Take these example strings, I want to split them such that the length is limited to X or less characters, a complete word is at the end of each string, and the remaining part is stored in another column. The words are always separated by space. I came across this partial solution in TSQL (doesn't create variable for extra words). However I need to do it in R. I was provided the first half solution in a previous question, this doesn't have the remaining words in new variables. I need help to create the new variable
{gsub(patt="(^.{2,100})([ ].+)", repl="\\1",y)}
For example:
XOVEW VJIEW NI **stays** XOVEW VJIEW NI (assuming X is 14)
XOVEW VJIEW NIGOI **becomes** XOVEW VJIEW (NIGOI goes to a new vector)
XOVEW VJIEWNIGOI **becomes** XOVEW (assuming X is 14)
So new variable will contain c("NIGOI","VJIEWNIGOI") coming from 2nd and 3rd row above.
v1 <- ifelse( nchar(vect) > 14, gsub( "(.*)\\s+(\\w+)", "\\1 - \\2", vect),vect);
values <- data.frame(do.call('rbind', lapply(strsplit(v1,split="-"), `length<-`,2)));
Output:
[,1] [,2]
[1,] "XOVEW VJIEW NI" NA
[2,] "XOVEW VJIEW " " NIGOI"
[3,] "XOVEW " " VJIEWNIGOI"
I have created a small vector which will check if your string length is greater or smaller than 14 (?nchar in case you want to understand it).
Then wherever, it is longer than 14 I have created a string seperated by a dash, This is just to segregate the two strings, where the first strings deptics any collection of word which is not the last one, the second string matches the last word of the statement.
To match these I used regex, a dot represents any character, a star zero or more matches(together it means any character with zero or more matches) , a \\s+ matches 1 or more spaces and \\w+ matches one or more words. Collectively the match is such that it should have last word segregated with rest of the string in cases where string length is more than 14 within ifelse. Also these characters are further captured into \\1 and \\2 with a dash separation. where \\1 matches the first non last word match and \\2 match the last word of the string.
At last do.call is used with with rbind(bind all the rows) and lapply(to get even number of columns across all the elements)
I hope this explains your query.
Related
I've solved 2022 advent of code day 6, but was wondering if there was a regex way to find the first occurance of 4 non-repeating characters:
From the question:
bvwbjplbgvbhsrlpgdmjqwftvncz
bvwbjplbgvbhsrlpgdmjqwftvncz
# discard as repeating letter b
bvwbjplbgvbhsrlpgdmjqwftvncz
# match the 5th character, which signifies the end of the first four character block with no repeating characters
in R I've tried:
txt <- "bvwbjplbgvbhsrlpgdmjqwftvncz"
str_match("(.*)\1", txt)
But I'm having no luck
You can use
stringr::str_extract(txt, "(.)(?!\\1)(.)(?!\\1|\\2)(.)(?!\\1|\\2|\\3)(.)")
See the regex demo. Here, (.) captures any char into consequently numbered groups and the (?!...) negative lookaheads make sure each subsequent . does not match the already captured char(s).
See the R demo:
library(stringr)
txt <- "bvwbjplbgvbhsrlpgdmjqwftvncz"
str_extract(txt, "(.)(?!\\1)(.)(?!\\1|\\2)(.)(?!\\1|\\2|\\3)(.)")
## => [1] "vwbj"
Note that the stringr::str_match (as stringr::str_extract) takes the input as the first argument and the regex as the second argument.
Regex and stringr newbie here. I have a data frame with a column from which I want to find 10-digit numbers and keep only the first three digits. Otherwise, I want to just keep whatever is there.
So to make it easy let's just pretend it's a simple vector like this:
new<-c("111", "1234567891", "12", "12345")
I want to write code that will return a vector with elements: 111, 123, 12, and 12345. I also need to write code (I'm assuming I'll do this iteratively) where I extract the first two digits of a 5-digit string, like the last element above.
I've tried:
gsub("\\d{10}", "", new)
but I don't know what I could put for the replacement argument to get what I'm looking for. Also tried:
str_replace(new, "\\d{10}", "")
But again I don't know what to put in for the replacement argument to get just the first x digits.
Edit: I disagree that this is a duplicate question because it's not just that I want to extract the first X digits from a string but that I need to do that with specific strings that match a pattern (e.g., 10 digit strings.)
If you are willing to use the library stringr from which comes the str_replace you are using. Just use str_extract
vec <- c(111, 1234567891, 12)
str_extract(vec, "^\\d{1,3}")
The regex ^\\d{1,3} matches at least 1 to a maximum of 3 digits occurring right in the beginning of the phrase. str_extract, as the name implies, extracts and returns these matches.
