PROBLEM STATEMENT

Payday Loans are high risk short term lending financial products and its very important to asses risk of payment default. Use dataset “paydayloan_collections.csv” to build a model whether repayment will be successful or not.

AIM: Break the dataset into test and train data.Use RandomForest and DecisionTress to build your model on train data and compare their performance on test data.

Initial Setup for using decision tree and random forest model:

library(tree)
## Warning: package 'tree' was built under R version 3.3.3
library(ISLR)
## Warning: package 'ISLR' was built under R version 3.3.3
library(randomForest)
## Warning: package 'randomForest' was built under R version 3.3.3
## randomForest 4.6-12
## Type rfNews() to see new features/changes/bug fixes.
library(dplyr)
## 
## Attaching package: 'dplyr'
## The following object is masked from 'package:randomForest':
## 
##     combine
## The following objects are masked from 'package:stats':
## 
##     filter, lag
## The following objects are masked from 'package:base':
## 
##     intersect, setdiff, setequal, union
loan=read.csv("paydayloan_collections.csv")
View(loan)
str(loan) #VERY IMP TO NOTE THAT PAYMENT IS OF FACTOR TYPE...WHILE REST OTHERS ARE ALSO OF FACTOR N NUM TYPE.
## 'data.frame':    30000 obs. of  31 variables:
##  $ payment: Factor w/ 2 levels "Denied","Success": 2 1 1 2 2 2 2 2 1 1 ...
##  $ var1   : Factor w/ 4 levels "kq","ma","qw",..: 3 3 3 4 2 1 4 3 4 3 ...
##  $ var2   : Factor w/ 9 levels "bq","hk","js",..: 2 7 9 3 8 5 5 7 7 7 ...
##  $ var3   : num  3.11 3.35 4.15 6.23 1.28 -2.45 1.05 5.41 7.29 3.13 ...
##  $ var4   : num  16.1 11.2 29.2 15.7 20.7 ...
##  $ var5   : num  -4.6 -18.55 18.91 2.81 14.98 ...
##  $ var6   : num  22.34 6.68 16.4 4.46 11.19 ...
##  $ var7   : num  13.53 12.78 3.67 5.13 17.66 ...
##  $ var8   : num  1.53 6.62 5.72 8.66 1.13 ...
##  $ var9   : Factor w/ 3 levels "ch","ja","nv": 3 3 1 2 3 1 1 1 1 1 ...
##  $ var10  : Factor w/ 3 levels "db","ld","pe": 2 2 2 2 2 1 2 2 1 2 ...
##  $ var11  : Factor w/ 2 levels "rl","te": 1 2 2 2 2 2 2 2 2 1 ...
##  $ var12  : num  4.46 4.04 -4.41 2.14 14.93 ...
##  $ var13  : Factor w/ 7 levels "cb","iz","kh",..: 1 3 1 2 6 2 1 1 3 6 ...
##  $ var14  : num  4.93 -0.76 1.21 3.56 2.2 0.37 -0.84 5.85 5.23 0.62 ...
##  $ var15  : num  26.5 16.2 29.1 23.6 -19.2 ...
##  $ var16  : num  10.48 -0.87 5.49 15.34 -3 ...
##  $ var17  : Factor w/ 5 levels "az","bw","ki",..: 4 3 2 2 3 3 5 2 2 2 ...
##  $ var18  : num  11.2 15.5 -10.8 -24.3 12 ...
##  $ var19  : Factor w/ 7 levels "bz","ev","fh",..: 2 7 5 2 5 2 2 7 6 5 ...
##  $ var20  : num  19.1 28.4 1.6 -11.9 18.5 ...
##  $ var21  : num  8.94 31.02 23.26 29.25 2.19 ...
##  $ var22  : num  -12.76 34.76 9.5 -1.53 10.24 ...
##  $ var23  : Factor w/ 10 levels "cz","da","fe",..: 9 1 6 6 9 6 8 9 8 8 ...
##  $ var24  : num  12.06 1.44 7.77 8.94 8.92 ...
##  $ var25  : num  2.46 9.44 8.7 19.33 5.48 ...
##  $ var26  : num  4.73 13.56 -1.75 23.73 -0.28 ...
##  $ var27  : num  -1.72 -2.24 5.96 5.54 4.01 6.65 1.22 9.08 0.87 -2.54 ...
##  $ var28  : num  0.91 0.24 1.91 0.85 1.21 -1.12 1.26 0.68 1.6 -1.43 ...
##  $ var29  : Factor w/ 2 levels "dg","ev": 2 2 2 2 2 2 2 2 1 2 ...
##  $ var30  : num  8 -2.9 22.7 36.3 11.3 ...
glimpse(loan)
