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

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

其他回答

如果你不介意顺序会改变,我建议你使用@job solution,否则,你可以使用这个:

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

    while last < len(seq):
        if len(seq) - (last + steps) < steps:
            until = len(seq)
            steps = len(seq) - last
        else:
            until = int(last + steps)
        out.append(seq[int(last): until])
        last += steps
return out

在这种情况下,我自己编写了代码:

def chunk_ports(port_start, port_end, portions):
    if port_end < port_start:
        return None

    total = port_end - port_start + 1

    fractions = int(math.floor(float(total) / portions))

    results = []

    # No enough to chuck.
    if fractions < 1:
        return None

    # Reverse, so any additional items would be in the first range.
    _e = port_end
    for i in range(portions, 0, -1):
        print "i", i

        if i == 1:
            _s = port_start
        else:
            _s = _e - fractions + 1

        results.append((_s, _e))

        _e = _s - 1

    results.reverse()

    return results

Divide_ports(1,10,9)将返回

[(1, 2), (3, 3), (4, 4), (5, 5), (6, 6), (7, 7), (8, 8), (9, 9), (10, 10)]

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)]

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])]

下面是一个添加None以使列表长度相等的列表

>>> from itertools import izip_longest
>>> def chunks(l, n):
    """ Yield n successive chunks from l. Pads extra spaces with None
    """
    return list(zip(*izip_longest(*[iter(l)]*n)))

>>> l=range(54)

>>> chunks(l,3)
[(0, 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, 51), (1, 4, 7, 10, 13, 16, 19, 22, 25, 28, 31, 34, 37, 40, 43, 46, 49, 52), (2, 5, 8, 11, 14, 17, 20, 23, 26, 29, 32, 35, 38, 41, 44, 47, 50, 53)]

>>> chunks(l,4)
[(0, 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52), (1, 5, 9, 13, 17, 21, 25, 29, 33, 37, 41, 45, 49, 53), (2, 6, 10, 14, 18, 22, 26, 30, 34, 38, 42, 46, 50, None), (3, 7, 11, 15, 19, 23, 27, 31, 35, 39, 43, 47, 51, None)]

>>> chunks(l,5)
[(0, 5, 10, 15, 20, 25, 30, 35, 40, 45, 50), (1, 6, 11, 16, 21, 26, 31, 36, 41, 46, 51), (2, 7, 12, 17, 22, 27, 32, 37, 42, 47, 52), (3, 8, 13, 18, 23, 28, 33, 38, 43, 48, 53), (4, 9, 14, 19, 24, 29, 34, 39, 44, 49, None)]