logit,GBM,knn,xgboost对一个信用卡数据的实现

Download the data

dataset = read.table("http://archive.ics.uci.edu/ml/machine-learning-databases/credit-screening/crx.data",
           sep = ",",
           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(0.7n))
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 = dataset
library(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

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(0.7n))
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 = 2:20
set.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 = 1:20
result =
  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