I have a data.frame like
a b c d
1 1 0 0 1
2 1 1 0 0
3 0 1 0 0
4 1 0 1 0
5 1 0 0 0
Which I generated using
df<- data.frame(a=sample(0:1,5,replace=T),b=sample(0:1,5,replace=T),c=sample(0:1,5,replace=T),d=sample(0:1,5,replace=T))
How can I get the result as 4, 2, 2, 3, 1 if I pass 1 to that function depicting to find the last index of 1 in each row.
One approach would be:
apply(df, 1, function(x) max(which(x == 1)))
If you wanted to be flexible about which element you're checking for and handle cases where the value is missing from a row:
max.row <- function(df, val) unname(apply(df, 1, function(x) tail(c(NA, which(x == val)), 1)))
max.row(df, 0)
# [1] 3 4 4 4
max.row(df, 1)
# [1] 4 2 2 3
max.row(df, 2)
# [1] NA NA NA NA
you can try max.col which is a little bit faster than apply
max.col(df, "last")
# [1] 2 4 4 2 4
Data
set.seed(1)
df <- data.frame(a=sample(0:1,5,replace=T),b=sample(0:1,5,replace=T),c=sample(0:1,5,replace=T),d=sample(0:1,5,replace=T))
df again, otherwise the results you are showing won't match with anybody else. +1 howeverAnother option is using pmax. We multiply the col(df) by 'df' and get the max value by row.
do.call(pmax,col(df)*df)
#[1] 4 2 2 3 1
col(df) is a convenient function to get the column index of the dataset.
col(df)
# [,1] [,2] [,3] [,4]
#[1,] 1 2 3 4
#[2,] 1 2 3 4
#[3,] 1 2 3 4
#[4,] 1 2 3 4
#[5,] 1 2 3 4
By doing the multiplication of 'df' with the col(df) of equal dimension, the '0' values will remain 0 while the places that are '1' will be replaced by the column index, i.e.
col(df)*df
# a b c d
#1 1 0 0 4
#2 1 2 0 0
#3 0 2 0 0
#4 1 0 3 0
#5 1 0 0 0
Now, we can get the max value per each row by do.call(pmax)
Seeing all the possible solutions and one from my side, here are the times taken by each replicated 10,000 times
apply(df,1,function(x){tail(which(x==1),1)})
user system elapsed
2.978 0.010 2.988
apply(df*col(df),1,function(x){max(x)})
user system elapsed
8.217 0.026 8.245
apply(df, 1, function(x) max(which(x == 1)))
user system elapsed
1.621 0.005 1.627
max.col(df, "last")
user system elapsed
1.348 0.004 1.352
Though @Mamoun Benghezal's answer is the most efficient, it doesn't solve my purpose of being flexible. The accepted answer does.