You can use system() or shell(), which adds "cmd.exe /c " to the front of
your command
so you can use DOS syntax. Remember to add double quotes when file names
have
spaces in them. E.g., I can call an old version of R with the following
and later read its
text output into my current session.
> i
?system
But this begs the question: WHY would you want to do this? More
specifically, what should R communicate to your other software, and
what should the other software communicate to R?
Cheers,
Bert
Bert Gunter
"The trouble with having an open mind is that people keep coming along
and stick
Jun:
"My problem is the pattern has to be dynamically constructed on the
input data of the function "
What does that mean? How can a pattern be "dynamically constructed"
when you have not made clear (at least to me, perhaps also to yourself
and/or others) *how* it is to be constructed?
Cheers,
B
Hi Ista,
Thanks for the suggestion. I didn't know mapply can be used this way! Let
me take one more step. Instead of defining a pattern for each string, I
would like to define a set of patterns from all the possible combination of
the unique values of those variables. Then I need each string to fi
Hi Jeff,
Thanks for the reply. I tried your suggestion and it doesn't seem to work
and I tried a simple pattern as follows and it works as expected
sub("(3\\.mg\\.kg)\\.(>50-70\\.kg)\\.(.*)", '\\1', "3.mg.kg.>50-70.kg.P05")
[1] "3.mg.kg"
sub("(3\\.mg\\.kg)\\.(>50-70\\.kg)\\.(.*)", '\\2', "3.mg.k
Hi all R users:
Does anybody have the experience of running an external software in R? I
try to use R to run ANSYS software, which is a engineering simulation
package. I ever have done this task in Matlab platform by executing
the following code line:
system('"C:\Program Files\Ansys Inc\v100\ANSY
Hi Bert,
In the final.pattern, there are ten patterns.
>sub(final.pattern, '\\1', test.string)
Expected results: "240.m.g" "3.mg.kg" "240.m.g"
Current results: "" "" "240.m.g"
>sub(final.pattern, '\\2', test.string)
Expected results: ">110.kg" ">110.kg" ">50-70.kg"
Current results: "" "" ">50-70
I am not near my computer today, but each parenthesis gets its own result
number, so you should put the parenthesis around the whole pattern of
alternatives instead of having many parentheses.
I recommend thinking in terms of what common information you expect to find in
these various strings,
Hi Bert,
I still couldn't make the multiple patterns to work. Here is an example. I
make the pattern as follows
final.pattern <-
"(240\\.m\\.g)\\.(>50-70\\.kg)\\.(.*)|(3\\.mg\\.kg)\\.(>50-70\\.kg)\\.(.*)|(240\\.m\\.g)\\.(>70-90\\.kg)\\.(.*)|(3\\.mg\\.kg)\\.(>70-90\\.kg)\\.(.*)|(240\\.m\\.g)\\.(>9
On Tue, Sep 6, 2016 at 2:00 AM, Rainer M Krug wrote:
> Please reply to the mailing list to keep the conversation available for
> everybody. I Send this mail to the mailing list as well.
>
> Simone Tenan writes:
>
>> Thanks Rainer,
>> it's very kind of you.
>> Unfortunately, I cannot save the (lar
Hi Sebastian,
here are examples with ggplot2 and basic graphic.
http://stackoverflow.com/questions/3777174/plotting-two-variables-as-lines-using-ggplot2-on-the-same-graph
http://stackoverflow.com/questions/17150183/r-plot-multiple-lines-in-one-graph
You may also impress your audience by using
For a linear model without an intercept term as in this example, neither
the usual permutation scheme for testing Ho: B1 = 0 nor usual definition of
R-squared apply. So you need to check what the developer of this code
chose to do. If I'm recalling correctly, in a linear model with an
intercept t
?lines
?points
to add to an existing base graphics graph. There are other ways to do
this in the other graph systems (ggplot, lattice,...) used in R.
Cheers,
Bert
Bert Gunter
"The trouble with having an open mind is that people keep coming along
and sticking things into it."
-- Opus (aka Berkele
That is contributed code. It could do anything the author felt like. I
recommend reading the source code.
--
Sent from my phone. Please excuse my brevity.
On September 5, 2016 11:52:15 PM PDT, Agustin Lobo wrote:
>Any reason why the R-square prob is not calculated by randomization in
>lmPerm::
Thank you for your reply. I am grouping citation and some social media
indictors ( number of tweets, mendeley readers, etc). The number of
citaions a paper recievs or the number of social media indicators that a
papers receives depends on time. For example a paper published in 2009 has
more time to
That step is easy, but context is hard. You really need to provide a
reproducible example. There are many models, many analysis tools, and many
timescales to choose from. In fact, this could easily be mistaken for a
question about statistics (not really on-topic here) since you have failed to
It is not the implementation of regex that requires double backslashes, but the
R string parser. You can use cat to see what the pattern looks like to the
parser. Try
cat( "\\(.*?\\)" )
--
Sent from my phone. Please excuse my brevity.
