列表中有很多列。我想使用一些功能来自动工作。
我有一个data.frame myData
没有myData$home_player_X
。我正在手动一个接一个地添加它。
如果我手动执行此操作,则代码如下所示:
myData$home_player_1 <- lDataFrames[[3]]$home_player_1
myData$home_player_2 <- lDataFrames[[3]]$home_player_2
...
myData$home_player_11 <- lDataFrames[[3]]$home_player_11
如果仅考虑<-
之后的部分,我可以将其转换为表达式:
eval(parse(text=paste("lDataFrames[[3]]$home_player_",i,sep="")))
但是我想转换整个字符串。整个字符串是这样的:
paste("myData$home_player_",i," <- lDataFrames[[3]]$home_player_", i,sep="")
我想将字符串转换为赋值语句,因此可以在for循环中完成
答案 0 :(得分:1)
您可以直接在mydata
中复制所需的列,而不必使用字符串。
cols <- grep("^home_player", names(lDataFrames[[3]]), value = TRUE)
mydata[cols] <- lDataFrames[[3]][cols]
使用可复制的示例,
df <- data.frame(home_player_1 = 1:5, home_player_2 = 6:10, home_player_3 = 11:15)
cols <- grep("^home_player", names(df), value = TRUE)
mydata <- data.frame(matrix(nrow = nrow(df), ncol = length(cols),
dimnames = list(NULL, cols)))
mydata[cols] <- df[cols]
mydata
# home_player_1 home_player_2 home_player_3
#1 1 6 11
#2 2 7 12
#3 3 8 13
#4 4 9 14
#5 5 10 15
答案 1 :(得分:0)
不要使用$
表示法,只需将变量名用作索引即可。我用Y
代替了lDataFrames[[3]]
,但翻译起来应该很容易。
myData = data.frame(Var1 = 1:10)
Y = data.frame(home_player_1 = 11:20,
home_player_2 = 21:30, home_player_3 = 31:40)
for(i in 1:3) {
VarName = paste0("home_player_", i)
myData[ ,VarName] = Y[ ,VarName]
}
myData
Var1 home_player_1 home_player_2 home_player_3
1 1 11 21 31
2 2 12 22 32
3 3 13 23 33
4 4 14 24 34
5 5 15 25 35
6 6 16 26 36
7 7 17 27 37
8 8 18 28 38
9 9 19 29 39
10 10 20 30 40