Sorting each row of a data frame

Question

I am trying to sort each row of a data frame using this line,

sapply(df, function(x) sort(x))

However, the columns are getting sorted instead of the rows.

For example, this data frame

5          10          7          1          5
6           3          9          2          4
4           5          1          3          3

is ending up like this:

4           3          1          1          3
5           5          7          2          4
6          10          9          3          5

And I want this:

1 5 5 7 10
2 3 4 6 9
1 3 3 4 5

Any recommendations? Thanks

Solution 1

You could use the plain apply function with MARGIN = 1 to apply over rows and then transpose the result.

t(apply(df, 1, sort))

Solution 2

You can transpose it (coverts it to matrix), and split by column and sort

t(sapply(split(t(df), col(t(df))), sort))
#   [,1] [,2] [,3] [,4] [,5]
# 1    1    5    5    7   10
# 2    2    3    4    6    9
# 3    1    3    3    4    5

Because a data.frame is a list of columns, when you sapply like that you are sorting the columns.

or apply by row

t(apply(df, 1, sort))

Comments

[Frank]

I had a similar idea: do.call(rbind, lapply(split(df, seq_len(nrow(df))), sort)) I think any approach here involves converting each row to an atomic vector (so it can be sorted), which is roughly the same as converting the whole thing to matrix.

[Rorschach]

@Frank yea that works too, very slow for larger data though

[Frank]

Hm, maybe so -- that t(sapply( beats do.call(rbind,lapply( -- though it's not obvious to me why. For the other part, transposing before splitting looks wasteful and inferior to split(df,seq_len(nrow(df))). If you have some benchmark in favor of transposing, I'd be surprised and interested.

[Rorschach]

@Frank yes, the first part is the slow part

[Varun Gawande]

If MARGIN was 2, the output would have been the Matrix, but sorted by the columns. Why do we have to transpose it while sorting according to Rows?