Re: [R] Substring replacement in string
Dear Hervé,
Many thanks for your suggestion. Gabor Grothendieck proposed a simple
one-liner that works perfectly for my purposes:
gsub("(\\b[a-oq-z][a-z0-9]*)", "1-\\U\\1", x, perl = TRUE)
where x is the respective string.
Best wishes,
Alrik
-Ursprüngliche Nachricht-
Von: Hervé Pagès [mailto:hpa...@fredhutch.org]
Gesendet: Samstag, 28. Februar 2015 23:29
An: Alrik Thiem; r-help@r-project.org
Betreff: Re: [R] Substring replacement in string
Hi Alrik,
With the Biostrings/IRanges infrastructure (Bioconductor packages), you
can do this with:
library(Biostrings)
x0 <- BString("pmin(pmax(pmin(x1, X2), pmin(X3, X4)) == Y, pmax(Z1,
z1))")
donttouch_words <- c("pmin", "pmax")
## Extract the substrings to modify (target substrings).
donttouch_regions <- reduce(do.call("c", lapply(donttouch_words,
matchPattern, x0)))
target_regions <- ranges(gaps(donttouch_regions))
target_substrings <- extractAt(x0, target_regions)
## Modify them.
old <- paste0(letters, collapse="")
new <- paste0(LETTERS, collapse="")
target_substrings <- chartr(old, new, target_substrings)
## Replace in original string.
x1 <- replaceAt(x0, target_regions, target_substrings)
Then:
> x1
57-letter "BString" instance
seq: pmin(pmax(pmin(X1, X2), pmin(X3, X4)) == Y, pmax(Z1, Z1))
> as.character(x1)
[1] "pmin(pmax(pmin(X1, X2), pmin(X3, X4)) == Y, pmax(Z1, Z1))"
Hope this helps,
H.
On 02/27/2015 02:19 PM, Alrik Thiem wrote:
> Dear R-help list,
>
> I would like to replace all lower-case letters in a string that are not
part
> of certain fixed expressions. For example, I have the string:
>
> "pmin(pmax(pmin(x1, X2), pmin(X3, X4)) == Y, pmax(Z1, z1))"
>
> Where I would like to replace all lower-case letters that do not belong to
> the functions "pmin" and "pmax" by 1 - toupper(...) to get
>
> "pmin(pmax(pmin(1 - X1, X2), pmin(X3, X4)) == Y, pmax(Z1, 1 - Z1))"
>
> Any ideas on how I could achieve that?
>
> Many thanks and best wishes,
>
> Alrik
>
>
>
> Alrik Thiem
> Post-Doctoral Researcher
>
> Department of Philosophy
> University of Geneva
> Rue de Candolle 2
> CH-1211 Geneva
>
> +41 76 527 80 83
>
> http://www.alrik-thiem.net
> http://www.compasss.org
>
> __
> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
--
Hervé Pagès
Program in Computational Biology
Division of Public Health Sciences
Fred Hutchinson Cancer Research Center
1100 Fairview Ave. N, M1-B514
P.O. Box 19024
Seattle, WA 98109-1024
E-mail: hpa...@fredhutch.org
Phone: (206) 667-5791
Fax:(206) 667-1319
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] r script in web
I cannot see how that would be possible. R must be installed in order to use it. --- Jeff NewmillerThe . . Go Live... DCN:Basics: ##.#. ##.#. Live Go... Live: OO#.. Dead: OO#.. Playing Research Engineer (Solar/BatteriesO.O#. #.O#. with /Software/Embedded Controllers) .OO#. .OO#. rocks...1k --- Sent from my phone. Please excuse my brevity. On February 28, 2015 12:29:17 PM PST, hamid-reza kadkhodazadeh wrote: >hi >i want to run r script in web without install r on server(my server is >windows). >is it possible?how? > >thank you > > [[alternative HTML version deleted]] > >__ >R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see >https://stat.ethz.ch/mailman/listinfo/r-help >PLEASE do read the posting guide >http://www.R-project.org/posting-guide.html >and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Firefox not showing R help.
See inline below.
On 01/03/15 10:59, Duncan Murdoch wrote:
On 28/02/2015 4:10 PM, Rolf Turner wrote:
Firefox recently updated itself on my laptop. Now when I ask for R help
--- e.g. ?plot --- I just get my home page. And no help. If I do "?plot"
again after Firefox has opened its window, I just get yet another
Firefox window, opened to my home page. (I have my preferences set to
"When Firefox starts Show my homepage"
--- as I always have had in the past.)
I would guess that browseURL() won't work for any URL. Is that right?
Yes. That is correct. E.g. if I do
browseURL("http://www.r-project.org/";)
I get taken to my home page, rather than to the R home page.
What does getOption("browser") give you in R?
"/usr/bin/firefox"
If it is just a character
string (e.g. "xdg-open" is what I get in Ubuntu), does it work from your
command line, outside of R, e.g. for me that test would be
xdg-open http://www.r-project.org
I tried
/usr/bin/firefox http://www.r-project.org/
from the Linux command line and was taken to the R home page,
seamlessly. I also tried
xdg-open http://www.r-project.org/
and that worked equally well.
Finally I tried
options(browser="xdg-open")
and then
?plot
and BINGO!!! the HTML help came up as requested.
So I have a working solution to my problem. But I *really* don't
understand why changing the browser from "/usr/bin/firefox" to
"xdg-open" made a difference. (Since there appears to be no difference
at the Linux command line.)
Anyway; thanks very much for solving my problem.
cheers,
Rolf
If that doesn't work, but you can figure out a command line way to open
a particular URL, change getOption("browser") to use that.
Duncan Murdoch
The Firefox that I am currently running is (according Firefox help
--> "About Firefox") is version 36.0.
Can anyone suggest to me how I can get my html R help back?