You may use
new<-c("111", "1234567891", "12")
sub("^(\\d{3})\\d{7}$", "\\1", new)
## => [1] "111" "123" "12"
See the R online demo and the regex demo.
Regex graph:
Details
^ - start of string anchor
(\d{3}) - Capturing group 1 (this value is accessed using \1 in the replacement pattern): three digit chars
\d{7} - seven digit chars
$ - end of string anchor.
So, the sub command only matches strings that are composed only of 10 digits, captures the first three into a separate group, and then replaces the whole string (as it is the whole match) with the three digits captured in Group 1.
You can use:
as.numeric(substring(my_vec,1,3))
#[1] 111 123 12
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).
For Example, lets say I have the following string
vec <- " #_Jim98 Did you turn off the stove #9am?"
I would like to count the number of # characters that contain only numbers,letters,#, and underscore symbol in the string. In the case above, it would only count 1 since #9am? contains the ? symbol, so it won't be counted.
Also, it could not be longer than 10 characters.
1) Search for # followed by any number of the allowed characters "\\w" followed by a whitespace character "\\s" or | end of string $. If zero word characters are allowable then change the + to *. The expression is vectorized, i.e. x can be a character vector. No packages are used.
x <- " #_Jim98 Did you turn off the stove #9am?" # test input
pat <- "#\\w+(\\s|$)"
lengths(regmatches(x, gregexpr(pat, x)))
## [1] 1
Note that the reason for regmatches is that gregexpr produces a -1 rather than a zero length vector for no matches whereas regmatches will produce a zero length vector. Thus it works for the edge case of no matches.
2) A slightly more compact solution would be this where pat is from above:
library(gsubfn)
lengths(strapplyc(x, pat))
## [1] 1
We can do this with a regular expression. I'm interpreting that you are counting words separated by space characters or occurring at the beginning or end of the string. This assumes the # is at the start of the word, and I match a # followed by some number of word characters \\w(letters and digits) or underscores. You can remove the first (^|\\s) if you don't care about having # at the beginning of the word and would like to count 3 words in, for example, " #_Jim98 Did the Latin#s or tom#domain turn off the stove #9am?"
stringr::str_count(" #_Jim98 Did you turn off the stove #9am?", "(^|\\s)#(\\w|_)*?($|\\s)")
#> [1] 1
Created on 2018-04-12 by the reprex package (v0.2.0).
I would like to remove the character 'V' (always the last one in the strings) from the following vector containing a large number of strings. They look similar to the following example:
str <- c("VDM 000 V2.1.1",
"ABVC 001 V10.15.0",
"ASDV 123 V1.20.0")
I know that it is always the last 'V', I would like to remove.
I also know that this character is either the sixth, seventh or eighth last character within these strings.
I was not really able to come up with a nice solution. I know that I have to use sub or gsub but I can only remove all V's rather than only the last one.
Has anyone got an idea?
Thank you!
This regex pattern is written to match a "V" that is then followed by 5 to 7 other non-"V" characters. The "[...]" construct is a "character-class" and within such constructs a leading "^" causes negation. The "{...} consturct allows two digits specifying minimum and maximum lengths, and the "$" matches the length-0 end-of-string which I think was desired when you wrote "sixth, seventh or eighth last character":
sub("(V)(.{5,7})$", "\\2", str)
[1] "VDM 000 2.1.1" "ABVC 001 10.15.0" "ASDV 123 1.20.0"
Since you only wanted a single substitution I used sub instead of gsub.
You can use:
gsub("V(\\d+.\\d+.\\d+)$","\\1",str)
##[1] "VDM 000 2.1.1" "ABVC 001 10.15.0" "ASDV 123 1.20.0"
The regex V(\\d+.\\d+.\\d+)$ matches the "version" consisting of the character "V" followed by three sets of digits (i.e., \\d+) separated by two "." at the end of the string (i.e., $). The parenthesis around the \\d+.\\d+.\\d+ provides a group within the match that can be referenced by \\1. Therefore, gsub will replace the whole match with the group, thereby removing that "V".
Since you know it's the last V you want to remove from the string, try this regex V(?=[^V]*$):
gsub("V(?=[^V]*$)", "", str, perl = TRUE)
# [1] "VDM 000 2.1.1" "ABVC 001 10.15.0" "ASDV 123 1.20.0"
The regex matches V before pattern [^V]*$ which consists of non V characters from the end of the String, which guarantees that the matched V is the last V in the string.