[#6810] which(.., arr.ind=TRUE) on lgRMatrix object produces result for transposed object

Date:
2023-06-07 06:24 Priority:
3

State:
Closed

Submitted by:
Hervé Pagès (hpages) Assigned to:
Martin Maechler (mmaechler)

Hardware:
None Product:
None

Operating System:
Linux Component:
None

Version:
None Severity:
normal

Resolution:
Fixed

URL:

Summary:
which(.., arr.ind=TRUE) on lgRMatrix object produces result for transposed object

Detailed description
Here is how to reproduce: library(Matrix) m <- matrix(c(FALSE, FALSE, TRUE, FALSE, TRUE, TRUE, FALSE, TRUE), ncol=2) lgr <- as(m, "RsparseMatrix") lgr # 4 x 2 sparse Matrix of class "lgRMatrix" # # [1,] . \| # [2,] . \| # [3,] \| . # [4,] . \| This is correct (same as which(m)): which(lgr) # [1] 3 5 6 8 This is not correct: which(lgr, arr.ind=TRUE) # row col # [1,] 1 2 # [2,] 2 2 # [3,] 3 1 # [4,] 4 2 The array-indices are correct but they are listed in the wrong order (they're listed row by row instead of column by column). The correct answer is: which(m, arr.ind=TRUE) # row col # [1,] 3 1 # [2,] 1 2 # [3,] 2 2 # [4,] 4 2 This is with Matrix 1.5-4.1 and R 4.3.0 on a 64-bit Intel machine running Ubuntu 23.04. Best, H.

Comments:

Message
Date: 2023-06-09 06:42 Sender: Hervé Pagès Great! Thanks
Date: 2023-06-09 05:28 Sender: Mikael Jagan Fixed now ... which(<Matrix>) and which(<matrix>) should now behave identically, i.e., consistently with help("which") ... Notably which(<[CRT]sparseMatrix>) now coerces to sparseVector to get a result in column-major order, and that coercion now happens in C, for speed.
Date: 2023-06-07 18:45 Sender: Hervé Pagès > it is NOT the result for the transposed matrix Sorry I meant to say "which(.., arr.ind=TRUE) on lgRMatrix object produces result in order of transposed object". I don't know how to edit the title of the issue. Anyways, it should not do that. As Mikael Jagan pointed out below, the contract as described in the man page is that 'which(m, arr.ind=TRUE)' should be equivalent to 'arrayInd(which(m, arr.ind=FALSE), dim(m), dimnames(m))'. This implies that 'arr.ind' should not affect the order in which the indices are returned. Thanks
Date: 2023-06-07 11:15 Sender: Mikael Jagan help("which") states: ======== Basically, the result is '(1:length(x))[x]' in typical cases; more generally, including when 'x' has 'NA''s, 'which(x)' is 'seq_along(x)[!is.na(x) & x]' plus 'names' when 'x' has. If 'arr.ind == TRUE' and 'x' is an 'array' (has a 'dim' attribute), the result is 'arrayInd(which(x), dim(x), dimnames(x))', namely a matrix whose rows each are the indices of one element of 'x'; see Examples below. ======== hence I am inclined to agree that we must change (somehow) ... I can look later ...
Date: 2023-06-07 10:34 Sender: Martin Maechler I can confirm what you see. However the result is just a row-ordered version of what we get for "base matrix". It is still correct and it is NOT the result for the transposed matrix. The main property that which() promises is satisfied: The resulting 2-column corresponds to cbind(i, j) where <matrix>[ i[k], j[k] ] is TRUE (or non-0). Why should we be more base-compatible here? Returning the result the way we do is most efficient for Rsparse matrices, and I don't think we promise anything about the ordering of the returned (i,j) indices?

Attached Files:

Changes

Field	Old Value	Date	By
Severity	major	2023-06-09 05:28	jaganmn
Resolution	None	2023-06-09 05:28	jaganmn
status_id	Open	2023-06-09 05:28	jaganmn
close_date	None	2023-06-09 05:28	jaganmn
assigned_to	none	2023-06-07 10:34	mmaechler