string = "ABC3JFD456"
Suppose I have the above string, and I wish to find what the first digit in the string is and store its value. In this case, I would want to store the value 3 (since it's the first-occuring digit in the string). grepl("\\d", string) only returns a logical value, but does not tell me anything about where or what the first digit is. Which regular expression should I use to find the value of the first digit?
Base R
regmatches(string, regexpr("\\d", string))
## [1] "3"
Or using stringi
library(stringi)
stri_extract_first(string, regex = "\\d")
## [1] "3"
Or using stringr
library(stringr)
str_extract(string, "\\d")
## [1] "3"
1) sub Try sub with the indicated regular expression which takes the shortest string until a digit, a digit and then everything following and replaces it with the digit:
sub(".*?(\\d).*", "\\1", string)
giving:
[1] "3"
This also works if string is a vector of strings.
2) strapplyc It would also be possible to use strapplyc from gsubfn in which case an even simpler regular expression could be used:
strapplyc(string, "\\d", simplify = TRUE)[1]
giving the same or use this which gives the same answer again but also works if string is a vector of strings:
sapply(strapplyc(string, "\\d"), "[[", 1)
Get the locations of the digits
tmp <- gregexpr("[0-9]", string)
iloc <- unlist(tmp)[1]
Extract the first digit
as.numeric(substr(string,iloc,iloc))
Using regexpr is simpler
tmp<-regexpr("[0-9]",string)
if(tmp[[1]]>=0) {
iloc <- tmp[1]
num <- as.numeric(substr(string,iloc,iloc))
}
Using rex may make this type of task a little simpler.
string = c("ABC3JFD456", "ARST4DS324")
re_matches(string,
rex(
capture(name = "first_number", digit)
)
)
#> first_number
#> 1 3
#> 2 4
> which( sapply( strsplit(string, ""), grepl, patt="[[:digit:]]"))[1]
[1] 4
Or
> gregexpr("[[:digit:]]", string)[[1]][1]
[1] 4
So:
> splstr[[1]][ which( sapply( splstr, grepl, patt="[[:digit:]]"))[1] ]
[1] "3"
Note that a full result from a gregexpr call is a list, hence the need to extract its first element with "[[":
> gregexpr("[[:digit:]]", string)
[[1]]
[1] 4 8 9 10
attr(,"match.length")
[1] 1 1 1 1
attr(,"useBytes")
[1] TRUE
A gsub solution that is based on replacing the substrings preceding and following the first digit with the empty string:
gsub("^\\D*(?=\\d)|(?<=\\d).*", "", string, perl = TRUE)
# [1] "3"
Related
I'm still a little confused by regex syntax. Can you please help me with these patterns:
_A00_A1234B_
_A00_A12345B_
_A1_A12345_
my approaches so far:
vapply(strsplit(files, "[_.]"), function(files) files[nchar(files) == 7][1], character(1))
or
str_extract(str2, "[A-Z][0-9]{5}[A-Z]")
The expected outputs are
A1234B
A12345B
A12345
Thanks!
You can try
library(stringr)
str_extract(str2, "[A-Z][0-9]{4,5}[A-Z]?")
#[1] "A1234B" "A12345B" "A12345"
Here, the pattern looks for a capital letter [A-Z], followed by 4 or 5 digits [0-9]{4,5}, followed by a capital letter [A-Z] ?
Or you can use stringi which would be faster
library(stringi)
stri_extract(str2, regex="[A-Z][0-9]{4,5}[A-Z]?")
#[1] "A1234B" "A12345B" "A12345"
Or a base R option would be
regmatches(str2,regexpr('[A-Z][0-9]{4,5}[A-Z]?', str2))
#[1] "A1234B" "A12345B" "A12345"
data
str2 <- c('_A00_A1234B_', '_A00_A12345B_', '_A1_A12345_')
vec <- c("_A00_A1234B_", "_A00_A12345B_", "_A1_A12345_")
You can use sub and this regex:
sub(".*([A-Z]\\d{4,5}[A-Z]?).*", "\\1", vec)
# [1] "A1234B" "A12345B" "A12345"
Using rex to construct the regular expression may make it more understandable.
x <- c("_A00_A1234B_", "_A00_A12345B_", "_A1_A12345_")
# approach #1, assumes always is between the second underscores.
re_matches(x,
rex(
"_",
anything,
"_",
capture(anything),
"_"
)
)
#> 1
#> 1 A1234B
#> 2 A12345B
#> 3 A12345
# approach #2, assumes an alpha, followed by 4 or 5 digits with a possible trailing alpha.
re_matches(x,
rex(
capture(
alpha,
between(digit, 4, 5),
maybe(alpha)
)
)
)
#> 1
#> 1 A1234B
#> 2 A12345B
#> 3 A12345
You can do this without using a regular expression ...