For what it's worth: I am using Linux, Fedora 17. (Yes, I know it's
elderly, but then so am I. :-) )
Also in case it has any relevance:
sessionInfo()
R version 3.1.2 (2014-10-31)
Platform: x86_64-unknown-linux-gnu (64-bit)
locale:
[1] LC_CTYPE=en_NZ.utf8 LC_NUMERIC=C
[3] LC_TIME=en_NZ.utf8LC_COLLATE=en_NZ.utf8
[5] LC_MONETARY=en_NZ.utf8LC_MESSAGES=en_NZ.utf8
[7] LC_PAPER=en_NZ.utf8 LC_NAME=C
[9] LC_ADDRESS=C LC_TELEPHONE=C
[11] LC_MEASUREMENT=en_NZ.utf8 LC_IDENTIFICATION=C
attached base packages:
[1] stats graphics grDevices utils datasets methods base
other attached packages:
[1] spatstat_1.40-0.064 misc_0.0-16
loaded via a namespace (and not attached):
[1] abind_1.4-0 deldir_0.1-7goftest_1.0-2 grid_3.1.2
[5] lattice_0.20-29 Matrix_1.1-4mgcv_1.8-3 nlme_3.1-118
[9] polyclip_1.3-1 tensor_1.5 tools_3.1.2
--
Rolf Turner
Technical Editor ANZJS
Department of Statistics
University of Auckland
Phone: +64-9-373-7599 ext. 88276
Home phone: +64-9-480-4619
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Lubridate and NameSpace
Not certain but perhaps: > lubridate::%m+% there is a subtle difference between what the two and three colons do. -Roy On Feb 28, 2015, at 9:25 AM, Glenn Schultz wrote: > Hello All, > > I am working on a package very near completion - roxygenizing it now. The > namespace imports from lubridate importFrom(lubridate,"%m+%"). The problem > is: > > Lubridate will not load with bondlab, I have to manually click on lubridate > in R Studio to load it > I tried calling function using lubridate:::%m+% within my function based on > some discussion I found online. This is likely the correct idea but I don't > think I am calling the function correctly since it does not work in the code > when I use the above syntax. > Any suggestions are appreciated> > > Best Glenn > > The description file is as follows: > > Package: BondLab > Type: Package > Title: A package for the analysis of structured products > Version: 0.0.0 > Date: 2013-12-08 > Author: Glenn Schultz, CFA > Maintainer: Glenn Schultz > Description: The package provides a suite of software utilities for the > analysis of Mortgage and Asset Backed securities > LazyLoad: yes > License: GPL(>=3.0) > Imports: termstrc, > lubridate, > methods, > optimx, > Suggests: knitr, > devtools, > testthat > VignetteBuilder: knitr > > The Namespace is as follows: > # Generated by roxygen2 (4.1.0): do not edit by hand > > export(CPR.To.SMM) > export(DollarRoll) > export(Effective.Convexity) > export(Effective.Duration) > export(EstimYTM) > export(MakeScenario) > export(Mortgage.Monthly.Payment) > export(MortgageCashFlow) > export(PPC.Ramp) > export(PassThroughAnalytics) > export(PassThroughOAS) > export(Rates) > export(Remain.Balance) > export(SMM.To.CPR) > export(SMMVector.To.CPR) > export(Sched.Prin) > export(TermStructure) > export(TimeValue) > export(bondprice) > import(methods) > import(optimx) > importFrom(lubridate,"%m+%") > importFrom(termstrc,create_cashflows_matrix) > importFrom(termstrc,create_maturities_matrix) > importFrom(termstrc,estim_cs) > importFrom(termstrc,estim_nss) > importFrom(termstrc,forwardrates) > importFrom(termstrc,spotrates) > > > __ > R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. ** "The contents of this message do not reflect any position of the U.S. Government or NOAA." ** Roy Mendelssohn Supervisory Operations Research Analyst NOAA/NMFS Environmental Research Division Southwest Fisheries Science Center ***Note new address and phone*** 110 Shaffer Road Santa Cruz, CA 95060 Phone: (831)-420-3666 Fax: (831) 420-3980 e-mail: roy.mendelss...@noaa.gov www: http://www.pfeg.noaa.gov/ "Old age and treachery will overcome youth and skill." "From those who have been given much, much will be expected" "the arc of the moral universe is long, but it bends toward justice" -MLK Jr. __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] r script in web
hi i want to run r script in web without install r on server(my server is windows). is it possible?how? thank you [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Lubridate and NameSpace
Hello All, I am working on a package very near completion - roxygenizing it now. The namespace imports from lubridate importFrom(lubridate,"%m+%"). The problem is: Lubridate will not load with bondlab, I have to manually click on lubridate in R Studio to load it I tried calling function using lubridate:::%m+% within my function based on some discussion I found online. This is likely the correct idea but I don't think I am calling the function correctly since it does not work in the code when I use the above syntax. Any suggestions are appreciated> Best Glenn The description file is as follows: Package: BondLab Type: Package Title: A package for the analysis of structured products Version: 0.0.0 Date: 2013-12-08 Author: Glenn Schultz, CFA Maintainer: Glenn Schultz Description: The package provides a suite of software utilities for the analysis of Mortgage and Asset Backed securities LazyLoad: yes License: GPL(>=3.0) Imports: termstrc, lubridate, methods, optimx, Suggests: knitr, devtools, testthat VignetteBuilder: knitr The Namespace is as follows: # Generated by roxygen2 (4.1.0): do not edit by hand export(CPR.To.SMM) export(DollarRoll) export(Effective.Convexity) export(Effective.Duration) export(EstimYTM) export(MakeScenario) export(Mortgage.Monthly.Payment) export(MortgageCashFlow) export(PPC.Ramp) export(PassThroughAnalytics) export(PassThroughOAS) export(Rates) export(Remain.Balance) export(SMM.To.CPR) export(SMMVector.To.CPR) export(Sched.Prin) export(TermStructure) export(TimeValue) export(bondprice) import(methods) import(optimx) importFrom(lubridate,"%m+%") importFrom(termstrc,create_cashflows_matrix) importFrom(termstrc,create_maturities_matrix) importFrom(termstrc,estim_cs) importFrom(termstrc,estim_nss) importFrom(termstrc,forwardrates) importFrom(termstrc,spotrates) __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] scatter plot matrix with multiple trend lines and p values
I am trying to create a scatter plot matrix in which the lower panels
contain
scatter plots colored by group with trend lines by group, and the upper
panels contain r and p values. Using pairs() I was able to get close (ie,
scatter plot matrix on bottom with one trend line and correlation
coefficients and p values in the upper panel) but was unable to add two
lines. Using spm() I was able to get the multiple trend lines, but not the
upper
panel statistics. Does anyone know of a way to accomplish this?
My code using pairs:
pairs(~Sepal.Length+Sepal.Width+Petal.Length+Petal.Width, data = iris,
bg=c("red","blue","green")[iris$Species],
pch = 22,
lower.panel =my_line <- function(x,y,...){
points(x,y,...)
abline(a = lm(y ~ x)$coefficients[1] , b = lm(y ~ x)$coefficients[2] ,
...)
},
upper.panel = panel.cor <- function(x, y, digits = 2, cex.cor, ...)
