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

这是另一种变体,它将“剩余”元素均匀地分布在所有块中,一次一个,直到一个都不剩。在这个实现中,较大的块出现在流程的开头。

def chunks(l, k):
  """ Yield k successive chunks from l."""
  if k < 1:
    yield []
    raise StopIteration
  n = len(l)
  avg = n/k
  remainders = n % k
  start, end = 0, avg
  while start < n:
    if remainders > 0:
      end = end + 1
      remainders = remainders - 1
    yield l[start:end]
    start, end = end, end+avg

例如,从14个元素的列表中生成4个块:

>>> list(chunks(range(14), 4))
[[0, 1, 2, 3], [4, 5, 6, 7], [8, 9, 10], [11, 12, 13]]
>>> map(len, list(chunks(range(14), 4)))
[4, 4, 3, 3]

其他回答

看到more_itertools.divide:

n = 2

[list(x) for x in mit.divide(n, range(5, 11))]
# [[5, 6, 7], [8, 9, 10]]

[list(x) for x in mit.divide(n, range(5, 12))]
# [[5, 6, 7, 8], [9, 10, 11]]

通过> pip Install more_itertools安装。

舍入linspace并将其用作索引是一种比amit12690建议的更简单的解决方案。

function chunks=chunkit(array,num)

index = round(linspace(0,size(array,2),num+1));

chunks = cell(1,num);

for x = 1:num
chunks{x} = array(:,index(x)+1:index(x+1));
end
end

这就是numpy.array_split*使用d'être的原因:

>>> import numpy as np
>>> print(*np.array_split(range(10), 3))
[0 1 2 3] [4 5 6] [7 8 9]
>>> print(*np.array_split(range(10), 4))
[0 1 2] [3 4 5] [6 7] [8 9]
>>> print(*np.array_split(range(10), 5))
[0 1] [2 3] [4 5] [6 7] [8 9]

*贷方是6号房间的零比雷埃夫斯

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]]

其他的解决方案似乎有点长。下面是一个使用列表理解和NumPy函数array_split的一行程序。Array_split (list, n)将简单地将列表分成n部分。

[x.tolist() for x in np.array_split(range(10), 3)]