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问题描述
可能的重复:
如何将列表平均拆分Python 中的大小块?
我很惊讶我找不到一个批处理"函数,它将一个可迭代对象作为输入并返回一个可迭代对象.
I am surprised I could not find a "batch" function that would take as input an iterable and return an iterable of iterables.
例如:
for i in batch(range(0,10), 1): print i
[0]
[1]
...
[9]
或:
for i in batch(range(0,10), 3): print i
[0,1,2]
[3,4,5]
[6,7,8]
[9]
现在,我写了一个我认为非常简单的生成器:
Now, I wrote what I thought was a pretty simple generator:
def batch(iterable, n = 1):
current_batch = []
for item in iterable:
current_batch.append(item)
if len(current_batch) == n:
yield current_batch
current_batch = []
if current_batch:
yield current_batch
但是上面的内容并没有给我我所期望的:
But the above does not give me what I would have expected:
for x in batch(range(0,10),3): print x
[0]
[0, 1]
[0, 1, 2]
[3]
[3, 4]
[3, 4, 5]
[6]
[6, 7]
[6, 7, 8]
[9]
所以,我错过了一些东西,这可能表明我完全缺乏对 python 生成器的理解.有人愿意为我指出正确的方向吗?
So, I have missed something and this probably shows my complete lack of understanding of python generators. Anyone would care to point me in the right direction ?
推荐答案
这可能更高效(更快)
def batch(iterable, n=1):
l = len(iterable)
for ndx in range(0, l, n):
yield iterable[ndx:min(ndx + n, l)]
for x in batch(range(0, 10), 3):
print x
使用列表的示例
data = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10] # list of data
for x in batch(data, 3):
print(x)
# Output
[0, 1, 2]
[3, 4, 5]
[6, 7, 8]
[9, 10]
它避免了构建新列表.
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