I'm trying to match a word that has 2 vowels in it (doesn't have to be consecutively) but the regex I've come up either matches nothing or not enough. This is the last iteration (dart).
final vowelRegex = new RegExp(r'[aeiouy]{2}');
Here's an example sentence being parsed and it should match, one, shoulder, their, and over. It's only matching shoulder and their. I understand why, because that's the expression I defined. How can the expression be defined to match on 2 vowels, regardless of position in the word?
one shoulder their the which over
The expression only needs to be tested on one word at a time so hopefully this simplifies things.
You can use :
new RegExp(r'(\w*[aeiouy]\w*){2}');
Both of the previous two answers are incorrect.
(\S*[aeiouy]\S*){2} can match substrings of non-whitespace characters even if they contain non-word characters (proof).
\S*[aeiouy]\S*[aeiouy]\S* has the same problem (proof).
Correct solution:
\b([^\Waeiou]*[aeiou]){2}\w*\b
And if you want only whitespace to count as the word boundary (rather than any non-word character), then use the following regex where the target word is in capture group \2.
(\s|^)(([^\Waeiou]*[aeiou]){2}\w*)(\s|$)
You can try this:
\S*[aeiouy]\S*[aeiouy]\S*
Explanation
\S* matches any non-whitespace character (equal to [^\r\n\t\f ])
* Quantifier — Matches between zero and unlimited times
[aeiou] Match a single character present in the list below [aeiou]
For input string : one shoulder their the which over
it will match four word: one shoulder their over
I'd do:
\b(?:\w*[aeiouy]+\w*){2,}\b
Explanation:
\b : word boundary
(?: : start non-capture group
\w* : 0 or more word characters
[aeiouy]+ : 1 or more vowels
\w* : 0 or more word characters
){2,} : end group repeated at least twice
\b : word boundary
Related
I've been trying to solve this problems for few hours but with no luck. The task is to write a regular expression that matches at least four words starting with the same letter. But! These words do not have to be one after another.
This regex should be able to match a line like this:
cat color coral chat
but also one like this:
cat take boom candle creepy drum cheek
Thank you!
So far I have got this regex but it only matches words when they are in order.
(\w)\w+\s+\1\w+\s+\1\w+\s+\1
If you have only words in the line that can be matched with \w:
\b(\w)\w*(?:(?:\s+\w+)*?\s+\1\w*){3}
Explanation
\b A word boundary to prevent a partial word match
(\w)\w* Capture a single word character in group 1 followed by matching optional word characters
(?: Non capture group to repeat as a whole part
(?:\s+\w+)*? Match 1+ whitespace chars and 1+ word chars in between in case the word does not start with the character captured in the back reference
\s+\1\w* Match 1+ whitespace chars, a backreference to the same captured character and optional word characters
){3} Close the non capture group and repeat 3 times
See a regex demo
Note that \s can also match a newline.
If the words that should with the same character should be at least 2 characters long (as (\w)\w+ matches 2 or more characters)
\b(\w)\w+(?:(?:\s+\w+)*?\s+\1\w+){3}
See another regex demo.
Another idea to match lines with at least 4 words starting with the same letter:
\b(\w)(?:.*?\b\1){3}
See this demo at regex101
This is not very accurate, it just checks if there are three \b word boundaries, each followed by \1 in the first group \b(\w) captured character to the right with .*? any characters in between.
I need a regular expression to match the first word with character 'a' in it for each line. For example my test string is this:
bbsc abcd aaaagdhskss
dsaa asdd aaaagdfhdghd
wwer wwww awww wwwd
Only the ones in BOLD fonts should be matched. How can I do that? I can match all the words with 'a' in it, but can't figure out how to only match the first occurrence.
Under the assumption that the only characters being used are word characters, i.e. \w characters, and white space then use:
/^(?:[^a ]+ +)*([^a ]*a\w*)\b/gm
^ Matches the start of the line
(?:[^a ]+ +)* Matches 0 or more occurrences of words composed of any character other than an a followed by one or more spaces in a non-capturing group.
([^a ]*a\w*)\b Matches a word ending on a word boundary (it is already guaranteed to begin on a word boundary) that contains an a. The word-boundary constraint allows for the word to be at the end of the line.
The first word with an a in it will be in group #1.
See demo
If we cannot assume that only word (\w) and white space characters are present, then use:
^(?:[^a ]+ +)*(\w*a\w*)\b
The difference is in scanning the first word with an a in it, (\w*a\w*), where we are guaranteed that we are scanning a string composed of only word characters.
What are you using? In many programs you can set limit. If possible: \b[b-z]*a[a-z]* with 1 limit.