{
usr <- par("usr"); on.exit(par(usr))
par(usr = c(0, 1, 0, 1))
# correlation coefficient
r <- cor(x, y)
txt <- format(c(r, 0.123456789), digits = digits)[1]
txt <- paste("r= ", txt, sep = "")
text(0.5, 0.6, txt)
# p-value calculation
p <- cor.test(x, y)$p.value
txt2 <- format(c(p, 0.123456789), digits = digits)[1]
txt2 <- paste("p= ", txt2, sep = "")
if(p<0.01) txt2 <- paste("p= ", "<0.01", sep = "")
text(0.5, 0.4, txt2)
})
--
View this message in context:
http://r.789695.n4.nabble.com/scatter-plot-matrix-with-multiple-trend-lines-and-p-values-tp4703996.html
Sent from the R help mailing list archive at Nabble.com.
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Substring replacement in string
Hi Alrik,
With the Biostrings/IRanges infrastructure (Bioconductor packages), you
can do this with:
library(Biostrings)
x0 <- BString("pmin(pmax(pmin(x1, X2), pmin(X3, X4)) == Y, pmax(Z1,
z1))")
donttouch_words <- c("pmin", "pmax")
## Extract the substrings to modify (target substrings).
donttouch_regions <- reduce(do.call("c", lapply(donttouch_words,
matchPattern, x0)))
target_regions <- ranges(gaps(donttouch_regions))
target_substrings <- extractAt(x0, target_regions)
## Modify them.
old <- paste0(letters, collapse="")
new <- paste0(LETTERS, collapse="")
target_substrings <- chartr(old, new, target_substrings)
## Replace in original string.
x1 <- replaceAt(x0, target_regions, target_substrings)
Then:
> x1
57-letter "BString" instance
seq: pmin(pmax(pmin(X1, X2), pmin(X3, X4)) == Y, pmax(Z1, Z1))
> as.character(x1)
[1] "pmin(pmax(pmin(X1, X2), pmin(X3, X4)) == Y, pmax(Z1, Z1))"
Hope this helps,
H.
On 02/27/2015 02:19 PM, Alrik Thiem wrote:
Dear R-help list,
I would like to replace all lower-case letters in a string that are not part
of certain fixed expressions. For example, I have the string:
"pmin(pmax(pmin(x1, X2), pmin(X3, X4)) == Y, pmax(Z1, z1))"
Where I would like to replace all lower-case letters that do not belong to
the functions "pmin" and "pmax" by 1 - toupper(...) to get
"pmin(pmax(pmin(1 - X1, X2), pmin(X3, X4)) == Y, pmax(Z1, 1 - Z1))"
Any ideas on how I could achieve that?
Many thanks and best wishes,
Alrik
Alrik Thiem
Post-Doctoral Researcher
Department of Philosophy
University of Geneva
Rue de Candolle 2
CH-1211 Geneva
+41 76 527 80 83
http://www.alrik-thiem.net
http://www.compasss.org
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Hervé Pagès
Program in Computational Biology
Division of Public Health Sciences
Fred Hutchinson Cancer Research Center
1100 Fairview Ave. N, M1-B514
P.O. Box 19024
Seattle, WA 98109-1024
E-mail: hpa...@fredhutch.org
Phone: (206) 667-5791
Fax:(206) 667-1319
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Firefox not showing R help.
On 28/02/2015 4:10 PM, Rolf Turner wrote:
>
> Firefox recently updated itself on my laptop. Now when I ask for R help
> --- e.g. ?plot --- I just get my home page. And no help. If I do "?plot"
> again after Firefox has opened its window, I just get yet another
> Firefox window, opened to my home page. (I have my preferences set to
>
> "When Firefox starts Show my homepage"
>
> --- as I always have had in the past.)
I would guess that browseURL() won't work for any URL. Is that right?
What does getOption("browser") give you in R? If it is just a character
string (e.g. "xdg-open" is what I get in Ubuntu), does it work from your
command line, outside of R, e.g. for me that test would be
xdg-open http://www.r-project.org
If that doesn't work, but you can figure out a command line way to open
a particular URL, change getOption("browser") to use that.
Duncan Murdoch
>
> The Firefox that I am currently running is (according Firefox help
> --> "About Firefox") is version 36.0.
>
> Can anyone suggest to me how I can get my html R help back?
>
> For what it's worth: I am using Linux, Fedora 17. (Yes, I know it's
> elderly, but then so am I. :-) )
>
> Also in case it has any relevance:
>
>>> sessionInfo()
>> R version 3.1.2 (2014-10-31)
>> Platform: x86_64-unknown-linux-gnu (64-bit)
>>
>> locale:
>> [1] LC_CTYPE=en_NZ.utf8 LC_NUMERIC=C
>> [3] LC_TIME=en_NZ.utf8LC_COLLATE=en_NZ.utf8
>> [5] LC_MONETARY=en_NZ.utf8LC_MESSAGES=en_NZ.utf8
>> [7] LC_PAPER=en_NZ.utf8 LC_NAME=C
>> [9] LC_ADDRESS=C LC_TELEPHONE=C
>> [11] LC_MEASUREMENT=en_NZ.utf8 LC_IDENTIFICATION=C
>>
>> attached base packages:
>> [1] stats graphics grDevices utils datasets methods base
>>
>> other attached packages:
>> [1] spatstat_1.40-0.064 misc_0.0-16
>>
>> loaded via a namespace (and not attached):
>> [1] abind_1.4-0 deldir_0.1-7goftest_1.0-2 grid_3.1.2
>> [5] lattice_0.20-29 Matrix_1.1-4mgcv_1.8-3 nlme_3.1-118
>> [9] polyclip_1.3-1 tensor_1.5 tools_3.1.2
>
> cheers,
>
> Rolf Turner
>
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] Firefox not showing R help.
Firefox recently updated itself on my laptop. Now when I ask for R help --- e.g. ?plot --- I just get my home page. And no help. If I do "?plot" again after Firefox has opened its window, I just get yet another Firefox window, opened to my home page. (I have my preferences set to "When Firefox starts Show my homepage" --- as I always have had in the past.) The Firefox that I am currently running is (according Firefox help --> "About Firefox") is version 36.0. Can anyone suggest to me how I can get my html R help back? For what it's worth: I am using Linux, Fedora 17. (Yes, I know it's elderly, but then so am I. :-) ) Also in case it has any relevance: sessionInfo() R version 3.1.2 (2014-10-31) Platform: x86_64-unknown-linux-gnu (64-bit) locale: [1] LC_CTYPE=en_NZ.utf8 LC_NUMERIC=C [3] LC_TIME=en_NZ.utf8LC_COLLATE=en_NZ.utf8 [5] LC_MONETARY=en_NZ.utf8LC_MESSAGES=en_NZ.utf8 [7] LC_PAPER=en_NZ.utf8 LC_NAME=C [9] LC_ADDRESS=C LC_TELEPHONE=C [11] LC_MEASUREMENT=en_NZ.utf8 LC_IDENTIFICATION=C attached base packages: [1] stats graphics grDevices utils datasets methods base other attached packages: [1] spatstat_1.40-0.064 misc_0.0-16 loaded via a namespace (and not attached): [1] abind_1.4-0 deldir_0.1-7goftest_1.0-2 grid_3.1.2 [5] lattice_0.20-29 Matrix_1.1-4mgcv_1.8-3 nlme_3.1-118 [9] polyclip_1.3-1 tensor_1.5 tools_3.1.2 cheers, Rolf Turner -- Rolf Turner Technical Editor ANZJS Department of Statistics University of Auckland Phone: +64-9-373-7599 ext. 88276 Home phone: +64-9-480-4619 __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Substring replacement in string
Dear Gabor,
That works perfectly!