x <- c('_A00_A1234B_', '_A00_A12345B_', '_A1_A12345_')
sapply(strsplit(x, '_', fixed=T), '[', 3)
# [1] "A1234B" "A12345B" "A12345"
If you insist on using a regular expression, the following will suffice.
regmatches(x, regexpr('[^_]+(?=_$)', x, perl=T))
I have a sequence like this in a list "MSGSRRKATPASRTRVGNYEMGRTLGEGSFAKVKYAKNTVTGDQAAIKILDREKVFRHKMVEQLKREISTMKLIKHPNVVEIIEVMASKTKIYIVLELVNGGELFDKIAQQGRLKEDEARRYFQQLINAVDYCHSRGVYHRDLKPENLILDANGVLKVSDFGLSAFSRQVREDGLLHTACGTPNYVAPEVLSDKGYDGAAADVWSCGVILFVLMAGYLPFDEPNLMTLYKRICKAEFSCPPWFSQGAKRVIKRILEPNPITRISIAELLEDEWFKKGYKPPSFDQDDEDITIDDVDAAFSNSKECLVTEKKEKPVSMNAFELISSSSEFSLENLFEKQAQLVKKETRFTSQRSASEIMSKMEETAKPLGFNVRKDNYKIKMKGDKSGRKGQLSVATEVFEVAPSLHVVELRKTGGDTLEFHKVCDSFYKNFSSGLKDVVWNTDAAAEEQKQ"
I would like to create a substring like wherever a "K" is present it needs to pull out 6 characters before and 6 characters after "K"
Ex : MSGSRRKATPASR , here -6..K..+6
for the whole sequence..I tried the substring function in R but we need to specify the start and end position. Here the positions are unknown
Thanks
.{6}K.{6}
Try this.This will give the desired result.
See demo.
http://regex101.com/r/dM0rS8/4
use this:
\w{7}(?<=K)\w{6}
this uses positive lookbehind to ensure that there are characters present before K.
demo here: http://regex101.com/r/pK3jK1/2
Using rex may make this type of task a little simpler.
x <- "MSGSRRKATPASRTRVGNYEMGRTLGEGSFAKVKYAKNTVTGDQAAIKILDREKVFRHKMVEQLKREISTMKLIKHPNVVEIIEVMASKTKIYIVLELVNGGELFDKIAQQGRLKEDEARRYFQQLINAVDYCHSRGVYHRDLKPENLILDANGVLKVSDFGLSAFSRQVREDGLLHTACGTPNYVAPEVLSDKGYDGAAADVWSCGVILFVLMAGYLPFDEPNLMTLYKRICKAEFSCPPWFSQGAKRVIKRILEPNPITRISIAELLEDEWFKKGYKPPSFDQDDEDITIDDVDAAFSNSKECLVTEKKEKPVSMNAFELISSSSEFSLENLFEKQAQLVKKETRFTSQRSASEIMSKMEETAKPLGFNVRKDNYKIKMKGDKSGRKGQLSVATEVFEVAPSLHVVELRKTGGDTLEFHKVCDSFYKNFSSGLKDVVWNTDAAAEEQKQ"
library(rex)
re_matches(x,
rex(
capture(name = "amino_acids",
n(any, 6),
"K",
n(any, 6)
)
),
global = TRUE)[[1]]
#> amino_acids
#>1 MSGSRRKATPASR
#>2 GEGSFAKVKYAKN
#>3 GDQAAIKILDREK
#>4 KMVEQLKREISTM
#>5 IEVMASKTKIYIV
#>6 GGELFDKIAQQGR
#>7 VYHRDLKPENLIL
#>8 DANGVLKVSDFGL
#>9 PEVLSDKGYDGAA
#>10 NLMTLYKRICKAE
#>11 WFSQGAKRVIKRI
#>12 LEDEWFKKGYKPP
#>13 AAFSNSKECLVTE
#>14 LENLFEKQAQLVK
#>15 ASEIMSKMEETAK
#>16 LGFNVRKDNYKIK
#>17 GDKSGRKGQLSVA
#>18 HVVELRKTGGDTL
#>19 VCDSFYKNFSSGL
However the above is greedy, each K will only appear in one result.