On September 6, 2016 6:33:15 AM PDT, Sarah Goslee
wrote:
You can ignore the message below. The maximizing routine buried within the frailty()
command buried with coxph() has a maximizer that is not the brightest. It sometimes gets
lost but then finds its way again. The message is from one of those. It likely took a
not-so-good update step, and too
On 06 Sep 2016, at 11:00 , Rainer M Krug wrote:
> Please reply to the mailing list to keep the conversation available for
> everybody. I Send this mail to the mailing list as well.
>
> Simone Tenan writes:
>
>> Thanks Rainer,
>> it's very kind of you.
>> Unfortunately, I cannot save the (lar
Dear R-users:
Let's see if you can help.
I have an matrix of class "ts" of 100 rows by 4 columns which called PP.
In each column I have the time series of quarterly GDP from 4 countries.
They applied the Hodrick -Prescott filter and now I want to plot
simultaneously cyclical component of the 4
R's implementation of regex requires double backslashes. Reading
?regex will tell you more.
cleanBetweenBrackets <- function(String)
{
return(gsub("\\(.*?\\)", "", String))
}
Str <- "The cat is crazy (but not too crazy)"
cleanBetweenBrackets(Str)
> cleanBetweenBrackets(Str)
[1] "The cat is c
Hi,
Is it possible to use time as an offset (exposure variable) in factor
analysis? If yes, would you please advise how?
Thanks,
Tahereh
Tahereh Dehdarirad
PhD Student of Library and Information Science
University of Barcelona, Spain
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__
Hi Rosa,
you may take advantage of the extremevalues package.
https://cran.r-project.org/web/packages/extremevalues/extremevalues.pdf
An example:
set.seed(1023)
v3 <- c(rnorm(100, 0, 0.2), rnorm(5, 4, 0.1), rnorm(5, -4, 0.1))
v4 <- sample(v3, length(v3))
nam <- as.character(1:length(v4))
df <-
Hello,
I have been able to figure this out using \\ (two back slashes for escape)
Working R code for what I wanted is...
cleanBetweenBrackets <- function(String)
{ return(gsub("\\(.*?\\)", "", String))}
I thought I had tried that (the \\) before I emailed the list. Please
ignore my previous em
?traceback
?debug
?trace
R has built-in debugging tools. Learn to use them.
Cheers,
Bert
Bert Gunter
"The trouble with having an open mind is that people keep coming along
and sticking things into it."
-- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
On Mon, Sep 5, 2016 at 1
Just noticed: My clumsy do.call() line in my previously posted code
below should be replaced with:
pat <- paste(pat,collapse = "|")
> pat <- c(pat1,pat2)
> paste(pat,collapse="|")
[1] "a+\\.*a+|b+\\.*b+"
replace this **
> pat <- do.call(paste,c(as.list(pat),
Hello,
I am trying to remove brackets and the text contained in brackets. I tried
to do a user defined formula... my attempt at this is pasted below.
cleanBetweenBrackets <- function(String)
{ return(gsub("\(.*?\)", "", String))}
I keep getting errors (namely that there is an unrecognised es
Any reason why the R-square prob is not calculated by randomization in
lmPerm::lmp? The help pages states "Either permutation test p-values
or the usual F-test p-values will be output", but I always get the F
test for R-square as with lm():
require(lmPerm)
x <- 1:1000
set.seed(1000)
y1 <- x*2+runi
Dear R-Team,
I have been trying to use the finegray routine that creates a special data
so that Fine and Gray model can be fit. However, it does not seem to work.
Could you please help me with this issue?
Thanks,
Ahalya.
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You need to *print* results explicitly within a for-loop.
E.g.
print(head(d_dataset[,
c(paste0("v_turnover_",year),
paste0("v_customer_", year))]))
See FAQ 7.16 for a related discussion.
cheers,
Rolf Turner
--
Technical Editor ANZJS
Department of Statistics
University of
Hi All,
I would like to write a program that iterates over a set of dynamically
generated variables and produces some stats or prints parts of the data.
# --- data
v_turnover_2011 <- c(10, 20, 30, 40 , 50)
v_customer_2011 <- c(0, 1, NA, 0, 1)
v_turnover_2012 <- c(10, 20, 30, 40 , 50)
v_customer_
Please reply to the mailing list to keep the conversation available for
everybody. I Send this mail to the mailing list as well.
Simone Tenan writes:
> Thanks Rainer,
> it's very kind of you.
> Unfortunately, I cannot save the (large) R object where the current R session
> is running. There is
Simone Tenan writes:
> Hi all,
> I'm using R remotely via ssh connection in linux. I need to save a large R
> object from the remote server to my laptop. How can I specify the path in
> the save() function?
You are working on the remote machine and there is no way that you can
specify "out of th
Hi all,
I'm using R remotely via ssh connection in linux. I need to save a large R
object from the remote server to my laptop. How can I specify the path in
the save() function?
Thanks much for your help,
Simone
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