element wise pattern recognition and string substitution
On Tue, Sep 6, 2016 at 11:59 PM, Jun Shen <jun.shen.ut at gmail.com> wrote:
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 find a pattern for itself.
Uh, humn, what?!? I have no idea what this means. Example? --Ista I know this is getting a little stretching. Thanks for all the
suggestion/comments from everyone. Jun On Tue, Sep 6, 2016 at 9:44 PM, Ista Zahn <istazahn at gmail.com> wrote:
If you want to mach each element of 'strings' to a different regex, do
it. Here are three ways, using your original example.
pattern1 <- "([^.]*)\\.([^.]*\\.[^.]*)\\.(.*)"
pattern2 <- "([^.]*)\\.([^.]*)\\.(.*)"
patterns <- c(pattern1,pattern2)
strings <- c('TX.WT.CUT.mean','mg.tx.cv')
for(i in seq(strings)) print(sub(patterns[i], "\\2", strings[i]))
mapply(sub, pattern = patterns, x = strings, MoreArgs=list(replacement =
"\\2"))
library(stringi)
stri_replace_all_regex(strings, patterns, "$2")
Best,
Ista
On Tue, Sep 6, 2016 at 9:20 PM, Jun Shen <jun.shen.ut at gmail.com> wrote:
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.kg.>50-70.kg.P05")
[1] ">50-70.kg"
sub("(3\\.mg\\.kg)\\.(>50-70\\.kg)\\.(.*)", '\\3',
"3.mg.kg.>50-70.kg.P05")
[1] "P05"
My problem is the pattern has to be dynamically constructed on the input
data of the function I am writing. It's actually not too difficult to
assemble the final.pattern with some code like the following
sort.var <- c('TX','WTCUT')
combn.sort.var <- do.call(expand.grid, lapply(sort.var,
function(x)paste('(',gsub('\\.','\\\\.',unlist(unique(all.exposure[x]))),
')', sep='')))
all.patterns <- do.call(paste, c(combn.sort.var, '(.*)', sep='\\.'))
final.pattern <- paste0(all.patterns, collapse='|')
You cannot run the code directly since the data object "all.exposure" is
not provided here.
Jun
On Tue, Sep 6, 2016 at 8:18 PM, Jeff Newmiller
<jdnewmil at dcn.davis.ca.us>
wrote:
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, and place your parentheses to capture that information. There is no other reason to put parentheses in the pattern... they are not grouping symbols. -- Sent from my phone. Please excuse my brevity. On September 6, 2016 5:01:04 PM PDT, Bert Gunter <bgunter.4567 at gmail.com> wrote:
Jun: 1. Tell us your desired result from your test vector and maybe someone will help. 2. As we played this game once already (you couldn't do it; I showed you how), this seems to be a function of your limitations with regular expressions. I'm probably not much better, but in any case, I don't intend to be your consultant. See if you can find someone locally to help you if you do not receive a satisfactory reply from the list. There are many people here who are pretty good at this sort of thing, but I don't know if they'll reply. Regex's are certainly complex. PERL people tend to be pretty good at them, I believe. There are numerous web sites and books on them if you need to acquire expertise for your work. 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 Tue, Sep 6, 2016 at 3:59 PM, Jun Shen <jun.shen.ut at gmail.com> wrote:
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)\\.(>90-110\\. kg)\\.(.*)|(3\\.mg\\.kg)\\.(>90-110\\.kg)\\.(.*)|(240\\.m\\ .g)\\.(50\\.kg\\.or\\.less)\\.(.*)|(3\\.mg\\.kg)\\.(50\\.kg\ \.or\\.less)\\.(.*)|(240\\.m\\.g)\\.(>110\\.kg)\\.(.*)|(3\\. mg\\.kg)\\.(>110\\.kg)\\.(.*)"
test.string <- c('240.m.g.>110.kg.geo.mean', '3.mg.kg.>110.kg.P05',
'240.m.g.>50-70.kg.geo.mean')
sub(final.pattern, '\\1', test.string)
sub(final.pattern, '\\2', test.string)
sub(final.pattern, '\\3', test.string)
Only the third string has been correctly parsed, which matches the
first
pattern. It seems the rest of the patterns are not called. Jun On Mon, Sep 5, 2016 at 10:21 PM, Bert Gunter <bgunter.4567 at gmail.com>
wrote:
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), sep="|"))
********************************************
sub(paste0("^[^b]*(",pat,").*$"),"\\1",z)
[1] "a.a" "bb" "b.bbb" -- 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 12:11 PM, Bert Gunter
<bgunter.4567 at gmail.com>
wrote:
Jun: You need to provide a clear specification via regular expressions
of
the patterns you wish to match -- at least for me to decipher it. Others may be smarter than I, though... Jeff: Thanks. I have now convinced myself that it can be done (a "proof" of sorts): If pat1, pat2,..., patn are m different
patterns
(in a vector of patterns) to be matched in a vector of n strings, where only one of the patterns will match in any string, then use paste() (probably via do.call()) or otherwise to paste them
together
separated by "|" to form the concatenated pattern, pat. Then
sub(paste0("^.*(",pat, ").*$"),"\\1",thevector)
should extract the matching pattern in each (perhaps with a
little
fiddling due to precedence rules); e.g.
