将清单大致等份的最佳方法是什么?例如,如果列表有7个元素,并将其分为2部分,我们希望其中一部分有3个元素,而另一部分应该有4个元素。

我正在寻找类似even_split(L, n)的东西,它将L分解为n部分。

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

上面的代码给出了3个块,而不是3个块。我可以简单地转置(遍历这个,取每列的第一个元素,称之为第一部分,然后取第二个元素,把它放在第二部分,等等),但这破坏了项目的顺序。


当前回答

这里有一个单独的函数,它处理了大多数不同的分裂情况:

def splitList(lst, into):
    '''Split a list into parts.

    :Parameters:
        into (str) = Split the list into parts defined by the following:
            '<n>parts' - Split the list into n parts.
                ex. 2 returns:  [[1, 2, 3, 5], [7, 8, 9]] from [1,2,3,5,7,8,9]
            '<n>parts+' - Split the list into n equal parts with any trailing remainder.
                ex. 2 returns:  [[1, 2, 3], [5, 7, 8], [9]] from [1,2,3,5,7,8,9]
            '<n>chunks' - Split into sublists of n size.
                ex. 2 returns: [[1,2], [3,5], [7,8], [9]] from [1,2,3,5,7,8,9]
            'contiguous' - The list will be split by contiguous numerical values.
                ex. 'contiguous' returns: [[1,2,3], [5], [7,8,9]] from [1,2,3,5,7,8,9]
            'range' - The values of 'contiguous' will be limited to the high and low end of each range.
                ex. 'range' returns: [[1,3], [5], [7,9]] from [1,2,3,5,7,8,9]
    :Return:
        (list)
    '''
    from string import digits, ascii_letters, punctuation
    mode = into.lower().lstrip(digits)
    digit = into.strip(ascii_letters+punctuation)
    n = int(digit) if digit else None

    if n:
        if mode=='parts':
            n = len(lst)*-1 // n*-1 #ceil
        elif mode=='parts+':
            n = len(lst) // n
        return [lst[i:i+n] for i in range(0, len(lst), n)]

    elif mode=='contiguous' or mode=='range':
        from itertools import groupby
        from operator import itemgetter

        try:
            contiguous = [list(map(itemgetter(1), g)) for k, g in groupby(enumerate(lst), lambda x: int(x[0])-int(x[1]))]
        except ValueError as error:
            print ('{} in splitList\n   # Error: {} #\n {}'.format(__file__, error, lst))
            return lst
        if mode=='range':
            return [[i[0], i[-1]] if len(i)>1 else (i) for i in contiguous]
        return contiguous

r = splitList([1, '2', 3, 5, '7', 8, 9], into='2parts')
print (r) #returns: [[1, '2', 3, 5], ['7', 8, 9]]

其他回答

由于舍入错误,此代码被破坏。不要使用它!!

assert len(chunkIt([1,2,3], 10)) == 10  # fails

这里有一个可行的方法:

def chunkIt(seq, num):
    avg = len(seq) / float(num)
    out = []
    last = 0.0

    while last < len(seq):
        out.append(seq[int(last):int(last + avg)])
        last += avg

    return out

测试:

>>> chunkIt(range(10), 3)
[[0, 1, 2], [3, 4, 5], [6, 7, 8, 9]]
>>> chunkIt(range(11), 3)
[[0, 1, 2], [3, 4, 5, 6], [7, 8, 9, 10]]
>>> chunkIt(range(12), 3)
[[0, 1, 2, 3], [4, 5, 6, 7], [8, 9, 10, 11]]

与job的答案相同,但考虑了大小小于块数量的列表。

def chunkify(lst,n):
    [ lst[i::n] for i in xrange(n if n < len(lst) else len(lst)) ]

如果n (chunk的数量)是7,LST(要分割的列表)是[1,2,3],则chunk为[[0],[1],[2]],而不是[[0],[1],[2],[],[],[],[],[]]

我尝试了大部分的解决方案,但它们并不适用于我的情况,所以我创建了一个新函数,适用于大多数情况和任何类型的数组:

import math

def chunkIt(seq, num):
    seqLen = len(seq)
    total_chunks = math.ceil(seqLen / num)
    items_per_chunk = num
    out = []
    last = 0

    while last < seqLen:
        out.append(seq[last:(last + items_per_chunk)])
        last += items_per_chunk

    return out
#!/usr/bin/python


first_names = ['Steve', 'Jane', 'Sara', 'Mary','Jack','Bob', 'Bily', 'Boni', 'Chris','Sori', 'Will', 'Won','Li']

def chunks(l, n):
for i in range(0, len(l), n):
    # Create an index range for l of n items:
    yield l[i:i+n]

result = list(chunks(first_names, 5))
print result

从这个链接中选择,这对我很有帮助。我有一个预先定义好的列表。

它提供长度<= n, >= 0的块

def

 chunkify(lst, n):
    num_chunks = int(math.ceil(len(lst) / float(n))) if n < len(lst) else 1
    return [lst[n*i:n*(i+1)] for i in range(num_chunks)]

例如

>>> chunkify(range(11), 3)
[[0, 1, 2], [3, 4, 5], [6, 7, 8], [9, 10]]
>>> chunkify(range(11), 8)
[[0, 1, 2, 3, 4, 5, 6, 7], [8, 9, 10]]