Reputation: 23
There are many columns in the list. I'd like to use some functions to work automatically.
I have a data.frame myData
There is no myData$home_player_X
. I'm adding it one by one manually.
If I do it manually, the code looks like this:
myData$home_player_1 <- lDataFrames[[3]]$home_player_1
myData$home_player_2 <- lDataFrames[[3]]$home_player_2
...
myData$home_player_11 <- lDataFrames[[3]]$home_player_11
If we only consider the part after <-
, I can convert it into an expression:
eval(parse(text=paste("lDataFrames[[3]]$home_player_",i,sep="")))
But I want to convert whole string. The whole string is this:
paste("myData$home_player_",i," <- lDataFrames[[3]]$home_player_", i,sep="")
I want to convert string into an assignment statement, so I can do it in a for loop
Upvotes: 2
Views: 120
Reputation: 388982
Instead of playing with strings, you can directly copy the required columns in mydata
.
cols <- grep("^home_player", names(lDataFrames[[3]]), value = TRUE)
mydata[cols] <- lDataFrames[[3]][cols]
Using reproducible example,
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
Upvotes: 1
Reputation: 37641
Instead of using the $
notation, just use the variable name as an index. I am substituting Y
for your lDataFrames[[3]]
, but it should be easy to translate.
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
Upvotes: 0