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
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]]
其他回答
看看numpy.split:
>>> a = numpy.array([1,2,3,4])
>>> numpy.split(a, 2)
[array([1, 2]), array([3, 4])]
def chunk_array(array : List, n: int) -> List[List]:
chunk_size = len(array) // n
chunks = []
i = 0
while i < len(array):
# if less than chunk_size left add the remainder to last element
if len(array) - (i + chunk_size + 1) < 0:
chunks[-1].append(*array[i:i + chunk_size])
break
else:
chunks.append(array[i:i + chunk_size])
i += chunk_size
return chunks
这是我的版本(灵感来自Max)
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]]
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]]
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)]
