I want to use a regex that looks for spaces with a minimum length of 2 in a row, and replaces the occurrence with another value for each occurrence of the space found.
For example:
I love to eat cake
There are 3 spaces after love and 4 spaces after eat. I want my regex to replace occurrences of a space more than 1, and to replace it with a value for each occurrence found.
The output I am trying to go for:
I love---to eat----cake
I tried something like
myStr.replace(/ {2,}/g, '-')
CodePudding user response:
You may use this code with a lookahead and a lookbehind:
const s = 'I love to eat cake'
var r = s.replace(/ (?= )|(?<= ) /g, '-');
console.log(r);
//=> 'I love---to eat----cake'
<iframe name="sif1" sandbox="allow-forms allow-modals allow-scripts" frameborder="0"></iframe>
RegEx Details:
(?= )
: Match a space only if that is followed by a space|
: OR(?<= )
: Match a space only if that is preceded by a space
CodePudding user response:
You can match two or more whitespaces and replace with the same amount of hyphens:
const s = 'I love to eat cake'
console.log(s.replace(/\s{2,}/g, (x) => '-'.repeat(x.length)) )
<iframe name="sif2" sandbox="allow-forms allow-modals allow-scripts" frameborder="0"></iframe>
The same approach can be used in Python (since you asked), re.sub(r'\s{2,}', lambda x: '-' * len(x.group()), s)
, see the Python demo.
Also, you may replace any whitespace that is followed with a whitespace char or is preceded with whitespace using
const s = 'I love to eat cake'
console.log(s.replace(/\s(?=\s|(?<=\s.))/gs, '-') )
console.log(s.replace(/\s(?=\s|(?<=\s\s))/g, '-') )
<iframe name="sif3" sandbox="allow-forms allow-modals allow-scripts" frameborder="0"></iframe>
See this regex demo. Here, s
flag makes .
match any char. g
makes the regex replace all occurrences. Also,
\s
- matches any whitespace(?=\s|(?<=\s.))
- a positive lookahead that matches a location that is immediately followed with a whitespace char (\s
), or (|
) if it is immediately preceded with a whitespace and any one char (which is the matched whitespace). If you use(?<=\s\s)
version, there is no need ofs
flag,\s\s
just makes sure the whitespace before the matched whitespace is checked.