Reputation: 936
I have a following table in R
df <- data.frame('a' = c(1,0,0,1,0),
'b' = c(1,0,0,1,0),
'c' = c(1,1,0,1,1))
df
a b c
1 1 1 1
2 0 0 1
3 0 0 0
4 1 1 1
4 0 0 1
What I want is to replace the row value with the column name whenever the row is equal to 1. The output would be this one:
a b c
1 a b c
2 0 0 c
3 0 0 0
4 a b c
4 0 0 c
How can I do this in R? Thanks.
Upvotes: 3
Views: 272
Reputation: 66834
You can do it with ifelse
, but you have to do some intermediate transposing to account for R's column-major order processing.
data.frame(t(ifelse(t(df)==1,names(df),0)))
a b c
1 a b c
2 0 0 c
3 0 0 0
4 a b c
5 0 0 c
Upvotes: 1
Reputation: 886948
We can use
df[] <- names(df)[(NA^!df) * col(df)]
df[is.na(df)] <- 0
df
# a b c
#1 a b c
#2 0 0 c
#3 0 0 0
#4 a b c
#4 0 0 c
Upvotes: 4
Reputation: 520918
We can try iterating over the names of the data frame, and then handling each column, for a base R option:
df <- data.frame(a=c(1,0,0,1,0), b=c(1,0,0,1,0), c=c(1,1,0,1,1))
df <- data.frame(sapply(names(df), function(x) {
y <- df[[x]]
y[y == 1] <- x
return(y)
}))
df
a b c
1 a b c
2 0 0 c
3 0 0 0
4 a b c
5 0 0 c
Upvotes: 2
Reputation: 70256
I would use Map
and replace
:
df[] <- Map(function(n, x) replace(x, x == 1, n), names(df), df)
df
# a b c
# 1 a b c
# 2 0 0 c
# 3 0 0 0
# 4 a b c
# 5 0 0 c
Upvotes: 6
Reputation: 323226
You can try stack
and unstack
a=stack(df)
a
values ind
1 1 a
2 0 a
3 0 a
4 1 a
5 0 a
6 1 b
7 0 b
8 0 b
9 1 b
10 0 b
11 1 c
12 1 c
13 0 c
14 1 c
15 1 c
a$values[a$values==1]=as.character(a$ind)[a$values==1]
unstack(a)
a b c
1 a b c
2 0 0 c
3 0 0 0
4 a b c
5 0 0 c
Upvotes: 2