I am trying to extract usernames tagged in a text-chat, such as "#Jack #Marie Hi there!"
I am trying to do it on the combination of # and whitespace but I cannot get the regex to match non-greedy (or at least this is what I think is wrong):
library(stringr)
str_extract(string = '#This is what I want to extract', pattern = "(?<=#)(.*)(?=\\s+)")
[1] "This is what I want to"
What I would like to extract instead is only This.
You could make your regex non greedy:
(?<=#)(.*?)(?=\s+)
Or if you want to capture only "This" after the # sign, you could try it like this using only a positive lookbehind:
(?<=#)\w+
Explanation
A positive lookbehind (?<=
That asserts that what is behind is an #
Close positive lookbehind )
Match one or more word characters \w+
The central part of your regex ((.*)) is a sequence of any chars.
Instead you shoud look for a sequence of chars other than white space
(\S+) or word chars (\w+).
Note also that I changed * to +, as you are probably not interested
in any empty sequence of chars.
To capture also a name which has "last" position in the source
string, the last part of your regex should match not only a sequence
of whitespace chars, but also the end of the string, so change
(?=\\s+) to (?=\\s+|$).
And the last remark: Actually you don't need the parentheses around
the "central" part.
So to sum up, the whole regex can be like this:
(?<=#)\w+(?=\s+|$)
(with global oprion).
Here is a non-regex approach or rather a minimal-regex approach since grep takes the detection of # through the regex engine
grep('#', strsplit(x, ' ')[[1]], value = TRUE)
#[1] "#This"
Or to avoid strsplit, we can use scan (taken from this answer), i.e.
grep('#', scan(textConnection(x), " "), value=TRUE)
#Read 7 items
#[1] "#This"
Related
My objective would be replacing a string by a symbol repeated as many characters as have the string, in a way as one can replace letters to capital letters with \\U\\1, if my pattern was "...(*)..." my replacement for what is captured by (*) would be something like x\\q1 or {\\q1}x so I would get so many x as characters captured by *.
Is this possible?
I am thinking mainly in sub,gsub but you can answer with other libraris like stringi,stringr, etc.
You can use perl = TRUE or perl = FALSE and any other options with convenience.
I assume the answer can be negative, since seems to be quite limited options (?gsub):
a replacement for matched pattern in sub and gsub. Coerced to character if possible. For fixed = FALSE this can include backreferences "\1" to "\9" to parenthesized subexpressions of pattern. For perl = TRUE only, it can also contain "\U" or "\L" to convert the rest of the replacement to upper or lower case and "\E" to end case conversion. If a character vector of length 2 or more is supplied, the first element is used with a warning. If NA, all elements in the result corresponding to matches will be set to NA.
Main quantifiers are (?base::regex):
?
The preceding item is optional and will be matched at most once.
*
The preceding item will be matched zero or more times.
+
The preceding item will be matched one or more times.
{n}
The preceding item is matched exactly n times.
{n,}
The preceding item is matched n or more times.
{n,m}
The preceding item is matched at least n times, but not more than m times.
Ok, but it seems to be an option (which is not in PCRE, not sure if in PERL or where...) (*) which captures the number of characters the star quantifier is able to match (I found it at https://www.rexegg.com/regex-quantifier-capture.html) so then it could be used \q1 (same reference) to refer to the first captured quantifier (and \q2, etc.). I also read that (*) is equivalent to {0,} but I'm not sure if this is really the fact for what I'm interested in.
EDIT UPDATE:
Since asked by commenters I update my question with an specific example provide by this interesting question. I modify a bit the example. Let's say we have a <- "I hate extra spaces elephant" so we are interested in keeping the a unique space between words, the 5 first characters of each word (till here as the original question) but then a dot for each other character (not sure if this is what is expected in the original question but doesn't matter) so the resulting string would be "I hate extra space. eleph..." (one . for the last s in spaces and 3 dots for the 3 letters ant in the end of elephant). So I started by keeping the 5 first characters with
gsub("(?<!\\S)(\\S{5})\\S*", "\\1", a, perl = TRUE)
[1] "I hate extra space eleph"
How should I replace the exact number of characters in \\S* by dots or any other symbol?
Quantifiers cannot be used in the replacement pattern, nor the information how many chars they match.
What you need is a \G base PCRE pattern to find consecutive matches after a specific place in the string:
a <- "I hate extra spaces elephant"
gsub("(?:\\G(?!^)|(?<!\\S)\\S{5})\\K\\S", ".", a, perl = TRUE)
See the R demo and the regex demo.
