Re: [R] Substring replacement in string
Hi Alrik,
On 02/28/2015 11:06 PM, Alrik Thiem wrote:
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.
Sounds good. I didn't realize that you also wanted to prefix the lower
case letters with 1 - so my solution was not doing the right thing
anyway. Here is the corrected solution, just in case:
library(Biostrings)
funnyReplace - function(x, protected_words)
{
x - BString(x)
## Extract the substrings to modify (target substrings).
protected_regions - reduce(do.call(c, lapply(protected_words,
matchPattern, x)))
target_regions - ranges(gaps(protected_regions))
target_substrings - extractAt(x, target_regions)
## Modify them (using a reg exp almost like Gabbor's except
## that p is not treated as an exception).
target_substrings - gsub((\\b[a-z][a-z0-9]*), 1 - \\U\\1,
target_substrings, perl=TRUE)
## Replace in original string.
x - replaceAt(x, target_regions, target_substrings)
as.character(x)
}
Then:
x - pmin(pmax(pmin(x1, X2), pmin(X3, X4)) == Y, pmax(Z1, z1))
funnyReplace(x, c(pmin, pmax))
[1] pmin(pmax(pmin(1 - X1, X2), pmin(X3, X4)) == Y, pmax(Z1, 1 - Z1))
It works even if a variable name starts with a p:
funnyReplace(pmin(p), c(pmin, pmax))
[1] pmin(1 - P)
and you can specify an arbitrary number of protected words.
Cheers,
H.
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] Substring replacement in string
On Fri, Feb 27, 2015 at 5:19 PM, Alrik Thiem alrik.th...@gmail.com 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.
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 alrik.th...@gmail.com 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
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 alrik.th...@gmail.com 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 alrik.th...@gmail.com
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]]
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R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
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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] 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 alrik.th...@gmail.com 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.
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 alrik.th...@gmail.com 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
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 alrik.th...@gmail.com 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 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 alrik.th...@gmail.com 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 alrik.th...@gmail.com 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
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 alrik.th...@gmail.com 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
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 alrik.th...@gmail.com 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 alrik.th...@gmail.com 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
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] Substring replacement in string
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 alrik.th...@gmail.com 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
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 alrik.th...@gmail.com 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.
[R] Substring replacement in string
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.
