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
看看numpy.split:
>>> a = numpy.array([1,2,3,4])
>>> numpy.split(a, 2)
[array([1, 2]), array([3, 4])]
其他回答
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)]
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
1>
import numpy as np
data # your array
total_length = len(data)
separate = 10
sub_array_size = total_length // separate
safe_separate = sub_array_size * separate
splited_lists = np.split(np.array(data[:safe_separate]), separate)
splited_lists[separate - 1] = np.concatenate(splited_lists[separate - 1],
np.array(data[safe_separate:total_length]))
splited_lists # your output
2>
splited_lists = np.array_split(np.array(data), separate)
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]]
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
