if elif with strings

Question

I'm refactoring some scripts that have a bunch of code with like this:

if 'mString1' in mStrVar:
    retVal = 1
elif 'mString2' in mStrVar:
    retVal = 2
elif 'mString3' in mStrVar:
    retVal = 3
elif 'mString4' in mStrVar:
    retVal = 4
else:
    retVal = 0

Personally I don't like this I always prefer the dict() approach, but in this specific case, I think, I cannot be done in that way.

is it possible to rewrite this in a more short way? mString goes around mString10 in some cases.

Any hints highly appreciated, and apologies if this is a duplicated I couldn't find any question related.

What if `mStrVar = "mString1mString2mString3mString4"`? IMHO the logic itself of mapping substrings to a single number is already fragile. — Has QUIT--Anony-Mousse, Jan 29 '13 at 12:13
Also, matching with regular expressions may be a bit more explicit, and may be able to extract the matching string in a single pass over the string. — Has QUIT--Anony-Mousse, Jan 29 '13 at 12:17
Agreed, but I don't control the input in this case `mStrVar`. That is passed by another system. We have to assume that `mString`s are mutually exclusive. — rhormaza, Jan 29 '13 at 12:17
@DavidHeffernan well, it is a bit long, just looking for a shorter alternative, if possible. — rhormaza, Jan 29 '13 at 12:20
`mString1` is in `mString10`, so you'd need to be careful about the order if you use this method (longest first should work, I think, but I haven't given it much thought.) — DSM, Jan 29 '13 at 12:31

score 1 · Accepted Answer · answered Jan 29 '13 at 12:24

You could create a mapping list and use a plain for loop to search for the strings in the mStrVar variable:

retVal_mapping = [
   ('mString1', 1),
   ('mString2', 2),
   ('mString3', 2),
   ('mString4', 2),
   ('', 0) # default as "'' in anyString" is always true.
]

for s, retVal in retVal_mapping:
    if s in mStrVar:
        break

# retVal now contains the right value

score 0 · Answer 2 · answered Jan 29 '13 at 12:19

If you would for example use the regular expression mString(\d+) you can do this:

# Precompile pattern into finite automaton:
_pattern = re.compile(r'mString(\d+)')

# Apply multiple times.
match = _pattern.match(mStrVar)
if match:
    retVal = int(match.group(1))
else:
    retVal = 0

The problem with your substring-code above is that it is not very well defined if e.g. mStrVar = "mString1mString2mString3mString4". The regular expression matching is actually much stricter, which may be desired.

You can do a dict style approach, too, for non-numeric patterns:

# Precompile pattern:
_pattern = re.compile(r'(abc|def|ghi)')
_map = { 'def' : 1, 'ghi' : 2, 'abc' : 3 }

match = _pattern.match(mStrVar)
if match:
    retVal = _map.get(match.group(1), 0)
else:
    retVal = 0

in some cases, you can just use the dict right away, too:

_map = { 'def' : 1, 'ghi' : 2, 'abc' : 3 }
retVal = _map.get(mStrVar, 0) # 0 is default!

score 0 · Answer 3 · answered Jan 29 '13 at 12:21

0

some_dict = dict(mString1=1, mString2=2, mString3=3)
return some_dict.get(string_value, 0)

Where string_value is one of mString1, mString2, mString3.

answered Jan 29 '13 at 12:21

UnLiMiTeD

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This is not equivalent to the original code if more than one of the terms can be found in `mStrVar`. – Tim Pietzcker Jan 29 '13 at 12:24
you forgot to check whether `mStringN` is in `mStrVar` – rhormaza Jan 29 '13 at 12:25
1

If its no in mStrVal it will return 0. – UnLiMiTeD Jan 29 '13 at 12:30

score 0 · Answer 4 · answered Jan 29 '13 at 12:25

0

mylist = ['mString1', 'mString2', 'mString3', 'mString4']
retVal = [i for i in range(len(mylist)) if mylist[i] = mStrVar][0] + 1

answered Jan 29 '13 at 12:25

f p

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score 0 · Answer 5 · answered Jan 29 '13 at 12:29

0

retval, = [i for i in xrange(1, 5) if 'mString%s' % i in mStrVar] or [0]

answered Jan 29 '13 at 12:29

Rich Tier

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if elif with strings

5 Answers5