# Partial Least Squares Regression in R

Partial least squares regression is a form of regression that involves the development of components of the original variables in a supervised way. What this means is that the dependent variable is used to help create the new components form the original variables. This means that when pls is used the linear combination of the new features helps to explain both the independent and dependent variables in the model.

In this post, we will use predict “income” in the “Mroz” dataset using pls. Below is some initial code.

``library(pls);library(Ecdat)``
``````data("Mroz")
str(Mroz)``````
``````## 'data.frame':    753 obs. of  18 variables:
##  \$ work      : Factor w/ 2 levels "yes","no": 2 2 2 2 2 2 2 2 2 2 ...
##  \$ hoursw    : int  1610 1656 1980 456 1568 2032 1440 1020 1458 1600 ...
##  \$ child6    : int  1 0 1 0 1 0 0 0 0 0 ...
##  \$ child618  : int  0 2 3 3 2 0 2 0 2 2 ...
##  \$ agew      : int  32 30 35 34 31 54 37 54 48 39 ...
##  \$ educw     : int  12 12 12 12 14 12 16 12 12 12 ...
##  \$ hearnw    : num  3.35 1.39 4.55 1.1 4.59 ...
##  \$ wagew     : num  2.65 2.65 4.04 3.25 3.6 4.7 5.95 9.98 0 4.15 ...
##  \$ hoursh    : int  2708 2310 3072 1920 2000 1040 2670 4120 1995 2100 ...
##  \$ ageh      : int  34 30 40 53 32 57 37 53 52 43 ...
##  \$ educh     : int  12 9 12 10 12 11 12 8 4 12 ...
##  \$ wageh     : num  4.03 8.44 3.58 3.54 10 ...
##  \$ income    : int  16310 21800 21040 7300 27300 19495 21152 18900 20405 20425 ...
##  \$ educwm    : int  12 7 12 7 12 14 14 3 7 7 ...
##  \$ educwf    : int  7 7 7 7 14 7 7 3 7 7 ...
##  \$ unemprate : num  5 11 5 5 9.5 7.5 5 5 3 5 ...
##  \$ city      : Factor w/ 2 levels "no","yes": 1 2 1 1 2 2 1 1 1 1 ...
##  \$ experience: int  14 5 15 6 7 33 11 35 24 21 ...``````

First, we must prepare our data by dividing it into a training and test set. We will do this by doing a 50/50 split of the data.

``````set.seed(777)
train<-sample(c(T,F),nrow(Mroz),rep=T) #50/50 train/test split
test<-(!train)``````

In the code above we set the “set.seed function in order to assure reduplication. Then we created the “train” object and used the “sample” function to make a vector with ‘T’ and ‘F’ based on the number of rows in “Mroz”. Lastly, we created the “test”” object base don everything that is not in the “train” object as that is what the exclamation point is for.

Now we create our model using the “plsr” function from the “pls” package and we will examine the results using the “summary” function. We will also scale the data since this the scale affects the development of the components and use cross-validation. Below is the code.

``````set.seed(777)
pls.fit<-plsr(income~.,data=Mroz,subset=train,scale=T,validation="CV")
summary(pls.fit)``````
``````## Data:    X dimension: 392 17
##  Y dimension: 392 1
## Fit method: kernelpls
## Number of components considered: 17
##
## VALIDATION: RMSEP
## Cross-validated using 10 random segments.
##        (Intercept)  1 comps  2 comps  3 comps  4 comps  5 comps  6 comps
## CV           11218     8121     6701     6127     5952     5886     5857
## adjCV        11218     8114     6683     6108     5941     5872     5842
##        7 comps  8 comps  9 comps  10 comps  11 comps  12 comps  13 comps
## CV        5853     5849     5854      5853      5853      5852      5852
## adjCV     5837     5833     5837      5836      5836      5835      5835
##        14 comps  15 comps  16 comps  17 comps
## CV         5852      5852      5852      5852
## adjCV      5835      5835      5835      5835
##
## TRAINING: % variance explained
##         1 comps  2 comps  3 comps  4 comps  5 comps  6 comps  7 comps
## X         17.04    26.64    37.18    49.16    59.63    64.63    69.13
## income    49.26    66.63    72.75    74.16    74.87    75.25    75.44
##         8 comps  9 comps  10 comps  11 comps  12 comps  13 comps  14 comps
## X         72.82    76.06     78.59     81.79     85.52     89.55     92.14
## income    75.49    75.51     75.51     75.52     75.52     75.52     75.52
##         15 comps  16 comps  17 comps
## X          94.88     97.62    100.00
## income     75.52     75.52     75.52``````

The printout includes the root mean squared error for each of the components in the VALIDATION section as well as the variance explained in the TRAINING section. There are 17 components because there are 17 independent variables. You can see that after component 3 or 4 there is little improvement in the variance explained in the dependent variable. Below is the code for the plot of these results. It requires the use of the “validationplot” function with the “val.type” argument set to “MSEP” Below is the code

``validationplot(pls.fit,val.type = "MSEP")``

We will do the predictions with our model. We use the “predict” function, use our “Mroz” dataset but only those index in the “test” vector and set the components to three based on our previous plot. Below is the code.

