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I'm going nuts trying to get a regex to detect spam of keywords in the user inputs. Usually there is some normal text at the start and the keyword spam at the end, separated by commas or other chars.
What I need is a regex to count the number of keywords to flag the text for a human to check it.
The text is usually like this:
[random text, with commas, dots and all]
keyword1, keyword2, keyword3, keyword4, keyword5,
Keyword6, keyword7, keyword8...
I've tried several regex to count the matches:
-This only gets one out of two keywords
[,-](\w|\s)+[,-]
-This also matches the random text
(?:([^,-]*)(?:[^,-]|$))
Can anyone tell me a regex to do this? Or should I take a different approach?
Thanks!
Pr your answer to my question, here is a regexp to match a string that occurs between two commas.
(?<=,)[^,]+(?=,)
This regexp does not match, and hence do not consume, the delimiting commas.
This regexp would match " and hence do not consume" in the previous sentence.
The fact that your regexp matched and consumed the commas was the reason why your attempted regexp only matched every other candidate.
Also if the whole input is a single string you will want to prevent linebreaks. In that case you will want to use;
(?<=,)[^,\n]+(?=,)
http://www.phpliveregex.com/p/1DJ
As others have said this is potentially a very tricky thing to do... It suffers from all of the same failures as general "word filtering" (e.g. people will "mask" the input). It is made even more difficult without plenty of example posts to test against...
Solution
Anyway, assuming that keywords will be on separate lines to the rest of the input and separated by commas you can match the lines with keywords in like:
Regex
#(?:^)((?:(?:[\w\.]+)(?:, ?|$))+)#m
Input
Taken from your question above:
[random text, with commas, dots and all]
keyword1, keyword2, keyword3, keyword4, keyword5,
Keyword6, keyword7, keyword8
Output
// preg_match_all('#(?:^)((?:(?:[\w]+)(?:, ?|$))+)#m', $string, $matches);
// var_dump($matches);
array(2) {
[0]=>
array(2) {
[0]=>
string(49) "keyword1, keyword2, keyword3, keyword4, keyword5,"
[1]=>
string(31) "Keyword6, keyword7, keyword8..."
}
[1]=>
array(2) {
[0]=>
string(49) "keyword1, keyword2, keyword3, keyword4, keyword5,"
[1]=>
string(31) "Keyword6, keyword7, keyword8"
}
}
Explanation
#(?:^)((?:(?:[\w]+)(?:, ?|$))+)#m
# => Starting delimiter
(?:^) => Matches start of line in a non-capturing group (you could just use ^ I was using |\n originally and didn't update)
( => Start a capturing group
(?: => Start a non-capturing group
(?:[\w]+) => A non-capturing group to match one or more word characters a-zA-Z0-9_ (Using a character class so that you can add to it if you need to....)
(?:, ?|$) => A non-capturing group to match either a comma (with an optional space) or the end of the string/line
)+ => End the non-capturing group (4) and repeat 5/6 to find multiple matches in the line
) => Close the capture group 3
# => Ending delimiter
m => Multi-line modifier
Follow up from number 2:
#^((?:(?:[\w]+)(?:, ?|$))+)#m
Counting keywords
Having now returned an array of lines only containing key words you can count the number of commas and thus get the number of keywords
$key_words = implode(', ', $matches[1]); // Join lines returned by preg_match_all
echo substr_count($key_words, ','); // 8
N.B. In most circumstances this will return NUMBER_OF_KEY_WORDS - 1 (i.e. in your case 7); it returns 8 because you have a comma at the end of your first line of key words.
Links
http://php.net/manual/en/reference.pcre.pattern.modifiers.php
http://www.regular-expressions.info/
http://php.net/substr_count
Why not just use explode and trim?
$keywords = array_map ('trim', explode (',', $keywordstring));
Then do a count() on $keywords.
If you think keywords with spaces in are spam, then you can iterate of the $keywords array and look for any that contain whitespace. There might be legitimate reasons for having spaces in a keyword though. If you're talking about superheroes on your system, for example, someone might enter The Tick or Iron Man as a keyword
I don't think counting keywords and looking for spaces in keywords are really very good strategies for detecting spam though. You might want to look into other bot protection strategies instead, or even use manual moderation.
How to match on the String of text between the commas?
This SO Post was marked as a duplicate to my posted question however since it is NOT a duplicate and there were no answers in THIS SO Post that answered my question on how to also match on the strings between the commas see below on how to take this a step further.
