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
还有一个解决方案
def make_chunks(data, chunk_size):
while data:
chunk, data = data[:chunk_size], data[chunk_size:]
yield chunk
>>> for chunk in make_chunks([1, 2, 3, 4, 5, 6, 7], 2):
... print chunk
...
[1, 2]
[3, 4]
[5, 6]
[7]
>>>
与任何可迭代的内部数据是生成器对象(不是列表)一个衬垫
In [259]: get_in_chunks = lambda itr,n: ( (v for _,v in g) for _,g in itertools.groupby(enumerate(itr),lambda (ind,_): ind/n)) In [260]: list(list(x) for x in get_in_chunks(range(30),7)) Out[260]: [[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]]
非常简单的事情:
def chunks(xs, n):
n = max(1, n)
return (xs[i:i+n] for i in range(0, len(xs), n))
对于Python 2,使用xrange()代替range()。
此时,我认为我们需要强制性的匿名递归函数。
Y = lambda f: (lambda x: x(x))(lambda y: f(lambda *args: y(y)(*args)))
chunks = Y(lambda f: lambda n: [n[0][:n[1]]] + f((n[0][n[1]:], n[1])) if len(n[0]) > 0 else [])
我非常喜欢tzot和J.F.Sebastian提出的Python文档版本,但它有两个缺点:
它不是很明确我通常不希望在最后一个块中有填充值
我在代码中经常使用这个:
from itertools import islice
def chunks(n, iterable):
iterable = iter(iterable)
while True:
yield tuple(islice(iterable, n)) or iterable.next()
更新:一个懒块版本:
from itertools import chain, islice
def chunks(n, iterable):
iterable = iter(iterable)
while True:
yield chain([next(iterable)], islice(iterable, n-1))