我想用随机森林为无监督学习准备数据。 程序如下:
最后它看起来像这样:
... Class
|1
Original |1
Data |1
|1
--------------
|2
Synthetic |2
Data |2
|2
我的R代码如下所示:
library(gtools) #for smartbind()
sample1 <- function(X) { sample(X, replace=T) }
g1 <- function(dat) { apply(dat,2,sample1) }
data$class <- rep(1, times=nrow(data)) #add attribute 'class' with value 1
synthData<-data.frame(g1(data[,1:ncol(data)])) #generate synthetic data with sampling from data
synthData$class <- rep(2, times=nrow(synthData)) #attribute 'class' is 2
colnames(synthData) <- colnames(data)
newData <- smartbind(data, synthData) #bind the data together
很明显,我是R的新手,但它确实有效 - 只有一个问题:合成数据中的属性类型与原始数据中的属性类型不同。如果原来他们是nums,现在他们成为因素。如何在生成合成数据时保留相同类型?
谢谢!
Data1(nums成为因素):
结构(列表(V2 = c(1.51793,1.51711,1.51645,1.51916,1.51131) ),V3 = c(13.21,12.89,13.44,14.15,13.69),V4 = c(3.48,3.62, 3.61,0,3.2),V5 = c(1.41,1.57,1.54,2.09,1.81),V6 = c(72.64, 72.96,72.39,72.74,72.81),V7 = c(0.59,0.61,0.66,0,1.76 ),V8 = c(8.43,8.11,8.03,10.88,5.43),V9 = c(0,0,0,0, 1.19),V10 = c(0,0,0,0,0),realClass =结构(c(1L,2L, 2L,5L,6L),. Label = c(“1”,“2”,“3”,“5”,“6”,“7”),class =“factor”)),。Name = c( “V2”, “V3”,“V4”,“V5”,“V6”,“V7”,“V8”,“V9”,“V10”,“realClass”),row.names = c(27L, 138L,77L,183L,186L),class =“data.frame”)
Data2(因子变为chrs):
结构(列表(realClass =结构(c(2L,2L,2L,1L,2L),. Label = c(“e”, “p”),class =“factor”),V2 =结构(c(6L,3L,4L,6L,6L),. Label = c(“b”, “c”,“f”,“k”,“s”,“x”),class =“factor”),V3 =结构(c(4L, 4L,3L,1L,1L),. Label = c(“f”,“g”,“s”,“y”),class =“factor”), V4 =结构(c(5L,5L,5L,3L,4L),。Label = c(“b”,“c”, “e”,“g”,“n”,“p”,“r”,“u”,“w”,“y”),class =“factor”), V5 =结构(c(1L,1L,1L,2L,1L),。Label = c(“f”,“t” ),class =“factor”),V6 =结构(c(3L,9L,3L,6L,3L) ),。Label = c(“a”,“c”,“f”,“l”,“m”,“n”,“p”,“s”,“y” ),class =“factor”),V7 =结构(c(2L,2L,2L,2L,2L) ),。Label = c(“a”,“f”),class =“factor”),V8 =结构(c(1L, 1L,1L,1L,1L),. Label = c(“c”,“w”),class =“factor”), V9 =结构(c(2L,2L,2L,1L,1L),。Label = c(“b”,“n” ),class =“factor”),V10 =结构(c(1L,1L,1L,10L, 4L),. Label = c(“b”,“e”,“g”,“h”,“k”,“n”,“o”,“p”,“r”, “u”,“w”,“y”),class =“factor”),V11 =结构(c(2L, 2L,2L,2L,1L),. Label = c(“e”,“t”),class =“factor”), V12 =结构(c(NA,NA,NA,1L,1L),. Label = c(“b”,“c”, “e”,“r”),class =“factor”),V13 =结构(c(3L,2L,3L, 3L,2L),. Label = c(“f”,“k”,“s”,“y”),class =“factor”), V14 =结构(c(3L,3L,2L,3L,2L),. Label = c(“f”,“k”, “s”,“y”),class =“factor”),V15 =结构(c(7L,8L,7L, 4L,7L),. Label = c(“b”,“c”,“e”,“g”,“n”,“o”,“p”,“w”, “y”),class =“factor”),V16 =结构(c(7L,7L,8L,4L, 1L),. Label = c(“b”,“c”,“e”,“g”,“n”,“o”,“p”,“w”,“y” ),class =“factor”),V17 =结构(c(1L,1L,1L,1L,1L) ),。