I am trying to add parentheses around certain book titles character strings and I want to be able to paste with the paste0 function. I want to take this string:
a <- c("I Like What I Know 1959 02e pdfDrama (amazon.com)", "My Liffe 1993 07e pdfDrama (amazon.com)")
wrap certain strings in parentheses:
a
[1] “I Like What I Know (1959) (02e) (pdfDrama) (amazon.com)”
[2] ”My Life (1993) (07e) (pdfDrama) (amazon.com)”
I have tried but can't figure out a way to replace them within the string:
paste0("(",str_extract(a, "\\d{4}"),")")
paste0("(",str_extract(a, ”[0-9]+.e”),”)”)
Help?
I can suggest a regex for a fixed number of words of specific type:
a <- c("I Like What I Know 1959 02e pdfDrama (amazon.com)","My Life 1993 07e pdfDrama (amazon.com)")
sub("\\b(\\d{4})(\\s+)(\\d+e)(\\s+)([a-zA-Z]+)(\\s+\\([^()]*\\))", "(\\1)\\2(\\3)\\4(\\5)\\6", a)
See the R demo
And here is the regex demo. In short,
\\b(\\d{4}) - captures 4 digits as a whole word into Group 1
(\\s+) - Group 2: one or more whitespaces
(\\d+e) - Group 3: one or more digits and e
(\\s+) - Group 4: ibid
([a-zA-Z]+) - Group 5: one or more letters
(\\s+\\([^()]*\\)) - Group 6: one or more whitespaces, (, zero or more chars other than ( and ), ).
The contents of the groups are inserted back into the result with the help of backreferences.
If there are more words, and you need to wrap words starting with a letter/digit/underscore after a 4-digit word in the string, use
gsub("(?:(?=\\b\\d{4}\\b)|\\G(?!\\A))\\s*\\K\\b(\\S+)", "(\\1)", a, perl=TRUE)
See the R demo and a regex demo
Details:
(?:(?=\\b\\d{4}\\b)|\\G(?!\\A)) - either the location before a 4-digit whole word (see the positive lookahead (?=\\b\\d{4}\\b)) or the end of the previous successful match
\\s* - 0+ whitespaces
\\K - omitting the text matched so far
\\b(\\S+) - Group 1 capturing 1 or more non-whitespace symbols that are preceded with a word boundary.
Related
I want to match binomials, that is, bisyllabic words, sometimes hyphenated, with slightly varied syllable reduplication; the variation always concerns the first (and, possibly, second) letter in the reduplicated syllable:
x <- c("pow-wow", "pickwick", "easy-peasy", "nitty-gritty", "bzzzzzzz", "mmmmmm", "shish", "wedged", "yaaaaaa")
Here, we have said syllable reduplication in pow-wow, pickwick, easy-peasy, and nitty-gritty (which are then the expected output) but not in bzzzzzzz, mmmmmm, shish, wedged and yaaaaa.
This regex does at least manage to get rid of wedged(which is pronounced as one syllable) as well as monosyllabic words by requiring the presence of a vowel in the capturing group:
grep("\\b\\w?((?!ed)(?=[aeiou])\\w{2,})-?\\w\\w?\\1\\b$", x, value = T, perl = T)
[1] "pow-wow" "pickwick" "easy-peasy" "nitty-gritty" "yaaaaa"
However, yaaaaa is getting matched too. To not match it my feeling is that the capturing group should be disallowed to contain two identical vowels in immediate succession but I don't know how to implement that restriction.
Any ideas?
It looks as though you want to match words that cannot contain ed after the initial chars and 2 or more repeated chars if the same chunk is not found farther in the string. Also, the allowed "difference" window at the start and middle is 0 to 2 characters.
