This question already has answers here:
in R, use gsub to remove all punctuation except period
(4 answers)
Closed 2 years ago.
I have a vector vec which has elements with a punctuation mark in it. I want to return all elements with punctuation mark except the one with asterisk.
vec <- c("a,","abc","ef","abc-","abc|","abc*01")
> vec[grepl("[^*][[:punct:]]", vec)]
[1] "a," "abc-" "abc|" "abc*01"
why does it return "abc*01" if there is a negation mark[^*] for it?
Maybe you can try grep like below
grep("\\*",grep("[[:punct:]]",vec,value = TRUE), value = TRUE,invert = TRUE) # nested `grep`s for double filtering
or
grep("[^\\*[:^punct:]]",vec,perl = TRUE, value = TRUE) # but this will fail for case `abc*01|` (thanks for feedback from #Tim Biegeleisen)
which gives
[1] "a," "abc-" "abc|"
You could use grepl here:
vec <- c("a,","abc-","abc|","abc*01")
vec[grepl("^(?!.*\\*).*[[:punct:]].*$", vec, perl=TRUE)]
[1] "a," "abc-" "abc|"
The regex pattern used ^(?!.*\\*).*[[:punct:]].*$ will only match contents which does not contain any asterisk characters, while also containing at least one punctuation character:
^ from the start of the string
(?!.*\*) assert that no * occurs anywhere in the string
.* match any content
[[:punct:]] match any single punctuation character (but not *)
.* match any content
$ end of the string
Related
I have the string in R
BLCU142-09|Apodemia_mejicanus
and I would like to get the result
Apodemia_mejicanus
Using the stringr R package, I have tried
str_replace_all("BLCU142-09|Apodemia_mejicanus", "[[A-Z0-9|-]]", "")
# [1] "podemia_mejicanus"
which is almost what I need, except that the A is missing.
You can use
sub(".*\\|", "", x)
This will remove all text up to and including the last pipe char. See the regex demo. Details:
.* - any zero or more chars as many as possible
\| - a | char (| is a special regex metacharacter that is an alternation operator, so it must be escaped, and since string literals in R can contain string escape sequences, the | is escaped with a double backslash).
See the R demo online:
x <- c("BLCU142-09|Apodemia_mejicanus", "a|b|c|BLCU142-09|Apodemia_mejicanus")
sub(".*\\|", "", x)
## => [1] "Apodemia_mejicanus" "Apodemia_mejicanus"
We can match one or more characters that are not a | ([^|]+) from the start (^) of the string followed by | in str_remove to remove that substring
library(stringr)
str_remove(str1, "^[^|]+\\|")
#[1] "Apodemia_mejicanus"
If we use [A-Z] also to match it will match the upper case letter and replace with blank ("") as in the OP's str_replace_all
data
str1 <- "BLCU142-09|Apodemia_mejicanus"
You can always choose to _extract rather than _remove:
s <- "BLCU142-09|Apodemia_mejicanus"
stringr::str_extract(s,"[[:alpha:]_]+$")
## [1] "Apodemia_mejicanus"
Depending on how permissive you want to be, you could also use [[:alpha:]]+_[[:alpha:]]+ as your target.
I would keep it simple:
substring(my_string, regexpr("|", my_string, fixed = TRUE) + 1L)
This question already has answers here:
How to delete everything after nth delimiter in R?
(2 answers)
Closed 3 years ago.
I would like to remove anything after the second comma in a string -including the second comma-. Here is an example:
x <- 'Day,Bobby,Jean,Gav'
gsub("(.*),.*", "\\1", x)
and it gives:
[1] "Day, Bobby, Jean"
while I want:
[1] "Day, Bobby
regardless of the number of names that may exist in x
Use
> x <- 'Day, Bobby, Jean, Gav'
> sub("^([^,]*,[^,]*),.*", "\\1", x)
[1] "Day, Bobby"
The ^([^,]*,[^,]*),.* pattern matches
^ - start of string
([^,]*,[^,]*) - Group 1: 0+ non-commas, a comma, and 0+ non-commas
,.* - a comma and the rest of the string.
The \1 in the replacement pattern will keep Group 1 value in the result.
We can also use strsplit and then paste
toString(head(strsplit(x, ",")[[1]], 2))
#[1] "Day, Bobby"
This question already has answers here:
Remove part of string after "."
(6 answers)
Closed 3 years ago.
