Chapter 6 Statistical models
In this chapter, we will not learn about all the models out there that you may or may not need. Instead, I will show you how can use what you have learned until now and how you can apply these concepts to modeling. Also, as you read in the beginning of the book, R has many many packages. So the model you need is most probably already implemented in some package and you will very likely not need to write your own from scratch.
In the first section, I will discuss the terminology used in this book. Then I will discuss linear regression; showing how linear regression works illsutrates very well how other models work too, without loss of generality. Then I will introduce the concepte of hyper-parameters with ridge regression. This chapter will then finish with an introduction to cross-validation as a way to tune the hyper-parameters of models that features them.
6.1 Terminology
Before continuing discussing about statistical models and model fitting it is worthwhile to discuss
terminology a little bit. Depending on your background, you might call an explanatory variable a
feature or the dependent variable the target. These are the same objects. The matrix of features
is usually called a design matrix, and the what statisticians call the intercept in a linear regression
machine learning engineers call the bias. Calling the intercept the bias is unfortunate, as bias
also has a very different meaning; bias is also what we call the error in a model that may cause
biased estimates. To finish up, the estimated parameters of the model may be called coefficients
or weights. Here again, I don’t like the using weight as weight as a very different meaning in
statistics.
So, in the remainder of this chapter, and book, I will use the terminology from the statistical
litterature, using dependent and explanatory variables (y and x), and calling the
estimated parameters coefficients and the intercept… well the intercept (the \(\beta\)s of the model).
However, I will talk of training a model, instead of estimating a model.
6.2 Fitting a model to data
Suppose you have a variable y that you wish to explain using a set of other variables x1, x2,
x3, etc. Let’s take a look at the Housing dataset from the Ecdat package:
library(Ecdat)
data(Housing)You can read a description of the dataset by running:
?HousingHousing package:Ecdat R Documentation
Sales Prices of Houses in the City of Windsor
Description:
a cross-section from 1987
_number of observations_ : 546
_observation_ : goods
_country_ : Canada
Usage:
data(Housing)
Format:
A dataframe containing :
price: sale price of a house
lotsize: the lot size of a property in square feet
bedrooms: number of bedrooms
bathrms: number of full bathrooms
stories: number of stories excluding basement
driveway: does the house has a driveway ?
recroom: does the house has a recreational room ?
fullbase: does the house has a full finished basement ?
gashw: does the house uses gas for hot water heating ?
airco: does the house has central air conditioning ?
garagepl: number of garage places
prefarea: is the house located in the preferred neighbourhood of the city ?
Source:
Anglin, P.M. and R. Gencay (1996) “Semiparametric estimation of
a hedonic price function”, _Journal of Applied Econometrics_,
*11(6)*, 633-648.
References:
Verbeek, Marno (2004) _A Guide to Modern Econometrics_, John Wiley
and Sons, chapter 3.
Journal of Applied Econometrics data archive : <URL:
http://qed.econ.queensu.ca/jae/>.
See Also:
‘Index.Source’, ‘Index.Economics’, ‘Index.Econometrics’,
‘Index.Observations’or by looking for Housing in the help pane of RStudio. Usually, you would take a look a the data
before doing any modeling:
glimpse(Housing)## Observations: 546
## Variables: 12
## $ price <dbl> 42000, 38500, 49500, 60500, 61000, 66000, 66000, 69000,…
## $ lotsize <dbl> 5850, 4000, 3060, 6650, 6360, 4160, 3880, 4160, 4800, 5…
## $ bedrooms <dbl> 3, 2, 3, 3, 2, 3, 3, 3, 3, 3, 3, 2, 3, 3, 2, 2, 3, 4, 1…
## $ bathrms <dbl> 1, 1, 1, 1, 1, 1, 2, 1, 1, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1…
## $ stories <dbl> 2, 1, 1, 2, 1, 1, 2, 3, 1, 4, 1, 1, 2, 1, 1, 1, 2, 3, 1…
## $ driveway <fct> yes, yes, yes, yes, yes, yes, yes, yes, yes, yes, yes, …
## $ recroom <fct> no, no, no, yes, no, yes, no, no, yes, yes, no, no, no,…
## $ fullbase <fct> yes, no, no, no, no, yes, yes, no, yes, no, yes, no, no…
## $ gashw <fct> no, no, no, no, no, no, no, no, no, no, no, no, no, no,…
## $ airco <fct> no, no, no, no, no, yes, no, no, no, yes, yes, no, no, …
## $ garagepl <dbl> 1, 0, 0, 0, 0, 0, 2, 0, 0, 1, 3, 0, 0, 0, 0, 0, 1, 0, 0…
## $ prefarea <fct> no, no, no, no, no, no, no, no, no, no, no, no, no, no,…Housing prices depend on a set of variables such as the number of bedrooms, the area it is located
and so on. If you believe that housing prices depend linearly on a set of explanatory variables,
you will want to estimate a linear model. To estimate a linear model, you will need to use the
built-in lm() function:
model1 <- lm(price ~ lotsize + bedrooms, data = Housing)lm() takes a formula as an argument, which defines the model you want to estimate. In this case,
I ran the following regression:
\[ \text{price} = \beta_0 + \beta_1 * \text{lotsize} + \beta_2 * \text{bedrooms} + \varepsilon \]
where \(\beta_0, \beta_1\) and \(\beta_2\) are three parameters to estimate. To take a look at the
results, you can use the summary() method (not to be confused with dplyr::summarise():
summary(model1)##
## Call:
## lm(formula = price ~ lotsize + bedrooms, data = Housing)
##
## Residuals:
## Min 1Q Median 3Q Max
## -65665 -12498 -2075 8970 97205
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 5.613e+03 4.103e+03 1.368 0.172
## lotsize 6.053e+00 4.243e-01 14.265 < 2e-16 ***
## bedrooms 1.057e+04 1.248e+03 8.470 2.31e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 21230 on 543 degrees of freedom
## Multiple R-squared: 0.3703, Adjusted R-squared: 0.3679
## F-statistic: 159.6 on 2 and 543 DF, p-value: < 2.2e-16if you wish to remove the intercept (\(alpha\)) from your model, you can do so with -1:
model2 <- lm(price ~ -1 + lotsize + bedrooms, data = Housing)
summary(model2)##
## Call:
## lm(formula = price ~ -1 + lotsize + bedrooms, data = Housing)
##
## Residuals:
## Min 1Q Median 3Q Max
## -67229 -12342 -1333 9627 95509
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## lotsize 6.283 0.390 16.11 <2e-16 ***
## bedrooms 11968.362 713.194 16.78 <2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 21250 on 544 degrees of freedom
## Multiple R-squared: 0.916, Adjusted R-squared: 0.9157
## F-statistic: 2965 on 2 and 544 DF, p-value: < 2.2e-16or if you want to use all the columns inside Housing:
model3 <- lm(price ~ ., data = Housing)
summary(model3)##
## Call:
## lm(formula = price ~ ., data = Housing)
##
## Residuals:
## Min 1Q Median 3Q Max
## -41389 -9307 -591 7353 74875
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -4038.3504 3409.4713 -1.184 0.236762