You may use
\b\w{0,2}(?!((.)\2+)(?!.*\1)|ed)([aeiou]\w+)-?\w\w?\3\b
See the regex demo
Details
\b - a word boundary (you may use ^ if your "words" are equal to whole string)
\w{0,2} - two or more word chars (replace with \p{L} to only match letters)
(?!((.)\2+)(?!.*\1)|ed) - no ed or two or more identical chars that do not repeat later in the string are allowed immediately to the right of the current location
([aeiou]\w+) - a vowel (captured in Group 3) and 1+ word chars (replace with \p{L} to only match letters)
-? - an optional hyphen
\w\w? - 1 or 2 word charsd
\3 - same value as captured in Group 3
\b - a word boundary (you may use $ if your "words" are equal to whole string)
I have some DNA sequences to process, they look like:
>KU508975.1 Acalypha australis maturase K (matK) gene, partial cds; chloroplast
TAAATTATGTGTCAGAGCTATTAATACCTTACCCCATCCATCTAGAAAAATGGGTTCAAATTCTTCGATA
TTGGCTGAAAGATCCCTCTTCTTTGCATTTATTACGACTCTTTCTTCATGAATATTGGAATTGGAACTGT
TTTCTTATTCCAAAGAAATCGATTGCTATTTTTACAAAAAGTAATCCAAGATTTTTCTTGTTTCTATATA
>KC747175.1 Achyranthes bidentata bio-material USDA:GRIN:PI613015 maturase K (matK) gene, partial cds; chloroplast
GATATATTAATACCTTACCCCGCTCATCTAGAAATCTTGGTTCAAACTCTCCGATACTGGTTGAAAGATG
CTTCTTCTTTGCATTTATTACGATTCTTTCTTTATGAGTGTCGTAATTGGATTAGTCTTATTACTCCAAA
AAAATCCATTTCCTTTTTGAAAAAAAGGAATCGAAGATTATTCTTGTTCCTATATAATTTCTATGTATGT
I coded a regex in order to detect the title line of each sequence:
(\>)([A-Z]{2}\d{6}\.?\d)\s([a-zA-Z]+\-?[a-zA-Z]+)\s([a-zA-Z]+\-?[a-zA-Z]+)\s(.*)\n
What function should I use to replace this whole match with its group3 + group4? In addition, I've got 72 matches in one txt file, how can I replace them in one run?
Your current regex won't work for lines where Group 3 or 4 contains a single letter word because [a-zA-Z]+\\-?[a-zA-Z]+ matches 1+ letters, then an optional hyphen, and then again 1+ letters (that means, there must be at least 2 letters). With [a-zA-Z]+(?:-[a-zA-Z]+)?, you can match 1+ letters followed with an optional sequence of - and then 1+ letters.
Also, \s also matches line breaks, and if your title lines are shorter than you assume then .* may grab a sequence line by mistake. You may use \h or [ \t] instead.
Note that \n is not necessary at the end of the pattern because .* matches any 0+ chars other than line break chars with an ICU regex library (it is used in your current code, str_replace_all).
In general, you should only capture with (...) what you need to keep, everything else can be just matched. Remove extra capturing parentheses, and it will save some performance.
If you add (?m)^ at the start, you will make sure you only match > that is at the start of a line.
You may use
"(?m)^>[A-Z]{2}\\d{6}\\.?\\d\\h+([a-zA-Z]+(?:-[a-zA-Z]+)?)\\h+([a-zA-Z]+(?:-[a-zA-Z]+)?).*"
See the regex demo.
Code:
Sequence <- str_replace_all(SequenceRaw,
"(?m)^>[A-Z]{2}\\d{6}\\.?\\d\\h+([a-zA-Z]+(?:-[a-zA-Z]+)?)\\h+([a-zA-Z]+(?:-[a-zA-Z]+)?).*",
"\\1 \\2")
I figured it out myself, with tidyverse packages:
library(tidyverse)
SequenceRaw <- read_file("PATH OF SEQUENCE FILE\\sequenceraw.fasta") ## e.g. sequenceraw.fasta
Sequence <- str_replace_all(SequenceRaw,
"(\\>)([A-Z]{2}\\d{6}\\.?\\d)\\s([a-zA-Z]+\\-?[a-zA-Z]+)\\s([a-zA-Z]+\\-?[a-zA-Z]+)\\s(.*)\\n",
">\\3 \\4\n") ## Keep '>' and add a new line with '\n'
write_file(Sequence, "YOUR PATH\\sequence.fasta")
Here is the result:
The original Title for this Question was : R Regex for word boundary excluding space.It reflected the manner I was approaching the problem in. However, this is a better solution to my particular problem. It should work as long as a particular delimiter is used to separate items within a 'cell'
This must be very simple, but I've hit a brick wall on it.
I have a dataframe column where each cell(row) is a comma separated list of items. I want to find the rows that have a specific item.
df<-data.frame( nms= c("XXXCAP,XXX CAPITAL LIMITED" , "XXX,XXX POLYMERS LIMITED, 3455" , "YYY,XXX REP LIMITED,999,XXX" ),
b = c('A', 'X', "T"))
nms b
1 XXXCAP,XXX CAPITAL LIMITED A
2 XXX,XXX POLYMERS LIMITED, 3455 X
3 YYY,XXX REP LIMITED,999,XXX T
I want to search for rows that have item XXX. Rows 2 and 3 should match. Row 1 has the string XXX as part of a larger string and obviously should not match.
