Good morning Ruud,
What sort of tasks are you going to be doing in R? Some tasks will be
faster on a single core extreme type processor, and other tasks can benefit
from a multi-core processor (which run at slower clock speeds than extreme
single-core). If you're working with large matrices,
Good morning,
You do not need to quote a forward slash / in R, but you do need to quote a
backslash when you're inputting it... so to get a string which actually
contains blah\/blah... you need to use blah\\/blah
Good afternoon,
The short answer is yes, the long answer is it depends.
It all depends on what you want to do with the data, I'm working with
dataframes of a couple of million lines, on this plain desktop machine and
for my purposes it works fine. I read in text files, manipulate them,
convert
What is the problem error message? I can say
fred - blah1\\/blah2\\/blah3
and then the string looks like...
cat(#, fred, '#\n', sep='')
#blah1\/blah2\/blah3#
If you just ask R to print it then it looks like...
fred
[1] blah1\\/blah2\\/blah3
when you're playing with strings and regular
Good morning,
Try expm() in the Matrix package by Douglas Bates and Martin Maechler
http://www.stats.bris.ac.uk/R/web/packages/Matrix/index.html
Note that there is a revised version of that paper, refer:
Cleve Moler and Charles Van Loan (2003) Nineteen dubious ways to compute the
exponential of
Neil Shephard wrote:
Another pro to consider is the cost, you can obtain R for free,
SAS/S-Plus/Stata all have licenses of some sort that require purchasing.
Neil
Which has the side effect of *not* restricting how many machines are
available for use or where; e.g. I was running big
Jenny,
Have a look at the R Newsletter Volume 3/2, October 2003
Regards,
Sean
Jenny Barnes wrote:
Dear R-help community,
I have searched through the archives and not been able ot find any advice
on how to plot a wind field with one arrow per grid square with the arrow
pointing in
or using the %in% operator...
?%in%
data[data$label %in% flist,]
regards,
Sean
Applejus wrote:
Hi,
You are right the == doesn't work, but there's a workaround using regular
expressions:
flist-fun|food
grep(flist, data$label)
will give you the vector [2 4] which are the numbers
Good morning,
Firstly I'd like to say that I'm a huge fan of R and I think it's great
system.
Part of the problem in searching for information is knowing what buzzwords /
keywords to use. I was recently caught out like this as I didn't see my
problem as a cumulative sum (keyword=cumsum) only
Good morning, I've searched high and low and I've tried many different ways
of doing this, but I can't seem to get it to work.
I'm looking for a way of vectorising a running balance; i.e. the value in
the first row of the dataframe is zero, and following rows add to this
running balance. This
Good afternoon Monica,
Relying on regular expressions, substituting nothing for everything
starting with a space until the end of the line (i.e. with a dollar sign)
str1 - sub( .*$, , str)
Regards,
Sean
Monica Pisica wrote:
Hi everyone,
I have a vector of strings, each string made
fyi On my machine match runs *much* faster...
t0 - Sys.time(); for (i in 1:reps) { match(Y,X) }; print(Sys.time() - t0)
Time difference of 0.1570001 secs
t0 - Sys.time(); for (i in 1:reps) { sapply(Y,function(Y){which(Y==X)})
}; print(Sys.time() - t0)
Time difference of 6.093 secs
6.09/.157
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