Label =“p”,class =“factor”),V18 =结构(c(3L, 3L,3L,3L,3L),. Label = c(“n”,“o”,“w”,“y”),class =“factor”), V19 =结构(c(2L,2L,2L,2L,2L),. Label = c(“n”,“o”, “t”),class =“factor”),V20 =结构(c(1L,1L,1L,5L, 3L),. Label = c(“e”,“f”,“l”,“n”,“p”),class =“factor”), V21 =结构(c(8L,8L,8L,4L,2L),. Label = c(“b”,“h”, “k”,“n”,“o”,“r”,“u”,“w”,“y”),class =“factor”),V22 =结构(c(5L, 5L,5L,5L,6L),. Label = c(“a”,“c”,“n”,“s”,“v”,“y”),class =“factor”), V23 =结构(c(3L,3L,5L,1L,2L),. Label = c(“d”,“g”, “l”,“m”,“p”,“u”,“w”),class =“factor”)),. Name = c(“realClass”, “V2”,“V3”,“V4”,“V5”,“V6”,“V7”,“V8”,“V9”,“V10”,“V11”, “V12”,“V13”,“V14”,“V15”,“V16”,“V17”,“V18”,“V19”,“V20”, “V21”,“V22”,“V23”),row.names = c(4105L,6207L,6696L,2736L, 3756L),class =“data.frame”)
答案 0 :(得分:2)
您始终可以使用此技巧来创建数字列
numcol <- as.numeric(as.character(factcol))
但我怀疑你的data.frame中有因子变量。
由于apply
会返回一个矩阵,如果您的数据中有一个因子,那么所有数字变量也会被强制为因子。
以下是使用玩具数据集
的示例set.seed(123)
toydat <- data.frame(A = 1:10, B = rnorm(10), C = LETTERS[1:10])
str(toydat)
## 'data.frame': 10 obs. of 3 variables:
## $ A: int 1 2 3 4 5 6 7 8 9 10
## $ B: num -0.5605 -0.2302 1.5587 0.0705 0.1293 ...
## $ C: Factor w/ 10 levels "A","B","C","D",..: 1 2 3 4 5 6 7 8 9 10
set.seed(1)
str(data.frame(apply(toydat[,1:2], 2, sample, replace = TRUE)))
## 'data.frame': 10 obs. of 2 variables:
## $ A: num 3 4 6 10 3 9 10 7 7 1
## $ B: num 1.5587 -0.2302 0.4609 0.0705 -1.2651 ...
# with the factor column C
set.seed(2)
str(data.frame(apply(toydat[,1:3], 2, sample, replace = TRUE)))
## 'data.frame': 10 obs. of 3 variables:
## $ A: Factor w/ 6 levels "10"," 2"," 5",..: 2 5 4 2 1 1 2 6 3 4
## $ B: Factor w/ 8 levels " 0.129288","-0.230177",..: 8 7 6 2 1 5 3 7 1 4
## $ C: Factor w/ 6 levels "B","D","E","G",..: 4 2 5 1 2 3 1 2 6 1
这是plyr
包变得有用的地方,因为你可以控制输出(使用** ply)。但在这种情况下,colwise
功能就足够了
require(plyr)
set.seed(2)
mysamplingfun <- colwise(function(x) sample(x, replace = TRUE))
str(mysamplingfun(toydat[,1:3]))
## 'data.frame': 10 obs. of 3 variables:
## $ A: int 2 8 6 2 10 10 2 9 5 6
## $ B: num 1.715 1.559 -1.265 -0.23 0.129 ...
## $ C: Factor w/ 10 levels "A","B","C","D",..: 7 4 9 2 4 5 2 4 10 2