Many thanks and best wishes,
Alrik
-Ursprüngliche Nachricht-
Von: Gabor Grothendieck [mailto:ggrothendi...@gmail.com]
Gesendet: Samstag, 28. Februar 2015 19:30
An: Alrik Thiem
Cc: r-help@r-project.org
Betreff: Re: [R] Substring replacement in string
Replace the + (i.e. 1 or more) in the pattern with a * (i.e. 0 or more):
x <- "pmin(pmax(pmin(a,B),pmin(a,C,d))==Y,pmax(E,e))"
gsub("(\\b[a-oq-z][a-z0-9]*)", "1-\\U\\1", x, perl = TRUE)
giving:
[1] "pmin(pmax(pmin(1-A,B),pmin(1-A,C,1-D))==Y,pmax(E,1-E))"
Here is a visualization of the regular expression:
https://www.debuggex.com/i/5ByOCQS2zIdPEf-f.png
On Sat, Feb 28, 2015 at 8:16 AM, Alrik Thiem wrote:
> Dear Gabor,
>
> Many thanks. Works like a charm, but I can't get it to work with
>
> "pmin(pmax(pmin(a,B),pmin(a,C,d))==Y,pmax(E,e))"
>
> i.e., with strings where there're no integers following the components in the
> pmin/pmax functions. Could this be generalized to handle both cases?
>
> Best wishes,
> Alrik
>
> -Ursprüngliche Nachricht-
> Von: Gabor Grothendieck [mailto:ggrothendi...@gmail.com]
> Gesendet: Samstag, 28. Februar 2015 13:35
> An: Alrik Thiem
> Cc: r-help@r-project.org
> Betreff: Re: [R] Substring replacement in string
>
> On Fri, Feb 27, 2015 at 5:19 PM, Alrik Thiem wrote:
>> I would like to replace all lower-case letters in a string that are not part
>> of certain fixed expressions. For example, I have the string:
>>
>> "pmin(pmax(pmin(x1, X2), pmin(X3, X4)) == Y, pmax(Z1, z1))"
>>
>> Where I would like to replace all lower-case letters that do not belong to
>> the functions "pmin" and "pmax" by 1 - toupper(...) to get
>>
>> "pmin(pmax(pmin(1 - X1, X2), pmin(X3, X4)) == Y, pmax(Z1, 1 - Z1))"
>>
>
> Assuming x is the input string:
>
> gsub("(\\b[a-oq-z][a-z0-9]+)", "1-\\U\\1", x, perl = TRUE)
> ## [1] "pmin(pmax(pmin(1-X1, X2), pmin(X3, X4)) == Y, pmax(Z1, 1-Z1))"
>
>
>
> --
> Statistics & Software Consulting
> GKX Group, GKX Associates Inc.
> tel: 1-877-GKX-GROUP
> email: ggrothendieck at gmail.com
>
--
Statistics & Software Consulting
GKX Group, GKX Associates Inc.
tel: 1-877-GKX-GROUP
email: ggrothendieck at gmail.com
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Substring replacement in string
Replace the + (i.e. 1 or more) in the pattern with a * (i.e. 0 or more):
x <- "pmin(pmax(pmin(a,B),pmin(a,C,d))==Y,pmax(E,e))"
gsub("(\\b[a-oq-z][a-z0-9]*)", "1-\\U\\1", x, perl = TRUE)
giving:
[1] "pmin(pmax(pmin(1-A,B),pmin(1-A,C,1-D))==Y,pmax(E,1-E))"
Here is a visualization of the regular expression:
https://www.debuggex.com/i/5ByOCQS2zIdPEf-f.png
On Sat, Feb 28, 2015 at 8:16 AM, Alrik Thiem wrote:
> Dear Gabor,
>
> Many thanks. Works like a charm, but I can't get it to work with
>
> "pmin(pmax(pmin(a,B),pmin(a,C,d))==Y,pmax(E,e))"
>
> i.e., with strings where there're no integers following the components in the
> pmin/pmax functions. Could this be generalized to handle both cases?
>
> Best wishes,
> Alrik
>
> -Ursprüngliche Nachricht-
> Von: Gabor Grothendieck [mailto:ggrothendi...@gmail.com]
> Gesendet: Samstag, 28. Februar 2015 13:35
> An: Alrik Thiem
> Cc: r-help@r-project.org
> Betreff: Re: [R] Substring replacement in string
>
> On Fri, Feb 27, 2015 at 5:19 PM, Alrik Thiem wrote:
>> I would like to replace all lower-case letters in a string that are not part
>> of certain fixed expressions. For example, I have the string:
>>
>> "pmin(pmax(pmin(x1, X2), pmin(X3, X4)) == Y, pmax(Z1, z1))"
>>
>> Where I would like to replace all lower-case letters that do not belong to
>> the functions "pmin" and "pmax" by 1 - toupper(...) to get
>>
>> "pmin(pmax(pmin(1 - X1, X2), pmin(X3, X4)) == Y, pmax(Z1, 1 - Z1))"
>>
>
> Assuming x is the input string:
>
> gsub("(\\b[a-oq-z][a-z0-9]+)", "1-\\U\\1", x, perl = TRUE)
> ## [1] "pmin(pmax(pmin(1-X1, X2), pmin(X3, X4)) == Y, pmax(Z1, 1-Z1))"
>
>
>
> --
> Statistics & Software Consulting
> GKX Group, GKX Associates Inc.
> tel: 1-877-GKX-GROUP
> email: ggrothendieck at gmail.com
>
--
Statistics & Software Consulting
GKX Group, GKX Associates Inc.
tel: 1-877-GKX-GROUP
email: ggrothendieck at gmail.com
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] evaluate logical expressions
Hi,
I assume input y to wrapper <- function(y) {function(x) {(y)}} is a
function. In the statement to assign gfunc[[i]]<-
gsub("(Gene)([[:digit:]])", "x[\\2]", func[[i]]) the mode of
gsub("(Gene)([[:digit:]])", "x[\\2]", func[[i]]) is character. Is this the
issue?
