I have a vector with the following elements:
myvec<- c("output.chr10.recalibrated", "output.chr11.recalibrated",
"output.chrY.recalibrated")
I want to selectively extract the value after chr and before .recalibrated and get the result.
Result:
10, 11, Y
You can do that with a mere sub:
> sub(".*?chr(.*?)\\.recalibrated.*", "\\1", myvec)
[1] "10" "11" "Y"
The pattern matches any symbols before the first chr, then matches and captures any characters up to the first .recalibrated, and then matches the rest of the characters. In the replacement pattern, we use a backreference \1 that inserts the captured value you need back into the resulting string.
See the regex demo
As an alternative, use str_match:
> library(stringr)
> str_match(myvec, "chr(.*?)\\.recalibrated")[,2]
[1] "10" "11" "Y"
It keeps all captured values and helps avoid costly unanchored lookarounds in the pattern that are necessary in str_extract.
The pattern means:
chr - match a sequence of literal characters chr
(.*?) - match any characters other than a newline (if you need to match newlines, too, add (?s) at the beginning of the pattern) up to the first
\\.recalibrated - .recalibrated literal character sequence.
Both answers failing in case of slightly different inputs like whatever.chr10.whateverelse.recalibrated here's my own approach only differing on the regex part with sub:
sub(".*[.]chr([^.]*)[.].*", "\\1", myvec)
what the regex does is:
.*[.]chr match as much as possible until finding '.chr' literraly
([^.]*) capture everything not a dot after chr (could be replaced by \\d+ to capture only numeric values, requiring at least one digit present
[.].* match the rest of the line after a literal dot
I prefer the character class escape of dots ([.]) on the backslash escape (\\.) as it's usually easier to read when you're back on the regex, that's my my opinion and not covered by any best practice I know of.
We can use str_extract to do this. We match one of more characters (.*) that follow 'chr' ((?<=chr)) and before the .recalibrated ((?=\\.recalibrated)).
library(stringr)
str_extract(myvec, "(?<=chr).*(?=\\.recalibrated)")
#[1] "10" "11" "Y"
Or use gsub to match the characters until chr or (|) that starts from .recalibrated to the end ($) of the string and replace it with ''.
gsub(".*\\.chr|\\.recalibrated.*$", "", myvec)
#[1] "10" "11" "Y"
Looks like XY problem. Why extract? If this is needed in further analysis steps, we could for example do this instead:
for(chrN in c(1:22, "X", "Y")) {
myVar <- paste0("output.chr", chrN, ".recalibrated")
#do some fun stuff with myVar
print(myVar)
}
Related
I have the following string from which I want to extract the content between the second pair of colons (in bold in the example):
"20160607181026_0000005:0607181026000000501:ES5206956802492:479"
I am using R and specifically the stringr package to manipulate strings.
The command I attempted to use is:
str_extract("20160607181026_0000005:0607181026000000501:ES5206956802492:479", ":(.*):")
where the regex pattern is expressed at the end of the command. This produces the following result:
":0607181026000000501:ES5206956802492:"
I know that there is a way of grouping results and back-reference them, which would allow me to select only the part I am interested in, but I don't seem to be able to figure out the right syntax.
How can I achieve this?
Also word from stringr,
library(stringr)
word(v1, 3, sep=':')
#[1] "ES5206956802492"
If the first character after the : starts with LETTERS, then we can use a compact regex. Here, we use regex lookaround ((?<=:)) and match a LETTERS ([A-Z]) that follows the : followed by one of more characters that are not a : ([^:]+).
str_extract(v1, "(?<=:)[A-Z][^:]+")
#[1] "ES5206956802492"
or if it is based on the position i.e. 2nd position, a base R option would be to match zero or more non : ([^:]*) followed by the first : followed by zero or more non : followed by the second : and then we capture the non : in a group ((...)) and followed by rest of the characters (.*). In the replacement, we use the backreference, i.e. \\1 (first capture group).
