Many thanks to Dimitris, William and David for very helpful answers which
solved my problem. Being a relatve newb, I am confused by something in the
solutions by Dimitris and David.
#Create a matrix A as follows:
> A <- matrix(sample(50, 21), 7, 3)
> A[sample(21, 5)] <- NA;A
[,1] [,2] [,3]
[1,] 36 38 24
[2,] 6 33 13
[3,] 12 42 10
[4,] 7 NA NA
[5,] 48 NA NA
[6,] 3 NA 47
[7,] 29 23 4
> B = row(A) - apply(is.na(A), 2, cumsum);B
[,1] [,2] [,3]
[1,] 1 1 1
[2,] 2 2 2
[3,] 3 3 3
[4,] 4 3 3
[5,] 5 3 3
[6,] 6 3 4
[7,] 7 4 5
#But:
> B = row(A) - apply(!is.na(A), 2, cumsum);B
[,1] [,2] [,3]
[1,] 0 0 0
[2,] 0 0 0
[3,] 0 0 0
[4,] 0 1 1
[5,] 0 2 2
[6,] 0 3 2
[7,] 0 3 2
This seems exactly backwards to me. The is.na(A) command should be
cumulatively summing the NA values and !is.na(A) should be doing so on the
non-NA values. But the opposite is the case. I'm glad I have a solution
but this apparent backwardness of expected logic has me worried.
I do have another, tougher question if anyone has the time, which is, given
a resulting matrix like B below:
> is.na(B) <- is.na(A);B
[,1] [,2] [,3]
[1,] 1 1 1
[2,] 2 2 2
[3,] 3 3 3
[4,] 4 NA NA
[5,] 5 NA NA
[6,] 6 NA 4
[7,] 7 4 5
how can I rearrange all the columns so that equal values are in the same
row, i.e. in the case above, the NA values are removed from columns 2 and 3
and all non-NA values that had been below them are moved up to replace them.
Thanks again for your help.
Jim
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