On 06/06/2018 6:28 AM, zListserv wrote:
Sorry. Here's how I re-defined print, print.default, and print.data.frame:
print = function(df, ..., right=FALSE, row.names=FALSE) base::print(df, ...,
right=right, row.names=row.names)
base::print doesn't have those arguments. It only has arguments
print(x, ...). You shouldn't redefine it, since it just dispatches to
one of the methods.
In fact, I think this redefinition is causing the problem way down
below: instead of your two methods applying to the base package
generic, they are applying only to your own generic defined here.
Auto-printing uses the base generic.
print.default = function(df, ..., right=FALSE, row.names=FALSE)
base::print.default(df, ..., right=right, row.names=row.names)
base::print.default doesn't have a row.names argument. It won't cause
an error, but will be ignored. It already has `right=FALSE` as a
default, so it seems pretty pointless to redefine it.
print.data.frame = function(df, ..., right=FALSE, row.names=FALSE)
base::print.data.frame(df, ..., right=right, row.names=row.names)
That definition makes sense if you want left justification and no row
names, but remember that some print methods may rely on the display of
row names for sensible output. (I can't think of any examples right
now, but I'd look at print methods for summary objects if I was
searching for them. There are several that rely on row names when they
print matrices, e.g. print.summary.lm.)
And as a general rule, you should use the same argument names as in the
generic, i.e. x instead of df. It's pretty rare, but someone might say
print(x = data.frame(1:10)), and your print.data.frame method would
absorb the argument into the ... , yielding an error
'argument "df" is missing, with no default'
and this is what it yields (I would like it to print without row names and with
text left-adjusted):
R> x <- as.data.frame(rep(c("a", "ab", "abc"), 7))
R> print(x)
rep(c("a", "ab", "abc"), 7)
a
ab
abc
a
ab
abc
a
ab
abc
a
ab
abc
a
ab
abc
a
ab
abc
a
ab
abc
R> head(x)
rep(c("a", "ab", "abc"), 7)
1 a
2 ab
3 abc
4 a
5 ab
6 abc
I don't get that, because I didn't redefine the generic, only the methods.
R> x
rep(c("a", "ab", "abc"), 7)
1 a
2 ab
3 abc
Or that.
Duncan Murdoch
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