I have a vector s of strings (or NAs), and would like to get a vector of same length of everything before first occurrence of punctionation (.).
s <- c("ABC1.2", "22A.2", NA)
I would like a result like:
[1] "ABC1" "22A" NA
You can remove all symbols (incl. a newline) from the first dot with the following Perl-like regex:
s <- c("ABC1.2", "22A.2", NA)
gsub("[.][\\s\\S]*$", "", s, perl=T)
## => [1] "ABC1" "22A" NA
See IDEONE demo
The regex matches
[.] - a literal dot
[\\s\\S]* - any symbols incl. a newline
$ - end of string.
All matched strings are removed from the input with "". As the regex engine analyzes the string from left to right, the first dot is matched with \\., and the greedy * quantifier with [\\s\\S] will match all up to the end of string.
If there are no newlines, a simpler regex will do: [.].*$:
gsub("[.].*$", "", s)
See another demo
Related
I have this string:
[1] "19980213" "19980214" "19980215" "19980216" "19980217" "iffi" "geometry"
[8] "date_consid"
and I want to match all the elements that are not dates and not "date_consid". I tried
res = grep("(?!\\d{8})|(?!date_consid)", vec, value=T)
But I just cant make it work...
You can use
vec <- c("19980213", "19980214", "19980215", "19980216","19980217", "iffi","geometry", "date_consid")
grep("^(\\d{8}|date_consid)$", vec, value=TRUE, invert=TRUE)
## => [1] "iffi" "geometry"
See the R demo
The ^(\d{8}|date_consid)$ regex matches a string that only consists of any eight digits or that is equal to date_consid.
The value=TRUE makes grep return values rather than indices and invert=TRUE inverses the regex match result (returns those that do not match).
The pattern that you tried gives all the matches because the lookaheads are unanchored.
Using separate statements with or | will still match all strings.
You can change to logic to asserting from the start of the string, what is directly to the right is not either 8 digits or date_consid in a single check.
Using a positive lookahead, you have to add perl=T and add an anchor ^ to assert the start of the string and add an anchor $ to assert the end of the string after the lookahead.
^(?!\\d{8}$|date_consid$)
^ Start of string
(?! Negative lookahead
\\d{8}$ Match 8 digits until end of string
| Or
date_consid$Match date_consid until end of string
) Close lookahead
For example
vec <- c("19980213", "19980214", "19980215", "19980216","19980217", "iffi","geometry", "date_consid")
grep("^(?!\\d{8}$|date_consid$)", vec, value=T, perl=T)
Output
[1] "iffi" "geometry"
I have the string in R
BLCU142-09|Apodemia_mejicanus
and I would like to get the result
Apodemia_mejicanus
Using the stringr R package, I have tried
str_replace_all("BLCU142-09|Apodemia_mejicanus", "[[A-Z0-9|-]]", "")
# [1] "podemia_mejicanus"
which is almost what I need, except that the A is missing.
You can use
sub(".*\\|", "", x)
This will remove all text up to and including the last pipe char. See the regex demo. Details:
.* - any zero or more chars as many as possible
\| - a | char (| is a special regex metacharacter that is an alternation operator, so it must be escaped, and since string literals in R can contain string escape sequences, the | is escaped with a double backslash).
See the R demo online:
x <- c("BLCU142-09|Apodemia_mejicanus", "a|b|c|BLCU142-09|Apodemia_mejicanus")
sub(".*\\|", "", x)
## => [1] "Apodemia_mejicanus" "Apodemia_mejicanus"
We can match one or more characters that are not a | ([^|]+) from the start (^) of the string followed by | in str_remove to remove that substring
library(stringr)
str_remove(str1, "^[^|]+\\|")
#[1] "Apodemia_mejicanus"
If we use [A-Z] also to match it will match the upper case letter and replace with blank ("") as in the OP's str_replace_all
data
str1 <- "BLCU142-09|Apodemia_mejicanus"
You can always choose to _extract rather than _remove:
s <- "BLCU142-09|Apodemia_mejicanus"
stringr::str_extract(s,"[[:alpha:]_]+$")
## [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)
I have the following character vector than I need to modify with gsub.
