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我想为随机森林的无监督学习准备数据。程序如下：

  • 获取数据并将值为 1 的属性“类”添加到所有示例
  • 从原始数据生成合成数据：
    • 虽然您没有与原始数据构建示例相同数量的示例：
      • 从原始数据中该属性的所有值中采样新的属性值
      • 对所有属性执行此操作并将它们组合到新示例中
  • 分配给合成数据值 2 的属性“类”
  • 将两个数据绑定在一起

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，我对 R 真的很陌生，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不同。如果原来它们是 nums，那么现在它们变成了因数。如何在生成合成数据时保留相同的类型？

谢谢！

Data1（数字成为因素）：

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 = structure(c(1L, 2L, 2L) , 5L, 6L), .Label = c("1", "2", "3", "5", "6", "7"), class = "factor")), .Names = c(" V2”、“V3”、“V4”、“V5”、“V6”、“V7”、“V8”、“V9”、“V10”、“realClass”），row.names = c(27L, 138L, 77L, 183L, 186L), 类 = "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”），类=“因子”），V2 =结构（c（6L， 3L, 4L, 6L, 6L), .Label = c("b", "c", "f", "k", "s", "x"), class = "factor"), V3 = structure( c(4L, 4L, 3L, 1L, 1L), .Label = c("f", "g", "s", "y"), class = "factor"), V4 = structure(c(5L, 5L, 5L, 3L, 4L), .Label = c("b", "c", "e", "g", "n", "p", "r", "u", "w", "y"), class = "factor"), V5 = structure(c(1L, 1L, 1L, 2L, 1L), .Label = c("f", "t"), class = "factor"), V6 = 结构(c(3L, 9L, 3L, 6L, 3L), .标签 = 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 = structure(c(1L, 1L, 1L, 1L, 1L), .Label = c("c", "w"), class = "factor"), V9 = structure(c(2L, 2L, 2L, 1L, 1L), .Label = c("b", "n"), class = "factor"), V10 = structure(c(1L, 1L, 1L, 10L, 4L), .Label = c("b", "e", "g", "h", “k”，“n”，“o”，“p”，“r”，“u”，“w”，“y”），类 =“因子”），V11 = 结构（c（2L，2L， 2L, 2L, 1L), .标签 = c("e", "t"), class = "factor"), V12 = structure(c(NA, NA, NA, 1L, 1L), .Label = c("b", "c", "e", "r"), class = "factor"), V13 = structure(c(3L, 2L, 3L, 3L, 2L), .Label = c("f", "k", "s", "y"), class = "factor"), V14 = structure(c(3L, 3L, 2L, 3L, 2L), .Label = c("f", "k", "s", "y") , class = "factor"), V15 = structure(c(7L, 8L, 7L, 4L, 7L), .Label = c("b", "c", "e", "g", "n", “o”、“p”、“w”、“y”）、类 =“因子”）、V16 = 结构（c（7L、7L、8L、4L、1L），.Label = c（“b”， “c”、“e”、“g”、“n”、“o”、“p”、“w”、“y”），类 = “因子”），V17 = 结构（c（1L，1L，1L，1L，1L），.标签 = “p”，类 = “因子”），V18 = 结构（c（3L，3L，3L，3L，3L），.Label = c（“n”，“o”，“w”，“y” ), 类 = "因子"), V19 = 结构 (c(2L, 2L, 2L, 2L, 2L), .Label = c("n", "o", "t"), 类 = "因子") , 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 = structure(c(5L, 5L, 5L, 5L, 6L), .Label = c("a", "c", "n", "s", "v", "y"), class = "factor"), V23 = structure(c(3L, 3L, 5L, 1L, 2L), .Label = c("d", "g", "l", “m”、“p”、“u”、“w”）、class =“因子”））、.Names = 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), 类 = "data.frame")

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1 回答 1

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你总是可以使用这个技巧来拥有数字列

numcol <- as.numeric(as.character(factcol))

但我怀疑你的 data.frame 中有因子变量。由于apply返回一个矩阵，如果您的数据中有一个因子，则所有数值变量也将被强制为因子。

这是一个示例，使用玩具数据集

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，因为您可以控制输出（使用 **ply）。但是在这种情况下，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
于 2012-08-05T22:55:35.937 回答