How do I create a regular expression to match a word at the beginning of a string?
We are looking to match stop at the beginning of a string and anything can follow it.
For example, the expression should match:
stop
stop random
stopping
If you wish to match only lines beginning with stop, use
^stop
If you wish to match lines beginning with the word stop followed by a space:
^stop\s
Or, if you wish to match lines beginning with the word stop, but followed by either a space or any other non-word character you can use (your regex flavor permitting)
^stop\W
On the other hand, what follows matches a word at the beginning of a string on most regex flavors (in these flavors \w matches the opposite of \W)
^\w
If your flavor does not have the \w shortcut, you can use
^[a-zA-Z0-9]+
Be wary that this second idiom will only match letters and numbers, no symbol whatsoever.
Check your regex flavor manual to know what shortcuts are allowed and what exactly do they match (and how do they deal with Unicode).
Try this:
/^stop.*$/
Explanation:
/ charachters delimit the regular expression (i.e. they are not part of the Regex per se)
^ means match at the beginning of the line
. followed by * means match any character (.), any number of times (*)
$ means to the end of the line
If you would like to enforce that stop be followed by a whitespace, you could modify the RegEx like so:
/^stop\s+.*$/
\s means any whitespace character
+ following the \s means there has to be at least one whitespace character following after the stop word
Note: Also keep in mind that the RegEx above requires that the stop word be followed by a space! So it wouldn't match a line that only contains: stop
If you want to match anything after a word, stop, and not only at the start of the line, you may use: \bstop.*\b - word followed by a line.
Or if you want to match the word in the string, use \bstop[a-zA-Z]* - only the words starting with stop.
Or the start of lines with stop - ^stop[a-zA-Z]* for the word only - first word only.
The whole line ^stop.* - first line of the string only.
And if you want to match every string starting with stop, including newlines, use: /^stop.*/s - multiline string starting with stop.
Like #SharadHolani said. This won't match every word beginning with "stop"
. Only if it's at the beginning of a line like "stop going".
#Waxo gave the right answer:
This one is slightly better, if you want to match any word beginning with "stop" and containing nothing but letters from A to Z.
\bstop[a-zA-Z]*\b
This would match all
stop (1)
stop random (2)
stopping (3)
want to stop (4)
please stop (5)
But
/^stop[a-zA-Z]*/
would only match (1) until (3), but not (4) & (5)
If you want to match anything that starts with "stop" including "stop going", "stop" and "stopping" use:
^stop
If you want to match the word stop followed by anything as in "stop going", "stop this", but not "stopped" and not "stopping" use:
^stop\W
/stop([a-zA-Z])+/
Will match any stop word (stop, stopped, stopping, etc)
However, if you just want to match "stop" at the start of a string
/^stop/
will do :D
If you want the word to start with "stop", you can use the following pattern.
"^stop.*"
This will match words starting with stop followed by anything.
/^stop*$/i
i - in case it is case sensitive.
I'd advise against a simple regular expression approach to this problem. There are too many words that are substrings of other unrelated words, and you'll probably drive yourself crazy trying to overadapt the simpler solutions already provided.
You'll want at least a naive stemming algorithm (try the Porter stemmer; there's available, free code in most languages) to process text first. Keep this processed text and the preprocessed text in two separate space-split arrays. Make sure each non-alphabetical character also gets its own index in this array. Whatever list of words you're filtering, stem them also.
The next step would be to find the array indices which match to your list of stemmed 'stop' words. Remove those from the unprocessed array, and then rejoin on spaces.
This is only slightly more complicated, but will be much more reliable an approach. If you've got any doubts on the value of a more NLP-oriented approach, you might want to do some research into clbuttic mistakes.
can you try this:
https://regex101.com/r/P3qfKG/1
reg = /stop(\w+| [^ ]+|$)/gm
it will select both stop and start with stop and next word;
Related
How do I create a regular expression to match a word at the beginning of a string?
We are looking to match stop at the beginning of a string and anything can follow it.
