# K Nearest Neighbor in R

K-nearest neighbor is one of many nonlinear algorithms that can be used in machine learning. By non-linear I mean that a linear combination of the features or variables is not needed in order to develop decision boundaries. This allows for the analysis of data that naturally does not meet the assumptions of linearity.

KNN is also known as a “lazy learner”. This means that there are known coefficients or parameter estimates. When doing regression we always had  coefficient outputs regardless of the type of regression (ridge, lasso, elastic net, etc.). What KNN does instead is used K nearest neighbors to give a label to an unlabeled example. Our job when using KNN is to determine the number of K neighbors to use that is most accurate based on the different criteria for assessing the models.

In this post, we will develop a KNN model using the “Mroz” dataset from the “Ecdat” package. Our goal is to predict if someone lives in the city based on the other predictor variables. Below is some initial code.

``library(class);library(kknn);library(caret);library(corrplot)``
`library(reshape2);library(ggplot2);library(pROC);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 ...``````

We need to remove the factor variable “work” as KNN cannot use factor variables. After this, we will use the “melt” function from the “reshape2” package to look at the variables when divided by whether the example was from the city or not.

``````Mroz\$work<-NULL
mroz.melt<-melt(Mroz,id.var='city')
Mroz_plots<-ggplot(mroz.melt,aes(x=city,y=value))+geom_boxplot()+facet_wrap(~variable, ncol = 4)
Mroz_plots`````` From the plots, it appears there are no differences in how the variable act whether someone is from the city or not. This may be a flag that classification may not work.

We now need to scale our data otherwise the results will be inaccurate. Scaling might also help our box-plots because everything will be on the same scale rather than spread all over the place. To do this we will have to temporarily remove our outcome variable from the data set because it’s a factor and then reinsert it into the data set. Below is the code.

``````mroz.scale<-as.data.frame(scale(Mroz[,-16]))
mroz.scale\$city<-Mroz\$city``````

We will now look at our box-plots a second time but this time with scaled data.

``````mroz.scale.melt<-melt(mroz.scale,id.var="city")
mroz_plot2<-ggplot(mroz.scale.melt,aes(city,value))+geom_boxplot()+facet_wrap(~variable, ncol = 4)
mroz_plot2`````` This second plot is easier to read but there is still little indication of difference.

We can now move to checking the correlations among the variables. Below is the code

``````mroz.cor<-cor(mroz.scale[,-17])
corrplot(mroz.cor,method = 'number')`````` There is a high correlation between husband’s age (ageh) and wife’s age (agew). Since this algorithm is non-linear this should not be a major problem.

We will now divide our dataset into the training and testing sets

``````set.seed(502)
ind=sample(2,nrow(mroz.scale),replace=T,prob=c(.7,.3))
train<-mroz.scale[ind==1,]
test<-mroz.scale[ind==2,]``````

Before creating a model we need to create a grid. We do not know the value of k yet so we have to run multiple models with different values of k in order to determine this for our model. As such we need to create a ‘grid’ using the ‘expand.grid’ function. We will also use cross-validation to get a better estimate of k as well using the “trainControl” function. The code is below.

``````grid<-expand.grid(.k=seq(2,20,by=1))
control<-trainControl(method="cv")``````

Now we make our model,

``````knn.train<-train(city~.,train,method="knn",trControl=control,tuneGrid=grid)
knn.train``````
``````## k-Nearest Neighbors
##
## 540 samples
##  16 predictors
##   2 classes: 'no', 'yes'
##
## No pre-processing
## Resampling: Cross-Validated (10 fold)
## Summary of sample sizes: 487, 486, 486, 486, 486, 486, ...
## Resampling results across tuning parameters:
##
##   k   Accuracy   Kappa
##    2  0.6000095  0.1213920
##    3  0.6368757  0.1542968
##    4  0.6424325  0.1546494
##    5  0.6386252  0.1275248
##    6  0.6329998  0.1164253
##    7  0.6589619  0.1616377
##    8  0.6663344  0.1774391
##    9  0.6663681  0.1733197
##   10  0.6609510  0.1566064
##   11  0.6664018  0.1575868
##   12  0.6682199  0.1669053
##   13  0.6572111  0.1397222
##   14  0.6719586  0.1694953
##   15  0.6571425  0.1263937
##   16  0.6664367  0.1551023
##   17  0.6719573  0.1588789
##   18  0.6608811  0.1260452
##   19  0.6590979  0.1165734
##   20  0.6609510  0.1219624
##
## Accuracy was used to select the optimal model using  the largest value.
## The final value used for the model was k = 14.``````

R recommends that k = 16. This is based on a combination of accuracy and the kappa statistic. The kappa statistic is a measurement of the accuracy of a model while taking into account chance. We don’t have a model in the sense that we do not use the ~ sign like we do with regression. Instead, we have a train and a test set a factor variable and a number for k. This will make more sense when you see the code. Finally, we will use this information on our test dataset. We will then look at the table and the accuracy of the model.

``````knn.test<-knn(train[,-17],test[,-17],train[,17],k=16) #-17 removes the dependent variable 'city
table(knn.test,test\$city)``````
``````##
## knn.test  no yes
##      no   19   8
##      yes  61 125``````
``````prob.agree<-(15+129)/213
prob.agree``````
``##  0.6760563``

Accuracy is 67% which is consistent with what we found when determining the k. We can also calculate the kappa. This done by calculating the probability and then do some subtraction and division. We already know the accuracy as we stored it in the variable “prob.agree” we now need the probability that this is by chance. Lastly, we calculate the kappa.

``````prob.chance<-((15+4)/213)*((15+65)/213)
kap<-(prob.agree-prob.chance)/(1-prob.chance)
kap``````
``##  0.664827``

A kappa of .66 is actually good.

The example we just did was with unweighted k neighbors. There are times when weighted neighbors can improve accuracy. We will look at three different weighing methods. “Rectangular” is unweighted and is the one that we used. The other two are “triangular” and “epanechnikov”. How these calculate the weights is beyond the scope of this post. In the code below the argument “distance” can be set to 1 for Euclidean and 2 for absolute distance.

``````kknn.train<-train.kknn(city~.,train,kmax = 25,distance = 2,kernel = c("rectangular","triangular",
"epanechnikov"))
plot(kknn.train)`````` ``kknn.train``
``````##
## Call:
## train.kknn(formula = city ~ ., data = train, kmax = 25, distance = 2,     kernel = c("rectangular", "triangular", "epanechnikov"))
##
## Type of response variable: nominal
## Minimal misclassification: 0.3277778
## Best kernel: rectangular
## Best k: 14``````

If you look at the plot you can see which value of k is the best by looking at the point that is the lowest on the graph which is right before 15. Looking at the legend it indicates that the point is the “rectangular” estimate which is the same as unweighted. This means that the best classification is unweighted with a k of 14. Although it recommends a different value for k our misclassification was about the same.

Conclusion

In this post, we explored both weighted and unweighted KNN. This algorithm allows you to deal with data that does not meet the assumptions of regression by ignoring the need for parameters. However, because there are no numbers really attached to the results beyond accuracy it can be difficult to explain what is happening in the model to people. As such, perhaps the biggest drawback is communicating results when using KNN.

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