用R语言对一个信用卡数据实现logit,GBM,knn,xgboost

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Prepare the data

数据来自UCIhttp://archive.ics.uci.edu/ml/machine-learning-databases/credit-screening,一个信a用卡的数据,具体各项变量名以及变量名代表的含义不明(应该是出于保护隐私的目的),本文会用logit,GBM,knn,xgboost来对数据进行分类预测,对比准确率

预计的准确率应该是:

xgboost > GBM > logit > knn

Download the data

dataset = read.table("http://archive.ics.uci.edu/ml/machine-learning-databases/credit-screening/crx.data", sep = ",", essay-header = F, na.strings = "?")

head(dataset)

  V1    V2    V3 V4 V5 V6 V7   V8 V9 V10 V11 V12 V13 V14 V15 V16
1  b 30.83 0.000  u  g  w  v 1.25  t   t   1   f   g 202   0   +
2  a 58.67 4.460  u  g  q  h 3.04  t   t   6   f   g  43 560   +
3  a 24.50 0.500  u  g  q  h 1.50  t   f   0   f   g 280 824   +
4  b 27.83 1.540  u  g  w  v 3.75  t   t   5   t   g 100   3   +
5  b 20.17 5.625  u  g  w  v 1.71  t   f   0   f   s 120   0   +
6  b 32.08 4.000  u  g  m  v 2.50  t   f   0   t   g 360   0   +

## save.csv(dataset,file = "creditCard.csv")

以上是数据的形式,接下来看下数据是否有缺失值和各个数据的类型

sapply(dataset,function(x) sum(is.na(x)))

 V1  V2  V3  V4  V5  V6  V7  V8  V9 V10 V11 V12 V13 V14 V15 V16 
 12  12   0   6   6   9   9   0   0   0   0   0   0  13   0   0 

sapply(dataset,class)

       V1        V2        V3        V4        V5        V6        V7        V8        V9       V10 
 "factor" "numeric" "numeric"  "factor"  "factor"  "factor"  "factor" "numeric"  "factor"  "factor" 
      V11       V12       V13       V14       V15       V16 
"integer"  "factor"  "factor" "integer" "integer"  "factor" 

Train and Test

分割数据的训练集和测试集,这里set.seed(123),设定70%的训练集,30%的测试集.

set.seed(123)dataset = na.omit(dataset)n = dim(dataset)[1]index = sample(n,round(0.7*n))train = dataset[index,]test = dataset[-index,]dim(train)

[1] 457  16

dim(test)

[1] 196  16

Change the variable into dummy variables

有时候,需要转化变量为哑变量,因为在一些挖掘场合,数据不能直接使用因子型的数据:

• knn

• glmnet

• svm

• xgboost

有些挖掘方法是可以使用因子变量的,比如:

• logistic regression

• raprt

• GBM

• randomforest

dataset2 = datasetlibrary(plyr)into_factor = function(x){
  
  if(class(x) == "factor"){
  n = length(x)
  data.fac = data.frame(x = x,y = 1:n)
  output = model.matrix(y~x,data.fac)[,-1]
  ## Convert factor into dummy variable matrix
  }else{
    output = x
  ## if x is numeric, output is x
  }
  output
  }into_factor(dataset$V4)[1:5,]

  xu xy
1  1  0
2  1  0
3  1  0
4  1  0
5  1  0

dataset2 = colwise(into_factor)(dataset2)dataset2 = do.call(cbind,dataset2)dataset2 = as.data.frame(dataset2)head(dataset2)

  V1    V2    V3 xu xy xgg xp xc xcc xd xe xff xi xj xk xm xq xr xw xx xdd xff xh xj xn xo xv xz
1  1 30.83 0.000  1  0   0  0  0   0  0  0   0  0  0  0  0  0  0  1  0   0   0  0  0  0  0  1  0
2  0 58.67 4.460  1  0   0  0  0   0  0  0   0  0  0  0  0  1  0  0  0   0   0  1  0  0  0  0  0
3  0 24.50 0.500  1  0   0  0  0   0  0  0   0  0  0  0  0  1  0  0  0   0   0  1  0  0  0  0  0
4  1 27.83 1.540  1  0   0  0  0   0  0  0   0  0  0  0  0  0  0  1  0   0   0  0  0  0  0  1  0
5  1 20.17 5.625  1  0   0  0  0   0  0  0   0  0  0  0  0  0  0  1  0   0   0  0  0  0  0  1  0
6  1 32.08 4.000  1  0   0  0  0   0  0  0   0  0  0  0  1  0  0  0  0   0   0  0  0  0  0  1  0
    V8 V9 V10 V11 V12 xp xs V14 V15 V16
1 1.25  1   1   1   0  0  0 202   0   1
2 3.04  1   1   6   0  0  0  43 560   1
3 1.50  1   0   0   0  0  0 280 824   1
4 3.75  1   1   5   1  0  0 100   3   1
5 1.71  1   0   0   0  0  1 120   0   1
6 2.50  1   0   0   1  0  0 360   0   1

dim(dataset2)