How to Match on single digit values in a CSV String
For example if the task is to search the string within the commas for a single 7, 8 or a single 9 but not match on combinations such as 17 or 77 or 78 but only the single 7s, 8s, or 9s see below...
The answer is to Use look arounds and place your search pattern within the look arounds:
(?<=^|,)[789](?=,|$)
See live demo.
The above Pattern is more concise however I've pasted below the Two Patterns provided as solutions to THIS this question of matching on Strings within the commas and they are:
(?<=^|,)[789](?=,|$) Provided by #Bohemian and chosen as the Correct Answer
(?:(?<=^)|(?<=,))[789](?:(?=,)|(?=$)) Provided in comments by #Ouroborus
Demo: https://regex101.com/r/fd5GnD/1
Your first regexp doesn't need a preceding comma
[\w\s]+[,-]
A regex that will match strings between two commas or start or end of string is
(?<=,|^)[^,]*(?=,|$)
Or, a bit more efficient:
(?<![^,])[^,]*(?![^,])
See the regex demo #1 and demo #2.
Details:
(?<=,|^) / (?<![^,]) - start of string or a position immediately preceded with a comma
[^,]* - zero or more chars other than a comma
(?=,|$) / (?![^,]) - end of string or a position immediately followed with a comma
If people still search for this in 2021
([^,\n])+
Match anything except new line and comma
regexr.com/60eme
I think the difficulty is that the random text can also contain commas.
If the keywords are all on one line and it is the last line of the text as a whole, trim the whole text removing new line characters from the end. Then take the text from the last new line character to the end. This should be your string containing the keywords. Once you have this part singled out, you can explode the string on comma and count the parts.
<?php
$string = " some gibberish, some more gibberish, and random text
keyword1, keyword2, keyword3
";
$lastEOL = strrpos(trim($string), PHP_EOL);
$keywordLine = substr($string, $lastEOL);
$keywords = explode(',', $keywordLine);
echo "Number of keywords: " . count($keywords);
I know it is not a regex, but I hope it helps nevertheless.
The only way to find a solution, is to find something that separates the random text and the keywords that is not present in the keywords. If a new line is present in the keywords, you can not use it. But are 2 consecutive new lines? Or any other characters.
$string = " some gibberish, some more gibberish, and random text
keyword1, keyword2, keyword3,
keyword4, keyword5, keyword6,
keyword7, keyword8, keyword9
";
$lastEOL = strrpos(trim($string), PHP_EOL . PHP_EOL); // 2 end of lines after random text
$keywordLine = substr($string, $lastEOL);
$keywords = explode(',', $keywordLine);
echo "Number of keywords: " . count($keywords);
(edit: added example for more new lines - long shot)
Behind this question is an effort to extract all references created by knitr and latex. Not finding another way, my thought was to read into R the .Rnw script and use a regular expression to find references -- where the latex syntax is \ref{caption referenced to}. My script has 250+ references, and some are very close to each other.
The text.1 example below works, but not the text example. I think it has to do with R chugging along to the final closing brace. How do I stop at the first closing brace and extract what preceded it to the opening brace?
library(stringr)
text.1 <- c(" \\ref{test}", "abc", "\\ref{test2}", " \\section{test3}", "{test3")
# In the regular expression below, look back and if find "ref{", grab everything until look behind for } at end
# braces are special characters and require escaping with double backslacs for R to recognize them as braces
# unlist converts the list returned by str_extract to a vector
unlist(str_extract_all(string = text.1, pattern = "(?<=ref\\{).*(?=\\}$)"))
[1] "test" "test2"
# a more complicated string, with more than one set of braces in an element
text <- c("text \ref{?bar labels precision} and more text \ref{?table column alignment}", "text \ref{?table space} }")
unlist(str_extract_all(string = text, pattern = "(?<=ref\\{).*(?=\\}$)"))
character(0)
The problem with text is the backslash in front of "ref" is being interpreted as a carriage return \r by the engine and R's parser; so you're trying to match "ref" but it's really (CR + "ef") ...
