Peter Otten wrote: > Wade Leftwich wrote: > > > from itertools import groupby > > > > def chunk(it, n=0): > > if n == 0: > > return iter([it]) > > def groupfun((x,y)): > > return int(x/n) > > grouped = groupby(enumerate(it), groupfun) > > counted = (y for (x,y) in grouped) > > return ((z for (y,z) in x) for x in counted) > > > >>>> [list(x) for x in chunk(range(10), 3)] > > [[0, 1, 2], [3, 4, 5], [6, 7, 8], [9]] > > > >>>> [x for x in chunk(range(10), 3)] > > [<generator object at 0xb7a34e4c>, > > <generator object at 0xb7a34dac>, > > <generator object at 0xb7a34d2c>, > > <generator object at 0xb7a34d6c>] > > Note that all but the last of these generators are useless: > > >>> chunks = [x for x in chunk(range(10), 3)] > >>> [list(x) for x in chunks] > [[], [], [], [9]] # did you expect that? > In [48]: chunkgen = chunk(range(10), 3)
In [49]: for x in chunkgen: ....: print list(x) ....: ....: [0, 1, 2] [3, 4, 5] [6, 7, 8] [9] > Peter That's an interesting gotcha that I've never run into when using this function, because in practice I never put the generator returned by chunk() inside a list comprehension. In [51]: chunkgen = chunk(range(10), 3) In [52]: [list(x) for x in chunkgen] Out[52]: [[0, 1, 2], [3, 4, 5], [6, 7, 8], [9]] But, as you pointed out -- In [57]: chunkgen = chunk(range(10), 3) In [58]: chunks = list(chunkgen) In [59]: [list(x) for x in chunks] Out[59]: [[], [], [], [9]] So apparently when we list(chunkgen), we are exhausting the generators that are its members. And if we do it again, we get rid of the [9] -- In [60]: [list(x) for x in chunks] Out[60]: [[], [], [], []] I'll admit that chunk() is needlessly tricky for most purposes. I wrote it to accept a really lengthy, possibly unbounded, iterator and write out 10,000-line files from it, and also to play with Python's new functional capabilities. -- Wade -- http://mail.python.org/mailman/listinfo/python-list
