如何将清单分成大小均匀的区块？

Question

I have a list of arbitrary length, and I need to split it up into equal size chunks and operate on it. There are some obvious ways to do this, like keeping a counter and two lists, and when the second list fills up, add it to the first list and empty the second list for the next round of data, but this is potentially extremely expensive.

I was wondering if anyone had a good solution to this for lists of any length, e.g. using generators.

I was looking for something useful in itertools but I couldn't find anything obviously useful. Might've missed it, though.

Related question: What is the most “pythonic” way to iterate over a list in chunks?

转载于:https://stackoverflow.com/questions/312443/how-do-you-split-a-list-into-evenly-sized-chunks

Answer 1

Here's a generator that yields the chunks you want:

def chunks(l, n):
    """Yield successive n-sized chunks from l."""
    for i in range(0, len(l), n):
        yield l[i:i + n]

---

import pprint
pprint.pprint(list(chunks(range(10, 75), 10)))
[[10, 11, 12, 13, 14, 15, 16, 17, 18, 19],
 [20, 21, 22, 23, 24, 25, 26, 27, 28, 29],
 [30, 31, 32, 33, 34, 35, 36, 37, 38, 39],
 [40, 41, 42, 43, 44, 45, 46, 47, 48, 49],
 [50, 51, 52, 53, 54, 55, 56, 57, 58, 59],
 [60, 61, 62, 63, 64, 65, 66, 67, 68, 69],
 [70, 71, 72, 73, 74]]

---

If you're using Python 2, you should use xrange() instead of range():

def chunks(l, n):
    """Yield successive n-sized chunks from l."""
    for i in xrange(0, len(l), n):
        yield l[i:i + n]

---

Also you can simply use list comprehension instead of writing a function. Python 3:

[l[i:i + n] for i in range(0, len(l), n)]

Python 2 version:

[l[i:i + n] for i in xrange(0, len(l), n)]