[R] Extending each element in a list, or rbind()-ing arrays of different length without recycling

markleeds at verizon.net markleeds at verizon.net
Thu Feb 12 21:06:21 CET 2009

```Hi Jason: below seems to work. you have to take the transpose because
the apply
returns the rows transposed. i'm also not sure how to make the NAs be
the last
ones but maybe someone can show us how to do that.

mat <- matrix(c(2,7,2,7,9,10,10,6,8,6,1,9,7,2,0),byrow=TRUE,nrow=3)
print(mat)

t(apply(mat,1, function(.row) {
.row[duplicated(.row)] <- NA
.row
}))

On Thu, Feb 12, 2009 at  2:31 PM, Jason Shaw wrote:

> Hi,
>
> I'm trying to take a matrix such as
>
>      [,1] [,2] [,3] [,4] [,5]
> [1,]    2    7    2    7    9
> [2,]   10   10    6    8    6
> [3,]    1    9    7    2    0
>
> and generate a new matrix which contains only the unique values in
> each row:
>
>      [,1] [,2] [,3] [,4] [,5]
> [1,]    2    7    9   NA   NA
> [2,]   10    6    8   NA   NA
> [3,]    1    9    7    2    0
>
> My problem is that I can use apply(matrix,MARGIN=1,FUN=unique) to find
> the unique values, but this leaves me with a list with arrays of
> different length:
>
>> x <- apply(peaks,MARGIN=1,FUN=unique)
> [[1]]
> [1] 2 7 9
>
> [[2]]
> [1] 10  6  8
>
> [[3]]
> [1] 1 9 7 2 0
>
> and using do.call("rbind",x) recycles the values of the shorter
> vectors instead of filling them with NA:
>
>> do.call("rbind",x)
>      [,1] [,2] [,3] [,4] [,5]
> [1,]    2    7    9    2    7
> [2,]   10    6    8   10    6
> [3,]    1    9    7    2    0
>
> So, I'd like to either take every element of the list and extend it
> with NAs to the length of the longest element, or rbind every element
> where missing places are filled with NAs instead of recycled values.
> Is this possible?  Of course, the solution is trivial using a loop,
> but I'm trying to avoid this.
>
> Thanks for any suggestions.
>
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```

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