如何在恒定大小的块中拆分可迭代对象 [英] how to split an iterable in constant-size chunks

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

<小时>

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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