[R] consecutive numbering of elements in a matrix

[David Winsemius]

On Nov 22, 2009, at 11:11 AM, Jim Bouldin wrote:

>
> 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.

Put the individual components together side by side with cbind and it  
will make more sense:

cbind( row(A), apply(is.na(A), 2, cumsum) )

And think about the fact that row(A) and apply(is.na(A), 2, cumsum)  
will be identical in the case where there are no NAs, so their  
difference would be a zero matrix. Double negativism strikes again....  
not(is.na) == "is"

> 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.

You cannot have unequal length columns in a matrix. Only a list is  
able to handle that task. So we need a more clear description of what  
you expect, preferably typed out in full so we can "see" it.

-- 
David.

>
> Thanks again for your help.
>
> Jim

David Winsemius, MD
Heritage Laboratories
West Hartford, CT