如何将任意长度的列表拆分为大小相等的块?
请参阅如果数据结果将直接用于循环,并且不需要存储,则如何以块形式遍历列表。
对于字符串输入的同一问题,请参见每n个字符拆分字符串?。相同的技术通常适用,但也有一些变化。
如何将任意长度的列表拆分为大小相等的块?
请参阅如果数据结果将直接用于循环,并且不需要存储,则如何以块形式遍历列表。
对于字符串输入的同一问题,请参见每n个字符拆分字符串?。相同的技术通常适用,但也有一些变化。
当前回答
我很好奇不同方法的性能,这里是:
在Python 3.5.1上测试
import time
batch_size = 7
arr_len = 298937
#---------slice-------------
print("\r\nslice")
start = time.time()
arr = [i for i in range(0, arr_len)]
while True:
if not arr:
break
tmp = arr[0:batch_size]
arr = arr[batch_size:-1]
print(time.time() - start)
#-----------index-----------
print("\r\nindex")
arr = [i for i in range(0, arr_len)]
start = time.time()
for i in range(0, round(len(arr) / batch_size + 1)):
tmp = arr[batch_size * i : batch_size * (i + 1)]
print(time.time() - start)
#----------batches 1------------
def batch(iterable, n=1):
l = len(iterable)
for ndx in range(0, l, n):
yield iterable[ndx:min(ndx + n, l)]
print("\r\nbatches 1")
arr = [i for i in range(0, arr_len)]
start = time.time()
for x in batch(arr, batch_size):
tmp = x
print(time.time() - start)
#----------batches 2------------
from itertools import islice, chain
def batch(iterable, size):
sourceiter = iter(iterable)
while True:
batchiter = islice(sourceiter, size)
yield chain([next(batchiter)], batchiter)
print("\r\nbatches 2")
arr = [i for i in range(0, arr_len)]
start = time.time()
for x in batch(arr, batch_size):
tmp = x
print(time.time() - start)
#---------chunks-------------
def chunks(l, n):
"""Yield successive n-sized chunks from l."""
for i in range(0, len(l), n):
yield l[i:i + n]
print("\r\nchunks")
arr = [i for i in range(0, arr_len)]
start = time.time()
for x in chunks(arr, batch_size):
tmp = x
print(time.time() - start)
#-----------grouper-----------
from itertools import zip_longest # for Python 3.x
#from six.moves import zip_longest # for both (uses the six compat library)
def grouper(iterable, n, padvalue=None):
"grouper(3, 'abcdefg', 'x') --> ('a','b','c'), ('d','e','f'), ('g','x','x')"
return zip_longest(*[iter(iterable)]*n, fillvalue=padvalue)
arr = [i for i in range(0, arr_len)]
print("\r\ngrouper")
start = time.time()
for x in grouper(arr, batch_size):
tmp = x
print(time.time() - start)
结果:
slice
31.18285083770752
index
0.02184295654296875
batches 1
0.03503894805908203
batches 2
0.22681021690368652
chunks
0.019841909408569336
grouper
0.006506919860839844
其他回答
没有魔力,但简单而正确:
def chunks(iterable, n):
"""Yield successive n-sized chunks from iterable."""
values = []
for i, item in enumerate(iterable, 1):
values.append(item)
if i % n == 0:
yield values
values = []
if values:
yield values
参见本参考
>>> orange = range(1, 1001)
>>> otuples = list( zip(*[iter(orange)]*10))
>>> print(otuples)
[(1, 2, 3, 4, 5, 6, 7, 8, 9, 10), ... (991, 992, 993, 994, 995, 996, 997, 998, 999, 1000)]
>>> olist = [list(i) for i in otuples]
>>> print(olist)
[[1, 2, 3, 4, 5, 6, 7, 8, 9, 10], ..., [991, 992, 993, 994, 995, 996, 997, 998, 999, 1000]]
>>>
蟒蛇3
下面是一个处理任意可迭代项的生成器:
def split_seq(iterable, size):
it = iter(iterable)
item = list(itertools.islice(it, size))
while item:
yield item
item = list(itertools.islice(it, size))
例子:
>>> import pprint
>>> pprint.pprint(list(split_seq(xrange(75), 10)))
[[0, 1, 2, 3, 4, 5, 6, 7, 8, 9],
[10, 11, 12, 13, 14, 15, 16, 17, 18, 19],
[20, 21, 22, 23, 24, 25, 26, 27, 28, 29],
[30, 31, 32, 33, 34, 35, 36, 37, 38, 39],
[40, 41, 42, 43, 44, 45, 46, 47, 48, 49],
[50, 51, 52, 53, 54, 55, 56, 57, 58, 59],
[60, 61, 62, 63, 64, 65, 66, 67, 68, 69],
[70, 71, 72, 73, 74]]
上面的答案(由koffein给出)有一个小问题:列表总是被分割成相等数量的分割,而不是每个分区的项目数相等。这是我的版本。“//chs+1”考虑到项目的数量可能不能完全除以分区大小,因此最后一个分区将仅被部分填充。
# Given 'l' is your list
chs = 12 # Your chunksize
partitioned = [ l[i*chs:(i*chs)+chs] for i in range((len(l) // chs)+1) ]
我知道这有点过时,但还没有人提到numpy.array_split:
import numpy as np
lst = range(50)
np.array_split(lst, 5)
结果:
[array([0, 1, 2, 3, 4, 5, 6, 7, 8, 9]),
array([10, 11, 12, 13, 14, 15, 16, 17, 18, 19]),
array([20, 21, 22, 23, 24, 25, 26, 27, 28, 29]),
array([30, 31, 32, 33, 34, 35, 36, 37, 38, 39]),
array([40, 41, 42, 43, 44, 45, 46, 47, 48, 49])]