Welcome to OStack Knowledge Sharing Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
157 views
in Technique[技术] by (71.8m points)

python - Flatten an irregular list of lists

Yes, I know this subject has been covered before (here, here, here, here), but as far as I know, all solutions, except for one, fail on a list like this:

L = [[[1, 2, 3], [4, 5]], 6]

Where the desired output is

[1, 2, 3, 4, 5, 6]

Or perhaps even better, an iterator. The only solution I saw that works for an arbitrary nesting is found in this question:

def flatten(x):
    result = []
    for el in x:
        if hasattr(el, "__iter__") and not isinstance(el, basestring):
            result.extend(flatten(el))
        else:
            result.append(el)
    return result

flatten(L)

Is this the best model? Did I overlook something? Any problems?

Question&Answers:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Answer

0 votes
by (71.8m points)

Using generator functions can make your example a little easier to read and probably boost the performance.

Python 2

def flatten(l):
    for el in l:
        if isinstance(el, collections.Iterable) and not isinstance(el, basestring):
            for sub in flatten(el):
                yield sub
        else:
            yield el

I used the Iterable ABC added in 2.6.

Python 3

In Python 3, the basestring is no more, but you can use a tuple of str and bytes to get the same effect there.

The yield from operator returns an item from a generator one at a time. This syntax for delegating to a subgenerator was added in 3.3

from collections.abc import Iterable

def flatten(l):
    for el in l:
        if isinstance(el, Iterable) and not isinstance(el, (str, bytes)):
            yield from flatten(el)
        else:
            yield el

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome to OStack Knowledge Sharing Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...