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如何在恒定大小的块中拆分可迭代

更新时间:2023-01-13 17:14:49

这可能更有效(更快)

  def批处理(iterable,n = 1):
l = len(可迭代)
对于ndx范围(0,l,n):
产生iterable [ndx: min(ndx + n,l)]

for x in batch(range(0,10),3):
print x






使用列表示例

  data = [0,1,2,3,4,5,6,7,8,9,10]#数据列表

批量x(数据,3):
打印(x)

#输出

[0,1,2]
[3,4 ,5]
[6,7,8]
[9,10]

它避免建立新列表。


Possible Duplicate:
How do you split a list into evenly sized chunks in Python?

I am surprised I could not find a "batch" function that would take as input an iterable and return an iterable of iterables.

For example:

for i in batch(range(0,10), 1): print i
[0]
[1]
...
[9]

or:

for i in batch(range(0,10), 3): print i
[0,1,2]
[3,4,5]
[6,7,8]
[9]

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

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]

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 ?

[Edit: I eventually realized that the above behavior happens only when I run this within ipython rather than python itself]

This is probably more efficient (faster)

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


Example using list

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]

It avoids building new lists.