I want to extract all substrings that begin with M and are terminated by a *
The string below as an example;
vec<-c("SHVANSGYMGMTPRLGLESLLE*A*MIRVASQ")
Would ideally return;
MGMTPRLGLESLLE
MTPRLGLESLLE
I have tried the code below;
regmatches(vec, gregexpr('(?<=M).*?(?=\\*)', vec, perl=T))[[1]]
but this drops the first M and only returns the first string rather than all substrings within.
"GMTPRLGLESLLE"
You can use
(?=(M[^*]*)\*)
See the regex demo. Details:
(?= - start of a positive lookahead that matches a location that is immediately followed with:
(M[^*]*) - Group 1: M, zero or more chars other than a * char
\* - a * char
) - end of the lookahead.
See the R demo:
library(stringr)
vec <- c("SHVANSGYMGMTPRLGLESLLE*A*MIRVASQ")
matches <- stringr::str_match_all(vec, "(?=(M[^*]*)\\*)")
unlist(lapply(matches, function(z) z[,2]))
## => [1] "MGMTPRLGLESLLE" "MTPRLGLESLLE"
If you prefer a base R solution:
vec <- c("SHVANSGYMGMTPRLGLESLLE*A*MIRVASQ")
matches <- regmatches(vec, gregexec("(?=(M[^*]*)\\*)", vec, perl=TRUE))
unlist(lapply(matches, tail, -1))
## => [1] "MGMTPRLGLESLLE" "MTPRLGLESLLE"
This could be done instead with a for loop on a char array converted from you string.
If you encounter a M you start concatenating chars to a new string until you encounter a *, when you do encounter a * you push the new string to an array of strings and start over from the first step until you reach the end of your loop.
It's not quite as interesting as using REGEX to do it, but it's failsafe.
It is not possible to use regular expressions here, because regular languages don't have memory states required for nested matches.
stringr::str_extract_all("abaca", "a[^a]*a") only gives you aba but not the sorrounding abaca.
The first M was dropped, because (?<=M) is a positive look behind which is by definition not part of the match, but just behind it.
Related
I have a list of string like so:
batch1, batch2, batch3, batch10, batch11
I am trying to add a 0 before the single digits batch01, batch02, batch03, batch10, batch11
I have found many similar questions and tried to write my own regex. I am very close, but I can't quite make it do what I want.
Batch <- gsub('(.{5})([0-9]{1}\\b)','\\10\\2', Batch)
outputs batch01, batch02, batch 03, batch100, batch110
\\s instead of \\b doesn't change any values
sampleNames$Batch <- gsub('(.{5})([0-9]{1})','\\10\\2', sampleNames$Batch) outputs bacth01, batch02, batch03, batch010, batch011
I've played around with a few other versions but I cannot seem to get it correct. I know this is a somewhat repetitive question, but I have not been able to alter previous solutions to do what I need to do.
We can capture the last digit and the lower case letter before it as two groups, then in the replacement specify the backreference of the groups and the 0 in between. Thus, it won't match the ones having two digits at the end of the string
sub("([a-z])(\\d)$", "\\10\\2", Batch)
[1] "batch01" "batch02" "batch03" "batch10" "batch11"
Or we may use sprintf/str_pad with str_replace
library(stringr)
str_replace(Batch, "\\d+$", function(x) sprintf("%02d", as.numeric(x)))
[1] "batch01" "batch02" "batch03" "batch10" "batch11"
data
Batch <- c("batch1", "batch2", "batch3", "batch10", "batch11")
Use
sampleNames$Batch <- sub("(\\D|^)(\\d)$", "\\10\\2", sampleNames$Batch, perl=TRUE)
See regex proof.
