使用numpy实现。linspace方法。

只需指定要将数组分成的部分的数量。各部门的规模将几乎相同。

例子:

import numpy as np   
a=np.arange(10)
print "Input array:",a 
parts=3
i=np.linspace(np.min(a),np.max(a)+1,parts+1)
i=np.array(i,dtype='uint16') # Indices should be floats
split_arr=[]
for ind in range(i.size-1):
    split_arr.append(a[i[ind]:i[ind+1]]
print "Array split in to %d parts : "%(parts),split_arr

给:

Input array: [0 1 2 3 4 5 6 7 8 9]
Array split in to 3 parts :  [array([0, 1, 2]), array([3, 4, 5]), array([6, 7, 8, 9])]

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]]

#!/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。

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]]

只要你不想要像连续块这样愚蠢的东西:

>>> def chunkify(lst,n):
...     return [lst[i::n] for i in xrange(n)]
... 
>>> chunkify(range(13), 3)
[[0, 3, 6, 9, 12], [1, 4, 7, 10], [2, 5, 8, 11]]

假设你想把一个列表[1、2、3、4、5、6、7、8]分成3个元素列表

如[[1,2,3]，[4,5,6]，[7,8]]，如果剩下的最后一个元素小于3，则将它们分组在一起。

my_list = [1, 2, 3, 4, 5, 6, 7, 8]
my_list2 = [my_list[i:i+3] for i in range(0, len(my_list), 3)]
print(my_list2)

输出:[[1,2,3]，[4,5,6]，[7,8]]

其中一部分的长度为3。用你自己的块大小替换3。