For example, the expression should match:
stop
stop random
stopping
If you wish to match only lines beginning with stop, use
^stop
If you wish to match lines beginning with the word stop followed by a space:
^stop\s
Or, if you wish to match lines beginning with the word stop, but followed by either a space or any other non-word character you can use (your regex flavor permitting)
^stop\W
On the other hand, what follows matches a word at the beginning of a string on most regex flavors (in these flavors \w matches the opposite of \W)
^\w
If your flavor does not have the \w shortcut, you can use
^[a-zA-Z0-9]+
Be wary that this second idiom will only match letters and numbers, no symbol whatsoever.
Check your regex flavor manual to know what shortcuts are allowed and what exactly do they match (and how do they deal with Unicode).
Try this:
/^stop.*$/
Explanation:
/ charachters delimit the regular expression (i.e. they are not part of the Regex per se)
^ means match at the beginning of the line
. followed by * means match any character (.), any number of times (*)
$ means to the end of the line
If you would like to enforce that stop be followed by a whitespace, you could modify the RegEx like so:
/^stop\s+.*$/
\s means any whitespace character
+ following the \s means there has to be at least one whitespace character following after the stop word
Note: Also keep in mind that the RegEx above requires that the stop word be followed by a space! So it wouldn't match a line that only contains: stop
If you want to match anything after a word, stop, and not only at the start of the line, you may use: \bstop.*\b - word followed by a line.
Or if you want to match the word in the string, use \bstop[a-zA-Z]* - only the words starting with stop.
Or the start of lines with stop - ^stop[a-zA-Z]* for the word only - first word only.
The whole line ^stop.* - first line of the string only.
And if you want to match every string starting with stop, including newlines, use: /^stop.*/s - multiline string starting with stop.
Like #SharadHolani said. This won't match every word beginning with "stop"
. Only if it's at the beginning of a line like "stop going".
#Waxo gave the right answer:
This one is slightly better, if you want to match any word beginning with "stop" and containing nothing but letters from A to Z.
\bstop[a-zA-Z]*\b
This would match all
stop (1)
stop random (2)
stopping (3)
want to stop (4)
please stop (5)
But
/^stop[a-zA-Z]*/
would only match (1) until (3), but not (4) & (5)
If you want to match anything that starts with "stop" including "stop going", "stop" and "stopping" use:
^stop
If you want to match the word stop followed by anything as in "stop going", "stop this", but not "stopped" and not "stopping" use:
^stop\W
/stop([a-zA-Z])+/
Will match any stop word (stop, stopped, stopping, etc)
However, if you just want to match "stop" at the start of a string
/^stop/
will do :D
If you want the word to start with "stop", you can use the following pattern.
"^stop.*"
This will match words starting with stop followed by anything.
/^stop*$/i
i - in case it is case sensitive.
I'd advise against a simple regular expression approach to this problem. There are too many words that are substrings of other unrelated words, and you'll probably drive yourself crazy trying to overadapt the simpler solutions already provided.
You'll want at least a naive stemming algorithm (try the Porter stemmer; there's available, free code in most languages) to process text first. Keep this processed text and the preprocessed text in two separate space-split arrays. Make sure each non-alphabetical character also gets its own index in this array. Whatever list of words you're filtering, stem them also.
The next step would be to find the array indices which match to your list of stemmed 'stop' words. Remove those from the unprocessed array, and then rejoin on spaces.
This is only slightly more complicated, but will be much more reliable an approach. If you've got any doubts on the value of a more NLP-oriented approach, you might want to do some research into clbuttic mistakes.
can you try this:
https://regex101.com/r/P3qfKG/1
reg = /stop(\w+| [^ ]+|$)/gm
it will select both stop and start with stop and next word;
How do I create a regular expression to match a word at the beginning of a string?
We are looking to match stop at the beginning of a string and anything can follow it.
For example, the expression should match:
stop
stop random
stopping
If you wish to match only lines beginning with stop, use
^stop
If you wish to match lines beginning with the word stop followed by a space:
^stop\s
Or, if you wish to match lines beginning with the word stop, but followed by either a space or any other non-word character you can use (your regex flavor permitting)
^stop\W
On the other hand, what follows matches a word at the beginning of a string on most regex flavors (in these flavors \w matches the opposite of \W)
^\w
If your flavor does not have the \w shortcut, you can use
^[a-zA-Z0-9]+
Be wary that this second idiom will only match letters and numbers, no symbol whatsoever.