z <-c(".fg.h.g.a.a", "bb..dd.ef.tgf.", "foo...b.bbb.tgy")
pat1 <- "a+\\.*a+" pat2 <-"b+\\.*b+" pat <- c(pat1,pat2)
pat <- do.call(paste,c(as.list(pat), sep="|")) pat
[1] "a+\\.*a+|b+\\.*b+"
sub(paste0("^[^b]*(",pat,").*$"), "\\1", z)
[1] "a.a" "bb" "b.bbb" 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 9:56 AM, Jun Shen <jun.shen.ut at gmail.com>
wrote:
Thanks for the reply, Bert. Your solution solves the example. I actually have a more general situation where I have this dot concatenated string from multiple
variables. The
problem is those variables may have values with dots in there.
The
number of dots are not consistent for all values of a variable. So I am
thinking
to define a vector of patterns for the vector of the string and
hopefully
to find a way to use a pattern from the pattern vector for each
value of
the string vector. The only way I can think of is "for" loop, which
can be
slow. Also these are happening in a function I am writing. Just wonder
if
there is another more efficient way. Thanks a lot. Jun On Mon, Sep 5, 2016 at 1:41 AM, Bert Gunter
<bgunter.4567 at gmail.com>
wrote:
Well, he did provide an example, and...
z <- c('TX.WT.CUT.mean','mg.tx.cv')
sub("^.+?\\.(.+)\\.[^.]+$","\\1",z)
[1] "WT.CUT" "tx" ## seems to do what was requested. Jeff would have to amplify on his initial statement however: do
you
mean that separate patterns can always be combined via "|" ? Or something deeper? 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 Sun, Sep 4, 2016 at 9:30 PM, Jeff Newmiller <jdnewmil at dcn.davis.ca.us> wrote:
Your opening assertion is false. Provide a reproducible example and someone will demonstrate. -- Sent from my phone. Please excuse my brevity. On September 4, 2016 9:06:59 PM PDT, Jun Shen <jun.shen.ut at gmail.com> wrote:
Dear list, I have a vector of strings that cannot be described by one
pattern.
So let's say I construct a vector of patterns in the same length
as the
vector of strings, can I do the element wise pattern recognition and
string
substitution.
For example,
pattern1 <- "([^.]*)\\.([^.]*\\.[^.]*)\\.(.*)"
pattern2 <- "([^.]*)\\.([^.]*)\\.(.*)"
patterns <- c(pattern1,pattern2)
strings <- c('TX.WT.CUT.mean','mg.tx.cv')
Say I want to extract "WT.CUT" from the first string and "tx"
from
the second string. If I do sub(patterns, '\\2', strings), only the first pattern will be
used.
looping the patterns doesn't work the way I want. Appreciate
any
comments.
Thanks.
Jun
[[alternative HTML version deleted]]
______________________________________________ R-help at 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 at 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 at 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.