extracting column headers

Question

I have the following data frame (data) from which I will like to extract and print column headers where ever there is a non-zero element:

  M1 M2 M3 M4
2  0  1  0 1
5  1 -1  0 0
7  0  1  1 0

Required output:

2: M2 M4
5: M1 M2
7: M2 M3 

So far my R code is not working:

colnames(data)[which(data[2] !=0),]

Help will be appreciated. Thanks

Answer 1

This will always return a list:

Map(`[`, list(names(df)), split(col(df)[df != 0],
                                row(df)[df != 0]))

# [[1]]
# [1] "M2" "M4"
# 
# [[2]]
# [1] "M1" "M2"
# 
# [[3]]
# [1] "M2" "M3"

(And you can wrap in inside setNames(..., rownames(df)) if you want the list to share df's row names.)

Answer 2

Let's take the more general case of irregular results:

dat <- structure(list(M1 = c(0L, 1L, 0L), M2 = c(1L, -1L, 1L), M3 = -1:1, 
M4 = c(1L, 0L, 0L)), .Names = c("M1", "M2", "M3", "M4"), 
  class = "data.frame", row.names = c("2", "5", "7"))

> dat
  M1 M2 M3 M4
2  0  1 -1  1
5  1 -1  0  0
7  0  1  1  0

The inner apply "loop" constructs a logical set of vectors. Because R is column oriented the second level of processing is done on columns. The outer apply "loop" extracts the appropriate items from colnames:

 apply( apply(dat,1, as.logical) , 2, function(ll) colnames(dat)[ll] ) 
$`2`
[1] "M2" "M3" "M4"

$`5`
[1] "M1" "M2"

$`7`
[1] "M2" "M3"

You could also have extracted the array indicator version of which() and then process the results:

 > which(dat != 0, arr.ind=TRUE)
  row col
5   2   1
2   1   2
5   2   2
7   3   2
2   1   3
7   3   3
2   1   4