I have a large list which contains expressed genes from many cell lines. Ensembl genes often come with version suffixes, but I need to remove them. I've found several references that describe this here or here, but they will not work for me, likely because of my data structure (I think its a nested array within a list?). Can someone help me with the particulars of the code and with my understanding of my own data structures?
Here's some example data
>listOfGenes_version <- list("cellLine1" = c("ENSG001.1", "ENSG002.1", "ENSG003.1"), "cellLine2" = c("ENSG003.1", "ENSG004.1"))
>listOfGenes_version
$cellLine1
[1] "ENSG001.1" "ENSG002.1" "ENSG003.1"
$cellLine2
[1] "ENSG003.1" "ENSG004.1"
And what I would like to see is
>listOfGenes_trimmed
$cellLine1
[1] "ENSG001" "ENSG002" "ENSG003"
$cellLine2
[1] "ENSG003" "ENSG004"
Here are some things I tried, but did not work
>listOfGenes_trimmed <- str_replace(listOfGenes_version, pattern = ".[0-9]+$", replacement = "")
Warning message:
In stri_replace_first_regex(string, pattern, fix_replacement(replacement), :
argument is not an atomic vector; coercing
>listOfGenes_trimmed <- lapply(listOfGenes_version, gsub('\\..*', '', listOfGenes_version))
Error in match.fun(FUN) :
'gsub("\\..*", "", listOfGenes_version)' is not a function, character or symbol
Thanks so much!
An option would be to specify the pattern as . (metacharacter - so escape) followeed by one or more digits (\\d+) at the end ($) of the string and replace with blank ('")
lapply(listOfGenes_version, sub, pattern = "\\.\\d+$", replacement = "")
#$cellLine1
#[1] "ENSG001" "ENSG002" "ENSG003"
#$cellLine2
#[1] "ENSG003" "ENSG004"
The . is a metacharacter that matches any character, so we need to escape it to get the literal value as the mode is by default regex
I'm trying to extract a string after a : or ; and before a ; if the 2nd punctuation is present, then to remove everything after a ; if present. Goal result is a number.
The current code is able to do between : and ; OR after : but cannot handle ; alone or : alone.
Also, gsub(|(OF 100); SEE NOTE) isn't working, and I'm not sure why the initial : isn't being excluded and needs the gsub at all.
test<-c("Score (ABC): 2 (of 100); see note","Amount of ABC; 30%","Presence of ABC: negative","ABC not tested")
#works for :/;
toupper((regmatches(toupper(test), gregexpr(":\\s* \\K.*?(?=;)", toupper(test), perl=TRUE))))
#works for :
test<-toupper((regmatches(toupper(test), gregexpr(":\\s* (.*)", toupper(test), perl=TRUE))))
#removes extra characters:
test<-gsub(": |(OF 100); SEE NOTE|%|; ","",test)
#Negative to numeric:
test[grepl("NEGATIVE|<1",test)]<-0
test
Expected result: 2 30 0
Here are some solutions.
The first two are base. The first only uses very simple regular expressions. The second is shorter and the regular expression is only a bit more complicated. In both cases we return NA if there is no match but you can replace NAs with 0 (using ifelse(is.na(x), 0, x) where x is the answer with NAs) afterwards if that is important to you.
The third is almost the same as the second but uses strapply in gsubfn. It returns 0 instead of NA.
1) read.table Replace all colons with semicolons and read it in as semicolon-separated fields. Pick off the second such field and remove the first non-digit and everything after it. Then convert what is left to numeric.
DF <- read.table(text = gsub(":", ";", test),
as.is = TRUE, fill = TRUE, sep = ";", strip.white = TRUE)
as.numeric(sub("\\D.*", "", DF$V2))
##[1] 2 30 NA
2) strcapture Match from the start characters which are not colon or semicolon and then match a colon or semicolon and then match a space and finally capture digits. Return the captured digits converted to numeric.
strcapture("^[^:;]+[;:] (\\d+)", test, list(num = numeric(0)))$num
##[1] 2 30 NA
3) strapply Using the same pattern as in (2) convert the match to numeric and return 0 if the match is empty.
library(gsubfn)
strapply(test, "^[^:;]+[;:] (\\d+)", as.numeric, simplify = TRUE, empty = 0)
## [1] 2 30 0
Another approach:
out <- gsub('(^.+?[;:][^0-9]+)(\\d+)(.*$)|^.+', '\\2', test)
out[out == ''] <- 0
as.numeric(out)
## [1] 2 30 0
Per the OP's description (italics is mine):
extract a string after a : or ; and before a ; if the 2nd punctuation is present, then to remove everything after a ; if present. Goal result is a number.