--
View this message in context:
http://r.789695.n4.nabble.com/evaluate-logical-expressions-tp4703964p4703970.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Substring replacement in string
Ah, I see what you mean. Thanks for suggesting. I'll try.
-Ursprüngliche Nachricht-
Von: Michael Dewey [mailto:i...@aghmed.fsnet.co.uk]
Gesendet: Samstag, 28. Februar 2015 17:31
An: Alrik Thiem
Cc: r-help@r-project.org
Betreff: Re: AW: [R] Substring replacement in string
Your original problem statement seemed to me to be one of wanting to
transform all the lower case identifiers to upper except for pmin and
pmax. My suggestion was not to bother and transform everything and then
define PMIN and PMAX.
On 28/02/2015 14:34, Alrik Thiem wrote:
> Dear Michael
>
> I'm not sure how you mean this. Maybe a more general description of my
> problem is helpful for clarifying.
>
> What I have to deal with are truth table output functions that always take,
> for example, the following form:
>
> Delta <- "(a*B+a*C*d<=>Y)*(E+e)"
>
> I.e. these functions will always have the structure (.*.+.*.+...<=>.)*(.+.),
> where the dots in the antecedent could be further conjunctions of unspecified
> complexity. I now need to filter all rows from the truth table that are
> compatible with this output function. To create the input part of the truth
> table "tt" for Delta above, I do:
>
> library(QCA) # createMatrix() function is part of this package
> Delta.upper <- toupper(Delta)
> f.names <- unique(unlist(strsplit(Delta.upper, "[(*+<=>)]")))
> f.names <- f.names[f.names != ""]
> tt <- data.frame(createMatrix(rep(2, length(f.names
> dimnames(tt) <- list(as.character(seq(2^length(f.names))), f.names)
> tt
>
>> tt
> A B C D Y E
> 1 0 0 0 0 0 0
> 2 0 0 0 0 0 1
> . . . . . . .
> 63 1 1 1 1 1 0
> 64 1 1 1 1 1 1
>
> I now need to transform Delta into a string of the following form in order to
> extract the subset of compatible rows from "tt":
>
> "pmin(pmax(pmin(1-tt$A,tt$B),pmin(1-tt$A,tt$C,1-tt$D))==tt$Y,pmax(tt$E,1-tt$E))==TRUE"
>
> With this string, I can then do:
>
>> tt[pmin(pmax(pmin(1-tt$A,tt$B),
>> pmin(1-tt$A,tt$C,1-tt$D))==tt$Y,pmax(tt$E,1-tt$E))==TRUE, ]
> A B C D Y E
> 1 0 0 0 0 0 0
> 2 0 0 0 0 0 1
> . . . . . . .
> 61 1 1 1 1 0 0
> 62 1 1 1 1 0 1
>
> -Ursprüngliche Nachricht-
> Von: Michael Dewey [mailto:i...@aghmed.fsnet.co.uk]
> Gesendet: Samstag, 28. Februar 2015 14:50
> An: Alrik Thiem
> Cc: r-help@r-project.org
> Betreff: Re: [R] Substring replacement in string
>
> Dear Alrik
>
> This may seem a silly suggestion but why not just define new functions
> PMIN and PMAX to call pmin and pmax. Obviously that does not solve your
> problem if it is more general than your example.
>
> On 28/02/2015 13:16, Alrik Thiem wrote:
>> Dear Gabor,
>>
>> Many thanks. Works like a charm, but I can't get it to work with
>>
>> "pmin(pmax(pmin(a,B),pmin(a,C,d))==Y,pmax(E,e))"
>>
>> i.e., with strings where there're no integers following the components in
>> the pmin/pmax functions. Could this be generalized to handle both cases?
>>
>> Best wishes,
>> Alrik
>>
>> -Ursprüngliche Nachricht-
>> Von: Gabor Grothendieck [mailto:ggrothendi...@gmail.com]
>> Gesendet: Samstag, 28. Februar 2015 13:35
>> An: Alrik Thiem
>> Cc: r-help@r-project.org
>> Betreff: Re: [R] Substring replacement in string
>>
>> On Fri, Feb 27, 2015 at 5:19 PM, Alrik Thiem wrote:
>>> I would like to replace all lower-case letters in a string that are not part
>>> of certain fixed expressions. For example, I have the string:
>>>
>>> "pmin(pmax(pmin(x1, X2), pmin(X3, X4)) == Y, pmax(Z1, z1))"
>>>
>>> Where I would like to replace all lower-case letters that do not belong to
>>> the functions "pmin" and "pmax" by 1 - toupper(...) to get
>>>
>>> "pmin(pmax(pmin(1 - X1, X2), pmin(X3, X4)) == Y, pmax(Z1, 1 - Z1))"
>
>>>
>>
>> Assuming x is the input string:
>>
>> gsub("(\\b[a-oq-z][a-z0-9]+)", "1-\\U\\1", x, perl = TRUE)
>> ## [1] "pmin(pmax(pmin(1-X1, X2), pmin(X3, X4)) == Y, pmax(Z1, 1-Z1))"
>>
>>
>>
>
--
Michael
http://www.dewey.myzen.co.uk
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Substring replacement in string
Your original problem statement seemed to me to be one of wanting to
transform all the lower case identifiers to upper except for pmin and
pmax. My suggestion was not to bother and transform everything and then
define PMIN and PMAX.
On 28/02/2015 14:34, Alrik Thiem wrote:
Dear Michael
I'm not sure how you mean this. Maybe a more general description of my problem
is helpful for clarifying.
What I have to deal with are truth table output functions that always take, for
example, the following form:
Delta <- "(a*B+a*C*d<=>Y)*(E+e)"
I.e. these functions will always have the structure (.*.+.*.+...<=>.)*(.+.), where the
dots in the antecedent could be further conjunctions of unspecified complexity. I now need to
filter all rows from the truth table that are compatible with this output function. To create
the input part of the truth table "tt" for Delta above, I do:
library(QCA) # createMatrix() function is part of this package
Delta.upper <- toupper(Delta)
f.names <- unique(unlist(strsplit(Delta.upper, "[(*+<=>)]")))
f.names <- f.names[f.names != ""]
tt <- data.frame(createMatrix(rep(2, length(f.names
dimnames(tt) <- list(as.character(seq(2^length(f.names))), f.names)
tt
tt
A B C D Y E
1 0 0 0 0 0 0
2 0 0 0 0 0 1
. . . . . . .
