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]]
其他回答
这将通过一个表达式将其拆分为相等的部分，同时保持顺序:
myList = list(range(18)) # given list
N = 5 # desired number of parts
[myList[(i*len(myList))//N:((i+1)*len(myList))//N] for i in range(N)]
# [[0, 1, 2], [3, 4, 5, 6], [7, 8, 9], [10, 11, 12, 13], [14, 15, 16, 17]]
各部分的差异不超过一个元素。把18分成5份，结果是3 + 4 + 3 + 4 + 4 = 18。
看看numpy.split:
>>> a = numpy.array([1,2,3,4])
>>> numpy.split(a, 2)
[array([1, 2]), array([3, 4])]
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
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])]
你可以把它简单地写成一个列表生成器:
def split(a, n):
k, m = divmod(len(a), n)
return (a[i*k+min(i, m):(i+1)*k+min(i+1, m)] for i in range(n))
例子:
>>> list(split(range(11), 3))
[[0, 1, 2, 3], [4, 5, 6, 7], [8, 9, 10]]