Details
(?:\G(?!^)|(?<!\S)\S{5}) - the end of the previous successful match or five non-whitespace chars not preceded with a non-whitespace char
\K - a match reset operator discarding text matched so far
\S - any non-whitespace char.
gsubfn is like gsub except the replacement string can be a function which inputs the match and outputs the replacement. The function can optionally be expressed a formula as we do here replacing each string of word characters with the output of the function replacing that string. No complex regular expressions are needed.
library(gsubfn)
gsubfn("\\w+", ~ paste0(substr(x, 1, 5), strrep(".", max(0, nchar(x) - 5))), a)
## [1] "I hate extra space. eleph..."
or almost the same except function is slightly different:
gsubfn("\\w+", ~ paste0(substr(x, 1, 5), substring(gsub(".", ".", x), 6)), a)
## [1] "I hate extra space. eleph..."
I have a vector of strings and I want to remove -es from all strings (words) ending in either -ses or -ces at the same time. The reason I want to do it at the same time and not consequitively is that sometimes it happens that after removing one ending, the other ending appears while I don't want to apply this pattern to a single word twice.
I have no idea how to use two patterns at the same time, but this is the best I could:
text <- gsub("[sc]+s$", "[sc]", text)
I know the replacement is not correct, but I wonder how can I show that I want to replace it with the letter I just detected (c or s in this case). Thank you in advance.
To remove es at the end of words, that is preceded with s or c, you may use
gsub("([sc])es\\b", "\\1", text)
gsub("(?<=[sc])es\\b", "", text, perl=TRUE)
To remove them at the end of strings, you can go on using your $ anchor:
gsub("([sc])es$", "\\1", text)
gsub("(?<=[sc])es$", "", text, perl=TRUE)
The first gsub TRE pattern is ([sc])es\b: a capturing group #1 that matches either s or c, and then es is matched, and then \b makes sure the next char is not a letter, digit or _. The \1 in the replacement is the backreference to the value stored in the capturing group #1 memory buffer.
In the second example with the PCRE regex (due to perl=TRUE), (?<=[sc]) positive lookbehind is used instead of the ([sc]) capturing group. Lookbehinds are not consuming text, the text they match does not land in the match value, and thus, there is no need to restore it anyhow. The replacement is an empty string.
Strings ending with "ces" and "ses" follow the same pattern, i.e. "*es$"
If I understand it correctly than you don't need two patterns.
Example:
x = c("ces", "ses", "mes)
gsub( pattern = "*([cs])es$", replacement = "\\1", x)
[1] "c" "s" "mes"
Hope it helps.
M
I have a column of texts look like below:
str1 = "ABCID 123456789 is what I'm looking for, could you help me to check this Item's status?"
I want to use gsub function in R to extract "ABCID 123456789" from there. The number might change with different numbers, but ABCID is a constant. Can someone know the solution with that please? Thanks very much!
We can use str_extract to select the fixed word followed by space and one or more numbers (\\d+)
library(stringr)
str_extract(df1$col1, "ABCID \\d+")
If there are multiple instances, use str_extract_all
str_extract_all(df1$col1, "ABCID \\d+")
NOTE: The OP states that to extract "ABCID 123456789" from there
If the number has constant length (9) you could you use positive lookbehind:
sub("(?<=ABCID \\d{9}).*", "", str1, perl = TRUE)
# [1] "ABCID 123456789"
Match the beginning of string (^) leading letters (ABCID), a space, digits (\d+) and everything else (.*) and replace it all with the captured portion, i.e. the portion within parentheses. Note that we want to use sub, not gsub, here because there is only one substitution.
sub("^(ABCID \\d+).*", "\\1", str1)
## [1] "ABCID 123456789"
I realize this is a rather simple question and I have searched throughout this site, but just can't seem to get my syntax right for the following regex challenges. I'm looking to do two things. First have the regex to pick up the first three characters and stop at a semicolon. For example, my string might look as follows:
Apt;House;Condo;Apts;
I'd like to go here
Apartment;House;Condo;Apartment
I'd also like to create a regex to substitute a word in between delimiters, while keep others unchanged. For example, I'd like to go from this:
feline;labrador;bird;labrador retriever;labrador dog; lab dog;
To this:
feline;dog;bird;dog;dog;dog;
Below is the regex I'm working with. I know ^ denotes the beginning of the string and $ the end. I've tried many variations, and am making substitutions, but am not achieving my desired out put. I'm also guessing one regex could work for both? Thanks for your help everyone.