## Observations: 30,000
## Variables: 31
## $ payment <fctr> Success, Denied, Denied, Success, Success, Success, S...
## $ var1    <fctr> qw, qw, qw, wv, ma, kq, wv, qw, wv, qw, qw, qw, wv, w...
## $ var2    <fctr> hk, rv, zg, js, xn, py, py, rv, rv, rv, py, bq, py, p...
## $ var3    <dbl> 3.11, 3.35, 4.15, 6.23, 1.28, -2.45, 1.05, 5.41, 7.29,...
## $ var4    <dbl> 16.06, 11.18, 29.19, 15.70, 20.71, 22.45, 23.02, 17.92...
## $ var5    <dbl> -4.60, -18.55, 18.91, 2.81, 14.98, 15.18, 17.59, -14.5...
## $ var6    <dbl> 22.34, 6.68, 16.40, 4.46, 11.19, -2.12, 6.65, 5.00, 13...
## $ var7    <dbl> 13.53, 12.78, 3.67, 5.13, 17.66, -8.24, -2.06, 1.34, 2...
## $ var8    <dbl> 1.53, 6.62, 5.72, 8.66, 1.13, 10.34, 12.20, -8.54, 4.4...
## $ var9    <fctr> nv, nv, ch, ja, nv, ch, ch, ch, ch, ch, ch, ch, ch, n...
## $ var10   <fctr> ld, ld, ld, ld, ld, db, ld, ld, db, ld, ld, ld, db, l...
## $ var11   <fctr> rl, te, te, te, te, te, te, te, te, rl, te, rl, te, t...
## $ var12   <dbl> 4.46, 4.04, -4.41, 2.14, 14.93, 8.64, 14.91, 13.97, 15...
## $ var13   <fctr> cb, kh, cb, iz, te, iz, cb, cb, kh, te, iz, np, te, i...
## $ var14   <dbl> 4.93, -0.76, 1.21, 3.56, 2.20, 0.37, -0.84, 5.85, 5.23...
## $ var15   <dbl> 26.48, 16.21, 29.06, 23.61, -19.16, 35.83, -8.38, 28.9...
## $ var16   <dbl> 10.48, -0.87, 5.49, 15.34, -3.00, 8.05, 15.66, 15.06, ...
## $ var17   <fctr> ov, ki, bw, bw, ki, ki, zk, bw, bw, bw, bw, bw, ki, z...
## $ var18   <dbl> 11.17, 15.50, -10.84, -24.26, 12.02, 32.12, 69.45, 15....
## $ var19   <fctr> ev, tg, me, ev, me, ev, ev, tg, qu, me, ev, ev, me, e...
## $ var20   <dbl> 19.15, 28.39, 1.60, -11.89, 18.47, 9.94, 27.55, 17.71,...
## $ var21   <dbl> 8.94, 31.02, 23.26, 29.25, 2.19, -0.26, 14.39, 4.09, 1...
## $ var22   <dbl> -12.76, 34.76, 9.50, -1.53, 10.24, 8.11, -4.73, 22.62,...
## $ var23   <fctr> ub, cz, ri, ri, ub, ri, tf, ub, tf, tf, tf, tf, ub, u...
## $ var24   <dbl> 12.06, 1.44, 7.77, 8.94, 8.92, 2.32, 8.57, 2.82, 15.00...
## $ var25   <dbl> 2.46, 9.44, 8.70, 19.33, 5.48, 6.89, 8.73, -11.05, 7.4...
## $ var26   <dbl> 4.73, 13.56, -1.75, 23.73, -0.28, -3.51, 2.87, 11.94, ...
## $ var27   <dbl> -1.72, -2.24, 5.96, 5.54, 4.01, 6.65, 1.22, 9.08, 0.87...
## $ var28   <dbl> 0.91, 0.24, 1.91, 0.85, 1.21, -1.12, 1.26, 0.68, 1.60,...
## $ var29   <fctr> ev, ev, ev, ev, ev, ev, ev, ev, dg, ev, dg, dg, ev, d...
## $ var30   <dbl> 8.00, -2.90, 22.67, 36.31, 11.33, 9.93, 21.75, 7.53, -...
table(loan$payment)
## 
##  Denied Success 
##   18755   11245