## lotsize 3.5463 0.3503 10.124 < 2e-16 ***
## bedrooms 1832.0035 1047.0002 1.750 0.080733 .
## bathrms 14335.5585 1489.9209 9.622 < 2e-16 ***
## stories 6556.9457 925.2899 7.086 4.37e-12 ***
## drivewayyes 6687.7789 2045.2458 3.270 0.001145 **
## recroomyes 4511.2838 1899.9577 2.374 0.017929 *
## fullbaseyes 5452.3855 1588.0239 3.433 0.000642 ***
## gashwyes 12831.4063 3217.5971 3.988 7.60e-05 ***
## aircoyes 12632.8904 1555.0211 8.124 3.15e-15 ***
## garagepl 4244.8290 840.5442 5.050 6.07e-07 ***
## prefareayes 9369.5132 1669.0907 5.614 3.19e-08 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 15420 on 534 degrees of freedom
## Multiple R-squared: 0.6731, Adjusted R-squared: 0.6664
## F-statistic: 99.97 on 11 and 534 DF, p-value: < 2.2e-16You can access different elements of model3 (for example) with $, because the result of lm()
is a list:
print(model3$coefficients)## (Intercept) lotsize bedrooms bathrms stories
## -4038.350425 3.546303 1832.003466 14335.558468 6556.945711
## drivewayyes recroomyes fullbaseyes gashwyes aircoyes
## 6687.778890 4511.283826 5452.385539 12831.406266 12632.890405
## garagepl prefareayes
## 4244.829004 9369.513239but I prefer to use the {broom} package, and more specifically the tidy() function, which
converts model3 into a neat data.frame:
results3 <- broom::tidy(model3)
glimpse(results3)## Observations: 12
## Variables: 5
## $ term <chr> "(Intercept)", "lotsize", "bedrooms", "bathrms", "stor…
## $ estimate <dbl> -4038.350425, 3.546303, 1832.003466, 14335.558468, 655…
## $ std.error <dbl> 3409.4713, 0.3503, 1047.0002, 1489.9209, 925.2899, 204…
## $ statistic <dbl> -1.184451, 10.123618, 1.749764, 9.621691, 7.086369, 3.…
## $ p.value <dbl> 2.367616e-01, 3.732442e-22, 8.073341e-02, 2.570369e-20…I explicitely write broom::tidy() because tidy() is a popular function name. For instance,
it is also a function from the {yardstick} package, which does not do the same thing at all. Since
I will also be using {yardstick} I prefer explicitely writing broom::tidy() to avoid conflicts.
Using broom::tidy() is useful, because you can then work on the results easily, for example if
you wish to only keep results that are significant at the 5% level:
results3 %>%
filter(p.value < 0.05)## # A tibble: 10 x 5
## term estimate std.error statistic p.value
## <chr> <dbl> <dbl> <dbl> <dbl>
## 1 lotsize 3.55 0.350 10.1 3.73e-22
## 2 bathrms 14336. 1490. 9.62 2.57e-20
## 3 stories 6557. 925. 7.09 4.37e-12
## 4 drivewayyes 6688. 2045. 3.27 1.15e- 3
## 5 recroomyes 4511. 1900. 2.37 1.79e- 2
## 6 fullbaseyes 5452. 1588. 3.43 6.42e- 4
## 7 gashwyes 12831. 3218. 3.99 7.60e- 5
## 8 aircoyes 12633. 1555. 8.12 3.15e-15
## 9 garagepl 4245. 841. 5.05 6.07e- 7
## 10 prefareayes 9370. 1669. 5.61 3.19e- 8You can even add new columns, such as the confidence intervals:
results3 <- broom::tidy(model3, conf.int = TRUE, conf.level = 0.95)
print(results3)## # A tibble: 12 x 7
## term estimate std.error statistic p.value conf.low conf.high
## <chr> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
## 1 (Intercept) -4038. 3409. -1.18 2.37e- 1 -10736. 2659.
## 2 lotsize 3.55 0.350 10.1 3.73e-22 2.86 4.23
## 3 bedrooms 1832. 1047. 1.75 8.07e- 2 -225. 3889.
## 4 bathrms 14336. 1490. 9.62 2.57e-20 11409. 17262.
## 5 stories 6557. 925. 7.09 4.37e-12 4739. 8375.
## 6 drivewayyes 6688. 2045. 3.27 1.15e- 3 2670. 10705.
## 7 recroomyes 4511. 1900. 2.37 1.79e- 2 779. 8244.
## 8 fullbaseyes 5452. 1588. 3.43 6.42e- 4 2333. 8572.
## 9 gashwyes 12831. 3218. 3.99 7.60e- 5 6511. 19152.
## 10 aircoyes 12633. 1555. 8.12 3.15e-15 9578. 15688.
## 11 garagepl 4245. 841. 5.05 6.07e- 7 2594. 5896.
## 12 prefareayes 9370. 1669. 5.61 3.19e- 8 6091. 12648.Going back to model estimation, you can of course use lm() in a pipe workflow:
Housing %>%
select(-driveway, -stories) %>%
lm(price ~ ., data = .) %>%
broom::tidy()## # A tibble: 10 x 5
## term estimate std.error statistic p.value
## <chr> <dbl> <dbl> <dbl> <dbl>
## 1 (Intercept) 3025. 3263. 0.927 3.54e- 1
## 2 lotsize 3.67 0.363 10.1 4.52e-22
## 3 bedrooms 4140. 1036. 3.99 7.38e- 5
## 4 bathrms 16443. 1546. 10.6 4.29e-24
## 5 recroomyes 5660. 2010. 2.82 5.05e- 3
## 6 fullbaseyes 2241. 1618. 1.38 1.67e- 1
## 7 gashwyes 13568. 3411. 3.98 7.93e- 5
## 8 aircoyes 15578. 1597. 9.75 8.53e-21
## 9 garagepl 4232. 883. 4.79 2.12e- 6
## 10 prefareayes 10729. 1753. 6.12 1.81e- 9The first . in the lm() function is used to indicate that we wish to use all the data from Housing
(minus driveway and stories which I removed using select() and the - sign), and the second . is
used to place the result from the two dplyr instructions that preceded is to be placed there.
The picture below should help you understand:
knitr::include_graphics("pics/pipe_to_second_position.png")
You have to specify this, because by default, when using %>% the left hand side argument gets
passed as the first argument of the function on the right hand side.
6.3 Diagnostics
Diagnostics are useful metrics to assess model fit. You can read some of these diagnostics, such as
the \(R^2\) at the bottom of the summary (when running summary(my_model)), but if you want to do
more than simply reading these diagnostics from RStudio, you can put those in a data.frame too,
using broom::glance():
glance(model3)## # A tibble: 1 x 11
## r.squared adj.r.squared sigma statistic p.value df logLik AIC
## <dbl> <dbl> <dbl> <dbl> <dbl> <int> <dbl> <dbl>
## 1 0.673 0.666 15423. 100.0 6.18e-122 12 -6034. 12094.
## # … with 3 more variables: BIC <dbl>, deviance <dbl>, df.residual <int>You can also plot the usual diagnostics plots using ggfortify::autoplot() which uses the
ggplot2 package under the hood:
library(ggfortify)
autoplot(model3, which = 1:6) + theme_minimal()
which=1:6 is an additional option that shows you all the diagnostics plot. If you omit this
option, you will only get 4 of them.