df$variable <- gsub("^apt$;", "Apartment;", df$variable, ignore.case = TRUE)
Here is an approach that uses look behind (so you need perl=TRUE):
> tmp <- c("feline;labrador;bird;labrador retriever;labrador dog; lab dog;",
+ "lab;feline;labrador;bird;labrador retriever;labrador dog; lab dog")
> gsub( "(?<=;|^) *lab[^;]*", "dog", tmp, perl=TRUE)
[1] "feline;dog;bird;dog;dog;dog;"
[2] "dog;feline;dog;bird;dog;dog;dog"
The (?<=;|^) is the look behind, it says that any match must be preceded by either a semi-colon or the beginning of the string, but what is matched is not included in the part to be replaced. The * will match 0 or more spaces (since your example string had one case where there was space between the semi-colon and the lab. It then matches a literal lab followed by 0 or more characters other than a semi-colon. Since * is by default greedy, this will match everything up to, but not including' the next semi-colon or the end of the string. You could also include a positive look ahead (?=;|$) to make sure it goes all the way to the next semi-colon or end of string, but in this case the greediness of * will take care of that.
You could also use the non-greedy modifier, then force to match to end of string or semi-colon:
> gsub( "(?<=;|^) *lab.*?(?=;|$)", "dog", tmp, perl=TRUE)
[1] "feline;dog;bird;dog;dog;dog;"
[2] "dog;feline;dog;bird;dog;dog;dog"
The .*? will match 0 or more characters, but as few as it can get away with, stretching just until the next semi-colon or end of line.
You can skip the look behind (and perl=TRUE) if you match the delimiter, then include it in the replacement:
> gsub("(;|^) *lab[^;]*", "\\1dog", tmp)
[1] "feline;dog;bird;dog;dog;dog;"
[2] "dog;feline;dog;bird;dog;dog;dog"
With this method you need to be careful that you only match the delimiter on one side (the first in my example) since the match consumes the delimiter (not with the look-ahead or look-behind), if you consume both delimiters, then the next will be skipped and only every other field will be considered for replacement.
I'd recommend doing this in two steps:
Split the string by the delimiters
Do the replacements
(optional, if that's what you gotta do) Smash the strings back together.
To split the string, I'd use the stringr library. But you can use base R too:
myString <- "Apt;House;Condo;Apts;"
# base R
splitString <- unlist(strsplit(myString, ";", fixed = T))
# with stringr
library(stringr)
splitString <- as.vector(str_split(myString, ";", simplify = T))
Once you've done that, THEN you can do the text substitution:
# base R
fixedApts <- gsub("^Apt$|^Apts$", "Apartment", splitString)
# with stringr
fixedApts <- str_replace(splitString, "^Apt$|^Apts$", "Apartment")
# then do the rest of your replacements
There's probabably a better way to do the replacements than regular expressions (using switch(), maybe?)
Use paste0(fixedApts, collapse = "") to collapse the vector into a single string at the end if that's what you need to do.
I'm looking for a regular expression to grep whole words, including words separated by digits or underscore. \\b considers digits and underscore as parts of words, not as boundaries.
For example, I'd like to catch MOUSE in "DOG MOUSE CAT", in "DOG MOUSE:CAT" but also in "DOG_MOUSE9CAT" and at the end or the beginning of an expression, as in "MOUSE9CAT" and "DOG_MOUSE". Basically, the boundary I'm looking for is any non-uppercase-alpha character plus beginning and end of line/expression (maybe missing some other cases caught by \\b here).
I've tried:
"[[0-9_]\\b]MOUSE[[0-9_]\\b]"
"[[0-9_]|\\b]MOUSE[[0-9_]|\\b]"
"[$|[^A-Z]]MOUSE[^|[^A-Z]]"
"[?<=^|[^A-Z]]MOUSE[?=$|[^A-Z]]"
None of them work.
I'm actually looking for several words (based on a long vector of values), so the final result should look something like
grep(paste("\\b", paste(searchwords, collapse = "\\b|\\b"), "\\b"), targettext)
(with a different delimiter because \\b is too restrictive for me).
(This is a similar question to the one asked by user Nick Sabbe in a comment here: Using grep in R to find strings as whole words (but not strings as part of words))
Use PCRE regex with lookarounds:
grep("(?<![A-Z])MOUSE(?![A-Z])", targettext, perl=TRUE)
See the regex demo
The (?<![A-Z]) negative lookbehind will fail the match if the word is preceded with an uppercase ASCII letter and the negative lookahead (?![A-Z]) will fail the match if the word is followed with an uppercase ASCII letter.
To apply the lookarounds to all the alternatives you have, use an outer grouping (?:...|...).
See the R online demo:
> targettext <- c("DOG MOUSE CAT","DOG MOUSE:CAT","DOG_MOUSE9CAT","MOUSE9CAT","DOG_MOUSE")
> searchwords <- c("MOUSE","FROG")
> grep(paste0("(?<![A-Z])(?:", paste(searchwords, collapse = "|"), ")(?![A-Z])"), targettext, perl=TRUE)
[1] 1 2 3 4 5