I think some of the other suggestions may miss that italicized criteria. So here is the OP's test set with one extra condition at the end to test that:
test<-c( "Score (ABC): 2 (of 100); see note",
"Amount of ABC; 30%",
"Presence of ABC: negative",
"...and before a ; if the second punctuation is present, then remove everything after a ; if present [so 666 should not be returned]")
One-liner to return results as requested:
sub( pattern='.+?[:;]\\D*?[^;](\\d*).*?;*.*',
replacement='\\1',
x=test, perl=TRUE)
Results matching OP's request:
[1] "2" "30" "" ""
If the OP really wants an integer with zeros where no match is found, set the sub() replacement = '0\\1' and wrap with as.integer() as follows:
as.integer( gsub( pattern='.+?[:;]\\D*?[^;](\\d*).*?;*.*',
replacement='0\\1',
x=test, perl=TRUE) )
Result:
[1] 2 30 0 0
Fully working online R (R 3.3.2) example:
https://ideone.com/TTuKzG
Regexp explanation
OP wants to find just one match in a string so the sub() function works just fine.
Technique for using sub() is to make a pattern that matches all strings, but use a capture group in the middle to capture zero or more digits if conditions around it are met.
The pattern .+?[:;]\\D*?[^;](\\d*).*?;*.* is read as follows
.+? Match any character (except for line terminators) + between one and unlimited times ? as few times as possible, expanding as needed (lazy)
[:;] Match a single character in the list between the square brackets, in this case : or ;
\\D Match any character that's NOT a digit (equal to [^0-9])
*? Quantifier * Matches between zero and unlimited times ? as few times as possible, expanding as needed (lazy)
[^;] The ^ hat as first character between square brackets means: Match a single character NOT present in the list between the square brackets, in this case match any character NOT ;
(\d*) Everything between curved brackets is a capturing group - this is the 1st capturing croup: \\d* matches a digit (equal to [0-9]) between zero and unlimited times, as many times as possible(greedy)
;* Match the ; character * between zero and unlimited times [so ; does not have to be present but is matched if it is there: This is the key to excluding anything after the second delimiter as the OP requested]
.* Match any character * between zero and unlimited times, as many times as possible (greedy) [so picks up everything to the end of the line]
The replacement = \\1 refers to the 1st capture group in our pattern. We replace everything that was matched by the pattern with what we found in the capture group. \\d* can match no digits, so will return an empty string if there is no number found where we are expecting it.
Suppose I have the following two strings and want to use grep to see which match:
business_metric_one
business_metric_one_dk
business_metric_one_none
business_metric_two
business_metric_two_dk
business_metric_two_none
And so on for various other metrics. I want to only match the first one of each group (business_metric_one and business_metric_two and so on). They are not in an ordered list so I can't index and have to use grep. At first I thought to do:
.*metric.*[^_dk|^_none]$
But this doesn't seem to work. Any ideas?
You need to use a PCRE pattern to filter the character vector:
x <- c("business_metric_one","business_metric_one_dk","business_metric_one_none","business_metric_two","business_metric_two_dk","business_metric_two_none")
grep("metric(?!.*_(?:dk|none))", x, value=TRUE, perl=TRUE)
## => [1] "business_metric_one" "business_metric_two"
See the R demo
The metric(?!.*(?:_dk|_none)) pattern matches
metric - a metric substring
(?!.*_(?:dk|none)) - that is not followed with any 0+ chars other than line break chars followed with _ and then either dk or none.
See the regex demo.
NOTE: if you need to match only such values that contain metric and do not end with _dk or _none, use a variation, metric.*$(?<!_dk|_none) where the (?<!_dk|_none) negative lookbehind fails the match if the string ends with either _dk or _none.
You can also do something like this:
grep("^([[:alpha:]]+_){2}[[:alpha:]]+$", string, value = TRUE)
# [1] "business_metric_one" "business_metric_two"
or use grepl to match dk and none, then negate the logical when you're indexing the original string:
string[!grepl("(dk|none)", string)]
# [1] "business_metric_one" "business_metric_two"
more concisely:
string[!grepl("business_metric_[[:alpha:]]+_(dk|none)", string)]
# [1] "business_metric_one" "business_metric_two"
Data:
string = c("business_metric_one","business_metric_one_dk","business_metric_one_none","business_metric_two","business_metric_two_dk","business_metric_two_none")