[1] 653  38

Logistic Regression

使用logistic回归来进行测试建模和预测,使用的函数是glm

logit.model = glm(V16~.,data = train,family = "binomial")logit.response = predict(logit.model,test,type = "response")logit.predict = ifelse(logit.response>0.5,"+","-")table(logit.predict,test$V16)

             logit.predict  -  +
            - 90 24
            + 13 69

accurancy1 = mean(logit.predict == test$V16)accurancy1

[1] 0.81122

GBM

使用GBM方法来进行预测,这里用的是caret,repeat-cv来选择最优树

library(caret)

ctrl = trainControl(method = "repeatedcv", number = 5, repeats = 5)set.seed(300)m_gbm = train(V16 ~ ., data=train, method = "gbm",  metric = "Kappa", trControl = ctrl)

gbm.predict = predict(m_gbm,test)table(gbm.predict,test$V16)

accurancy2 = mean(gbm.predict == test$V16)accurancy2

[1] 0.85714

knn method for classification

knn set k = 5

This is a model without cross-validation

首先测试一个knn模型,不做CV,不做标准化,不做数据类型转换得到的结果,这里,不转换数据类型会把因子类型的变量舍弃,仅保留数值变量

library(caret)knn.model1 = knn3(V16 ~ .,data = train, k = 5) 
 knn.response1 = predict(knn.model1,test,class = "response") 
 knn.predict1 = ifelse(knn.response1[,1]<0.5,"+","-")

table(knn.predict1,test$V16)

            knn.predict1  -  +
           - 78 48
           + 25 45

mean(knn.predict1 == test$V16)

[1] 0.62755

knn after scale

After scaling and convert into dummy variables:

经过标准化和数据转换之后的准确率:

knn.dataset = cbind(
                colwise(scale)(dataset2[,-38]),V16 = as.factor(dataset2$V16)
                )

set.seed(123)

index = sample(n,round(0.7*n))

train.knn = knn.dataset[index,]

test.knn = knn.dataset[-index,]

knn.model1 = knn3(V16 ~ .,data = train.knn, k = 5)

  knn.predict1 = predict(knn.model1,test.knn,,type = "class") table(knn.predict1,test.knn$V16)

            knn.predict1  0  1
           0 89 32
           1 14 61

mean(knn.predict1 == test.knn$V16)

[1] 0.76531

knn CV for k

my-try

不管是我的这个程序函数caret,总算出来应该是k=2的时候误差最小,但是实际情况不是这样

library(class)cv.knn = function(data,n=5,k){
  index = sample(1:5,nrow(data),replace = T)
  acc=0
  for ( i in 1:5){
    ind = index == i
    train = data[-ind,]
    test = data[ind,]
    knn.model1 = knn3(V16 ~ .,data = train, k = k)  
    knn.predict= predict(knn.model1,test,type = "class") 
    acc[i] = mean(knn.predict == test$V16)
  }
    mean(acc)}cv.knn(train.knn,3,5)

[1] 0.8533

k = 2:20set.seed(123)acc = sapply(k,function(x) cv.knn(train.knn,3,x))plot(k,acc,type = "b")

k.final = which.max(acc)knn.model.f = knn3(V16 ~ .,data = train.knn, k = k.final)  knn.predict.f = predict(knn.model.f,test.knn,type = "class") 
table(knn.predict.f,test.knn$V16)

             knn.predict.f  0  1
            0 81 31
            1 22 62

mean(knn.predict.f == test.knn$V16)

[1] 0.72959

library(caret)

fitControl <- trainControl(method = "cv", number = 10)

knnTune <- train(x = dataset2[1:37], y = dataset2[,38], method = "knn", preProc = c("center", "scale"),tuneGrid = data.frame(.k = 1:20), trControl = fitControl)

直接train,test来看:

效果是k=5最好

knn_train_test = function(train,test,k =5){
    knn.model.f = knn3(V16 ~ .,data = train, k = k)  
    knn.predict.f = predict(knn.model.f,test,type = "class") 
    mean(knn.predict.f == test$V16)}x = 1:20result = 
  sapply(x,         function(x) knn_train_test(train.knn,test.knn,k = x))  plot(x,result,type = "b")

k.final = which.max(result)accurancy3 = knn_train_test(train.knn,test.knn,k = k.final)accurancy3

[1] 0.75

xgboost

Install:

##　devtools::install_github(‘dmlc/xgboost‘,subdir=‘R-package‘)

require(xgboost)

require(methods)

require(plyr)

set.seed(123)

set.seed(123)

index = sample(n,round(0.7*n))

train.xg = dataset2[index,]

test.xg = dataset2[-index,]

label <- as.matrix(train.xg[,38,drop =F])

data <- as.matrix(train.xg[,-38,drop =F])

data2 <-  as.matrix(test.xg[,-38,drop =F])

label2 =  as.matrix(test.xg[,38,drop =F])

# weight <- as.numeric(dtrain[[32]]) * testsize / length(label)

xgmat <- xgb.DMatrix(data, label = label, missing = -10000)

param <- list("objective" = "binary:logistic","bst:eta" = 1,"bst:max_depth" = 2,"eval_metric" = "logloss","silent" = 1,"nthread" = 16 ,"min_child_weight" =1.45)

nround =275

bst = xgb.train(param, xgmat, nround )

res1 = predict(bst,data2)

pre1 = ifelse(res1>0.5,1,0)

table(pre1,label2)

    label2
pre1  0  1
   0 91 15
   1 12 78

accurancy4 = mean(pre1 ==label2)

accurancy4 

[1] 0.86224

Final Results

用R语言对一个信用卡数据实现logit,GBM,knn,xgboost

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原文地址：http://www.cnblogs.com/payton/p/5340538.html