I know you have a solution, but I would have suggested using print() with
quote=FALSE as a better way to illuminate what is going on, as in this
example:
foo - 'bahbah'
foo
[1] bah\bah
print(foo)
[1] bah\bah
print(foo, quote=FALSE)
[1] bahbah
As others said, the backslash isn't really there.
Dear R People:
I'm sure that this is a very simple problem, but I have been wresting with
it for some time.
I have the following file that has the following one line:
CRS(+init=epsg:28992)
Fair enough. I scan it into R and get the following:
u
[1] CRS(\+init=epsg:28992\)
(1) The backslashes are not really there; they are an artefact of the R
print() function.
Try cat(u,\n). I think this might be an FAQ.
(2) Is not your problem the fact that your are setting replacement
equal to the
thing you are trying to get rid of? I.e. don't you want
v -
They're not actually there so don't try too hard to rid yourself of them:
x - \'
length(x)
print(x)
cat(x, \n)
Make sure you're clear on the difference between what's stored by the computer
and what it print()s. Rarely the same, though cat() is often slightly more
honest.
On Nov 17, 2013,
I actually solved the problem in a backhanded (backslashed?) sort of
way...took out the quotation marks in the original file. All is well.
Thanks!
On Sun, Nov 17, 2013 at 4:38 PM, Rolf Turner r.tur...@auckland.ac.nzwrote:
(1) The backslashes are not really there; they are an artefact of the
On Sun, Nov 17, 2013 at 10:42 PM, R. Michael Weylandt
michael.weyla...@gmail.com michael.weyla...@gmail.com wrote:
They're not actually there so don't try too hard to rid yourself of them:
x - \'
length(x)
print(x)
cat(x, \n)
Did you mean to do 'nchar(x)' to show that \ was one character?
On Sun, Nov 17, 2013 at 6:24 PM, Barry Rowlingson
b.rowling...@lancaster.ac.uk wrote:
On Sun, Nov 17, 2013 at 10:42 PM, R. Michael Weylandt
michael.weyla...@gmail.com michael.weyla...@gmail.com wrote:
Did you mean to do 'nchar(x)' to show that \ was one character?
'length' gives the number of
7 matches
Mail list logo
