Reputation: 633
I would like to add a new column D to data.frames in a list that contains the first part of column B. But I'm not sure how to adress within lists down to the column level?
create some data
df1 <- data.frame(A = "hey", B = "wass.7", C = "up")
df2 <- data.frame(A = "how", B = "are.1", C = "you")
dfList <- list(df1,df2)
desired output:
# a new column removing the last part of column B
[[1]]
A B C D
1 hey wass.7 up wass
[[2]]
A B C D
1 how are.1 you are
for each data frame I did this, which worked
df1$D<-sub('\\..*', '', df1$B)
in a function I tried this, which is probably not correctly addressing the columns and returns "unexpected symbol..."
dfList <- lapply(rapply(dfList, function(x)
x$D<-sub('\\..*', '', x$B) how = "list"),
as.data.frame)
the lapply(rapply)
part is copied from Using gsub in list of dataframes with R
Upvotes: 1
Views: 61
Reputation: 107767
The rapply
solution does work. However, you needed a comma before the how argument to resolve the error. Additionally, you will NOT be able to assign one new column only replace existing ones. Since rapply
is a recursive call, it will run the gsub
across every element in nested list so across ALL columns of ALL dataframes.
Otherwise use a simple lapply
per @JilberUrbina's answer.
df1 <- data.frame(A = "hey", B = "wass.7", C = "up", stringsAsFactors = F)
df2 <- data.frame(A = "how", B = "are.1", C = "you", stringsAsFactors = F)
dfList <- list(df1,df2)
dfList <- lapply(rapply(dfList, function(x)
sub('\\..*', '', x), how = "list"),
as.data.frame)
dfList
# [[1]]
# A B C
# 1 hey wass up
# [[2]]
# A B C
# 1 how are you
Upvotes: 1
Reputation: 61214
Check this out
lapply(dfList, function(x){
x$D <-sub('\\..*', '', x$B);
x
})
[[1]]
A B C D
1 hey wass.7 up wass
[[2]]
A B C D
1 how are.1 you are
Upvotes: 2