Check your regex flavor manual to know what shortcuts are allowed and what exactly do they match (and how do they deal with Unicode).
Try this:
/^stop.*$/
Explanation:
/ charachters delimit the regular expression (i.e. they are not part of the Regex per se)
^ means match at the beginning of the line
. followed by * means match any character (.), any number of times (*)
$ means to the end of the line
If you would like to enforce that stop be followed by a whitespace, you could modify the RegEx like so:
/^stop\s+.*$/
\s means any whitespace character
+ following the \s means there has to be at least one whitespace character following after the stop word
Note: Also keep in mind that the RegEx above requires that the stop word be followed by a space! So it wouldn't match a line that only contains: stop
If you want to match anything after a word, stop, and not only at the start of the line, you may use: \bstop.*\b - word followed by a line.
Or if you want to match the word in the string, use \bstop[a-zA-Z]* - only the words starting with stop.
Or the start of lines with stop - ^stop[a-zA-Z]* for the word only - first word only.
The whole line ^stop.* - first line of the string only.
And if you want to match every string starting with stop, including newlines, use: /^stop.*/s - multiline string starting with stop.
Like #SharadHolani said. This won't match every word beginning with "stop"
. Only if it's at the beginning of a line like "stop going".
#Waxo gave the right answer:
This one is slightly better, if you want to match any word beginning with "stop" and containing nothing but letters from A to Z.
\bstop[a-zA-Z]*\b
This would match all
stop (1)
stop random (2)
stopping (3)
want to stop (4)
please stop (5)
But
/^stop[a-zA-Z]*/
would only match (1) until (3), but not (4) & (5)
If you want to match anything that starts with "stop" including "stop going", "stop" and "stopping" use:
^stop
If you want to match the word stop followed by anything as in "stop going", "stop this", but not "stopped" and not "stopping" use:
^stop\W
/stop([a-zA-Z])+/
Will match any stop word (stop, stopped, stopping, etc)
However, if you just want to match "stop" at the start of a string
/^stop/
will do :D
If you want the word to start with "stop", you can use the following pattern.
"^stop.*"
This will match words starting with stop followed by anything.
/^stop*$/i
i - in case it is case sensitive.
I'd advise against a simple regular expression approach to this problem. There are too many words that are substrings of other unrelated words, and you'll probably drive yourself crazy trying to overadapt the simpler solutions already provided.
You'll want at least a naive stemming algorithm (try the Porter stemmer; there's available, free code in most languages) to process text first. Keep this processed text and the preprocessed text in two separate space-split arrays. Make sure each non-alphabetical character also gets its own index in this array. Whatever list of words you're filtering, stem them also.
The next step would be to find the array indices which match to your list of stemmed 'stop' words. Remove those from the unprocessed array, and then rejoin on spaces.
This is only slightly more complicated, but will be much more reliable an approach. If you've got any doubts on the value of a more NLP-oriented approach, you might want to do some research into clbuttic mistakes.
can you try this:
https://regex101.com/r/P3qfKG/1
reg = /stop(\w+| [^ ]+|$)/gm
it will select both stop and start with stop and next word;
I am trying to make a swearing prevention system, so far I have ignored the whitespace (with "\s*") and I've ignored the case("(?i)"). How would I ignore repeated characters ? e.g heeeello.
There is no flag that you can turn on to simply ignore any duplicate characters. However, you can use the 'one or more' quantifier (+) to match one or more occurrence of any character, character class, or group. For example the pattern he+l+o will match all of the following:
helo
heelo
hello
heeeello
Assuming you want a general solution to remove repeated characters, you'll want to replace (.)\1 with \1 repeatedly as long as it succeeds.
Use + to catch as many repetition of a sequence in () as there are. e+ will catch all the e's in heeeeello.
Check out rubular.com, very simple way to learn, practice and test regex.
