[R] element wise pattern recognition and string substitution
Jun Shen
jun.shen.ut at gmail.com
Sat Sep 10 06:06:01 CEST 2016
Hi Jeff,
I have been trying different methods and found your approach is the most
efficient. I am able to resolve the string-parsing problem. Let me report
back to the group.
This following example explains what I was trying to achieve.
melt.results is where the strings reside, testdata is a snippet of data
where the unique values are derived. replace.metaChar is a function I
defined. Thanks for the help from everyone and appreciate any comment.
Jun
################################################################
melt.results <- structure(list(param = c("Cmin1", "Cminss", "Cmaxss",
"Cmin1",
"Cminss", "Cmin1", "Cminss", "Cmaxss", "Cmin1", "Cminss"), variable =
structure(c(1L,
5L, 9L, 14L, 18L, 21L, 25L, 29L, 34L, 38L), .Label =
c("240.mg.>110.kg.geo.mean",
"240.mg.>110.kg.cv", "240.mg.>110.kg.P05", "240.mg.>110.kg.P95",
"3.mg.kg.>110.kg.geo.mean", "3.mg.kg.>110.kg.cv", "3.mg.kg.>110.kg.P05",
"3.mg.kg.>110.kg.P95", "240.mg.>50-70.kg.geo.mean", "240.mg.>50-70.kg.cv",
"240.mg.>50-70.kg.P05", "240.mg.>50-70.kg.P95", "3.mg.kg.>50-70.kg.geo.mean",
"3.mg.kg.>50-70.kg.cv", "3.mg.kg.>50-70.kg.P05", "3.mg.kg.>50-70.kg.P95",
"240.mg.50.kg.or.less.geo.mean", "240.mg.50.kg.or.less.cv",
"240.mg.50.kg.or.less.P05",
"240.mg.50.kg.or.less.P95", "3.mg.kg.50.kg.or.less.geo.mean",
"3.mg.kg.50.kg.or.less.cv", "3.mg.kg.50.kg.or.less.P05",
"3.mg.kg.50.kg.or.less.P95",
"240.mg.>70-90.kg.geo.mean", "240.mg.>70-90.kg.cv", "240.mg.>70-90.kg.P05",
"240.mg.>70-90.kg.P95", "3.mg.kg.>70-90.kg.geo.mean", "3.mg.kg.>70-90.kg.cv",
"3.mg.kg.>70-90.kg.P05", "3.mg.kg.>70-90.kg.P95", "240.mg.>90-110.kg.geo.mean",
"240.mg.>90-110.kg.cv", "240.mg.>90-110.kg.P05", "240.mg.>90-110.kg.P95",
"3.mg.kg.>90-110.kg.geo.mean", "3.mg.kg.>90-110.kg.cv",
"3.mg.kg.>90-110.kg.P05",
"3.mg.kg.>90-110.kg.P95"), class = "factor"), value = c(97L,
144L, 76L, 137L, 18L, 104L, 92L, 87L, 111L, 41L)), .Names = c("param",
"variable", "value"), row.names = c(1L, 14L, 27L, 40L, 53L, 61L,
74L, 87L, 100L, 113L), class = "data.frame")
testdata <- structure(list(TX = c("240.mg", "3.mg.kg", "240.mg", "3.mg.kg",
"240.mg", "3.mg.kg", "240.mg", "3.mg.kg", "240.mg", "3.mg.kg"
), WTCUT = c(">50-70.kg", ">50-70.kg", ">70-90.kg", ">70-90.kg",
">90-110.kg", ">90-110.kg", "50.kg.or.less", "50.kg.or.less",
">110.kg", ">110.kg")), .Names = c("TX", "WTCUT"), row.names = c(1L,
2L, 7L, 8L, 19L, 20L, 21L, 22L, 129L, 130L), class = "data.frame")
replace.metaChar <- function(string) {
metaChar <-
c("\\$","\\*","\\+","\\.","\\?","\\[","\\]","\\^","\\{","\\}","\\|","\\(","\\)","\\\\")
metaReplace <- paste('\\',metaChar, sep='')
for(r in seq(metaChar)) gsub(metaChar[r], metaReplace[r], string) ->
string
return(string)
}
sort.var <- c('TX','WTCUT')
one.pattern <- paste('\\b',paste(sapply(sapply(sort.var,
function(x)replace.metaChar(unique(testdata[[x]]))), function(y)
paste('(',paste(y,collapse='|'),')', sep='')), collapse='\\.'), '\\.(.*)',
sep='')
n.sort.var <- length(sort.var)
one.replacement <- paste('\\', seq(n.sort.var+1), collapse='\t', sep='')
one.results <- strsplit(sub(one.pattern, one.replacement,
melt.results$variable), split='\t')
melt.results[c(sort.var,'STATS')] <- as.data.frame(do.call(rbind,
one.results))
On Wed, Sep 7, 2016 at 3:04 AM, Jeff Newmiller <jdnewmil at dcn.davis.ca.us>
wrote:
> Here are some suggestions:
>
> test.string <- c( '240.m.g.>110.kg.geo.mean'
> , '3.mg.kg.>110.kg.P05'
> , '240.m.g.>50-70.kg.geo.mean'
> )
> # based on your literal idea
> suggested.pattern1 <-
> "(240\\.m\\.g|3\\.mg\\.kg)\\.(>50-70\\.kg|>70-90\\.kg|>90-11
> 0\\.kg|50\\.kg\\.or\\.less|>110\\.kg)\\.(.*)"
>
> resultL <- strsplit( sub( suggested.pattern1
> , "\\1\t\\2\t\\3"
> , test.string )
> , split = "\t"
> )
>
> # equivalent based on apparent repetitive patterns in your sample data
> suggested.pattern2 <- "(.*?m\\.g|kg)\\.(.*?kg|.*?less)\\.(.*)"
>
> resultL2 <- strsplit( sub( suggested.pattern2
> , "\\1\t\\2\t\\3"
> , test.string
> )
> , split = "\t"
> )
>
> # put results into an organized table
> DF <- setNames( data.frame( do.call( rbind, resultL ) )
> , c( "First", "Second", "Third" )
> )
>
> By the way... please aim to make your examples reproducible. It would have
> been easy for you to define the necessary variables in example form
> rather than sending a non-reproducible example.
>
>
> On Tue, 6 Sep 2016, Jun Shen 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\\.k
>> g)\\.(.*)|(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.
>> >>> >>
>> >>> >>
>> >>
>> >>
>>
>>
>>
>>
>>
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