You can also get the residuals of the regression in two ways; either you grab them directly from the model fit:
resi3 <- residuals(model3)or you can augment the original data with a residuals column, using broom::augment():
housing_aug <- augment(model3)Let’s take a look at housing_aug:
glimpse(housing_aug)## Observations: 546
## Variables: 19
## $ price <dbl> 42000, 38500, 49500, 60500, 61000, 66000, 66000, 6900…
## $ lotsize <dbl> 5850, 4000, 3060, 6650, 6360, 4160, 3880, 4160, 4800,…
## $ bedrooms <dbl> 3, 2, 3, 3, 2, 3, 3, 3, 3, 3, 3, 2, 3, 3, 2, 2, 3, 4,…
## $ bathrms <dbl> 1, 1, 1, 1, 1, 1, 2, 1, 1, 2, 2, 1, 1, 1, 1, 1, 1, 1,…
## $ stories <dbl> 2, 1, 1, 2, 1, 1, 2, 3, 1, 4, 1, 1, 2, 1, 1, 1, 2, 3,…
## $ driveway <fct> yes, yes, yes, yes, yes, yes, yes, yes, yes, yes, yes…
## $ recroom <fct> no, no, no, yes, no, yes, no, no, yes, yes, no, no, n…
## $ fullbase <fct> yes, no, no, no, no, yes, yes, no, yes, no, yes, no, …
## $ gashw <fct> no, no, no, no, no, no, no, no, no, no, no, no, no, n…
## $ airco <fct> no, no, no, no, no, yes, no, no, no, yes, yes, no, no…
## $ garagepl <dbl> 1, 0, 0, 0, 0, 0, 2, 0, 0, 1, 3, 0, 0, 0, 0, 0, 1, 0,…
## $ prefarea <fct> no, no, no, no, no, no, no, no, no, no, no, no, no, n…
## $ .fitted <dbl> 66037.98, 41391.15, 39889.63, 63689.09, 49760.43, 663…
## $ .se.fit <dbl> 1790.507, 1406.500, 1534.102, 2262.056, 1567.689, 236…
## $ .resid <dbl> -24037.9757, -2891.1515, 9610.3699, -3189.0873, 11239…
## $ .hat <dbl> 0.013477335, 0.008316321, 0.009893730, 0.021510891, 0…
## $ .sigma <dbl> 15402.01, 15437.14, 15431.98, 15437.02, 15429.89, 154…
## $ .cooksd <dbl> 2.803214e-03, 2.476265e-05, 3.265481e-04, 8.004787e-0…
## $ .std.resid <dbl> -1.56917096, -0.18823924, 0.62621736, -0.20903274, 0.…A few columns have been added to the original data, among them .resid which contains the
residuals. Let’s plot them:
ggplot(housing_aug) +
geom_density(aes(.resid))
Fitted values are also added to the original data, under the variable .fitted. It would also have
been possible to get the fitted values with:
fit3 <- fitted(model3)but I prefer using augment(), because the columns get merged to the original data, which then
makes it easier to find specific individuals, for example, you might want to know for which housing
units the model underestimates the price:
total_pos <- housing_aug %>%
filter(.resid > 0) %>%
summarise(total = n()) %>%
pull(total)we find 261 individuals where the residuals are positive. It is also easier to extract outliers:
housing_aug %>%
mutate(prank = cume_dist(.cooksd)) %>%
filter(prank > 0.99) %>%
glimpse()## Observations: 6
## Variables: 20
## $ price <dbl> 163000, 125000, 132000, 175000, 190000, 174500
## $ lotsize <dbl> 7420, 4320, 3500, 9960, 7420, 7500
## $ bedrooms <dbl> 4, 3, 4, 3, 4, 4
## $ bathrms <dbl> 1, 1, 2, 2, 2, 2
## $ stories <dbl> 2, 2, 2, 2, 3, 2
## $ driveway <fct> yes, yes, yes, yes, yes, yes
## $ recroom <fct> yes, no, no, no, no, no
## $ fullbase <fct> yes, yes, no, yes, no, yes
## $ gashw <fct> no, yes, yes, no, no, no
## $ airco <fct> yes, no, no, no, yes, yes
## $ garagepl <dbl> 2, 2, 2, 2, 2, 3
## $ prefarea <fct> no, no, no, yes, yes, yes
## $ .fitted <dbl> 94826.68, 77688.37, 85495.58, 108563.18, 115125.03, 1…
## $ .se.fit <dbl> 2520.691, 3551.954, 3544.961, 2589.680, 2185.603, 258…
## $ .resid <dbl> 68173.32, 47311.63, 46504.42, 66436.82, 74874.97, 559…
## $ .hat <dbl> 0.02671105, 0.05303793, 0.05282929, 0.02819317, 0.020…
## $ .sigma <dbl> 15144.70, 15293.34, 15298.27, 15159.14, 15085.99, 152…
## $ .cooksd <dbl> 0.04590995, 0.04637969, 0.04461464, 0.04616068, 0.041…
## $ .std.resid <dbl> 4.480428, 3.152300, 3.098176, 4.369631, 4.904193, 3.6…
## $ prank <dbl> 0.9963370, 1.0000000, 0.9945055, 0.9981685, 0.9926740…prank is a variable I created with cume_dist() which is a dplyr function that returns the
proportion of all values less than or equal to the current rank. For example:
example <- c(5, 4.6, 2, 1, 0.8, 0, -1)
cume_dist(example)## [1] 1.0000000 0.8571429 0.7142857 0.5714286 0.4285714 0.2857143 0.1428571by filtering prank > 0.99 we get the top 1% of outliers according to Cook’s distance.
6.4 Interpreting models
Model interpretation is essential in the social sciences. If one wants to know the effect of
variable x on the dependent variable y, marginal effects have to be computed. This is easily
done in R with the {margins} package, which aims to provide the same functionality as the
margins command in STATA:
library(margins)
effects_model3 <- margins(model3)
summary(effects_model3)## factor AME SE z p lower upper
## aircoyes 12632.8904 1555.0329 8.1239 0.0000 9585.0819 15680.6989
## bathrms 14335.5585 1482.9885 9.6667 0.0000 11428.9545 17242.1624
## bedrooms 1832.0035 1045.6558 1.7520 0.0798 -217.4442 3881.4512
## drivewayyes 6687.7789 2045.2636 3.2699 0.0011 2679.1359 10696.4219
## fullbaseyes 5452.3855 1587.9782 3.4335 0.0006 2340.0054 8564.7657
## garagepl 4244.8290 847.2173 5.0103 0.0000 2584.3136 5905.3444
## gashwyes 12831.4063 3217.6211 3.9879 0.0001 6524.9848 19137.8277
## lotsize 3.5463 0.3503 10.1250 0.0000 2.8598 4.2328
## prefareayes 9369.5132 1669.1034 5.6135 0.0000 6098.1307 12640.8957
## recroomyes 4511.2838 1899.9255 2.3745 0.0176 787.4982 8235.0694
## stories 6556.9457 924.4211 7.0930 0.0000 4745.1137 8368.7777It is also possible to plot the results:
plot(effects_model3)
This uses the basic R plotting capabilities, which is useful because it is a simple call to the
function plot() but if you’ve been using ggplot2 and want this graph to have the same feel as
the others made with ggplot2 you first need to save the summary in a variable.
summary(effects_model3) is a data.frame with many more details. Let’s overwrite this
effects_model3 with its summary:
effects_model3 <- summary(effects_model3)And now it is possible to use ggplot2 to have the same plot:
ggplot(data = effects_model3) +
geom_point(aes(factor, AME)) +
geom_errorbar(aes(x = factor, ymin = lower, ymax = upper)) +
geom_hline(yintercept = 0) +
theme_minimal() +
theme(axis.text.x = element_text(angle = 45))
Of course for model3, the marginal effects are the same as the coefficients, so let’s estimate a
logit model and compute the marginal effects. Logit models can be estimated using the glm() function.
