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Suppose I have the following string:

trend  = '(A|B|C)_STRING'

I want to expand this to:

A_STRING
B_STRING
C_STRING

The OR condition can be anywhere in the string. i.e STRING_(A|B)_STRING_(C|D)

would expand to

STRING_A_STRING_C
STRING_B_STRING C
STRING_A_STRING_D
STRING_B_STRING_D

I also want to cover the case of an empty conditional:

(|A_)STRING would expand to:

A_STRING
STRING

Here's what I've tried so far:

def expandOr(trend):
    parenBegin = trend.index('(') + 1
    parenEnd = trend.index(')')
    orExpression = trend[parenBegin:parenEnd]
    originalTrend = trend[0:parenBegin - 1]
    expandedOrList = []

    for oe in orExpression.split("|"):
        expandedOrList.append(originalTrend + oe)

But this is obviously not working.

Is there any easy way to do this using regex?

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1  
You realize you're discarding everything after the closing parenthesis, right? Do you not see a way to fix that? –  jwodder 1 hour ago
 
Not sure what you mean. The code works for the case where the parentheses come at the end the of the string. i.e. STRING_(A|B) –  Mark Kennedy 53 mins ago
 
Right, the code works there because there's nothing after the parentheses to discard, but if you input FOO_(A|B)_BAR, you get FOO_A and FOO_B, with the _BAR being discarded. Do you not realize that this is what's wrong with your code? Do you not see how you forgot to handle the substring after the )? –  jwodder 48 mins ago

1 Answer

up vote 0 down vote

I would do this to extract the groups:

def extract_groups(trend):
    l_parens = [i for i,c in enumerate(trend) if c == '(']
    r_parens = [i for i,c in enumerate(trend) if c == ')']
    assert len(l_parens) == len(r_parens)
    return [trend[l+1:r].split('|') for l,r in zip(l_parens,r_parens)]

And then you can evaluate the product of those extracted groups using itertools.product:

expr = 'STRING_(A|B)_STRING_(C|D)'
from itertools import product
list(product(*extract_groups(expr)))
Out[92]: [('A', 'C'), ('A', 'D'), ('B', 'C'), ('B', 'D')]

Now it's just a question of splicing those back onto your original expression. I'll use re for that :)

#python3.3+
def _gen(it):
    yield from it

p = re.compile('\(.*?\)')

for tup in product(*extract_groups(trend)):
    gen = _gen(tup)
    print(p.sub(lambda x: next(gen),trend))

STRING_A_STRING_C
STRING_A_STRING_D
STRING_B_STRING_C
STRING_B_STRING_D

There's probably a more readable way to get re.sub to sequentially substitute things from an iterable, but this is what came off the top of my head.

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