63 1 1 1 1 1 0
64 1 1 1 1 1 1
I now need to transform Delta into a string of the following form in order to extract the
subset of compatible rows from "tt":
"pmin(pmax(pmin(1-tt$A,tt$B),pmin(1-tt$A,tt$C,1-tt$D))==tt$Y,pmax(tt$E,1-tt$E))==TRUE"
With this string, I can then do:
tt[pmin(pmax(pmin(1-tt$A,tt$B),
pmin(1-tt$A,tt$C,1-tt$D))==tt$Y,pmax(tt$E,1-tt$E))==TRUE, ]
A B C D Y E
1 0 0 0 0 0 0
2 0 0 0 0 0 1
. . . . . . .
61 1 1 1 1 0 0
62 1 1 1 1 0 1
-Ursprüngliche Nachricht-
Von: Michael Dewey [mailto:i...@aghmed.fsnet.co.uk]
Gesendet: Samstag, 28. Februar 2015 14:50
An: Alrik Thiem
Cc: r-help@r-project.org
Betreff: Re: [R] Substring replacement in string
Dear Alrik
This may seem a silly suggestion but why not just define new functions
PMIN and PMAX to call pmin and pmax. Obviously that does not solve your
problem if it is more general than your example.
On 28/02/2015 13:16, Alrik Thiem wrote:
Dear Gabor,
Many thanks. Works like a charm, but I can't get it to work with
"pmin(pmax(pmin(a,B),pmin(a,C,d))==Y,pmax(E,e))"
i.e., with strings where there're no integers following the components in the
pmin/pmax functions. Could this be generalized to handle both cases?
Best wishes,
Alrik
-Ursprüngliche Nachricht-
Von: Gabor Grothendieck [mailto:ggrothendi...@gmail.com]
Gesendet: Samstag, 28. Februar 2015 13:35
An: Alrik Thiem
Cc: r-help@r-project.org
Betreff: Re: [R] Substring replacement in string
On Fri, Feb 27, 2015 at 5:19 PM, Alrik Thiem wrote:
I would like to replace all lower-case letters in a string that are not part
of certain fixed expressions. For example, I have the string:
"pmin(pmax(pmin(x1, X2), pmin(X3, X4)) == Y, pmax(Z1, z1))"
Where I would like to replace all lower-case letters that do not belong to
the functions "pmin" and "pmax" by 1 - toupper(...) to get
"pmin(pmax(pmin(1 - X1, X2), pmin(X3, X4)) == Y, pmax(Z1, 1 - Z1))"
Assuming x is the input string:
gsub("(\\b[a-oq-z][a-z0-9]+)", "1-\\U\\1", x, perl = TRUE)
## [1] "pmin(pmax(pmin(1-X1, X2), pmin(X3, X4)) == Y, pmax(Z1, 1-Z1))"
--
Michael
http://www.dewey.myzen.co.uk
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] title of r plots
Use help("plotmath") to see all the details - you can use tildes for spaces,
asterisks for no spaces.
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Fri, Feb 27, 2015 at 8:52 PM, li li wrote:
>
> Thanks very much.
> Also How do add an empty space when using expression()?
> When I do the following, it returns an error message.
>
>
> plot(1,1,main=expression(-70*degree*C%+-%10*degree*C/Ambient Condition))
>
>
>
>Hanna
>
>
> 2015-02-27 23:03 GMT-05:00 William Dunlap :
>
>> plot(1,1,main=expression(-70*degree*C%+-%10*degree*C/Ambient))
>>
>> Bill Dunlap
>> TIBCO Software
>> wdunlap tibco.com
>>
>> On Fri, Feb 27, 2015 at 7:27 PM, li li wrote:
>>
>>> Hi all,
>>> I would like to add "-70°C ± 10°C/Ambient" as the title of my plot.
>>> Could anyone give some help on this?
>>> Thanks.
>>>Hanna
>>>
>>> [[alternative HTML version deleted]]
>>>
>>> __
>>> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
>>> https://stat.ethz.ch/mailman/listinfo/r-help
>>> PLEASE do read the posting guide
>>> http://www.R-project.org/posting-guide.html
>>>
>>> and provide commented, minimal, self-contained, reproducible code.
>>
>>
>>
>
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Substring replacement in string
string <- "pmin(1, x)"
expr <- parse(text=string)[[1]]
will convert the string to an unevaluated language object.
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Fri, Feb 27, 2015 at 11:25 PM, Alrik Thiem wrote:
> Many thanks. Unfortunately, I cannot work directly on these expressions
> since they’re only created from other strings. Would I first have to
> transform these strings to unevaluated expressions?
>
>
>
> *Von:* William Dunlap [mailto:wdun...@tibco.com]
> *Gesendet**:* Freitag, 27. Februar 2015 23:39
> *An:* Alrik Thiem
> *Cc:* r-help@r-project.org
> *Betreff:* Re: [R] Substring replacement in string
>
>
>
> If your string will always represent an R expression, you could work with
>
> the expression directly with functions like all.names() and substitute().
>
>
>
> f <- function (expr)
>
> {
>
> toReplace <- setdiff(all.names(expr), c("pmin", "pmax"))
>
> toReplace <- grep(value = TRUE, "[a-z]", toReplace)
>
> names(toReplace) <- toReplace
>
> replacementList <- lapply(toReplace, function(name) call("-",
>
> 1, as.name(toupper(name
>
> do.call(substitute, list(expr, replacementList))
>
> }
>
>
>
> > In <- quote(pmin(pmax(pmin(x1, X2), pmin(X3, X4)) == Y, pmax(Z1, z1)))
>
> > Desired <- quote(pmin(pmax(pmin(1 - X1, X2), pmin(X3, X4)) == Y,
> pmax(Z1, 1 - Z1)))
>
> > all.equal(Desired, f(In))
>
> [1] TRUE
>
>
>
>
>
>
>
>
>
>
> Bill Dunlap
> TIBCO Software
> wdunlap tibco.com
>
>
>
> On Fri, Feb 27, 2015 at 2:19 PM, Alrik Thiem
> wrote:
>
> Dear R-help list,
>
> I would like to replace all lower-case letters in a string that are not
> part
> of certain fixed expressions. For example, I have the string:
>
> "pmin(pmax(pmin(x1, X2), pmin(X3, X4)) == Y, pmax(Z1, z1))"
>
> Where I would like to replace all lower-case letters that do not belong to
> the functions "pmin" and "pmax" by 1 - toupper(...) to get
>
> "pmin(pmax(pmin(1 - X1, X2), pmin(X3, X4)) == Y, pmax(Z1, 1 - Z1))"
>
> Any ideas on how I could achieve that?
>
> Many thanks and best wishes,
>
> Alrik
>
>
>
> Alrik Thiem
> Post-Doctoral Researcher
>
> Department of Philosophy
> University of Geneva
> Rue de Candolle 2
> CH-1211 Geneva
>
> +41 76 527 80 83
>
> http://www.alrik-thiem.net
> http://www.compasss.org
>
> __
> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
>
>
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Substring replacement in string
Dear Michael
I'm not sure how you mean this. Maybe a more general description of my problem
is helpful for clarifying.