strings <- c("x", "pm2.5.median", "rmin.10000m", "rmin.2500m", "rmax.5000m")
Desired output of filtered strings:
"x", "pm2.5.median", "rmin", "rmin", "rmax"
My current attempt works for everything except the pm2.5.median string which has dots that need to be preserved. I'm really just trying to remove the buffer size that is appended to the end of each variable, e.g. 1000m, 2500m, 5000m, 7500m, and 10000m.
gsub("\\..*m$", "", strings)
"x", "pm2", "rmin", "rmin", "rmax"
Match a dot, any number of digits, m and the end of string and replace that with the empty string. Note that we prefer sub to gsub here because we are only interested in one replacement per string.
sub("\\.\\d+m$", "", strings)
## [1] "x" "pm2.5.median" "rmin" "rmin" "rmax"
The .* pattern matches any 0 or more chars, as many as possible. The \..*m$ pattern matches the first (leftmost) . in the string and then grab all the text after it if it ends with m.
You need
> sub("\\.[^.]*m$", "", strings)
[1] "x" "pm2.5.median" "rmin" "rmin" "rmax"
Here, \.[^.]*m$ matches ., then 0 or more chars other than a dot and then m at the end of the string.
See the regex demo.
Details
\. - a dot (must be escaped since it is a special regex char otherwise)
[^.]* - a negated character class matching any char but . 0 or more times
m - an m char
$ - end of string.
I have the following string in R: "xxx, yyy. zzz"
I want to get the yyy part only, which are in between "," and "."
I don't want to use regex.
I searched half a day, found many string functions in R but none which deal with "cut before/after a character" function.
Is there such?
We can use gsub to match zero or more characters that are not a , ([^,]*) from the start (^) of the string followed by a , followed by zero or more spaces (\\s*) or (!) a dot (\\. - it is a metacharacter meaning any character so it is escaped) followed by other characters (.*) until the end of the string ($) and replace it with blank ("")
gsub("^[^,]*,\\s*|\\..*$", "", str1)
#[1] "yyy"
If we don't need regex then strsplit the string by , followed by zero or more spaces or with a . and select the second entry after converting the list output to vector ([[1]])
strsplit(str1, ",\\s*|\\.")[[1]][2]
#[1] "yyy"
data
str1 <- "xxx, yyy. zzz"
It could be that this suffices:
unlist(strsplit("xxx, yyy. zzz","[,.]"))[2] # get yyy with space, or:
gsub(" ","",unlist(strsplit("xxx, yyy. zzz","[,.]")))[2] # remove space
I would like to extract parts of strings. The string is:
> (x <- 'ab/cd efgh "xyz xyz"')
> [1] "ab/cd efgh \"xyz xyz\""
Now, I would like first to extract the first part:
> # get "ab/cd efgh"
> sub(" \"[/A-Za-z ]+\"","",x)
[1] "ab/cd efgh"
But I don't succeed in extracting the second part:
> # get "xyz xyz"
> sub("(\"[A-Za-z ]+\")$","\\1",x, perl=TRUE)
[1] "ab/cd efgh \"xyz xyz\""
What is wrong with this code?
Thanks for help.
Your last snippet does not work because you reinsert the whole match back into the result: (\"[A-Za-z ]+\")$ matches and captures ", 1+ letters and spaces, " into Group 1 and \1 in the replacement puts it back.
You may actually get the last part inside quotes by removing all chars other than " at the start of the string:
x <- 'ab/cd efgh "xyz xyz"'
sub('^[^"]+', "", x)
See the R demo
The sub here will find and replace just once, and it will match the string start (with ^) followed with 1+ chars other than " with [^"]+ negated character class.
To get this to work with sub, you have to match the whole string. The help file says
For sub and gsub return a character vector of the same length and with the same attributes as x (after possible coercion to character). Elements of character vectors x which are not substituted will be returned unchanged (including any declared encoding).
So to get this to work with your regex, pre-pend the sometimes risky catchall ".*"
sub(".*(\"[A-Za-z ]+\")$","\\1",x, perl=TRUE)
[1] "\"xyz xyz\""