As an example, we are going to use the Participation data, also from the Ecdat package:
data(Participation)?ParticpationParticipation package:Ecdat R Documentation
Labor Force Participation
Description:
a cross-section
_number of observations_ : 872
_observation_ : individuals
_country_ : Switzerland
Usage:
data(Participation)
Format:
A dataframe containing :
lfp labour force participation ?
lnnlinc the log of nonlabour income
age age in years divided by 10
educ years of formal education
nyc the number of young children (younger than 7)
noc number of older children
foreign foreigner ?
Source:
Gerfin, Michael (1996) “Parametric and semiparametric estimation
of the binary response”, _Journal of Applied Econometrics_,
*11(3)*, 321-340.
References:
Davidson, R. and James G. MacKinnon (2004) _Econometric Theory
and Methods_, New York, Oxford University Press, <URL:
http://www.econ.queensu.ca/ETM/>, chapter 11.
Journal of Applied Econometrics data archive : <URL:
http://qed.econ.queensu.ca/jae/>.
See Also:
‘Index.Source’, ‘Index.Economics’, ‘Index.Econometrics’,
‘Index.Observations’The variable of interest is lfp: whether the individual participates in the labour force. To know which variables are relevant in the decision to participate in the labour force, one could estimate a logit model, using glm().
logit_participation <- glm(lfp ~ ., data = Participation, family = "binomial")
broom::tidy(logit_participation)## # A tibble: 7 x 5
## term estimate std.error statistic p.value
## <chr> <dbl> <dbl> <dbl> <dbl>
## 1 (Intercept) 10.4 2.17 4.79 1.69e- 6
## 2 lnnlinc -0.815 0.206 -3.97 7.31e- 5
## 3 age -0.510 0.0905 -5.64 1.72e- 8
## 4 educ 0.0317 0.0290 1.09 2.75e- 1
## 5 nyc -1.33 0.180 -7.39 1.51e-13
## 6 noc -0.0220 0.0738 -0.298 7.66e- 1
## 7 foreignyes 1.31 0.200 6.56 5.38e-11From the results above, one can only interpret the sign of the coefficients. To know how much a
variable influences the labour force participation, one has to use margins():
effects_logit_participation <- margins(logit_participation) %>%
summary()
print(effects_logit_participation)## factor AME SE z p lower upper
## age -0.1064 0.0176 -6.0494 0.0000 -0.1409 -0.0719
## educ 0.0066 0.0060 1.0955 0.2733 -0.0052 0.0185
## foreignyes 0.2834 0.0399 7.1102 0.0000 0.2053 0.3615
## lnnlinc -0.1699 0.0415 -4.0994 0.0000 -0.2512 -0.0887
## noc -0.0046 0.0154 -0.2981 0.7656 -0.0347 0.0256
## nyc -0.2775 0.0333 -8.3433 0.0000 -0.3426 -0.2123We can use the previous code to plot the marginal effects:
ggplot(data = effects_logit_participation) +
geom_point(aes(factor, AME)) +
geom_errorbar(aes(x = factor, ymin = lower, ymax = upper)) +
geom_hline(yintercept = 0) +
theme_minimal() +
theme(axis.text.x = element_text(angle = 45))
So an infinitesimal increase, in say, non-labour income (lnnlinc) of 0.001 is associated with a decrease of the probability of labour force participation by 0.001*17 percentage points.
You can also extract the marginal effects of a single variable:
head(dydx(Participation, logit_participation, "lnnlinc"))## dydx_lnnlinc
## 1 -0.15667764
## 2 -0.20014487
## 3 -0.18495109
## 4 -0.05377262
## 5 -0.18710476
## 6 -0.19586986Which makes it possible to extract the effect for a list of individuals that you can create yourself:
my_subjects = tribble(
~lfp, ~lnnlinc, ~age, ~educ, ~nyc, ~noc, ~foreign,
"yes", 10.780, 7.0, 4, 1, 1, "yes",
"no", 1.30, 9.0, 1, 4, 1, "yes"
)
dydx(my_subjects, logit_participation, "lnnlinc")## dydx_lnnlinc
## 1 -0.09228119
## 2 -0.17953451I used the tribble() function from the tibble package to create this test data set, row by row.
Then, using dydx(), I get the marginal effect of variable lnnlinc for these two individuals.
6.5 Comparing models
Let’s estimate another model on the same data; prices are only positive, so a linear regression might not be the best model, because the model could predict negative prices. Let’s look at the distribution of prices:
ggplot(Housing) +
geom_density(aes(price))
it looks like modeling the log of price might provide a better fit:
model_log <- lm(log(price) ~ ., data = Housing)
result_log <- broom::tidy(model_log)
print(result_log)## # A tibble: 12 x 5
## term estimate std.error statistic p.value
## <chr> <dbl> <dbl> <dbl> <dbl>
## 1 (Intercept) 10.0 0.0472 212. 0.
## 2 lotsize 0.0000506 0.00000485 10.4 2.91e-23
## 3 bedrooms 0.0340 0.0145 2.34 1.94e- 2
## 4 bathrms 0.168 0.0206 8.13 3.10e-15
## 5 stories 0.0923 0.0128 7.20 2.10e-12
## 6 drivewayyes 0.131 0.0283 4.61 5.04e- 6
## 7 recroomyes 0.0735 0.0263 2.79 5.42e- 3
## 8 fullbaseyes 0.0994 0.0220 4.52 7.72e- 6
## 9 gashwyes 0.178 0.0446 4.00 7.22e- 5
## 10 aircoyes 0.178 0.0215 8.26 1.14e-15
## 11 garagepl 0.0508 0.0116 4.36 1.58e- 5
## 12 prefareayes 0.127 0.0231 5.50 6.02e- 8Let’s take a look at the diagnostics:
glance(model_log)## # A tibble: 1 x 11
## r.squared adj.r.squared sigma statistic p.value df logLik AIC
## <dbl> <dbl> <dbl> <dbl> <dbl> <int> <dbl> <dbl>
## 1 0.677 0.670 0.214 102. 3.67e-123 12 73.9 -122.
## # … with 3 more variables: BIC <dbl>, deviance <dbl>, df.residual <int>Let’s compare these to the ones from the previous model:
diag_lm <- glance(model3)
diag_lm <- diag_lm %>%
mutate(model = "lin-lin model")
diag_log <- glance(model_log)
diag_log <- diag_log %>%
mutate(model = "log-lin model")
diagnostics_models <- full_join(diag_lm, diag_log)## Joining, by = c("r.squared", "adj.r.squared", "sigma", "statistic", "p.value", "df", "logLik", "AIC", "BIC", "deviance", "df.residual", "model")print(diagnostics_models)## # A tibble: 2 x 12
## r.squared adj.r.squared sigma statistic p.value df logLik AIC
## <dbl> <dbl> <dbl> <dbl> <dbl> <int> <dbl> <dbl>
## 1 0.673 0.666 1.54e+4 100.0 6.18e-122 12 -6034. 12094.
## 2 0.677 0.670 2.14e-1 102. 3.67e-123 12 73.9 -122.
## # … with 4 more variables: BIC <dbl>, deviance <dbl>, df.residual <int>,
## # model <chr>I saved the diagnostics in two different data.frame objects using the glance() function and added a
model column to indicate which model the diagnostics come from. Then I merged both datasets using
full_join(), a {dplyr} function.