You need to capture a single character then check for any repetition of it with using a backreference to the lately captured group:
(.)\1+
If string is matched then it has repetition.
Live demo
The problem is harder than you think. Let's assume that you want to match "no fewer than this number of characters" for each word in your dictionary. Then you would have to create a dictionary of regexes with a + after each character…
Initial dictionary:
boom
smurf
tree
cannibals
Process the dictionary with a text editor:
sed -e 's/\(.\)/\1\+/g' dictionary.txt > regex.txt
This puts a + between all characters:
b+o+o+m+
s+m+u+r+f+
t+r+e+e+
c+a+n+n+i+b+a+l+s+
And now you can match your "repeated" words:
bom : no match
smuuurf : match
trees : no match
canibals : no match
cannnibalssss : match
You might want to add "word boundaries" - so that smurfette doesn't get caught by smurf. This would mean adding \b before and after each expression ("word boundary").
Note - it's not enough to remove all duplicate letters from both the dictionary, and the words to be matched - otherwise you risk banning "pop" because you had "poop" on your list (and how would you know to stop when "pooop" had reached exactly two characters). This is why I prefer this solution over some of the others that recommend stripping repeats.
I am trying to match lines in a configuration that start with the word "deny" but do not end with the word "log". This seems terribly elementary but I can not find my solution in any of the numerous forums I have looked. My beginners mindset led me to try "^deny.* (?!log$)" Why wouldn't this work? My understanding is that it would find any strings that begin with "deny" followed by any character for 0 or more digits where the end of line is something other than log.
When given a line like deny this log, your ^deny.*(?!log$) regex (I'm omitting the space that was in your sample question) is evaluated as follows:
^deny matches "deny".
.* means "match 0 or more of any character", so it can match " this log".
^(?!log$) means "make sure that the next characters aren't 'log' then the end of the line." In this case, they're not - they're just the end of the line - so the regex matches.
Try this regex instead:
^deny.*$(?<!log)
"Match deny at the beginning of the string, then match to the end of the line, then use a zero-width negative look-behind assertion to check that whatever we just matched at the end of the line is not 'log'."
With all of that said...
Regexes aren't necessarily the best tool for the job. In this case, a simple Boolean operator like
if (/^deny/ and not /log$/)
is probably clearer than a more advanced regex like
if (/^deny.*$(?<!log)/)
(?!log$) is a zero-width negative look-ahead assertion that means don't match if immediately ahead at this point in the string is log and the end of the string, but the .* in your regex has already greedily consumed all the characters right up to the end of the string so there is no way the log could then match.
If your regular expression implementation supports look-behinds you could use a regex such as in Josh Kelley's answer, if you were using javascript you could use
/^deny(?:.{0,2}|.*(?!log)...)$/m
The m flag means multiline mode, which makes ^ and $ match the start and end of every line rather than just the start and end of the string.
Note that three . are positioned after the negative look-ahead so that it has space to match log if it is there. Including these three dots meant it was also necessary to add the .{0,2} option so that strings with from zero to two characters after deny would also match. The (?:a|b) means a non-capturing group where a or b has to match.
I was wondering how to match a line without either of two words?
For example, I would like to match a line without neither Chapter nor Part. So neither of these two lines is a match:
("Chapter 2 The Economic Problem 31" "#74")
("Part 2 How Markets Work 51" "#94")
while this is a match
("Scatter Diagrams 21" "#64")
My python-style regex will be like (?<!(Chapter|Part)).*?\n. I know it is not right and will appreciate your help.
Try this:
^(?!.*(Chapter|Part)).*
#MRAB's solution will work, but here's another option:
(?m)^(?:(?!\b(?:Chapter|Part)\b).)*$
The . matches one character at a time, after the lookahead checks that it's not the first character of Chapter or Part. The word boundaries (\b) make sure it doesn't incorrectly match part of a longer word, like Partition.
The ^ and $ are start- and end anchors; they ensure that you match a whole line. $ is better than \n because it also matches the end of the last line, which won't necessarily have a linefeed at the end. The (?m) at the beginning modifies the meaning of the anchors; without that, they only match at the beginning and end of the whole input, not of individual lines.