What I have to deal with are truth table output functions that always take, for
example, the following form:
Delta <- "(a*B+a*C*d<=>Y)*(E+e)"
I.e. these functions will always have the structure (.*.+.*.+...<=>.)*(.+.),
where the dots in the antecedent could be further conjunctions of unspecified
complexity. I now need to filter all rows from the truth table that are
compatible with this output function. To create the input part of the truth
table "tt" for Delta above, I do:
library(QCA) # createMatrix() function is part of this package
Delta.upper <- toupper(Delta)
f.names <- unique(unlist(strsplit(Delta.upper, "[(*+<=>)]")))
f.names <- f.names[f.names != ""]
tt <- data.frame(createMatrix(rep(2, length(f.names
dimnames(tt) <- list(as.character(seq(2^length(f.names))), f.names)
tt
> tt
A B C D Y E
1 0 0 0 0 0 0
2 0 0 0 0 0 1
. . . . . . .
63 1 1 1 1 1 0
64 1 1 1 1 1 1
I now need to transform Delta into a string of the following form in order to
extract the subset of compatible rows from "tt":
"pmin(pmax(pmin(1-tt$A,tt$B),pmin(1-tt$A,tt$C,1-tt$D))==tt$Y,pmax(tt$E,1-tt$E))==TRUE"
With this string, I can then do:
> tt[pmin(pmax(pmin(1-tt$A,tt$B),
> pmin(1-tt$A,tt$C,1-tt$D))==tt$Y,pmax(tt$E,1-tt$E))==TRUE, ]
A B C D Y E
1 0 0 0 0 0 0
2 0 0 0 0 0 1
. . . . . . .
61 1 1 1 1 0 0
62 1 1 1 1 0 1
-Ursprüngliche Nachricht-
Von: Michael Dewey [mailto:i...@aghmed.fsnet.co.uk]
Gesendet: Samstag, 28. Februar 2015 14:50
An: Alrik Thiem
Cc: r-help@r-project.org
Betreff: Re: [R] Substring replacement in string
Dear Alrik
This may seem a silly suggestion but why not just define new functions
PMIN and PMAX to call pmin and pmax. Obviously that does not solve your
problem if it is more general than your example.
On 28/02/2015 13:16, Alrik Thiem wrote:
> Dear Gabor,
>
> Many thanks. Works like a charm, but I can't get it to work with
>
> "pmin(pmax(pmin(a,B),pmin(a,C,d))==Y,pmax(E,e))"
>
> i.e., with strings where there're no integers following the components in the
> pmin/pmax functions. Could this be generalized to handle both cases?
>
> Best wishes,
> Alrik
>
> -Ursprüngliche Nachricht-
> Von: Gabor Grothendieck [mailto:ggrothendi...@gmail.com]
> Gesendet: Samstag, 28. Februar 2015 13:35
> An: Alrik Thiem
> Cc: r-help@r-project.org
> Betreff: Re: [R] Substring replacement in string
>
> On Fri, Feb 27, 2015 at 5:19 PM, Alrik Thiem wrote:
>> I would like to replace all lower-case letters in a string that are not part
>> of certain fixed expressions. For example, I have the string:
>>
>> "pmin(pmax(pmin(x1, X2), pmin(X3, X4)) == Y, pmax(Z1, z1))"
>>
>> Where I would like to replace all lower-case letters that do not belong to
>> the functions "pmin" and "pmax" by 1 - toupper(...) to get
>>
>> "pmin(pmax(pmin(1 - X1, X2), pmin(X3, X4)) == Y, pmax(Z1, 1 - Z1))"
>>
>
> Assuming x is the input string:
>
> gsub("(\\b[a-oq-z][a-z0-9]+)", "1-\\U\\1", x, perl = TRUE)
> ## [1] "pmin(pmax(pmin(1-X1, X2), pmin(X3, X4)) == Y, pmax(Z1, 1-Z1))"
>
>
>
--
Michael
http://www.dewey.myzen.co.uk
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Substring replacement in string
Dear Alrik
This may seem a silly suggestion but why not just define new functions
PMIN and PMAX to call pmin and pmax. Obviously that does not solve your
problem if it is more general than your example.
On 28/02/2015 13:16, Alrik Thiem wrote:
Dear Gabor,
Many thanks. Works like a charm, but I can't get it to work with
"pmin(pmax(pmin(a,B),pmin(a,C,d))==Y,pmax(E,e))"
i.e., with strings where there're no integers following the components in the
pmin/pmax functions. Could this be generalized to handle both cases?
Best wishes,
Alrik
-Ursprüngliche Nachricht-
Von: Gabor Grothendieck [mailto:ggrothendi...@gmail.com]
Gesendet: Samstag, 28. Februar 2015 13:35
An: Alrik Thiem
Cc: r-help@r-project.org
Betreff: Re: [R] Substring replacement in string
On Fri, Feb 27, 2015 at 5:19 PM, Alrik Thiem wrote:
I would like to replace all lower-case letters in a string that are not part
of certain fixed expressions. For example, I have the string:
"pmin(pmax(pmin(x1, X2), pmin(X3, X4)) == Y, pmax(Z1, z1))"
Where I would like to replace all lower-case letters that do not belong to
the functions "pmin" and "pmax" by 1 - toupper(...) to get
"pmin(pmax(pmin(1 - X1, X2), pmin(X3, X4)) == Y, pmax(Z1, 1 - Z1))"
Assuming x is the input string:
gsub("(\\b[a-oq-z][a-z0-9]+)", "1-\\U\\1", x, perl = TRUE)
## [1] "pmin(pmax(pmin(1-X1, X2), pmin(X3, X4)) == Y, pmax(Z1, 1-Z1))"
--
Michael
http://www.dewey.myzen.co.uk
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] Using a text file as a removeWord dictionary in tm_map
Hi list
Although this query applies specifically to the tm package, perhaps it's
something that others might be able to lend a thought to.
Using tm to do some initial text mining, I want to include an external
(to R) generated dictionary of words that I want removed from the corpus.
I have created a comma separated list of terms in " " marks in a
stopList.txt plain UTF-8 file. I want to read this into R, so do:
> stopDict <- read.table('~/path/to/file/stopList.txt', sep=',')
When I want to load it as part of the removeWords function in tm, I do:
> docs <- tm_map(docs, removeWords, stopDict)
which has no effect. Neither does:
> docs <- tm_map(docs, removeWords, c(stopDict))
What am I not seeing/ doing?
How do I pass a text file with pre-defined terms to the removeWords
transform of tm?
Thanks for any ideas.
