How to capture multiple repeating patterns with regular expression?

Question

I get some string like this: \input{{whatever}{1}}\mypath{{path1}{path2}{path3}...{pathn}}\shape{{0.2}{0.3}} I would like to capture all the paths: path1, path2, ... pathn. I tried the re module in python. However, it does not support multiple capture. For example: r"\\mypath\{(\{[^\{\}\[\]]*\})*\}" will only return the last matched group. Applying the pattern to search(r"\mypath{{path1}{path2}})" will only return groups() as ("{path2}",)

Then I found an alternative way to do this:

    gpathRegexPat=r"(?:\\mypath\{)((\{[^\{\}\[\]]*\})*)(?:\})"
    gpathRegexCp=re.compile(gpathRegexPat)
    strpath=gpathRegexCp.search(r'\mypath{{sadf}{ad}}').groups()[0]
    >>> strpath
    '{sadf}{ad}'
    p=re.compile('\{([^\{\}\[\]]*)\}')
    >>> p.findall(strpath)
    ['sadf', 'ad']

or:

    >>> gpathRegexPat=r"\\mypath\{(\{[^{}[\]]*\})*\}"
    >>> gpathRegexCp=re.compile(gpathRegexPat, flags=re.I|re.U)
    >>> strpath=gpathRegexCp.search(r'\input{{whatever]{1}}\mypath{{sadf}{ad}}\shape{{0.2}{0.1}}').group()
    >>> strpath
    '\\mypath{{sadf}{ad}}'
    >>> p.findall(strpath)
    ['sadf', 'ad']

At this point, I thought, why not just use the findall on the original string? I may use: gpathRegexPat=r"(?:\\mypath\{)(?:\{[^\{\}\[\]]*\})*?\{([^\{\}\[\]]*)\}(?:\{[^\{\}\[\]]*\})*?(?:\})": if the first (?:\{[^\{\}\[\]]*\})*? matches 0 time and the 2nd (?:\{[^\{\}\[\]]*\})*? matches 1 time, it will capture sadf; if the first (?:\{[^\{\}\[\]]*\})*? matches 1 time, the 2nd one matches 0 time, it will capture ad. However, it will only return ['sadf'] with this regex.

With out all those extra patterns ((?:\\mypath\{) and (?:\})), it actually works:

    >>> p2=re.compile(r'(?:\{[^\{\}\[\]]*\})*?\{([^\{\}\[\]]*)\}(?:\{[^\{\}\[\]]*\})*?')
    >>> p2.findall(strpath)
    ['sadf', 'ad']
    >>> p2.findall('{adadd}{dfada}{adafadf}')
    ['adadd', 'dfada', 'adafadf']

Can anyone explain this behavior to me? Is there any smarter way to achieve the result I want?

Joran Beasley · Accepted Answer · 2012-09-28 14:14:01Z

2

re.findall("{([^{}]+)}",text)

should work

returns

['path1', 'path2', 'path3', 'pathn']

finally

my_path = r"\input{{whatever}{1}}\mypath{{path1}{path2}{path3}...{pathn}}\shape{{0.2}{0.3}}"
#get the \mypath part
my_path2 = [p for p in my_path.split("\\") if p.startswith("mypath")][0]
print re.findall("{([^{}]+)}",my_path2)

or even better

re.findall("{(path\d+)}",text) #will only return things like path<num> inside {}

edited Sep 28, 2012 at 14:14

answered Sep 28, 2012 at 13:38

Joran Beasley

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5 Comments

Matthias Over a year ago

To respect Wang's path specification, the regular expression should be r"\{([^{}[\]]*)\}".

Wang Over a year ago

@Joran, thanks for reply, but consider a string buried inside other charaters : \input{{whatever}{1}}\mypath{{path1}{path2}{path3}...{pathn}}\shape{{0.1}{0.2}}. It will capture the parameters in input{} and \shape{} also. It still need to use 2-step capture I guess.

Wang Over a year ago

@ Joran Beasley, Thanks! So it is also 2-step. Can you explain a little bit more why the regex r"(?:\\mypath\{)(?:\{[^\{\}\[\]]*\})*?\{([^\{\}\[\]]*)\}(?:\{[^\{\}\[\]]*\})*?(?‌:\})" can not return all paths with findall() either?

Joran Beasley Over a year ago

naw with regex you want to keep them short ... long regex's take a long time for me to decode ...

jfs Over a year ago

@Wang: to extract the sequence (step 1): L = re.findall(r"\\mypath{({.*?})}", input_text) (assume \mypath occurs multiple times). Step 2: for each item in the list L call re.findall("{([^{}]+)}", item)

Hans Then · Accepted Answer · 2012-09-28 14:34:10Z

1

You are right. It is not possible to return repeated subgroups inside a group. To do what you want, you can use a regular expression to capture the group and then use a second regular expression to capture the repeated subgroups.

In this case that would be something like: \\mypath{(?:\{.*?\})}. This will return {path1}{path2}{path3}

Then to find the repeating patterns of {pathn} inside that string, you can simply use \{(.*?)\}. This will match anything withing the braces. The .*? is a non-greedy version of .*, meaning it will return the shortest possible match instead of the longest possible match.

edited Sep 28, 2012 at 14:34

answered Sep 28, 2012 at 13:38

Hans Then

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4 Comments

jfs Over a year ago

note: regex module supports repeated captures. Though they are not necessary in this case

Wang Over a year ago

@Sebastian. Thanks! This is not a build-in module I guess.

Wang Over a year ago

@Hans Then. Thanks! Can you explain a little bit more why the regex r"(?:\\mypath\{)(?:\{[^\{\}\[\]]*\})*?\{([^\{\}\[\]]*)\}(?:\{[^\{\}\[\]]*\})*?(?:\})" can not return all paths with findall() either?

Hans Then Over a year ago

That is because findall will return only the patterns that match your regular expression. Since you start with \mypath, findall will only return matches that start with that string, of which you have only one.

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