Also * is greedy by default, meaning it will match as much as it can and still allow the remainder of the regular expression to match. Use *? or a negated character class to prevent greediness.
unlist(str_extract_all(text, '(?<=\ref\\{)[^}]*'))
# [1] "?bar labels precision" "?table column alignment" "?table space"
As you can see, you can use a character class to match either (\r or r + "ef") ...
x <- c(' \\ref{test}', 'abc', '\\ref{test2}', ' \\section{test3}', '{test3',
'text \ref{?bar labels precision} and more text \ref{?table column alignment}',
'text \ref{?table space} }')
unlist(str_extract_all(x, '(?<=[\rr]ef\\{)[^}]*'))
# [1] "test" "test2" "?bar labels precision"
# [4] "?table column alignment" "?table space"
EDITED
The reason why it didn't capture what is before the closing brace } is because you added an end of line anchor $. Remove $ and it would work.
Therefore, you new code should be like this
unlist(str_extract_all(string = text, pattern = "(?<=ref\\{)[^}]*(?=\\})"))
See DEMO
I have a dataframe that consists of web-scraped data. One of the fields scraped was a time in clock time, but the scraping process wasn't perfect. Most of the 'good' data look something like '4:33, or '103:20 (so a leading single quote, and two fields, minutes and seconds). Also, there is some bad data, the most common one being '],, but also some containing text. I'd like a new string that is something like 4:33, and for bad data, just blank.
So my plan of attack is to match my good data form, and then replace everything else with a blank space. Sometime like time <- gsub('[0-9]+:[0-9]+', '', time). I know this would replace my pattern with a blank, and I want the opposite, but I'm unsure as to how to negate this whole pattern. A simple carat doesn't seem to work, nor applying it to a group. I tried something like gsub("(.)+([0-9]+)(:)([0-9]+)", "\\2\\3\\4", time) but that isn't working either.
Sample:
dput(sample)
c("'], ", "' Ling (2-0)vsThe Dragon(2-0)", "'8:18", "'13:33",
"'43:33")
Expected output:
c("", "", "8:18", "13:33", "43:33")
We can use grep to replace the elements that do not follow the pattern to '' and then replace the quotes (') with ''. Here, the pattern is the strings that start (^) with ' followed by numbers, :, numbers in that order to the end ($) of the string. So, all other string elements (by negating i.e. !) are assigned to '' using the logical index from grepl and we use sub to replace the '.
sample[!grepl("^'\\d+:\\d+$", sample)] <- ''
sub("'", '', sample)
#[1] "" "" "8:18" "13:33" "43:33"
Or we can also do this in one step using gsub by replacing all those characters (.) that do not follow the pattern \\d+:\\d+ with ''.
gsub("(\\d+:\\d+)(*SKIP)(*F)|.", '', sample, perl=TRUE)
#[1] "" "" "8:18" "13:33" "43:33"
Or another option is str_extract from library(stringr). It is not clear whether there are other patterns such as "some text '08:20 value" in the OP's original dataset or not. The str_extract will also extract those time values, if present.
library(stringr)
str_extract(sample, '\\d+:\\d+')
#[1] NA NA "8:18" "13:33" "43:33"
It will give NA instead of '' for those that doesn't follow the pattern.
You can use sub:
sub('.+?(?=[0-9]+:[0-9]+)|.+', '', sample, perl = TRUE)
[1] "" "" "8:18" "13:33" "43:33"
The regex consists of two parts that are combined with a logical or (|).
.+?(?=[0-9]+:[0-9]+)
This regex matches a positive number of characters followed by the target pattern.
.+ This regex matches a positive number of characters.
The logic: Replace everything preceding thte target pattern with an empty string (''). If there is no target pattern, replace everything with the empty string.
I am trying to extract all of the words in the string below contained within the brackets following the word 'tokens' only if the 'tokens' occurs after 'tag(noun)'.
For example, I have the string:
m<- "phrase('The New York State Department',[det([lexmatch(['THE']),
inputmatch(['The']),tag(det),tokens([the])]),mod([lexmatch(['New York State']),
inputmatch(['New','York','State']),tag(noun),tokens([new,york,state])]),
head([lexmatch([department]),inputmatch(['Department']),tag(noun),
tokens([department])])],0/29,[])."
I want to get a list of all of the words that occur within the brackets after the word 'tokens' only when the word tokens occurs after 'tag(noun)'.
Therefore, I want my output to be a vector of the following:
[1] new, york, state, department
How do I do this? I'm assuming I have to use a regular expression, but I'm lost on how to write this in R.
Thanks!
