Question

How can I yield multiple items at a time from an iterable object?

For example, with a sequence of arbitrary length, how can I iterate through the items in the sequence, in groups of X consecutive items per iteration?

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Solution

Your question is a bit vague, but check out the grouper recipe in the itertools documentation.

def grouper(n, iterable, fillvalue=None):
    "grouper(3, 'ABCDEFG', 'x') --> ABC DEF Gxx"
    args = [iter(iterable)] * n
    return izip_longest(fillvalue=fillvalue, *args)

(Zipping the same iterator several times with [iter(iterable)]*n is an old trick, but encapsulating it in this function avoids confusing code, and it is the same exact form and interface many people will use. It's a somewhat common need and it's a bit of a shame it isn't actually in the itertools module.)

OTHER TIPS

Here's another approach that works on older version of Python that don't have izip_longest:

def grouper(n, seq):
  result = []
  for x in seq:
    result.append(x)
    if len(result) >= n:
      yield tuple(result)
      del result[:]
  if result:
    yield tuple(result)

No filler, so the last group might have fewer than n elements.

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