EXPLANATION
--------------------------------------------------------------------------------
( group and capture to \1:
--------------------------------------------------------------------------------
\D non-digits (all but 0-9)
--------------------------------------------------------------------------------
| OR
--------------------------------------------------------------------------------
^ the beginning of the string
--------------------------------------------------------------------------------
) end of \1
--------------------------------------------------------------------------------
( group and capture to \2:
--------------------------------------------------------------------------------
\d digits (0-9)
--------------------------------------------------------------------------------
) end of \2
--------------------------------------------------------------------------------
$ before an optional \n, and the end of the
string
You can also use the following solution:
sapply(vec, function(x) {
d <- gsub("([[:alpha:]]+)(\\d)", "\\2", x)
if(nchar(d) == 1) {
gsub("([[:alpha:]]+)(\\d)", "\\10\\2", x)
} else {
x
}
})
batch1 batch2 batch3 batch10 batch11
"batch01" "batch02" "batch03" "batch10" "batch11"
I have phonemic transcriptions of English words such as these:
test <- c("ˈsɜːtnli", "ˈtwɛnti", "ˈfɒksi", "kɑːnt", "ʧeɪnʤd", "vɪkˈtɔːrɪə", "wɒznt", "ðeər", "dɪdnt",
"ˈdɪzni", "ˈəʊnli", "ˈfæbrɪks", "sɪˈkjʊərɪti", "ˈnjuːzˌpeɪpər", "ɑhɑː")
I'd like to match mono-syllabic words, i.e., words that contain a single vowel. My set of phonemic vowels is this:
vowel <- "iː|aɪ|ɔː|ɔɪ|əʊ|ɛə|eɪ|aʊ|eə|uː|ɑː|ɪə|ɜː|ʊə|ə|ɪ|ɒ|ʊ|ʌ|æ|e|ɑ|ɛ|i"
Using str_count and the vector vowel as pattern, I'm able to match a fairly good set of words:
library(stringr)
test[str_count(test, vowel) == 1]
[1] "kɑːnt" "ʧeɪnʤd" "wɒznt" "ðeər" "dɪdnt"
However, wɒznt and dɪdntcan be seen as bi-syllabic (as the nsound can replace a vowel so that nt counts as a second vowel). So the question is, how can I match mono-syllabic words except those that end in nt?
What I've tried so far is this set operation, which works well but looks clumsy:
setdiff(test[str_count(test, vowel) == 1], test[str_count(test, paste0("[^", vowel, "]nt$")) == 1])
[1] "kɑːnt" "ʧeɪnʤd" "ðeər"
I'd much rather have a single more concise regex. Any ideas?
You can use
test <- c("ˈsɜːtnli", "ˈtwɛnti", "ˈfɒksi", "kɑːnt", "ʧeɪnʤd", "vɪkˈtɔːrɪə", "wɒznt", "ðeər", "dɪdnt",
"ˈdɪzni", "ˈəʊnli", "ˈfæbrɪks", "sɪˈkjʊərɪti", "ˈnjuːzˌpeɪpər", "ɑhɑː")
vowel <- "iː|aɪ|ɔː|ɔɪ|əʊ|ɛə|eɪ|aʊ|eə|uː|ɑː|ɪə|ɜː|ʊə|ə|ɪ|ɒ|ʊ|ʌ|æ|e|ɑ|ɛ|i"
library(stringr)
p <- paste0("^(?!.*(?<!",vowel,")nt$)(?:(?!",vowel,").)*(?:",vowel,")(?:(?!",vowel,").)*$")
test[str_detect(test, p)]
## => [1] "kɑːnt" "ʧeɪnʤd" "ðeər"
See the online R demo. See the regex demo. The pattern means
^ - start of string
(?!.*(?<!",vowel,")nt$) - immediately to the right, there must not be any 0+ chars other than line break chars as many as possible followed with nt (not preceded with any of the specified vowel sound sequences) and end of string
(?:(?!",vowel,").)* - any char but a line break char, zero or more times as many as possible, that does not start a vowel char sequence
(?:",vowel,") - any of the specified vowel sound sequences
(?:(?!",vowel,").)* - any char but a line break char, zero or more times as many as possible, that does not start a vowel char sequence
$ - end of string.
This is a somewhat concise solution (thanks to #G5W for the decisive hint):
vowel_cc <- paste0(unique(unlist(strsplit(gsub("\\|", "", vowel), ""))), collapse = "")
vowel_cc
[1] "iːaɪɔəʊɛeuɑɜɒʌæ"
test[str_count(test, paste0(vowel, "|[^", vowel_cc, "]+nt$")) == 1]
[1] "kɑːnt" "ʧeɪnʤd" "ðeər"
This solution uses a vector vowel_cc consisting of all unique characters in vowels. These serve as input for a negated character class. The pattern specifies nt as one of the vowel alternatives on the condition that it be preceded by one or more non-vowel_ccs and occur at string end.
My string patterns looks like this:
UNB+UNOC:3+4399945681577+_GLN_Company__+180101:0050+10870 and I am trying to extract everything after the second last +, i.e. 180101:0050+10870.
Thus far, I managed to address the second last block 180101:0050 with this expression (?<=\+)[^\+]+(?=\+[^\+]*$) but fail to include the last block including the last +. Here is my sample: regex101
The expression is meant for R and I still need to escape the characters later on. This format it just for testing purposes in Regex101.