sub("[^:]*:[^:]*:([^:]+).*", "\\1", v1)
#[1] "ES5206956802492"
Or the repeating part can be captured to make it compact
sub("([^:]*:){2}([^:]+).*", "\\2", v1)
#[1] "ES5206956802492"
Or with strsplit, we split at delimiter : and extract the 3rd element.
strsplit(v1, ":")[[1]][3]
#[1] "ES5206956802492"
data
v1 <- "20160607181026_0000005:0607181026000000501:ES5206956802492:479"
I want to replace all the single character in my string with a blank. My idea is that there should be a space before and after the single character. So i have put spaces before and after the character but that doesn't seem to work. I also wanted to replace string with more than 1 char. i.e if i want to replace all char with length 2 or so, then how would the code change.
str="I have a cat of white color"
str=gsub("([[:space:]][[a-z]][[:space:]])", "", str)
I want to replace all the single character in my string with a blank. My idea is that there should be a space before and after the single character.
The idea is not correct, a word is not always surrounded with spaces. What if the words is at the beginning of the string? Or at the end? Or is followed with a punctuation?
Use \b word boundary:
There are three different positions that qualify as word boundaries:
- 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.
NOTE that in R, when you use gsub, it is best to use it with the PCRE regex (pass perl=T):
POSIX 1003.2 mode of gsub and gregexpr does not work correctly with repeated word-boundaries (e.g., pattern = "\b"). Use perl = TRUE for such matches (but that may not work as expected with non-ASCII inputs, as the meaning of ‘word’ is system-dependent).
So, to match all 1-letter words, you need to use
gsub("(?i)\\b[a-z]\\b", "REPLACEMENT", input, perl=T) ## To replace 1 ASCII letter words
Note that (?i) is a case-insensitive modifier (making a match both a and A).
Now, you need to match 2 letter words:
gsub("(?i)\\b[a-z]{2}\\b", "REPLACEMENT", input, perl=T) ## To replace 2 ASCII letter words
Here, we are using a limiting quantifier {min, max} / {max} to specify how many times the pattern quantified with this construct can be repeated.
See IDEONE demo:
> input = "I am a football fan"
> gsub("(?i)\\b[a-z]\\b", "REPLACEMENT", input, perl=T) ## To replace 1 ASCII letter words
[1] "REPLACEMENT am REPLACEMENT football fan"
gsub("(?i)\\b[a-z]{2}\\b", "REPLACEMENT", input, perl=T) ## To replace 2 ASCII letter words
[1] "I REPLACEMENT a football fan"
You need to use the quantifier regex property, e.g. [a-z]{2} which matches the letters a to z twice together. The regex pattern you want is something along the lines of this:
\\s[a-z]{2}\\s
You can build this regex dynamically in R using an input number of characters. Here is a code snippet which demonstrates this:
str <- "I have a cat of white color"
nchars <- 2
exp <- paste0("\\s[a-z]{", nchars, "}\\s")
> gsub(exp, "", str)
[1] "I have a catwhite color"
I'm looking for a regular expression to catch all digits in the first 7 characters in a string.
This string has 12 characters:
A12B345CD678
I would like to remove A and B only since they are within the first 7 chars (A12B345) and get
12345CD678
So, the CD678 should not be touched. My current solution in R:
paste(paste(str_extract_all(substr("A12B345CD678",1,7), "[0-9]+")[[1]],collapse=""),substr("A12B345CD678",8,nchar("A12B345CD678")),sep="")
It seems too complicated. I split the string at 7 as described, match any digits in the first 7 characters and bind it with the rest of the string.
Looking for a general answer, my current solution is to split the first 7 characters and just match all digits in this sub string.
Any help appreciated.
You can use the known SKIP-FAIL regex trick to match all the rest of the string beginning with the 8th character, and only match non-digit characters within the first 7 with a lookbehind:
s <- "A12B345CD678"
gsub("(?<=.{7}).*$(*SKIP)(*F)|\\D", "", s, perl=T)
## => [1] "12345CD678"
See IDEONE demo
The perl=T is required for this regex to work. The regex breakdown:
(?<=.{7}).*$(*SKIP)(*F) - matches any character but a newline (add (?s) at the beginning if you have newline symbols in the input), as many as possible (.*) up to the end ($, also \\z might be required to remove final newlines), but only if preceded with 7 characters (this is set by the lookbehind (?<=.{7})). The (*SKIP)(*F) verbs make the engine omit the whole matched text and advance the regex index to the position at the end of that text.
| - or...