If it is not possible, use group to capture and match latter: ([b-z]*a[a-z]*).*
Try:
^(?:[^a ]+ )*(\w*a\w*) .*$
Basically what it says is: capture a bunch of words that are composed of anything but the letter a (or <space>) then capture a word that must include the letter a.
Group 1 should hold the first word with a.
The words' length could be 2 or 6-10 and could be separated by space or comma. The word only include alphabet, not case sensitive.
Here is the groups of words that should be matched:
RE,re,rereRE
Not matching groups:
RE,rere,rel
RE,RERE
Here is the pattern that I have tried
((([a-zA-Z]{2})|([a-zA-Z]{6,10}))(,|\s+)?)
But unfortunately this pattern can match string like this: RE,RERE
Look like the word boundary has not been set.
You could match chars a-z either 2 or 6 - 10 times using an alternation
Then repeat that pattern 0+ times preceded by a comma or a space [ ,].
^(?:[A-Za-z]{6,10}|[A-Za-z]{2})(?:[, ](?:[A-Za-z]{6,10}|[A-Za-z]{2}))*$
Explanation
^ Start of string
(?:[A-Za-z]{6,10}|[A-Za-z]{2}) Match chars a-z 6 -10 or 2 times
(?: Non capturing group
[, ](?:[A-Za-z]{6,10}|[A-Za-z]{2}) Match comma or space and repeat previous pattern
)* Close non capturing group and repeat 0+ times
$ End of string
Regex demo
If lookarounds are supported, you might also assert what is directly on the left and on the right is not a non whitespace character \S.
(?<!\S)(?:[A-Za-z]{6,10}|[A-Za-z]{2})(?:[ ,](?:[A-Za-z]{6,10}|[A-Za-z]{2}))*(?!\S)
Regex demo
([a-zA-Z]{2}(,|\s)|[a-zA-Z]{6,10}|(,|\s))
This one will get only the words who have 2 letter, or between 6 and 10
\b,?([a-zA-Z]{6,10}|[a-zA-Z]{2}),?\b
You can use this
^(?!.*\b[a-z]{4}\b)(?:(?:[a-z]{2}|[a-z]{6,10})(?:,|[ ]+)?)+$
Regex Demo
This regex will match your first case, but neither of your two other cases:
^((([a-zA-Z]{2})|([a-zA-Z]{6,10}))(,|[ ]+|$))+$
I'm making the assumption here that each line should be a single match.
Here it is in action.
I have a text:
'1 2 3 ab AB úá awindow BCwindow'
Currently to get only words I use this regex: [a-zA-Zá-ú]+ and this is the result:
['ab', 'awindow', 'bcwindow', 'úá']
I would like to remove 'window' string of mathes words to get this:
['ab','a','bc','úá']
Thanks.
If word window always appears at the end of a matching word, you could do:
(?<!\S)[a-zA-Zá-ú]+?(?:(?!\S)|(?=window))
This assures you don't have extra non-whitespace characters preceding a word (prevents a match to begin from middle of a longer string) or following it. You may use word boundaries \b instead:
\b[a-zA-Zá-ú]+?(?:\b|(?=window))
Live demo
Breakdown:
\b Match a word boundary position (where a word begins)
[a-zA-Zá-ú]+? Match characters in class at least one time, ungreedily
(?: Start of non-capturing group
\b Match a word boundary (here we mean end of word)
| Or
(?=window) A positive lookahead, assert following characters are window
) End of non-capturing group
Whenever second word boundary is matched or positive lookahead asserts then engine is satisfied and every thing up to that point is returned as a match.
I thought [^0-9a-zA-Z]* excludes all alpha-numeric letters, but allows for special characters, spaces, etc.
With the search string [^0-9a-zA-Z]*ELL[^0-9A-Z]* I expect outputs such as
ELL
ELLs
The ELL
Which ELLs
However I also get following outputs
Ellis Island
Bellis
How to correct this?
You may use
(?:\b|_)ELLs?(?=\b|_)
See the regex demo.
It will find ELL or ELLs if it is surrounded with _ or non-word chars, or at the start/end of the string.
Details:
(?:\b|_) - a non-capturing alternation group matching a word boundary position (\b) or (|) a _
ELLs? - matches ELL or ELLs since s? matches 1 or 0 s chars
(?=\b|_) - a positive lookahead that requires the presence of a word boundary or _ immediately to the right of the current location.
change the * to +
a * means any amount including none. A + means one or more. What you probably want though is a word boundry:
\bELL\b
A word boundry is a position between \w and \W (non-word char), or at the beginning or end of a string if it begins or ends (respectively) with a word character ([0-9A-Za-z_]). More here about that:
What is a word boundary in regexes?