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
其他回答
另一种方法是这样的，这里的想法是用石斑鱼，但要去掉None。在本例中，所有的'small_parts'都由列表第一部分的元素组成，'larger_parts'则由列表的后一部分组成。' bigger parts'的长度为len(small_parts) + 1。我们需要把x看成两个不同的子部分。
from itertools import izip_longest
import numpy as np
def grouper(n, iterable, fillvalue=None): # This is grouper from itertools
"grouper(3, 'ABCDEFG', 'x') --> ABC DEF Gxx"
args = [iter(iterable)] * n
return izip_longest(fillvalue=fillvalue, *args)
def another_chunk(x,num):
extra_ele = len(x)%num #gives number of parts that will have an extra element
small_part = int(np.floor(len(x)/num)) #gives number of elements in a small part
new_x = list(grouper(small_part,x[:small_part*(num-extra_ele)]))
new_x.extend(list(grouper(small_part+1,x[small_part*(num-extra_ele):])))
return new_x
我设置它的方式返回一个元组列表:
>>> x = range(14)
>>> another_chunk(x,3)
[(0, 1, 2, 3), (4, 5, 6, 7, 8), (9, 10, 11, 12, 13)]
>>> another_chunk(x,4)
[(0, 1, 2), (3, 4, 5), (6, 7, 8, 9), (10, 11, 12, 13)]
>>> another_chunk(x,5)
[(0, 1), (2, 3, 4), (5, 6, 7), (8, 9, 10), (11, 12, 13)]
>>>
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]]
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)]
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
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]]