As you can see, the model with the logarithm of the prices as the dependent variable has a higher likelihood (and thus lower AIC and BIC) than the simple linear model. Let’s take a look at the diagnostics plots:
autoplot(model_log, which = 1:6) + theme_minimal()
6.6 Using a model for prediction
Once you estimated a model, you might want to use it for prediction. This is easily done using the
predict() function that works with most models. Prediction is also useful as a way to test the
accuracy of your model: split your data into a training set (used for estimation) and a testing
set (used for the pseudo-prediction) and see if your model overfits the data. We are going to see
how to do that in a later section; for now, let’s just get acquainted with predict() and other
functions. I insist, keep in mind that this section is only to get acquainted with these functions.
We are going to explore prediction, overfitting and tuning of models in a later section.
Let’s go back to the models we estimated in the previous section, model3 and model_log. Let’s also
take a subsample of data, which we will be using for prediction:
set.seed(1234)
pred_set <- Housing %>%
sample_n(20)so that we get always the same pred_set I set the random seed first. Let’s take a look at the
data:
print(pred_set)## price lotsize bedrooms bathrms stories driveway recroom fullbase gashw
## 1 52000 4280 2 1 1 yes no no no
## 2 62500 3900 3 1 2 yes no no no
## 3 175000 8960 4 4 4 yes no no no
## 4 141000 8100 4 1 2 yes yes yes no
## 5 54000 2856 3 1 3 yes no no no
## 6 52000 4130 3 2 2 yes no no no
## 7 66000 4160 3 1 1 yes yes yes no
## 8 95000 4260 4 2 2 yes no no yes
## 9 97000 12090 4 2 2 yes no no no
## 10 70000 6300 3 1 1 yes no no no
## 11 85000 6420 3 1 1 yes no yes no
## 12 39000 4000 3 1 2 yes no no no
## 13 59900 3450 3 1 2 yes no no no
## 14 53000 4050 2 1 1 yes no no no
## 15 60000 2610 4 3 2 no no no no
## 16 72500 5720 2 1 2 yes no no no
## 17 35500 3000 3 1 2 no no no no
## 18 40000 2650 3 1 2 yes no yes no
## 19 35000 3500 2 1 1 yes yes no no
## 20 37000 4400 2 1 1 yes no no no
## airco garagepl prefarea
## 1 yes 2 no
## 2 no 0 no
## 3 yes 3 no
## 4 yes 2 yes
## 5 no 0 yes
## 6 no 2 no
## 7 yes 0 no
## 8 no 0 no
## 9 no 2 yes
## 10 yes 2 no
## 11 yes 0 yes
## 12 no 1 no
## 13 no 1 no
## 14 no 0 no
## 15 no 0 no
## 16 yes 0 yes
## 17 no 0 no
## 18 no 1 no
## 19 no 0 no
## 20 no 0 noIf we wish to use it for prediction, this is easily done with predict():
predict(model3, pred_set)## 1 2 3 4 5 6 7
## 63506.66 49425.47 150689.71 106607.68 61649.59 73066.34 66387.12
## 8 9 10 11 12 13 14
## 79701.11 112496.42 72502.20 79260.00 54024.93 52074.46 41568.47
## 15 16 17 18 19 20
## 68666.08 76050.14 39546.02 54689.81 44129.28 42809.67This returns a vector of predicted prices. This can then be used to compute the Root Mean Squared Error
for instance. Let’s do it within a tidyverse pipeline:
rmse <- pred_set %>%
mutate(predictions = predict(model3, .)) %>%
summarise(sqrt(sum(predictions - price)**2/n()))The root mean square error of model3 is 1666.1312666.
I also used the n() function which returns the number of observations in a group (or all the
observations, if the data is not grouped). Let’s compare model3 ’s RMSE with the one from
model_log:
rmse2 <- pred_set %>%
mutate(predictions = exp(predict(model_log, .))) %>%
summarise(sqrt(sum(predictions - price)**2/n()))Don’t forget to exponentiate the predictions, remember you’re dealing with a log-linear model! model_log’s
RMSE is 1359.0392252 which is lower than model3’s. However, keep in mind that the model was estimated
on the whole data, and then the prediction quality was assessed using a subsample of the data the
model was estimated on… so actually we can’t really say if model_log’s predictions are very useful.
Of course, this is the same for model3.
In a later section we are going to learn how to do cross validation to avoid this issue.
Also another problem of what I did before, unrelated to statistics per se, is that I wanted to compute the same quantity for two different models, and did so by copy and pasting 3 lines of code. That’s not much, but if I wanted to compare 10 models, copy and paste mistakes could have sneaked in. Instead, it would have been nice to have a function that computes the RMSE and then use it on my models. We are going to learn how to write our own function and use it just like if it was another built-in R function.
6.7 Beyond linear regression
R has a lot of other built-in functions for regression, such as glm() (for Generalized Linear
Models) and nls() for (for Nonlinear Least Squares). There are also functions and additional
packages for time series, panel data, machine learning, bayesian and nonparametric methods.
Presenting everything here would take too much space, and would be pretty useless as you can find
whatever you need using an internet search engine. What you have learned until now is quite general
and should work on many type of models. To help you out, here is a list of methods and the
recommended packages that you can use:
| Model | Package | Quick example |
|---|---|---|
| Robust Linear Regression | MASS |
rlm(y ~ x, data = mydata) |
| Nonlinear Least Squares | stats2 |
nls(y ~ x1 / (1 + x2), data = mydata)3 |
| Logit | stats |
glm(y ~ x, data = mydata, family = "binomial") |
| Probit | stats |
glm(y ~ x, data = mydata, family = binomial(link = "probit")) |
| K-Means | stats |
kmeans(data, n)4 |
| PCA | stats |
prcomp(data, scale = TRUE, center = TRUE)5 |
| Multinomial Logit | mlogit |
Requires several steps of data pre-processing and formula definition, refer to the Vignette for more details. |
| Cox PH | survival |
coxph(Surv(y_time, y_status) ~ x, data = mydata)6 |
| Time series | Several, depending on your needs. | Time series in R is a vast subject that would require a very thick book to cover. You can get started with the following series of blog articles, Tidy time-series, part 1, Tidy time-series, part 2, Tidy time-series, part 3 and Tidy time-series, part 3 |
| Panel data | plm |
plm(y ~ x, data = mydata, model = "within|random") |
| Neural Networks | Several, depending on your needs. | R is a very popular programming language for machine learning. This blog post lists and compares some of the most useful packages for Neural nets and deep learning. |
| Nonparametric regression | np |
Several functions and options available, refer to the Vignette for more details. |
I put neural networks in the table, but you can also find packages for regression trees, naive
bayes, and pretty much any machine learning method out there! The same goes for Bayesian methods.
Popular packages include rstan, rjags which link R to STAN and JAGS (two other pieces of software
that do the Gibbs sampling for you) which are tools that allow you to fit very general models. It
is also possible to estimate models using Bayesian inference without the need of external tools,
with the bayesm package which estimates the usual micro-econometric models.
There really are a lot of packages available for Bayesian inference, and you can find them all in the
related CRAN Task View.