\\D - a non-digit character.
See the regex demo.
The regex solution is cool, but I'd use something easier to read for maintainability. E.g.
library(stringr)
str_sub(s, 1, 7) = gsub('[A-Z]', '', str_sub(s, 1, 7))
You can also use a simple negative lookbehind:
s <- "A12B345CD678"
gsub("(?<!.{7})\\D", "", s, perl=T)
I need to find a regexp that allows me to find strings in which i have all the required numbers but only once.
For example:
a <- c("12","13","112","123","113","1123","23","212","223","213","2123","312","323","313","3123","1223","1213","12123","2313","23123","13123")
I want to get:
"123" "213" "312"
The pattern 123 only once and in any order and in any position of the string
I tried a lot of things and this seemed to be the closer while it's still very far from what I want :
grep('[1:3][1:3][1:3]', a, value=TRUE)
[1] "113" "313" "2313" "13123"
What i exactly need is to find all 3 digit numbers containing 1 2 AND 3 digits
Then you can safely use
grep('^[123]{3}$', a, value=TRUE)
##=> [1] "112" "123" "113" "212" "223" "213" "312" "323" "313"
The regex matches:
^ - start of string
[123]{3} - Exactly 3 characters that are either 1, or 2 or 3
$ - assert the position at the end of string.
Also, if you only need unique values, use unique.
If you do not need to allow the same digit more than once, you need a Perl-based regex:
grep('^(?!.*(.).*\\1)[123]{3}$', a, value=TRUE, perl=T)
## => [1] "123" "213" "312"
Note the double escaped back-reference. The (?!.*(.).*\\1) negative look-ahead will check if the string has no repeated symbols with the help of a capturing group (.) and a back-reference that forces the same captured text to appear in the string. If the same characters are found, there will be no match. See IDEONE demo.
The (?!.*(.).*\\1) is a negative look-ahead. It only asserts the absence of some pattern after the current regex engine position, i.e. it checks and returns true if there is no match, otherwise it returns false. Thus, it does not not "consume" characters, it does not "match" the pattern inside the look-ahead, the regex engine stays at the same location in the input string. In this regex, it is the beginning of string (^). So, right at the beginning of the string, the regex engine starts looking for .* (any character but a newline, 0 or more repetitions), then captures 1 character (.) into group 1, again matches 0 or more characters with .*, and then tries to match the same text inside group 1 with \\1. Thus, if there is 121, there will be no match since the look-ahead will return false as it will find two 1s.
you can as well use this
grep('^([123])((?!\\1)\\d)(?!\\2|\\1)\\d', a, value=TRUE, perl=T)
see demo
I have a string titled thisLine and I'd like to remove all characters before the first integer. I can use the command
regexpr("[0123456789]",thisLine)[1]
to determine the position of the first integer. How do I use that index to split the string?
The short answer:
sub('^\\D*', '', thisLine)
where
^ matches the beginning of the string
\\D matches any non-digit (it is the opposite of \\d)
\\D* tries to match as many consecutive non-digits as possible
My personal preference, skipping regexp altogether:
sub("^.*?(\\d)","\\1",thisLine)
#breaking down the regex
#^ beginning of line
#. any character
#* repeated any number of times (including 0)
#? minimal qualifier (match the fewest characters possible with *)
#() groups the digit
#\\d digit
#\\1 backreference to first captured group (the digit)
You want the substring function.
Or use gsub to do work in one shot:
> gsub('^[^[:digit:]]*[[:digit:]]', '', 'abc1def')
[1] "def"
You may want to include that first digit, which can be done with a capture:
> gsub('^[^[:digit:]]*([[:digit:]])', '\\1', 'abc1def')
[1] "1def"
Or as flodel and Alan indicate, simply replace "all leading digits" with a blank. See flodel's answer.