Re: [R] diag() problem

From: Duncan Murdoch <murdoch_at_stats.uwo.ca>
Date: Thu 20 Oct 2005 - 01:22:05 EST

> On 10/19/2005 10:55 AM, Robin Hankin wrote:
>

>>> Hi
>>> 
>>> I have a matrix "u", for which diag() gives an error:
>>> 
>>> u <- structure(c(5.42334674128216, -2.31319389204264, -5.83059042218476,
>>>   -1.64112369640695, -2.31319389212801, 3.22737617646609,  
>>> 1.85200668021569,
>>>   -0.57102273078531, -5.83059042231881, 1.85200668008156,  
>>> 11.9488923894962,
>>>   -3.5525537165941, -1.64112369587405, -0.571022730886046,  
>>> -3.55255371755604,
>>>   10.0989829379577), .Dim = c(4, 4), .Dimnames = list(c("constant",
>>>   NA, NA, NA), c("constant", NA, NA, NA)))
>>> 
>>>  > u
>>>            constant       <NA>      <NA>       <NA>
>>> constant  5.423347 -2.3131939 -5.830590 -1.6411237
>>> <NA>     -2.313194  3.2273762  1.852007 -0.5710227
>>> <NA>     -5.830590  1.8520067 11.948892 -3.5525537
>>> <NA>     -1.641124 -0.5710227 -3.552554 10.0989829
>>>  > is.matrix(u)
>>> [1] TRUE
>>>  > diag(u)
>>> Error in if (is.list(nms) && !any(sapply(nms, is.null)) && all((nm <-  
>>> nms[[1]][1:m]) ==  :
>>>      missing value where TRUE/FALSE needed
>>> 
>>>  >
>>> 
>>> 
>>> What's going on here?

>
>
> It's trying to check whether the row names match the column names, in
> which case it will assign those names to the diagonal elements. But the
> writer didn't figure someone would have NA names, so the test
>
> all((nm <- nms[[1]][1:m]) == nms[[2]][1:m])
>
> fails.
>
> It could be "fixed" by putting "na.rm=TRUE" into the all(), but that's
> probably not right:

>
> > all(c(1, NA) == c(1, 2), na.rm = TRUE)
> [1] TRUE
>
> I think we want to wrap the values in "paste", to convert to non-missing
> characters. That would be
>
> all(paste((nm <- nms[[1]][1:m])) == paste(nms[[2]][1:m]))
>
> and would give
>
> > diag(u)
> constant <NA> <NA> <NA>
> 5.423347 3.227376 11.948892 10.098983
>
> Any objections to me committing this change?

I object: it can't tell the difference between the name "NA" and a missing name. A better fix is to wrap the all() in isTRUE(). This leaves the names off the result (since we don't know if the rownames and colnames match).

Duncan Murdoch



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