6.8 Hyper-parameters
Hyper-parameters are parameters of the model that cannot be directly learned from the data. A linear regression does not have any hyper-parameters, but a random forest for instance has several. You might have heard of ridge regression, lasso and elasticnet. These are extensions to linear models that avoid over-fitting by penalizing large models. These extensions of the linear regression have hyper-parameters that the practitioner has to tune. There are several ways one can tune these parameters, for example, by doing a grid-search, or a random search over the grid or using more elaborate methods. To introduce hyper-parameters, let’s get to know ridge regression, also called Tikhonov regularization.
6.8.1 Ridge regression
Ridge regression is used when the data you are working with has a lot of explanatory variables, or when there is a risk that a simple linear regression might overfit to the training data, because, for example, your explanatory variables are collinear. If you are training a linear model and then you notice that it generalizes very badly to new, unseen data, it is very likely that the linear model you trained overfit the data. In this case, ridge regression might prove useful. The way ridge regression works might seem counter-intuititive; it boils down to fitting a worse model to the training data, but in return, this worse model will generalize better to new data.
The closed form solution of the ordinary least squares estimator is defined as:
\[ \widehat{\beta} = (X'X)^{-1}X'Y \]
where \(X\) is the design matrix (the matrix made up of the explanatory variables) and \(Y\) is the dependent variable. For ridge regression, this closed form solution changes a little bit:
\[ \widehat{\beta} = (X'X + \lambda I_p)^{-1}X'Y \]
where \(\lambda \in \mathbb{R}\) is an hyper-parameter and \(I_p\) is the identity matrix of dimension \(p\) (\(p\) is the number of explanatory variables). This formula above is the closed form solution to the following optimisation program:
\[ \sum_{i=1}^n \left(y_i - \sum_{j=1}^px_{ij}\beta_j\right)^2 \]
such that:
\[ \sum_{j=1}^p(\beta_j)^2 < c \]
for any strictly positive \(c\).
The glmnet() function from the {glmnet} package can be used for ridge regression, by setting
the alpha argument to 0 (setting it to 1 would do LASSO, and setting it to a number between
0 and 1 would do elasticnet). But in order to compare linear regression and ridge regression,
let me first divide the data into a training set and a testing set:
index <- 1:nrow(Housing)
set.seed(12345)
train_index <- sample(index, round(0.90*nrow(Housing)), replace = FALSE)
test_index <- setdiff(index, train_index)
train_x <- Housing[train_index, ] %>%
select(-price)
train_y <- Housing[train_index, ] %>%
pull(price)
test_x <- Housing[test_index, ] %>%
select(-price)
test_y <- Housing[test_index, ] %>%
pull(price)I do the train/test split this way, because glmnet() requires a design matrix as input, and not
a formula. Design matrices can be created using the model.matrix() function:
library("glmnet")
train_matrix <- model.matrix(train_y ~ ., data = train_x)
test_matrix <- model.matrix(test_y ~ ., data = test_x)Let’s now run a linear regression, by setting the penalty to 0:
model_lm_ridge <- glmnet(y = train_y, x = train_matrix, alpha = 0, lambda = 0)The model above provides the same result as a linear regression, because I set lambda to 0. Let’s
compare the coefficients between the two:
coef(model_lm_ridge)## 13 x 1 sparse Matrix of class "dgCMatrix"
## s0
## (Intercept) -3247.030393
## (Intercept) .
## lotsize 3.520283
## bedrooms 1745.211187
## bathrms 14337.551325
## stories 6736.679470
## drivewayyes 5687.132236
## recroomyes 5701.831289
## fullbaseyes 5708.978557
## gashwyes 12508.524241
## aircoyes 12592.435621
## garagepl 4438.918373
## prefareayes 9085.172469and now the coefficients of the linear regression (because I provide a design matrix, I have to use
lm.fit() instead of lm() which requires a formula, not a matrix.)
coef(lm.fit(x = train_matrix, y = train_y))## (Intercept) lotsize bedrooms bathrms stories
## -3245.146665 3.520357 1744.983863 14336.336858 6737.000410
## drivewayyes recroomyes fullbaseyes gashwyes aircoyes
## 5686.394123 5700.210775 5709.493884 12509.005265 12592.367268
## garagepl prefareayes
## 4439.029607 9085.409155as you can see, the coefficients are the same. Let’s compute the RMSE for the unpenalized linear regression:
preds_lm <- predict(model_lm_ridge, test_matrix)
rmse_lm <- sqrt(mean(preds_lm - test_y)^2)The RMSE for the linear unpenalized regression is equal to 2077.4197343.
Let’s now run a ridge regression, with lambda equal to 100, and see if the RMSE is smaller:
model_ridge <- glmnet(y = train_y, x = train_matrix, alpha = 0, lambda = 100)and let’s compute the RMSE again:
preds <- predict(model_ridge, test_matrix)
rmse <- sqrt(mean(preds - test_y)^2)The RMSE for the linear penalized regression is equal to 2072.6117757, which is smaller than before.
But which value of lambda gives smallest RMSE? To find out, one must run model over a grid of
lambda values and pick the model with lowest RMSE. This procedure is available in the cv.glmnet()
function, which picks the best value for lambda:
best_model <- cv.glmnet(train_matrix, train_y)
# lambda that minimises the MSE
best_model$lambda.min## [1] 66.07936According to cv.glmnet() the best value for lambda is 66.0793576. In the
next section, we will implement cross validation ourselves, in order to find the hyper-parameters
of a random forest.
6.9 Training, validating, and testing models
Cross-validation is an important procedure which is used to compare models but also to tune the hyper-parameters of a model.
In this section, we are going to use several packages from the
{tidymodels} collection of packages, namely
{recipes},
{rsample} and
{parsnip} to train a random forest the tidy way. I will
also use {mlrMBO} to tune the hyper-parameters of the random forest.
6.9.1 Set up
Let’s load the needed packages:
library("tidyverse")
library("tidymodels")
library("parsnip")
library("brotools")
library("mlbench")Load the data, included in the {mlrbench} package:
data("BostonHousing2")I will train a random forest to predict the housing price, which is the cmedv column:
head(BostonHousing2)## town tract lon lat medv cmedv crim zn indus chas nox
## 1 Nahant 2011 -70.9550 42.2550 24.0 24.0 0.00632 18 2.31 0 0.538
## 2 Swampscott 2021 -70.9500 42.2875 21.6 21.6 0.02731 0 7.07 0 0.469
## 3 Swampscott 2022 -70.9360 42.2830 34.7 34.7 0.02729 0 7.07 0 0.469
## 4 Marblehead 2031 -70.9280 42.2930 33.4 33.4 0.03237 0 2.18 0 0.458
## 5 Marblehead 2032 -70.9220 42.2980 36.2 36.2 0.06905 0 2.18 0 0.458
## 6 Marblehead 2033 -70.9165 42.3040 28.7 28.7 0.02985 0 2.18 0 0.458
## rm age dis rad tax ptratio b lstat
## 1 6.575 65.2 4.0900 1 296 15.3 396.90 4.98
## 2 6.421 78.9 4.9671 2 242 17.8 396.90 9.14
## 3 7.185 61.1 4.9671 2 242 17.8 392.83 4.03
## 4 6.998 45.8 6.0622 3 222 18.7 394.63 2.94
## 5 7.147 54.2 6.0622 3 222 18.7 396.90 5.33
## 6 6.430 58.7 6.0622 3 222 18.7 394.12 5.21Only keep relevant columns:
boston <- BostonHousing2 %>%
select(-medv, -tract, -lon, -lat) %>%
rename(price = cmedv)I remove tract, lat and lon because the information contained in the column town is enough.