However, because XXX in row 1 is separated by spaces in each side, I am having trouble filtering it out with \\b or [[:<:]]
grep("\\bXXX\\b",df$nms, value = F) #matches 1,2,3
The easiest way to do this of course is strsplit() but I'd like to avoid it.Any suggestions on performance are welcome.
When \b does not "work", the problem usually lies in the definition of the "whole word".
A word boundary can occur in one of three positions:
Before the first character in the string, if the first character is a word character.
After the last character in the string, if the last character is a word character.
Between two characters in the string, where one is a word character and the other is not a word character.
It seems you want to only match a word in between commas or start/end of the string).
You may use a PCRE regex (note the perl=TRUE argument) like
(?<![^,])XXX(?![^,])
See the regex demo (the expression is "converted" to use positive lookarounds due to the fact it is a demo with a single multiline string).
Details
(?<![^,]) (equal to (?<=^|,)) - either start of the string or a comma
XXX - an XXX word
(?![^,]) (equal to (?=$|,)) - either end of the string or a comma
R demo:
> grep("(?<![^,])XXX(?![^,])",df$nms, value = FALSE, perl=TRUE)
## => [1] 2 3
The equivalent TRE regex will look like
> grep("(?:^|,)XXX(?:$|,)",df$nms, value = FALSE)
Note that here, non-capturing groups are used to match either start of string or , (see (?:^|,)) and either end of string or , (see ((?:$|,))).
This is perhaps a somewhat simplistic solution, but it works for the examples which you've provided:
library(stringr)
df$nms %>%
str_replace_all('\\s', '') %>% # Removes all spaces, tabs, newlines, etc
str_detect('(^|,)XXX(,|$)') # Detects string XXX surrounded by comma or beginning/end
[1] FALSE TRUE TRUE
Also, have a look at this cheatsheet made by RStudio on Regular Expressions - it is very nicely made and very useful (I keep going back to it when I'm in doubt).
In my data (which is text), there are abbreviations.
Is there any functions or code that search for abbreviations in text? For example, detecting 3-4-5 capital letter abbreviations and letting me count how often they happen.
Much appreciated!
detecting 3-4-5 capital letter abbreviations
You may use
\b[A-Z]{3,5}\b
See the regex demo
Details:
\b - a word boundary
[A-Z]{3,5} - 3, 4 or 5 capital letters (use [[:upper:]] to match letters other than ASCII, too)
\b - a word boundary.
R demo online (leveraging the regex occurrence count code from #TheComeOnMan)
abbrev_regex <- "\\b[A-Z]{3,5}\\b";
x <- "XYZ was seen at WXYZ with VWXYZ and did ABCDEFGH."
sum(gregexpr(abbrev_regex,x)[[1]] > 0)
## => [1] 3
regmatches(x, gregexpr(abbrev_regex, x))[[1]]
## => [1] "XYZ" "WXYZ" "VWXYZ"
You can use the regular expression [A-Z] to match any ocurrence of acapital letter. If you want this pattern to be repeated 3 times you can add \1{3} to your regex. Consider using variables and a loop to get the job done for 3 to 5 repetition times.
In R, I have a string where it contains repeated groups of characters:
testString <- "Hi hi missing u lollol hahahahalol sillybilly haaaaa!"
I'm trying to use a gsub regex to replace repeated groups of characters within each word to produce the following output:
"Hi hi missing u lol halol sillybilly haaaaa!"
I've tried the following line but it isn't producing the right output:
gsub("[[:blank:]](.+?){2,}[[blank]]\\1",
replacement="\\1", testString, perl=TRUE)
What have I done wrong?
You may match repeated consecutive word chars and skip them, and then handle all other repeated consecutive chars with a solution like
x <- "Hi hi missing u lollol hahahahalol sillybilly haaaaa!"
gsub("(\\w)\\1+(*SKIP)(*F)|(\\w+?)\\2+", "\\2", x, perl=TRUE)
See the regex demo and an online R demo
Details:
(\\w)\\1+(*SKIP)(*F) - match and capture a word char (with (\\w), this can be adjusted) and then 1+ ocurrences of this same char (with \\1+) and then the whole text is discarded and the engine goes on to search for another match after the end of the match (with the PCRE (*SKIP)(*FAIL) verbs sequence)
| - or
(\\w+?)\\2+ - 1 or more word chars, as few as possible, are captured into Group 2 (with (\\w+?)) and then 1+ occurrences of the same value are matched (with \\2+).
The replacement is just the Group 2 value.