We could capture group based on the occurrence of + from the end ($) of the string.
sub(".*\\+([^+]+\\+[^+]+$)", "\\1", str1)
#[1] "180101:0050+10870"
data
str1 <- "UNB+UNOC:3+4399945681577+_GLN_Company__+180101:0050+10870"
You may use
\+\K[^+]+\+[^+]*$
Or, if you would like to use it with stringr::str_extract:
(?<=\+)[^+]+\+[^+]*$
See the regex demo. Details:
\+ - a + char
\K - match reset operator
(?<=\+) - location right after a + symbol
[^+]+ - one or more chars other than +
\+ - a +
[^+]+ - one or more chars other than +
$ - end of string.
See R demo online:
x <- "UNB+UNOC:3+4399945681577+_GLN_Company__+180101:0050+10870"
regmatches(x, regexpr("\\+\\K[^+]+\\+[^+]*$", x, perl=TRUE))
## => [1] "180101:0050+10870"
library(stringr)
str_extract(x, "(?<=\\+)[^+]+\\+[^+]*$")
## => [1] "180101:0050+10870"
Another way you can do in this case:
library(stringr)
str_extract("UNB+UNOC:3+4399945681577+_GLN_Company__+180101:0050+10870", "\\d+:\\d+\\+\\d+")
#"180101:0050+10870"
How can I match all words starting with plan_ and not ending with template without using invert = TRUE? In the below example, I'd like to match only the second string. I tried with negative lookahead but it does not work, maybe because of greediness?
names <- c("plan_x_template", "plan_x")
grep("^plan.*(?!template)$",
names,
value = TRUE, perl = TRUE
)
#> [1] "plan_x_template" "plan_x"
I mean one can also solve the problem with two regex calls but I'd like to see how it works the other way :-)
is_plan <- grepl("^plan_", names)
is_template <- grepl("_template$", names)
names[is_plan & !is_template]
#> [1] "plan_x"
You may use
names <- c("plan_x_template", "plan_x")
grep("^plan(?!.*template)",
names,
value = TRUE, perl = TRUE
)
See the R online demo
The ^plan(?!.*template) pattern matches:
^ - a start of string
plan - a plan substring
(?!.*template) - a negative lookahead that fails the match if, immediately to the left of the current location, there are 0+ chars other than line break chars (since perl = TRUE is used and the pattern is processed with a PCRE engine, the . does not match all possible chars as opposed to the default grep TRE regex engine), as many as possible, followed with template substring.
NOTE: In case of multiline strings, you need to use a DOTALL modifier in the regex, "(?s)^plan(?!.*template)".
Lest's say I have a string:
test <- "(pop+corn)-bread+salt"
I want to replace the plus sign that is only between parenthesis by '|', so I get:
"(pop|corn)-bread+salt"
I tried:
gsub("([+])","\\|",test)
But it replaces all the plus signs of the string (obviously)
If you want to replace all + symbols that are inside parentheses (if there may be 1 or more), you can use any of the following solutions:
gsub("\\+(?=[^()]*\\))", "|", x, perl=TRUE)
See the regex demo. Here, the + is only matched when it is followed with any 0+ chars other than ( and ) (with [^()]*) and then a ). It is only good if the input is well-formed and there is no nested parentheses as it does not check if there was a starting (.
gsub("(?:\\G(?!^)|\\()[^()]*?\\K\\+", "|", x, perl=TRUE)
This is a safer solution since it starts matching + only if there was a starting (. See the regex demo. In this pattern, (?:\G(?!^)|\() matches the end of the previous match (\G(?!^)) or (|) a (, then [^()]*? matches any 0+ chars other than ( and ) chars, and then \K discards all the matched text and \+ matches a + that will be consumed and replaced. It still does not handle nested parentheses.
Also, see an online R demo for the above two solutions.
library(gsubfn)
s <- "(pop(+corn)+unicorn)-bread+salt+malt"
gsubfn("\\((?:[^()]++|(?R))*\\)", ~ gsub("+", "|", m, fixed=TRUE), s, perl=TRUE, backref=0)
## => [1] "(pop(|corn)|unicorn)-bread+salt+malt"
This solves the problem of matching nested parentheses, but requires the gsubfn package. See another regex demo. See this regex description here.
Note that in case you do not have to match nested parentheses, you may use "\\([^()]*\\)" regex with the gsubfn code above. \([^()]*\) regex matches (, then any zero or more chars other than ( and ) (replace with [^)]* to match )) and then a ).
We can try
sub("(\\([^+]+)\\+","\\1|", test)
#[1] "(pop|corn)-bread+salt"