To train and evaluate the model’s performance, I split the data in two. One data set, called the training set, will be further split into two down below. I won’t touch the second data set, the test set, until the very end, to finally assess the model’s performance.
train_test_split <- initial_split(boston, prop = 0.9)
housing_train <- training(train_test_split)
housing_test <- testing(train_test_split)initial_split(), training() and testing() are functions from the {rsample} package.
I will train a random forest on the training data, but the question, is which random forest? Because random forests have several hyper-parameters, and as explained in the intro these hyper-parameters cannot be directly learned from the data, which one should we choose? We could train 6 random forests for instance and compare their performance, but why only 6? Why not 16?
In order to find the right hyper-parameters, the practitioner can use values from the literature that seemed to have worked well (like is done in Macro-econometrics) or you can further split the train set into two, create a grid of hyperparameter, train the model on one part of the data for all values of the grid, and compare the predictions of the models on the second part of the data. You then stick with the model that performed the best, for example, the model with lowest RMSE. The thing is, you can’t estimate the true value of the RMSE with only one value. It’s like if you wanted to estimate the height of the population by drawing one single observation from the population. You need a bit more observations. To approach the true value of the RMSE for a give set of hyperparameters, instead of doing one split, let’s do 30. Then we compute the average RMSE, which implies training 30 models for each combination of the values of the hyperparameters.
First, let’s split the training data again, using the mc_cv() function from {rsample} package.
This function implements Monte Carlo cross-validation:
validation_data <- mc_cv(housing_train, prop = 0.9, times = 30)What does validation_data look like?
validation_data## # # Monte Carlo cross-validation (0.9/0.1) with 30 resamples
## # A tibble: 30 x 2
## splits id
## <list> <chr>
## 1 <split [411/45]> Resample01
## 2 <split [411/45]> Resample02
## 3 <split [411/45]> Resample03
## 4 <split [411/45]> Resample04
## 5 <split [411/45]> Resample05
## 6 <split [411/45]> Resample06
## 7 <split [411/45]> Resample07
## 8 <split [411/45]> Resample08
## 9 <split [411/45]> Resample09
## 10 <split [411/45]> Resample10
## # … with 20 more rowsLet’s look further down:
validation_data$splits[[1]]## <411/45/456>The first value is the number of rows of the first set, the second value of the second, and the third was the original amount of values in the training data, before splitting again.
How should we call these two new data sets? The author of {rsample}, Max Kuhn, talks about
the analysis and the assessment sets:
Now, in order to continue I need pre-process the data. I will do this in three steps. The first and the second step are used to center and scale the numeric variables and the third step converts character and factor variables to dummy variables. This is needed because I will train a random forest, which cannot handle factor variables directly. Let’s define a recipe to do that, and start by pre-processing the testing set. I write a wrapper function around the recipe, because I will need to apply this recipe to various data sets:
simple_recipe <- function(dataset){
recipe(price ~ ., data = dataset) %>%
step_center(all_numeric()) %>%
step_scale(all_numeric()) %>%
step_dummy(all_nominal())
}Once the recipe is defined, I can use the prep() function, which estimates the parameters from
the data which are needed to process the data. For example, for centering, prep() estimates
the mean which will then be subtracted from the variables. With bake() the estimates are then
applied on the data:
testing_rec <- prep(simple_recipe(housing_test), testing = housing_test)
test_data <- bake(testing_rec, new_data = housing_test)It is important to split the data before using prep() and bake(), because if not, you will
use observations from the test set in the prep() step, and thus introduce knowledge from the test
set into the training data. This is called data leakage, and must be avoided. This is why it is
necessary to first split the training data into an analysis and an assessment set, and then also
pre-process these sets separately. However, the validation_data object cannot now be used with
recipe(), because it is not a dataframe. No worries, I simply need to write a function that extracts
the analysis and assessment sets from the validation_data object, applies the pre-processing, trains
the model, and returns the RMSE. This will be a big function, at the center of the analysis.
But before that, let’s run a simple linear regression, as a benchmark. For the linear regression, I will not use any CV, so let’s pre-process the training set:
trainlm_rec <- prep(simple_recipe(housing_train), testing = housing_train)
trainlm_data <- bake(trainlm_rec, new_data = housing_train)
linreg_model <- lm(price ~ ., data = trainlm_data)
broom::augment(linreg_model, newdata = test_data) %>%
rmse(price, .fitted)## # A tibble: 1 x 3
## .metric .estimator .estimate
## <chr> <chr> <dbl>
## 1 rmse standard 0.345broom::augment() adds the predictions to the test_data in a new column, .fitted. I won’t
use this trick with the random forest, because there is no augment() method for random forests
from the {ranger} which I’ll use. I’ll add the predictions to the data myself.
Ok, now let’s go back to the random forest and write the big function:
my_rf <- function(mtry, trees, split, id){
analysis_set <- analysis(split)
analysis_prep <- prep(simple_recipe(analysis_set), training = analysis_set)
analysis_processed <- bake(analysis_prep, new_data = analysis_set)
model <- rand_forest(mtry = mtry, trees = trees) %>%
set_engine("ranger", importance = 'impurity') %>%
fit(price ~ ., data = analysis_processed)
assessment_set <- assessment(split)
assessment_prep <- prep(simple_recipe(assessment_set), testing = assessment_set)
assessment_processed <- bake(assessment_prep, new_data = assessment_set)
tibble::tibble("id" = id,
"truth" = assessment_processed$price,
"prediction" = unlist(predict(model, new_data = assessment_processed)))
}The rand_forest() function is available from the {parsnip} package. This package provides an
unified interface to a lot of other machine learning packages. This means that instead of having to
learn the syntax of range() and randomForest() and, and… you can simply use the rand_forest()
function and change the engine argument to the one you want (ranger, randomForest, etc).
Let’s try this function:
results_example <- map2_df(.x = validation_data$splits,
.y = validation_data$id,
~my_rf(mtry = 3, trees = 200, split = .x, id = .y))head(results_example)## # A tibble: 6 x 3
## id truth prediction
## <chr> <dbl> <dbl>
## 1 Resample01 -0.518 -0.259
## 2 Resample01 -1.03 -0.354
## 3 Resample01 -0.913 -0.345
## 4 Resample01 -0.874 -0.407
## 5 Resample01 0.430 0.241
## 6 Resample01 0.259 0.301I can now compute the RMSE when mtry = 3 and trees = 200:
results_example %>%
group_by(id) %>%
rmse(truth, prediction) %>%
summarise(mean_rmse = mean(.estimate)) %>%
pull## [1] 0.6096097The random forest has already lower RMSE than the linear regression. The goal now is to lower this
RMSE by tuning the mtry and trees hyperparameters. For this, I will use Bayesian Optimization
methods implemented in the {mlrMBO} package.