Remove newlines and then extract the portion matched to the part between parentheses in pattern pat. Then split apart such strings by commas and simplify into a character vector:
library(gsubfn)
pat <- "tag.noun.,tokens..(.*?)\\]"
strapply(gsub("\\n", "", m), pat, ~ unlist(strsplit(x, ",")), simplify = c)
giving:
[1] "new" "york" "state" "department"
Visualization: Here is the debuggex representation of the regular expression in pat. (Note that we need to double the backslash when put within R's double quotes):
tag.noun.,tokens..(.*?)\]
Debuggex Demo
Note that .*? means match the shortetst string of any characters such that the entire pattern matches - without the ? it would try to match the longest string.
How about something like this. Here i'll use the regcatputedmatches helper function to make it easier to extract the captured matches.
m<- "phrase('The New York State Department',[det([lexmatch(['THE']),inputmatch(['The']),tag(det),tokens([the])]),mod([lexmatch(['New York State']),inputmatch(['New','York','State']),tag(noun),tokens([new,york,state])]),head([lexmatch([department]),inputmatch(['Department']),tag(noun),tokens([department])])],0/29,[])."
rx <- gregexpr("tag\\(noun\\),tokens\\(\\[([^]]+)\\]\\)", m, perl=T)
lapply(regcapturedmatches(m,rx), function(x) {
unlist(strsplit(c(x),","))
})
# [[1]]
# [1] "new" "york" "state" "department"
The regular expression is a bit messy because your desired match contains many special regular expression symbols so we need to properly escape them.
Here is a one liner if you like:
paste(unlist(regmatches(m, gregexpr("(?<=tag\\(noun\\),tokens\\(\\[)[^\\]]*", m, perl=T))), collapse=",")
[1] "new,york,state,department"
Broken down:
# Get match indices
indices <- gregexpr("(?<=tag\\(noun\\),tokens\\(\\[)[^\\]]*", m, perl=T)
# Extract the matches
matches <- regmatches(m, indices)
# unlist and paste together
paste(unlist(matches), collapse=",")
[1] "new,york,state,department"
I am trying to extract a leading string by stripping off an optional trailing string, where the trailing strings are a subset of possible leading strings but not vice versa. Suppose the leading string is like [a-z]+ and the trailing string is like c. Thus from "abc" I want to extract "ab", and from "ab" I also want to get "ab". Something like this:
^([a-z]+)(?:c|)
The problem is that the [a-z]+ matches the entire string, using the empty option in the alternative, so the grabbed value is "abc" or "ab". (The (?: tells it not to grab the second part.) I want some way to make it take the longer option, or the first option, in the alternative, and use that to determine what matches the first part.
I have also tried putting the desired target inside both of the alternatives:
^([a-z]+)c|^([a-z]+)
I think that it should prefer to match the first one of the two possible alternatives, but I get the same results as above.
I am doing this in R, so I can use either the POSIX or the Perl regex library.
(The actual problem involves futures trading symbols. These have a root "instrument name" like [A-Z0-9]+, followed by an "expiration code" like [FGHJKMNQUVXZ][0-9]{1,2}. Given a symbol like "ZNH3", I want to strip the "H3" to get "ZN". But if I give it "ZN" I also want to get back "ZN".)
Try this:
> library(gsubfn)
> strapplyc(c("abc", "abd"), "^(\\w+?)c?$", simplify = TRUE)
[1] "ab" "abd"
and even easier:
> sub("c$", "", c("abc", "abd"))
[1] "ab" "abd"
Here's a working regular expression:
vec <- c("ZNH3", "ZN", "ZZZ33", "ABF")
sub("(\\w+)[FGHJKMNQUVXZ]\\d{1,2}", "\\1", vec)
# [1] "ZN" "ZN" "ZZ" "ABF"
A variation on the non-greedy answers using base code only.
codes <- c("ZNH3", "CLZ4")
matched <- regmatches(codes, regexec("^([A-Z0-9]+?)[FGHJKMNQUVXZ][0-9]{1,2}$", codes))
# [[1]]
# [1] "ZNH3" "ZN"
#
# [[2]]
# [1] "CLZ4" "CL"
sapply(matched, `[[`, 2) # extract just codes
# [1] "ZN" "CL"
Use a 'non-greedy' match for the first part of the regex, followed by the definitions of your 'optional allowed suffixes' anchored by the 'end-of-string'...
This regex (.+?)([FGHJKMNQUVXZ][0-9]{1,2})?$matches...
(.+?) as few characters as possible
([FGHJKMNQUVXZ][0-9]{1,2})? followed by an allowable (but optional) suffix
$ followed by the end of string
The required result is in the first captured element of the match (however that may be referenced in 'r') :-)