Cheers
Sun
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Substring replacement in string
Dear Gabor,
Many thanks. Works like a charm, but I can't get it to work with
"pmin(pmax(pmin(a,B),pmin(a,C,d))==Y,pmax(E,e))"
i.e., with strings where there're no integers following the components in the
pmin/pmax functions. Could this be generalized to handle both cases?
Best wishes,
Alrik
-Ursprüngliche Nachricht-
Von: Gabor Grothendieck [mailto:ggrothendi...@gmail.com]
Gesendet: Samstag, 28. Februar 2015 13:35
An: Alrik Thiem
Cc: r-help@r-project.org
Betreff: Re: [R] Substring replacement in string
On Fri, Feb 27, 2015 at 5:19 PM, Alrik Thiem wrote:
> I would like to replace all lower-case letters in a string that are not part
> of certain fixed expressions. For example, I have the string:
>
> "pmin(pmax(pmin(x1, X2), pmin(X3, X4)) == Y, pmax(Z1, z1))"
>
> Where I would like to replace all lower-case letters that do not belong to
> the functions "pmin" and "pmax" by 1 - toupper(...) to get
>
> "pmin(pmax(pmin(1 - X1, X2), pmin(X3, X4)) == Y, pmax(Z1, 1 - Z1))"
>
Assuming x is the input string:
gsub("(\\b[a-oq-z][a-z0-9]+)", "1-\\U\\1", x, perl = TRUE)
## [1] "pmin(pmax(pmin(1-X1, X2), pmin(X3, X4)) == Y, pmax(Z1, 1-Z1))"
--
Statistics & Software Consulting
GKX Group, GKX Associates Inc.
tel: 1-877-GKX-GROUP
email: ggrothendieck at gmail.com
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] Package survMisc
Hi R users. I have some problems with the package “survMisc”. When I am
loading it I am getting the following
> library(survMisc)
Loading required package: survival
Loading required package: splines
Loading required package: km.ci
Loading required package: ggplot2
Loading required package: data.table
data.table 1.9.4 For help type: ?data.table
*** NB: by=.EACHI is now explicit. See README to restore previous behaviour.
Loading required package: gridExtra
Loading required package: grid
Loading required package: rpart
Attaching package: ‘survMisc’
The following objects are masked from ‘package:stats’:
AIC, BIC, median, quantile
In the above output I noticed the line with the three stars (*). In
order to restore the data.table in its previous behavior I tried to locate
the README file but I couldn’t.
I ignored that NB in the previous output and I continue to run the
example given in the above mentioned package for the routine comp(). The
commands and the output are given below.
> ### 2 curves
> data(kidney,package="KMsurv")
> s1 <- survfit(Surv(time=time, event=delta) ~ type, data=kidney )
> comp(s1)
$tne
t n e n_type=1 e_type=1 n_type=2 e_type=2
1: 1.586 2 431
431
2: 3.580 2 401
401
3: 4.572 4 362
362
4: 5.566 2 331
331
5: 8.560 4 302 30
2
6: 9.554 2 271
271
7: 10.5 50 2 251
251
8: 11.544 2 22 1
221
9: 15.528 4 14 2
142
10: 16.5 26 2 13 1
131
11: 18.5 22 2 11 1
111
12: 23.5 8 24 1
41
13: 26.5 6 23 1
31
$tests
$tests$lrTests
ChiSq df p
Log-rank0 1 1
Gehan-Breslow (mod~ Wilcoxon) 0 1 1
Tarone-Ware 0 1 1
Peto-Peto 0 1 1
Mod~ Peto-Peto (Andersen)0 1 1
Flem~-Harr~ with p=1, q=1 0 1 1
$tests$supTests
Q p
Log-rank 0 1
Gehan-Breslow (mod~ Wilcoxon) 0 1
Tarone-Ware 0 1
Peto-Peto 0 1
Mod~ Peto-Peto (Andersen) 0 1
Renyi Flem~-Harr~ with p=1, q=1 0 1
Notice the zeros (0) that corresponds to the test statistics. (To my
opinion those zeros are strongly related to the NB above).
Next I noticed the following strange, to my opinion, thing. More
precisely I have written the following
routine
proc<-function(){
rm(list=ls())
library(survMisc)
d<-read.table("C:\\Program
Files\\R\\Data\\Survival\\HosmLem.txt",fill=TRUE,header=TRUE)
d4<-as.factor(d[,4])
s<-survfit(Surv(d[,2], d[,5])~d4)
ctest<-comp(s)$tests
print(ctest)
}
The data used are those of Hosmer and Lemeshow book on Applied Survival
Analysis. The first rows of this data set follow.
id Time Age Drug Censorentdateenddate
15 460 1 05/15/1990 10/14/1990
26 351 0 09/19/1989 03/20/1990
38 301 1 04/21/1991 12/20/1991
43 301 1 01/03/1991 04/04/1991
5 22 360 1 09/18/1989 07/19/1991
61 321 003/18/1991 04/17/1991
When I run the function proc() I am getting the answer
> proc()
Error in Surv(d[, 2], d[, 5]) : object 'd' not found
In contrast when I run the same routine command-by-command I am getting the
following output
$lrTests
ChiSq df p
Log-rank 0 1 1
Gehan-Breslow (mod~ Wilcoxon) 0 1 1
Tarone-Ware 0 1 1
Peto-Peto 0 1 1
Mod~ Peto-Peto (Andersen) 0 1 1
Flem~-Harr~ with p=1, q=1 0 1 1
$supTests
Q p
Log-rank 0 1
Gehan-Breslow (mod~ Wilcoxon) 0 1
Tarone-Ware0 1
Peto-Peto0 1
Mod~ Peto-Peto (Andersen)
Re: [R] Substring replacement in string
On Fri, Feb 27, 2015 at 5:19 PM, Alrik Thiem wrote:
> I would like to replace all lower-case letters in a string that are not part
> of certain fixed expressions. For example, I have the string:
>
> "pmin(pmax(pmin(x1, X2), pmin(X3, X4)) == Y, pmax(Z1, z1))"
>
> Where I would like to replace all lower-case letters that do not belong to
> the functions "pmin" and "pmax" by 1 - toupper(...) to get
>
> "pmin(pmax(pmin(1 - X1, X2), pmin(X3, X4)) == Y, pmax(Z1, 1 - Z1))"
>
Assuming x is the input string:
gsub("(\\b[a-oq-z][a-z0-9]+)", "1-\\U\\1", x, perl = TRUE)
## [1] "pmin(pmax(pmin(1-X1, X2), pmin(X3, X4)) == Y, pmax(Z1, 1-Z1))"
--
Statistics & Software Consulting
GKX Group, GKX Associates Inc.
tel: 1-877-GKX-GROUP
email: ggrothendieck at gmail.com
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