6.9.2 Bayesian hyperparameter optimization
I will re-use the code from above, and define a function that does everything from pre-processing to returning the metric I want to minimize by tuning the hyperparameters, the RMSE:
tuning <- function(param, validation_data){
mtry <- param[1]
trees <- param[2]
results <- purrr::map2_df(.x = validation_data$splits,
.y = validation_data$id,
~my_rf(mtry = mtry, trees = trees, split = .x, id = .y))
results %>%
group_by(id) %>%
rmse(truth, prediction) %>%
summarise(mean_rmse = mean(.estimate)) %>%
pull
}This is exactly the code from before, but it now returns the RMSE. Let’s try the function with the values from before:
tuning(c(3, 200), validation_data)## [1] 0.6085931I now follow the code that can be found in the arxiv paper to run the optimization. I think I got the gist of the paper, but I did not understand everything yet. For now, I am still experimenting with the library at the moment, but from what I understand, a simpler model, called the surrogate model, is used to look for promising points and to evaluate the value of the function at these points. This seems somewhat similar (in spirit) to the Indirect Inference method as described in Gourieroux, Monfort, Renault.
Let’s first load the package and create the function to optimize:
library("mlrMBO")fn <- makeSingleObjectiveFunction(name = "tuning",
fn = tuning,
par.set = makeParamSet(makeIntegerParam("x1", lower = 3, upper = 8),
makeIntegerParam("x2", lower = 100, upper = 500)))This function is based on the function I defined before. The parameters to optimize are also
defined as are their bounds. I will look for mtry between the values of 3 and 8, and trees
between 50 and 500.
Now comes the part I didn’t quite get.
# Create initial random Latin Hypercube Design of 10 points
library(lhs)# for randomLHS
des <- generateDesign(n = 5L * 2L, getParamSet(fn), fun = randomLHS)I think this means that these 10 points are the points used to start the whole process. I did not understand why they have to be sampled from a hypercube, but ok. Then I choose the surrogate model, a random forest too, and predict the standard error. Here also, I did not quite get why the standard error can be an option.
# Specify kriging model with standard error estimation
surrogate <- makeLearner("regr.ranger", predict.type = "se", keep.inbag = TRUE)Here I define some options:
# Set general controls
ctrl <- makeMBOControl()
ctrl <- setMBOControlTermination(ctrl, iters = 10L)
ctrl <- setMBOControlInfill(ctrl, crit = makeMBOInfillCritEI())And this is the optimization part:
# Start optimization
result <- mbo(fn, des, surrogate, ctrl, more.args = list("validation_data" = validation_data))result## Recommended parameters:
## x1=8; x2=314
## Objective: y = 0.484
##
## Optimization path
## 10 + 10 entries in total, displaying last 10 (or less):
## x1 x2 y dob eol error.message exec.time ei
## 11 8 283 0.4855415 1 NA <NA> 7.353 -3.276847e-04
## 12 8 284 0.4852047 2 NA <NA> 7.321 -3.283713e-04
## 13 8 314 0.4839817 3 NA <NA> 7.703 -3.828517e-04
## 14 8 312 0.4841398 4 NA <NA> 7.633 -2.829713e-04
## 15 8 318 0.4841066 5 NA <NA> 7.692 -2.668354e-04
## 16 8 314 0.4845221 6 NA <NA> 7.574 -1.382333e-04
## 17 8 321 0.4843018 7 NA <NA> 7.693 -3.828924e-05
## 18 8 318 0.4868457 8 NA <NA> 7.696 -8.692828e-07
## 19 8 310 0.4862687 9 NA <NA> 7.594 -1.061185e-07
## 20 8 313 0.4878694 10 NA <NA> 7.628 -5.153015e-07
## error.model train.time prop.type propose.time se mean
## 11 <NA> 0.011 infill_ei 0.450 0.0143886864 0.5075765
## 12 <NA> 0.011 infill_ei 0.427 0.0090265872 0.4971003
## 13 <NA> 0.012 infill_ei 0.443 0.0062693960 0.4916927
## 14 <NA> 0.012 infill_ei 0.435 0.0037308971 0.4878950
## 15 <NA> 0.012 infill_ei 0.737 0.0024446891 0.4860699
## 16 <NA> 0.013 infill_ei 0.442 0.0012713838 0.4850705
## 17 <NA> 0.012 infill_ei 0.444 0.0006371109 0.4847248
## 18 <NA> 0.013 infill_ei 0.467 0.0002106381 0.4844576
## 19 <NA> 0.014 infill_ei 0.435 0.0002182254 0.4846214
## 20 <NA> 0.013 infill_ei 0.748 0.0002971160 0.4847383So the recommended parameters are 8 for mtry and 314 for trees. The
user can access these recommended parameters with result$x$x1 and result$x$x2.
The value of the RMSE is lower than before, and equals 0.4839817. It can be accessed with
result$y.
Let’s now train the random forest on the training data with this values. First, I pre-process the
training data
training_rec <- prep(simple_recipe(housing_train), testing = housing_train)
train_data <- bake(training_rec, new_data = housing_train)Let’s now train our final model and predict the prices:
final_model <- rand_forest(mtry = result$x$x1, trees = result$x$x2) %>%
set_engine("ranger", importance = 'impurity') %>%
fit(price ~ ., data = train_data)
price_predict <- predict(final_model, new_data = select(test_data, -price))Let’s transform the data back and compare the predicted prices to the true ones visually:
cbind(price_predict * sd(housing_train$price) + mean(housing_train$price),
housing_test$price)## .pred housing_test$price
## 1 30.59078 33.4
## 2 20.45857 22.1
## 3 20.65227 15.0
## 4 21.00041 18.9
## 5 21.85504 21.0
## 6 22.22871 19.7
## 7 25.50695 23.5
## 8 23.37194 22.8
## 9 23.10205 22.9
## 10 35.32846 43.8
## 11 19.75544 19.4
## 12 20.62800 18.5
## 13 19.32962 20.5
## 14 17.47344 19.2
## 15 17.85629 23.0
## 16 37.12904 50.0
## 17 19.92393 17.4
## 18 25.12932 29.9
## 19 28.58319 29.6
## 20 27.52733 30.5
## 21 36.23437 48.5
## 22 37.18469 50.0
## 23 22.24399 24.4
## 24 25.13665 27.5
## 25 36.78475 44.8
## 26 31.46111 31.5
## 27 32.45295 44.0
## 28 29.72116 36.5
## 29 29.56234 35.1
## 30 23.19542 23.9
## 31 23.80056 26.4
## 32 19.83584 17.8
## 33 22.48759 19.4
## 34 20.94058 18.7
## 35 24.33038 23.1
## 36 22.50212 20.6
## 37 15.94037 17.8
## 38 21.41992 50.0
## 39 12.89569 12.3
## 40 12.40893 12.7
## 41 15.62015 11.9
## 42 19.64325 15.0
## 43 14.30708 11.7
## 44 12.09624 8.3
## 45 13.70672 14.4
## 46 17.14228 19.1
## 47 15.79025 13.3
## 48 16.86834 21.4
## 49 20.49379 16.8
## 50 21.41435 22.4Let’s now compute the RMSE:
tibble::tibble("truth" = test_data$price,
"prediction" = unlist(price_predict)) %>%
rmse(truth, prediction)## # A tibble: 1 x 3
## .metric .estimator .estimate
## <chr> <chr> <dbl>
## 1 rmse standard 0.481This package gets installed with R, no need to add it↩
The formula in the example is shown for illustration purposes.↩
datamust only contain numeric values, andnis the number of clusters.↩datamust only contain numeric values, or a formula can be provided.↩Surv(y_time, y_status)creates a survival object, wherey_timeis the time to eventy_status. It is possible to create more complex survival objects depending on exactly which data you are dealing with.↩