In R, using `unique()` with extra conditions to extract submatrices: easy solution without plyr

Question

In R, let M be the matrix

     [,1] [,2] [,3]
[1,]    1    2    3
[2,]    1    3    3
[3,]    2    4    5
[4,]    6    7    8

I would like to select the submatrix m

     [,1] [,2] [,3]
[1,]    1    3    3
[2,]    2    4    5
[3,]    6    7    8

using unique on M[,1], specifying to keep the row with the maximal value in the second columnM. At the end, the algorithm should keep row [2,] from the set \{[1,], [2,]\}. Unfortunately unique() returns me a vector with actual values, and not row numbers, after elimination of duplicates.

Is there a way to get the asnwer without the package plyr? Thanks a lot, Avitus

I am not a programmer, so I would like to learn step by step as much as possible :-). plyr is great, btw. — Avitus, May 22 '13 at 13:28

score 1 · Answer 1 · answered May 22 '13 at 13:28

1

You're looking for duplicated.

m <- as.matrix(read.table(text="1    2    3
1    3    3
2    4    5
6    7    8"))
m <- m[order(m[,2], decreasing=TRUE), ]
m[!duplicated(m[,1]),]

#      V1 V2 V3
# [1,]  6  7  8
# [2,]  2  4  5
# [3,]  1  3  3

answered May 22 '13 at 13:28

Matthew Plourde

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but this will not preserve the original row order. – flodel May 22 '13 at 13:32
If OP specifies that as a requirement, I'll remove my answer. – Matthew Plourde May 22 '13 at 13:32
@flodel and Matthew Plourde: the row order is not a problem. Both answers are much appreciated; I was searching for a short but smart code to get the result. Many thanks! Avitus – Avitus May 22 '13 at 13:40

score 1 · Accepted Answer · answered May 22 '13 at 13:30

1

Here's how:

is.first.max <- function(x) seq_along(x) == which.max(x)

M[as.logical(ave(M[, 2], M[, 1], FUN = is.first.max)), ]
#      [,1] [,2] [,3]
# [1,]    1    3    3
# [2,]    2    4    5
# [3,]    6    7    8

answered May 22 '13 at 13:30

flodel

87,577
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score 1 · Answer 3 · answered May 22 '13 at 13:38

1

Not the most efficient:

M <- matrix(c(1,1,2,6,2,3,4,7,3,3,5,8),4)

t(sapply(unique(M[,1]),function(i) {temp <- M[M[,1]==i,,drop=FALSE]
                                    temp[which.max(temp[,2]),]                                  
         }))

#     [,1] [,2] [,3]
#[1,]    1    3    3
#[2,]    2    4    5
#[3,]    6    7    8

answered May 22 '13 at 13:38

Roland

127,288
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I never used `sapply` before :-). Thanks for your code. It is interesting the use of `t()` at the very end, though. – Avitus May 22 '13 at 13:50

In R, using `unique()` with extra conditions to extract submatrices: easy solution without plyr

3 Answers3