Question

我是新的R用户。目前我正在研究一个数据集，其中我必须将多个二进制列转换为单因子列

以下是示例：

当前数据集如：

$ Property.RealEstate                   : num  1 1 1 0 0 0 0 0 1 0 ...
$ Property.Insurance                    : num  0 0 0 1 0 0 1 0 0 0 ...
$ Property.CarOther                     : num  0 0 0 0 0 0 0 1 0 1 ...
$ Property.Unknown                      : num  0 0 0 0 1 1 0 0 0 0 ...

Property.RealEstate  Property.Insurance  Property.CarOther  Property.Unknown
                  1                   0                  0                 0
                  0                   1                  0                 0
                  1                   0                  0                 0
                  0                   1                  0                 0
                  0                   0                  1                 0
                  0                   0                  0                 1

重新编码的列应为：

   Property
1  Real estate
2  Insurance
3  Real estate
4  Insurance
5  CarOther
6  Unknown

它基本上与melt.matrix函数相反。

感谢大家的宝贵意见。它确实有效。但是有一个问题，我有一些值为：

的行

Property.RealEstate Property.Insurance Property.CarOther Property.Unknown

         0                      0                      0               0

我希望将这些标记为NA或Null

如果您也建议，也会有所帮助。

谢谢

Answer 1

> mat <- matrix(c(0,1,0,0,0,
+                 1,0,0,0,0,
+                 0,0,0,1,0,
+                 0,0,1,0,0,
+                 0,0,0,0,1), ncol = 5, byrow = TRUE)
> colnames(mat) <- c("Level1","Level2","Level3","Level4","Level5")
> mat
     Level1 Level2 Level3 Level4 Level5
[1,]      0      1      0      0      0
[2,]      1      0      0      0      0
[3,]      0      0      0      1      0
[4,]      0      0      1      0      0
[5,]      0      0      0      0      1

根据每行中每个1的索引创建一个新因子使用矩阵列名称作为每个级别的标签

NewFactor <- factor(apply(mat, 1, function(x) which(x == 1)), 
                    labels = colnames(mat)) 

> NewFactor 
[1] Level2 Level1 Level4 Level3 Level5 
Levels: Level1 Level2 Level3 Level4 Level5

你也可以尝试：

factor(mat%*%(1:ncol(mat)), labels = colnames(mat))

也使用Tomas解决方案 - ifounf在某处SO

as.factor(colnames(mat)[mat %*% 1:ncol(mat)])

Answer 2

融化无疑是一种解决方案。我建议使用reshape2熔化如下：

library(reshape2)

df=data.frame(Property.RealEstate=c(0,0,1,0,0,0),
              Property.Insurance=c(0,1,0,1,0,0),
              Property.CarOther=c(0,0,0,0,1,0),
              Property.Unknown=c(0,0,0,0,0,1))

#add id column (presumably you have ids more meaningful than row numbers)
df$row=1:nrow(df)

#melt to "long" format
long=melt(df,id="row")

#only keep 1's
long=long[which(long$value==1),]

#merge in ids for NA entries
long=merge(df[,"row",drop=F],long,all.x=T)

#clean up to match example output
long=long[order(long$row),"variable",drop=F]
names(long)="Property"
long$Property=gsub("Property.","",long$Property,fixed=T)

#results
long

Answer 3

不同的东西：

获取数据：

dat <- data.frame(Property.RealEstate=c(1,0,1,0,0,0),Property.Insurance=c(0,1,0,1,0,0),Property.CarOther=c(0,0,0,0,1,0),Property.Unknown=c(0,0,0,0,0,1))

重塑它：

names(dat)[row(t(dat))[t(dat)==1]]
#[1] "Property.RealEstate" "Property.Insurance"  "Property.RealEstate"
#[4] "Property.Insurance"  "Property.CarOther"   "Property.Unknown"

如果您想要清理它，请执行以下操作：

gsub("Property\\.","",names(dat)[row(t(dat))[t(dat)==1]])
#[1] "RealEstate" "Insurance"  "RealEstate" "Insurance"  "CarOther"   "Unknown"

如果您更喜欢因子输出：

factor(row(t(dat))[t(dat)==1],labels=names(dat))

......并清理完毕：

factor(row(t(dat))[t(dat)==1],labels=gsub("Property\\.","",names(dat)) )

Answer 4

或者，你可以用天真的方式做到这一点。我认为它比任何其他建议（包括我的其他建议）更透明。

df=data.frame(Property.RealEstate=c(0,0,1,0,0,0),
              Property.Insurance=c(0,1,0,1,0,0),
              Property.CarOther=c(0,0,0,0,1,0),
              Property.Unknown=c(0,0,0,0,0,1))

propcols=c("Property.RealEstate", "Property.Insurance", "Property.CarOther", "Property.Unknown")

df$Property=NA

for(colname in propcols)({
  coldata=df[,colname]
  df$Property[which(coldata==1)]=colname
})

df$Property=gsub("Property.","",df$Property,fixed=T)

R：将多个二进制列转换为一个因子变量，其因子是二进制列名

4 个答案: