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
def chunks(iterable,n):
"""assumes n is an integer>0
"""
iterable=iter(iterable)
while True:
result=[]
for i in range(n):
try:
a=next(iterable)
except StopIteration:
break
else:
result.append(a)
if result:
yield result
else:
break
g1=(i*i for i in range(10))
g2=chunks(g1,3)
print g2
'<generator object chunks at 0x0337B9B8>'
print list(g2)
'[[0, 1, 4], [9, 16, 25], [36, 49, 64], [81]]'
其他回答
不调用len(),这对大型列表很有用:
def splitter(l, n):
i = 0
chunk = l[:n]
while chunk:
yield chunk
i += n
chunk = l[i:i+n]
这是可迭代的:
def isplitter(l, n):
l = iter(l)
chunk = list(islice(l, n))
while chunk:
yield chunk
chunk = list(islice(l, n))
上述产品的功能风味:
def isplitter2(l, n):
return takewhile(bool,
(tuple(islice(start, n))
for start in repeat(iter(l))))
OR:
def chunks_gen_sentinel(n, seq):
continuous_slices = imap(islice, repeat(iter(seq)), repeat(0), repeat(n))
return iter(imap(tuple, continuous_slices).next,())
OR:
def chunks_gen_filter(n, seq):
continuous_slices = imap(islice, repeat(iter(seq)), repeat(0), repeat(n))
return takewhile(bool,imap(tuple, continuous_slices))
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
这是一个生成大小均匀的块的生成器:
def chunks(lst, n):
"""Yield successive n-sized chunks from lst."""
for i in range(0, len(lst), n):
yield lst[i:i + n]
import pprint
pprint.pprint(list(chunks(range(10, 75), 10)))
[[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]]
对于Python 2,使用xrange代替range:
def chunks(lst, n):
"""Yield successive n-sized chunks from lst."""
for i in xrange(0, len(lst), n):
yield lst[i:i + n]
下面是一行理解列表。不过,上面的方法更可取,因为使用命名函数使代码更容易理解。对于Python 3:
[lst[i:i + n] for i in range(0, len(lst), n)]
对于Python 2:
[lst[i:i + n] for i in xrange(0, len(lst), n)]
考虑使用matplotlib.cbook片段
例如:
import matplotlib.cbook as cbook
segments = cbook.pieces(np.arange(20), 3)
for s in segments:
print s
您可以使用更多的intertools.chunked_甚至与math.eil一起使用。这可能是最容易理解的吗?
from math import ceil
import more_itertools as mit
from pprint import pprint
pprint([*mit.chunked_even(range(19), ceil(19 / 5))])
# [[0, 1, 2, 3], [4, 5, 6, 7], [8, 9, 10, 11], [12, 13, 14, 15], [16, 17, 18]]
pprint([*mit.chunked_even(range(20), ceil(20 / 5))])
# [[0, 1, 2, 3], [4, 5, 6, 7], [8, 9, 10, 11], [12, 13, 14, 15], [16, 17, 18, 19]]
pprint([*mit.chunked_even(range(21), ceil(21 / 5))])
# [[0, 1, 2, 3, 4],
# [5, 6, 7, 8],
# [9, 10, 11, 12],
# [13, 14, 15, 16],
# [17, 18, 19, 20]]
pprint([*mit.chunked_even(range(3), ceil(3 / 5))])
# [[0], [1], [2]]