Iterate over a python sequence in multiples of n?

Question

How do I process the elements of a sequence in batches, idiomatically?

For example, with the sequence \"abcdef\" and a batch size of 2, I would like to do something

Answer 1

From the docs of more_itertools: more_itertools.chunked()

more_itertools.chunked(iterable, n)

Break an iterable into lists of a given length:

>>> list(chunked([1, 2, 3, 4, 5, 6, 7], 3))
[[1, 2, 3], [4, 5, 6], [7]]

If the length of iterable is not evenly divisible by n, the last returned list will be shorter.

Answer 2

Adapted from this answer for Python 3:

def groupsgen(seq, size):
    it = iter(seq)
    iterating = True
    while iterating:
        values = ()
        try:
            for n in range(size):
                values += (next(it),)
        except StopIteration:
            iterating = False
            if not len(values):
                return None
        yield values

It will safely terminate and won't discard values if their number is not divisible by size.

Answer 3

One solution, although I challenge someone to do better ;-)

a = 'abcdef'
b = [[a[i-1], a[i]] for i in range(1, len(a), 2)]

for x, y in b:
  print "%s%s\n" % (x, y)

Answer 4

Don't forget about the zip() function:

a = 'abcdef'
for x,y in zip(a[::2], a[1::2]):
  print '%s%s' % (x,y)

Answer 5

How about itertools?

from itertools import islice, groupby

def chunks_islice(seq, size):
    while True:
        aux = list(islice(seq, 0, size))
        if not aux: break
        yield "".join(aux)

def chunks_groupby(seq, size):
    for k, chunk in groupby(enumerate(seq), lambda x: x[0] / size):
        yield "".join([i[1] for i in chunk])