Python: cleaner list comprehension

Is there a cleaner way to write this:

fo开发者_开发知识库r w in [w for w in words if w != '']:

I want to loop over a dictionary words, but only words that != ''. Thanks!

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You don't need a listcomp here. Just write:

for w in words:
    if w != '':
        # ...

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Assuming that you are after the keys, why not try:

[w for w in words if w]

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Testing that an element does not equal '' isn't going to filter out whitespace elements. If that's what you're after, you probably want to use str.isspace (or a regular expression).

If you use a list comprehension, you'll make an extra copy of the list as an intermediary object. Probably not a big deal, but a generator won't use the extra memory.

I'd do it like this, with a generator:

for word in (w for w in words if not w.isspace()):
    # do stuff

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I think your solution is sub optimal. You're iterating over the list words twice - once in the list comprehension to create the non-null terms and again in the loop to do the processing. It would be better if you used a genexp like so.

for w in (x for x in words if x): process(w)

That way, the genexp will lazily return a list of non-nulls.

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filter(lambda w: w != '', words) or filter(None, words)

this is suggestion, it may not be the best solution for your problem.

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What does the body of the outer for loop do?

If it's a function call you could potentially just do:

[f(w) for w in words if w != '']