too many indentation when checking a string with many regular expressions serially in python

Question

I get deep indentation when I write code like below

match = re.search(some_regex_1, s)
if match:
    # do something with match data
else:
    match = re.search(some_regex_2, s)
    if match:
        # do something with match data
    else:
        match = re.search(soem_regex_3, s)
        if match:
            # do something with match data
        else:
            # ...
            # and so on

I tried to rewrite as:

if match = re.search(some_regex_1, s):
    # ...
elif match = re.search(some_regex_2, s):
    # ...
elif ....
    # ...
...

but Python doesn't allow that syntax. What should I do to avoid deep indentation in this case?

The answers below address this specific case, but, generally, it's important to note that you need `if a == b: ... ` in Python, not just one equals sign as you have above. Or, depending on the scenario, `if a is b: ...` The `elif` statements above work in theory, but you need the equals-equals. — Karmel, May 21 '12 at 18:45
@Karmel, I think he's trying to assign and test `match` for truthness at the same time, like folks might do in C: `while (data=fread(fp)) {` — JoeFish, May 21 '12 at 18:51
Good point. I thought is was a more general assumption. My apologies, @Le Curious :) — Karmel, May 21 '12 at 18:55

Silas Ray · Accepted Answer · 2012-05-21T18:52:59.727

6

regexes = (regex1, regex2, regex3)
for regex in regexes:
    match = re.search(regex, s)
    if match:
        #do stuff
        break

Alternatively (more advanced):

def process1(match_obj):
    #handle match 1

def process2(match_obj):
    #handle match 2

def process3(match_obj):
    #handle match 3
.
.
.
handler_map = ((regex1, process1), (regex2, process2), (regex3, process3))
for regex, handler in handler_map:
    match = re.search(regex, s)
    if match:
        result = handler(match)
        break
else:
    #else condition if no regex matches

edited May 21 '12 at 18:52

answered May 21 '12 at 18:38

Silas Ray

25,682
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Nice! You probably want to enumerate the regexes so you can keep track of which regex was matched though. `for i, regex in enumerate(regexes): ...` – Junuxx May 21 '12 at 18:39
@Junuxx, wasn't in the spec, but sure, could be useful in some cases. You might also just want to know that something was matched to handle the final else condition. – Silas Ray May 21 '12 at 18:40
I assumed that the 'do stuff' could be different depending on which regex it matched and you don't want to test it again :p – Junuxx May 21 '12 at 18:41
I suggest to add variable i, so you would be able to easily check which regex has matched. After iteration if i == num of regex, then no regex has matched. – kravemir May 21 '12 at 18:42
Oh yeah, I see your points. Though then you end up with another if/elif/else construct... let me put in something better. – Silas Ray May 21 '12 at 18:42
3

Instead of 'if not result' you can use an 'else' block in the for loop. – Kamil Kisiel May 21 '12 at 18:51
1

@KamilKisiel Good call, changed. – Silas Ray May 21 '12 at 18:53

pillmuncher · Answer 2 · 2012-09-17T03:19:41.503

If you can use finditer() instead of search() (most of the time you can), you could join all your regexes into one and use symbolic group names. Here is an example:

import re

regex = """
   (?P<number> \d+ ) |
   (?P<word> \w+ ) |
   (?P<punctuation> \. | \! | \? | \, | \; | \: ) |
   (?P<whitespace> \s+ ) |
   (?P<eof> $ ) |
   (?P<error> \S )
"""

scan = re.compile(pattern=regex, flags=re.VERBOSE).finditer

for match in scan('Hi, my name is Joe. I am 1 programmer.'):
    token_type = match.lastgroup
    if token_type == 'number':
        print 'found number "%s"' % match.group()
    elif token_type == 'word':
        print 'found word "%s"' % match.group()
    elif token_type == 'punctuation':
        print 'found punctuation character "%s"' % match.group()
    elif token_type == 'whitespace':
        print 'found whitespace'
    elif token_type == 'eof':
        print 'done parsing'
        break
    else:
        raise ValueError('String kaputt!')

score 0 · Answer 3 · answered May 21 '12 at 18:40

0

if re.search(some_regex_1, s) is not None:
    # ...
elif re.search(some_regex_2, s) is not None:
    # ...
elif ....
    # ...
...

search() returns None if there is no match found, so in your if statement it will proceed to the next test.

answered May 21 '12 at 18:40

Andrew Sledge

10,163
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4

But then you can't actually use the match data for anything without re-running the search. – Silas Ray May 21 '12 at 18:41
+ ` is not None` isn't required. If i were in you, i would delete this answer, i think it gets downvoted :) – kravemir May 21 '12 at 18:45

too many indentation when checking a string with many regular expressions serially in python

3 Answers3