[R] Printing left-justified character strings

[Duncan Murdoch]

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