If you want to output an AA for each K
library(rex)
locs <- re_matches(x,
rex(
"K" %if_prev_is% n(any, 6) %if_next_is% n(any, 6)
),
global = TRUE, locations = TRUE)[[1]]
substring(x, locs$start - 6, locs$end + 6)
#> [1] "MSGSRRKATPASR" "GEGSFAKVKYAKN" "GSFAKVKYAKNTV" "AKVKYAKNTVTGD"
#> [5] "GDQAAIKILDREK" "KILDREKVFRHKM" "EKVFRHKMVEQLK" "KMVEQLKREISTM"
#> [9] "REISTMKLIKHPN" "STMKLIKHPNVVE" "IEVMASKTKIYIV" "VMASKTKIYIVLE"
#>[13] "GGELFDKIAQQGR" "AQQGRLKEDEARR" "VYHRDLKPENLIL" "DANGVLKVSDFGL"
#>[17] "PEVLSDKGYDGAA" "NLMTLYKRICKAE" "LYKRICKAEFSCP" "WFSQGAKRVIKRI"
#>[21] "GAKRVIKRILEPN" "LEDEWFKKGYKPP" "EDEWFKKGYKPPS" "WFKKGYKPPSFDQ"
#>[25] "AAFSNSKECLVTE" "ECLVTEKKEKPVS" "CLVTEKKEKPVSM" "VTEKKEKPVSMNA"
#>[29] "LENLFEKQAQLVK" "KQAQLVKKETRFT" "QAQLVKKETRFTS" "ASEIMSKMEETAK"
#>[33] "KMEETAKPLGFNV" "LGFNVRKDNYKIK" "VRKDNYKIKMKGD" "KDNYKIKMKGDKS"
#>[37] "NYKIKMKGDKSGR" "IKMKGDKSGRKGQ" "GDKSGRKGQLSVA" "HVVELRKTGGDTL"
#>[41] "DTLEFHKVCDSFY" "VCDSFYKNFSSGL" "NFSSGLKDVVWNT"
I am trying to use the indexes that were returned from searching through a string for every instance of a character. When I use gregexp (pattern, text),
lookfor<-"n"
string<-"ATTnGGCnATTn"
gregexpr(pattern=lookfor,text=string)
I get the following:
[[1]]
[1] 4 8 12
attr(,"match.length")
[1] 1 1 1
attr(,"useBytes")
[1] TRUE
How do I index through the first line to be able to use those locations? Thank you in advance for your help!
Addition (2) : After thinking about this for a while, I came to the conclusion that you could have simply used unlist on your original gregexpr call
> unlist(gregexpr("n", string))
# [1] 4 8 12
From your comment
I am looking for the position of each letter n
it follows that you could do any of these:
> which(strsplit(string, "")[[1]] == "n")
# [1] 4 8 12
> cumsum(nchar(strsplit(string, "n")[[1]])+1)
# [1] 4 8 12
> nc <- 1:nchar(string)
> which(substring(string, nc, nc) == "n")
# [1] 4 8 12
Addition (1) in regards to the similar strings (comment in another answer) : You could use strsplit again, and locate those values with one of the methods above
> string2 <- "ATTTGGCCATTG"
> w <- which(strsplit(string, "")[[1]] == "n")
> strsplit(string2, "")[[1]][w]
[1] "T" "C" "G"
If you want to extract all the matches, you can use the builtin function regmatches()
m <- gregexpr(regexp,string)
regmatches(string,m)
This will return a list of character vectors because string can be greater than length 1. If you're only passing one string in, you can get at the vector of matches bypassing the list with
regmatches(string,m)[[1]]
Here is a step-by-step method to find the indices. I suspect there are more efficient ways to achieve the same result. The argument fixed = TRUE tells R to look for the literal lower case "n" rather than treat it as a regular expression.
Having done so, the [[1]] portion at the end retains only the indices element of the list
To show all indices, use the length function.
string="ATTnGGCnATTn"
index <- gregexpr(pattern = "n", text = string, fixed = TRUE)[[1]]
first.index <- index[1:length(index)]
I think I figured out my own answer!
Using the Biostrings package in Bioconductor:
string<-"ATTnGGCnATTn"
matches<-matchPattern(pattern="n", subject=string)
m<-as.vector(ranges(matches))
Now I can call up each position of of "n" in all similar strings. Thank you for those that took the time to answer.
A comment on my answer to this question which should give the desired result using strsplit does not, even though it seems to correctly match the first and last commas in a character vector. This can be proved using gregexpr and regmatches.
