How to write in more pythonic way?

Question

I want to write this code in more pythonic way. Have you any idea how I can do this?

def counter():
    for x in self.get_xs():
        total_x = 0
        result = (re.sub('[^0-9]', '', x))
        for number in result:
            total_x += int(number)
        yield(total_x)

Where did result came from? Perhaps you meant: re.sub('[^0-9]', '', x). — Ashwini Chaudhary
– Ashwini Chaudhary, Commented Jan 21, 2015 at 8:51

Ashwini Chaudhary · Accepted Answer · 2015-01-21 09:03:22Z

2

You can calculate the sum using sum() and a generator expression. If the iterable passed to sum() is empty(i.e re.sub() returned '') then it will simply return the default start value 0.

def counter():
    r = re.compile('[^0-9]')
    for x in self.get_xs():
        yield sum(int(number) for number in r.sub('', x))

In Python 3.3+ you can use yield from:

def counter():
    r = re.compile('[^0-9]')
    yield from (sum(map(int, r.sub('', x))) for x in self.get_xs())

edited Jan 21, 2015 at 9:03

answered Jan 21, 2015 at 8:57

Ashwini Chaudhary

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6 Comments

tobias_k Over a year ago

I think, in this case, where there's just one yield from, you could also do return (<generator>) in Python 2.

poke Over a year ago

It would be a good idea to move the compiled regular expression out of the function so it’s reused instead of recompiled every time. Also, the function likely needs a self argument.

Ashwini Chaudhary Over a year ago

@poke IIRC Python caches regular expressions.

Matthias Over a year ago

The docs say: "The compiled versions of the most recent patterns passed to re.match(), re.search() or re.compile() are cached, so programs that use only a few regular expressions at a time needn’t worry about compiling regular expressions." re.sub is not mentioned.

poke Over a year ago

@AshwiniChaudhary Interesting, I didn’t know that!

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