``````set.seed(777)
pls.pred<-predict(pls.fit,Mroz[test,],ncomp=3)``````

After this, we will calculate the mean squared error. This is done by subtracting the results of our predicted model from the dependent variable of the test set. We then square this information and calculate the mean. Below is the code

``mean((pls.pred-Mroz\$income[test])^2)``
``## [1] 63386682``

As you know, this information is only useful when compared to something else. Therefore, we will run the data with a tradition least squares regression model and compare the results.

``````set.seed(777)
lm.fit<-lm(income~.,data=Mroz,subset=train)
lm.pred<-predict(lm.fit,Mroz[test,])
mean((lm.pred-Mroz\$income[test])^2)``````
``## [1] 59432814``

The least squares model is slightly better then our partial least squares model but if we look at the model we see several variables that are not significant. We will remove these see what the results are

``summary(lm.fit)``
``````##
## Call:
## lm(formula = income ~ ., data = Mroz, subset = train)
##
## Residuals:
##    Min     1Q Median     3Q    Max
## -20131  -2923  -1065   1670  36246
##
## Coefficients:
##               Estimate Std. Error t value Pr(>|t|)
## (Intercept) -1.946e+04  3.224e+03  -6.036 3.81e-09 ***
## workno      -4.823e+03  1.037e+03  -4.651 4.59e-06 ***
## hoursw       4.255e+00  5.517e-01   7.712 1.14e-13 ***
## child6      -6.313e+02  6.694e+02  -0.943 0.346258
## child618     4.847e+02  2.362e+02   2.052 0.040841 *
## agew         2.782e+02  8.124e+01   3.424 0.000686 ***
## educw        1.268e+02  1.889e+02   0.671 0.502513
## hearnw       6.401e+02  1.420e+02   4.507 8.79e-06 ***
## wagew        1.945e+02  1.818e+02   1.070 0.285187
## hoursh       6.030e+00  5.342e-01  11.288  < 2e-16 ***
## ageh        -9.433e+01  7.720e+01  -1.222 0.222488
## educh        1.784e+02  1.369e+02   1.303 0.193437
## wageh        2.202e+03  8.714e+01  25.264  < 2e-16 ***
## educwm      -4.394e+01  1.128e+02  -0.390 0.697024
## educwf       1.392e+02  1.053e+02   1.322 0.186873
## unemprate   -1.657e+02  9.780e+01  -1.694 0.091055 .
## cityyes     -3.475e+02  6.686e+02  -0.520 0.603496
## experience  -1.229e+02  4.490e+01  -2.737 0.006488 **
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 5668 on 374 degrees of freedom
## Multiple R-squared:  0.7552, Adjusted R-squared:  0.744
## F-statistic: 67.85 on 17 and 374 DF,  p-value: < 2.2e-16``````
``````set.seed(777)
lm.fit<-lm(income~work+hoursw+child618+agew+hearnw+hoursh+wageh+experience,data=Mroz,subset=train)
lm.pred<-predict(lm.fit,Mroz[test,])
mean((lm.pred-Mroz\$income[test])^2)``````
``## [1] 57839715``

As you can see the error decreased even more which indicates that the least squares regression model is superior to the partial least squares model. In addition, the partial least squares model is much more difficult to explain because of the use of components. As such, the least squares model is the favored one.