So why does strsplit split on each comma in this example, even though regmatches only returns two matches for the same regex?
# We would like to split on the first comma and
# the last comma (positions 4 and 13 in this string)
x <- "123,34,56,78,90"
# Splits on every comma. Must be wrong.
strsplit( x , '^\\w+\\K,|,(?=\\w+$)' , perl = TRUE )[[1]]
#[1] "123" "34" "56" "78" "90"
# Ok. Let's check the positions of matches for this regex
m <- gregexpr( '^\\w+\\K,|,(?=\\w+$)' , x , perl = TRUE )
# Matching positions are at
unlist(m)
[1] 4 13
# And extracting them...
regmatches( x , m )
[[1]]
[1] "," ","
Huh?! What is going on?
The theory of #Aprillion is exact, from R documentation:
The algorithm applied to each input string is
repeat {
if the string is empty
break.
if there is a match
add the string to the left of the match to the output.
remove the match and all to the left of it.
else
add the string to the output.
break.
}
In other words, at each iteration ^ will match the begining of a new string (without the precedent items.)
To simply illustrate this behavior:
> x <- "12345"
> strsplit( x , "^." , perl = TRUE )
[[1]]
[1] "" "" "" "" ""
Here, you can see the consequence of this behavior with a lookahead assertion as delimiter (Thanks to #JoshO'Brien for the link.)
I would like to capture the first match, and return NA if there is no match.
regexpr("a+", c("abc", "def", "cba a", "aa"), perl=TRUE)
# [1] 1 -1 3 1
# attr(,"match.length")
# [1] 1 -1 1 2
x <- c("abc", "def", "cba a", "aa")
m <- regexpr("a+", x, perl=TRUE)
regmatches(x, m)
# [1] "a" "a" "aa"
So I expected "a", NA, "a", "aa"
Staying with regexpr:
r <- regexpr("a+", x)
out <- rep(NA,length(x))
out[r!=-1] <- regmatches(x, r)
out
#[1] "a" NA "a" "aa"
use regexec instead, since it returns a list which will allow you to catch the character(0)'s before unlisting
R <- regmatches(x, regexec("a+", x))
unlist({R[sapply(R, length)==0] <- NA; R})
# [1] "a" NA "a" "aa"
In R 3.3.0, it is possible to pull out both the matches and the non-matched results using the invert=NA argument. From the help file, it says
if invert is NA, regmatches extracts both non-matched and matched substrings, always starting and ending with a non-match (empty if the match occurred at the beginning or the end, respectively).
The output is a list, typically, in most cases of interest, (matching a single pattern), regmatches with this argument will return a list with elements of either length 3 or 1. 1 is the case of where no matches are found and 3 is the case with a match.
myMatch <- regmatches(x, m, invert=NA)
myMatch
[[1]]
[1] "" "a" "bc"
[[2]]
[1] "def"
[[3]]
[1] "cb" "a" " a"
[[4]]
[1] "" "aa" ""
So to extract what you want (with "" in place of NA), you can use sapply as follows:
myVec <- sapply(myMatch, function(x) {if(length(x) == 1) "" else x[2]})
myVec
[1] "a" "" "a" "aa"
At this point, if you really want NA instead of "", you can use
is.na(myVec) <- nchar(myVec) == 0L
myVec
[1] "a" NA "a" "aa"
Some revisions:
Note that you can collapse the last two lines into a single line:
myVec <- sapply(myMatch, function(x) {if(length(x) == 1) NA_character_ else x[2]})
The default data type of NA is logical, so using it will result in additional data conversions. Using the character version NA_character_, avoids this.
An even slicker extraction method for the final line is to use [:
sapply(myMatch, `[`, 2)
[1] "a" NA "a" "aa"
So you can do the whole thing in a fairly readable single line:
sapply(regmatches(x, m, invert=NA), `[`, 2)
Using more or less the same construction as yours -
chars <- c("abc", "def", "cba a", "aa")
chars[
regexpr("a+", chars, perl=TRUE) > 0
][1] #abc
chars[
regexpr("q", chars, perl=TRUE) > 0
][1] #NA
#vector[
# find all indices where regexpr returned positive value i.e., match was found
#][return the first element of the above subset]
Edit - Seems like I misunderstood the question. But since two people have found this useful I shall let it stay.
You can use stringr::str_extract(string, pattern). It will return NA if there is no matches. It has simpler function interface than regmatches() as well.