R devel - May 2002 - merge.data.frame can coerce character vectors to factor in some circumstances (PR#1608)

If the following two conditions are met:

1) all.x is TRUE

2) at least 1 row in y does not have a match in x

then any character vectors in y will be coerced to be factors. Here is a simple
example (previously provided on r-devel):
> x <- data.frame(a = 1:4)
> y <- data.frame(b = LETTERS[1:3]) 
> y$b <- as.character(y$b) 
> z <- merge(x, y, by = 0, all.x = TRUE)
> z
  Row.names a    b
1         1 1    A
2         2 2    B
3         3 3    C
4         4 4   <NA>
> sapply(z, data.class)
Row.names         a         b 
 "factor" "numeric"  "factor"
> 
This problem could be fixed by changing the line in merge.data.frame:

for (i in seq(along = y)) is.na(y[[i]]) <- (lxy + 1):(lxy + nxx) 

to:

for (i in seq(along = y)) y[((lxy + 1):(lxy + nxx)), i] <- NA 

To the extent that this is a feature rather than a bug (if so, I would like to
know why), then I would suggest that the following sentence be added to the
documentation for merge at the end of the section on all.x

"Be aware that, if all.x equals `TRUE', character vectors in `y'
will be
converted to factors if any rows in y have no matching row in `x'."

Thanks,

Dave Kane

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On Wed, 29 May 2002 a296180@agate.fmr.com wrote:
> If the following two conditions are met:
>
> 1) all.x is TRUE
>
> 2) at least 1 row in y does not have a match in x
>
> then any character vectors in y will be coerced to be factors. Here is a
simple
> example (previously provided on r-devel):
>
> > x <- data.frame(a = 1:4)
> > y <- data.frame(b = LETTERS[1:3])
> > y$b <- as.character(y$b)
> > z <- merge(x, y, by = 0, all.x = TRUE)
> > z
>   Row.names a    b
> 1         1 1    A
> 2         2 2    B
> 3         3 3    C
> 4         4 4   <NA>
> > sapply(z, data.class)
> Row.names         a         b
>  "factor" "numeric"  "factor"
> >
>
> This problem could be fixed by changing the line in merge.data.frame:
>
> for (i in seq(along = y)) is.na(y[[i]]) <- (lxy + 1):(lxy + nxx)
>
> to:
>
> for (i in seq(along = y)) y[((lxy + 1):(lxy + nxx)), i] <- NA
But other problems would be introduced, as the two operations are
not equivalent (and the right one has been used).
> To the extent that this is a feature rather than a bug (if so, I would like
to
> know why),
I have already patiently explained it to you.  It is a side issue of
subscripting of data frames converting character columns to factor.
I have also given you a workaround.
>  then I would suggest that the following sentence be added to the
> documentation for merge at the end of the section on all.x
>
> "Be aware that, if all.x equals `TRUE', character vectors in
`y' will be
> converted to factors if any rows in y have no matching row in
`x'."
As I said before, this is a consequence of the general rules.  Data frames
are not designed to have character columns, and those who insist on using
them must make themselves aware of the consequences.

-- 
Brian D. Ripley,                  ripley@stats.ox.ac.uk
Professor of Applied Statistics,  http://www.stats.ox.ac.uk/~ripley/
University of Oxford,             Tel:  +44 1865 272861 (self)
1 South Parks Road,                     +44 1865 272860 (secr)
Oxford OX1 3TG, UK                Fax:  +44 1865 272595

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