[python]将列表分组成包含多个子列表的列表

想取得的效果是： [1, 2, 3, 4, 5, 6, 7] --> [[1, 2, 3], [4, 5, 6], [7]]

下面三种方法取自博文 http://sandrotosi.blogspot.com/2011/04/python-group-list-in-sub-lists-of-n.html 及评论。

第一种方法

def group_iter(iterator, n=2):
    """ Given an iterator, it returns sub-lists made of n items
    (except the last that can have len < n)
    inspired by http://countergram.com/python-group-iterator-list-function"""
    accumulator = []
    for item in iterator:
        accumulator.append(item)
        if len(accumulator) == n: # tested as fast as separate counter
            yield accumulator
            accumulator = [] # tested faster than accumulator[:] = []
            # and tested as fast as re-using one list object
    if len(accumulator) != 0:
        yield accumulator

效果：

>>> group_iter([1, 2, 3, 4, 5, 6], 3)
<generator object group_iter at 0x02A43418>
>>> list(group_iter([1, 2, 3, 4, 5, 6], 3))
[[1, 2, 3], [4, 5, 6]]
>>> list(group_iter([1, 2, 3, 4, 5, 6], 4))
[[1, 2, 3, 4], [5, 6]]

第二种方法：

>>> original_list = [1, 2, 3, 4, 5, 6]
>>> list_size = 5
>>> [original_list[i:i+list_size] for i in xrange(0, len(original_list), list_size)]
[[1, 2, 3, 4, 5], [6]]
>>>

第三种方法：

def splitarray(array, gsize):
    arraylen = len(array)
    for i in range(arraylen / gsize):
        yield array[i * gsize:(i * gsize) + gsize]
    if arraylen % gsize != 0:
        yield array[-(arraylen % gsize):]

效果：

>>> original_list = [1, 2, 3, 4, 5, 6]
>>> list_size = 5
>>> [original_list[i:i+list_size] for i in xrange(0, len(original_list), list_size)]
[[1, 2, 3, 4, 5], [6]]
>>>

参考

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