Philippe Grosjean wrote:
> I know this topic has been discussed already several times. Is it a
> workable solution that emerged? I would like to place R graph in vector
> format in an OpenOffice Writer document (solution working in Linux AND
> Mac OS X AND Windows). I have tried to play with pstoed
Hi Henning,
Henning Wildhagen wrote:
i tried your code, but it results in an error:
"error in var(x) : `x` is empty"
Do you have an idea what the reason might be?
If you ask questions on this list, please provide some code snippets,
which the others can copy and paste to reproduce your error
Hi Henning,
Henning Wildhagen wrote:
Given a dataframe of three columns, where col1 is of type factor, col 2 and
col3 are numeric and pairs of observations i would like to perform a paired
t-test for each level of col1.
I would like to avoid specifying the levels of col1 manually because it
se
Hi Charles,
Charles C. Berry wrote:
> Try this:
>
> f2 <- function(...) sapply( substitute(list(...)), deparse )[-1]
Wow, this works. Thanks a lot!
> p.s. Why do you want this as mode "character"?
For the usual purpose of 'substitute': to label something. The use case
is that I have a
an see, the missing behaviour is marked NA, because it is
uncomparable. To get the number of TRUE values, use
> sum( r1==r2, na.rm=TRUE )
[1] 1
And to get the number of comparable observations, i.e. those without NA,
use e.g.
> length( na.omit( r1==r2 ) )
[1] 3
I hope this helps you to wor
o\"", "3 + 5" )
My third try seems to be wrong as well:
> f3 <- function(...) lapply( list(...), substitute )
> f3( 2+7, "foo", 3+5 )
Error in lapply(list(...), substitute) :
'...' used in an incorrect context
Does anybody has a good hi
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