the reponse variable is of classification type.ie.payment=denied/sucess lets make decision tree first and then random forest method

Step 1:divide the data into train and test(50:50%)

set.seed(2)
s=sample(1:nrow(loan),15000)
loan_train=loan[s,]
loan_test=loan[-s,]

step 2: making tree model for train data

tree.loan=tree(payment~.,data=loan_train)

lets plot it

plot(tree.loan) #model with 7 nodes.
text(tree.loan,pretty=0)

tree.loan #text format output
## node), split, n, deviance, yval, (yprob)
##       * denotes terminal node
## 
##  1) root 15000 19790.0 Denied ( 0.62847 0.37153 )  
##    2) var25 < 9.105 7041  9332.0 Success ( 0.37722 0.62278 )  
##      4) var17: bw 3197  4338.0 Denied ( 0.58555 0.41445 )  
##        8) var25 < 1.345 1355  1661.0 Success ( 0.30258 0.69742 ) *
##        9) var25 > 1.345 1842  1875.0 Denied ( 0.79370 0.20630 ) *
##      5) var17: az,ki,ov,zk 3844  3889.0 Success ( 0.20395 0.79605 ) *
##    3) var25 > 9.105 7959  6708.0 Denied ( 0.85074 0.14926 )  
##      6) var25 < 15.195 3510  3703.0 Denied ( 0.77949 0.22051 )  
##       12) var23: cz,da,fe,po,qu,ri,sy,tf 2778  2408.0 Denied ( 0.84377 0.15623 ) *
##       13) var23: ub,yv 732  1011.0 Denied ( 0.53552 0.46448 )  
##         26) var17: bw 327   262.0 Denied ( 0.86239 0.13761 ) *
##         27) var17: az,ki,ov,zk 405   473.7 Success ( 0.27160 0.72840 ) *
##      7) var25 > 15.195 4449  2754.0 Denied ( 0.90695 0.09305 ) *
summary(tree.loan)
## 
## Classification tree:
## tree(formula = payment ~ ., data = loan_train)
## Variables actually used in tree construction:
## [1] "var25" "var17" "var23"
## Number of terminal nodes:  7 
## Residual mean deviance:  0.8886 = 13320 / 14990 
## Misclassification error rate: 0.1718 = 2577 / 15000

output:for train dataset having 7 nodes: residual mean dev=0.88 error rate=0.1718

step 3:predict for test dataset

tree.pred=predict(tree.loan,newdata = loan_test,type ="class")

step 4:check the predictions made on test data and calculate error.

table(tree.pred,loan_test$payment)
##          
## tree.pred Denied Success
##   Denied    8076    1373
##   Success   1252    4299
(1252+1373)/15000
## [1] 0.175

error on test data=0.175

lets find error on train data:

tree.pred.train=predict(tree.loan,newdata = loan_train,type ="class")
table(tree.pred.train,loan_train$payment)
##                
## tree.pred.train Denied Success
##         Denied    8123    1273
##         Success   1304    4300
(1304+1273)/15000
## [1] 0.1718

CONCLUSION:ERROR FOUND ON TRAIN DATA IS 0.1718 WHILE ON TEST DATA 0.175.(GOOD MODEL)

Step 5: STILL LETS TRY AND DO PRUNING FOR OPTIMUM MODEL OF TREE. WE HERE USE CROSS VALIDATION FUNCTION AND PRUNE FUNCTION
r set.seed(2) cv.loan=cv.tree(tree.loan,FUN = prune.misclass)
lets plot the pruned tree
r plot(cv.loan$size,cv.loan$dev,type="b")
plot shows that error is minimum on 7 nodes ONLY where deviance is minimum
Hence the tree itself is best size with 7 nodes which already exist.

Lets do random forest method

Here response is clasification type(payment=success/denied)

step 1:build rf model on train dataset.

rf.loan=randomForest(payment~.,data = loan_train)#U CAN USE ARGUMENT do.trace=T
rf.loan
## 
## Call:
##  randomForest(formula = payment ~ ., data = loan_train) 
##                Type of random forest: classification
##                      Number of trees: 500
## No. of variables tried at each split: 5
## 
##         OOB estimate of  error rate: 13.43%
## Confusion matrix:
##         Denied Success class.error
## Denied    8537     890  0.09440967
## Success   1125    4448  0.20186614

output: interpretatn:ntree=500 no. of variables tried at each split=5 OOB estimate of error rate=13.43% i.e.(1125+890)/15000=13.43%

step 2:predicting rf model created on test dataset and find error

rf.loan.pred=predict(rf.loan,newdata = loan_test)
t=table(loan_test$payment,rf.loan.pred)
t
##          rf.loan.pred
##           Denied Success
##   Denied    8496     832
##   Success   1149    4523

interpretation:

(1149+832)/15000
## [1] 0.1320667

error=13.20% by random forest model.

step 3:Plot variablesImpPlot(i.e. variableswhich are important):

importance(rf.loan)
##       MeanDecreaseGini
## var1          40.32600
## var2         128.59517
## var3         168.25258
## var4         167.97693
## var5         163.46701
## var6         163.03284
## var7         252.18814
## var8         165.88244
## var9          25.67926
## var10         26.78148
## var11         12.40307
## var12        160.85998
## var13         98.65424
## var14        164.22960
## var15        168.69391
## var16        340.11406
## var17        517.84024
## var18        167.12299
## var19         96.80594
## var20        162.52858
## var21        168.95544
## var22        168.19927
## var23        265.28208
## var24        168.26335
## var25       2360.36785
## var26        157.98775
## var27        162.64079
## var28        154.84962
## var29         39.65420
## var30        167.07499
varImpPlot(rf.loan)

variables 25,17,16,23 are most imp variables contributing to the output.

Conclusion:error generated on PREDICTING TEST data by random forest is 13.20% as against 17.5% by decision tree model.

Hence random forest is better than decision tree model .