Wacek Kusnierczyk wrote:
But can you be sure that there is no legitimate reason for expecting
the current behaviour?
you surely know the answer.
Actually, I don't. I was just pointing out the generic risk of fixing
something that isn't broken by breaking something that works. There's a
Peter Dalgaard wrote:
> Wacek Kusnierczyk wrote:
>
>>>
>>> (a) do not descend recursively into the function part (first element)
>>> of a call
>>> (b) do descend, unless it is a name
>>
>> if it is a name, how would you descend?
>
> By calling a recursive function which has it as the argument. It
Wacek Kusnierczyk wrote:
(a) do not descend recursively into the function part (first element)
of a call
(b) do descend, unless it is a name
if it is a name, how would you descend?
By calling a recursive function which has it as the argument. It's not a
problem unless you want it to be (
Peter Dalgaard wrote:
> Wacek Kusnierczyk wrote:
>> Gabor Grothendieck wrote:
>>> This comes from the all.vars function and would indicate
>>> a bug in that base R function.
>>>
>>>
>>
>> hush! a user bug, i presume? but indeed,
>>
>> all.vars(expression(foo(bar)()))
>> # character(0)
>> all.n
Wacek Kusnierczyk wrote:
Gabor Grothendieck wrote:
This comes from the all.vars function and would indicate
a bug in that base R function.
hush! a user bug, i presume? but indeed,
all.vars(expression(foo(bar)()))
# character(0)
all.names(expression(foo(bar)()))
# "foo" "bar"
Semantic
Gabor Grothendieck wrote:
> This comes from the all.vars function and would indicate
> a bug in that base R function.
>
>
hush! a user bug, i presume? but indeed,
all.vars(expression(foo(bar)()))
# character(0)
all.names(expression(foo(bar)()))
# "foo" "bar"
vQ
>> f = function(a) function(
This comes from the all.vars function and would indicate
a bug in that base R function.
> f = function(a) function() paste(a, a, sep="")
> all.vars(~ fo(o)())
character(0)
On Tue, Feb 3, 2009 at 8:24 AM, Wacek Kusnierczyk
wrote:
> Peter Dalgaard wrote:
>> Gabor Grothendieck wrote:
>>> On Sat, J
Peter Dalgaard wrote:
> Gabor Grothendieck wrote:
>> On Sat, Jan 31, 2009 at 6:01 PM, Wacek Kusnierczyk
> th some additional boring pedantry wrt. ?gsubfn, which says:
>>>
>>> " If 'replacement' is a formula instead of a function then a one
>>> line function is created whose body is the right ha
Wacek Kusnierczyk wrote:
> Gabor Grothendieck wrote:
>
>> On Sat, Jan 31, 2009 at 4:46 PM, Wacek Kusnierczyk
>> wrote:
>>
>>
>>>
>>> to extend the context, if you were to solve the problem in perl, the
>>> regex below would work in perl 5.10, but not in earlier versions of
>>> perl; an
Peter Dalgaard wrote:
> Gabor Grothendieck wrote:
>> On Sat, Jan 31, 2009 at 6:01 PM, Wacek Kusnierczyk
> th some additional boring pedantry wrt. ?gsubfn, which says:
>>>
>>> " If 'replacement' is a formula instead of a function then a one
>>> line function is created whose body is the right ha
On Sun, Feb 1, 2009 at 6:44 AM, Peter Dalgaard wrote:
> Gabor Grothendieck wrote:
>>
>> On Sat, Jan 31, 2009 at 6:01 PM, Wacek Kusnierczyk
>
> th some additional boring pedantry wrt. ?gsubfn, which says:
>>>
>>> " If 'replacement' is a formula instead of a function then a one
>>>line function
Gabor Grothendieck wrote:
On Sat, Jan 31, 2009 at 6:01 PM, Wacek Kusnierczyk
th some additional boring pedantry wrt. ?gsubfn, which says:
" If 'replacement' is a formula instead of a function then a one
line function is created whose body is the right hand side of the
formula and whose
On Sat, Jan 31, 2009 at 6:01 PM, Wacek Kusnierczyk
wrote:
> Gabor Grothendieck wrote:
>> On Sat, Jan 31, 2009 at 4:46 PM, Wacek Kusnierczyk
>> wrote:
>>
>>> David Hajage wrote:
>>>
Thank you, it's perfect.
>>> to extend the context, if you were to solve the problem in perl, the
>>>
Gabor Grothendieck wrote:
> On Sat, Jan 31, 2009 at 4:46 PM, Wacek Kusnierczyk
> wrote:
>
>> David Hajage wrote:
>>
>>> Thank you, it's perfect.
>>>
>>>
>> to extend the context, if you were to solve the problem in perl, the
>> regex below would work in perl 5.10, but not in earlier
On Sat, Jan 31, 2009 at 4:46 PM, Wacek Kusnierczyk
wrote:
> David Hajage wrote:
>> Thank you, it's perfect.
>>
>
> to extend the context, if you were to solve the problem in perl, the
> regex below would work in perl 5.10, but not in earlier versions of
> perl; another approach is to replace the
David Hajage wrote:
> Thank you, it's perfect.
>
to extend the context, if you were to solve the problem in perl, the
regex below would work in perl 5.10, but not in earlier versions of
perl; another approach is to replace the unwanted leading characters
with equally many replacement character
Thank you, it's perfect.
david
2009/1/30 Wacek Kusnierczyk
> David Hajage wrote:
> > Hello R users,
> >
> > I have a string, for example:
> > x <- "\t\tabc\t def"
> >
> > This string can contain any number of tabulations. I want to replace each
> > tabulation of the begining of the string by th
David Hajage wrote:
> Hello R users,
>
> I have a string, for example:
> x <- "\t\tabc\t def"
>
> This string can contain any number of tabulations. I want to replace each
> tabulation of the begining of the string by the same number of space:
> " abc\t def"
>
> I'm trying to do this with gsub :
>
Hello R users,
I have a string, for example:
x <- "\t\tabc\t def"
This string can contain any number of tabulations. I want to replace each
tabulation of the begining of the string by the same number of space:
" abc\t def"
I'm trying to do this with gsub :
> gsub("\t", " ", x) # replace every \
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