split a generator/iterable every n items in python (splitEvery)

Question

I'm trying to write the Haskell function 'splitEvery' in Python. Here is it's definition:

splitEvery :: Int -> [e] -> [[e]]     @'splitEvery' n@ splits a list into length-n pieces.  The last     piece will be shorter if @n@ does not evenly divide the length of     the list. 

The basic version of this works fine, but I want a version that works with generator expressions, lists, and iterators. And, if there is a generator as an input it should return a generator as an output!

Tests

# should not enter infinite loop with generators or lists splitEvery(itertools.count(), 10) splitEvery(range(1000), 10)  # last piece must be shorter if n does not evenly divide assert splitEvery(5, range(9)) == [[0, 1, 2, 3, 4], [5, 6, 7, 8]]  # should give same correct results with generators tmp = itertools.islice(itertools.count(), 10) assert list(splitEvery(5, tmp)) == [[0, 1, 2, 3, 4], [5, 6, 7, 8]] 

Current Implementation

Here is the code I currently have but it doesn't work with a simple list.

def splitEvery_1(n, iterable):     res = list(itertools.islice(iterable, n))     while len(res) != 0:         yield res         res = list(itertools.islice(iterable, n)) 

This one doesn't work with a generator expression (thanks to jellybean for fixing it):

def splitEvery_2(n, iterable):      return [iterable[i:i+n] for i in range(0, len(iterable), n)] 

There has to be a simple piece of code that does the splitting. I know I could just have different functions but it seems like it should be and easy thing to do. I'm probably getting stuck on an unimportant problem but it's really bugging me.

---

It is similar to grouper from http://docs.python.org/library/itertools.html#itertools.groupby but I don't want it to fill extra values.

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

It does mention a method that truncates the last value. This isn't what I want either.

The left-to-right evaluation order of the iterables is guaranteed. This makes possible an idiom for clustering a data series into n-length groups using izip(*[iter(s)]*n).

list(izip(*[iter(range(9))]*5)) == [[0, 1, 2, 3, 4]] # should be [[0, 1, 2, 3, 4], [5, 6, 7, 8]] 

Answer 1

from itertools import islice  def split_every(n, iterable):     i = iter(iterable)     piece = list(islice(i, n))     while piece:         yield piece         piece = list(islice(i, n)) 

Some tests:

>>> list(split_every(5, range(9))) [[0, 1, 2, 3, 4], [5, 6, 7, 8]]  >>> list(split_every(3, (x**2 for x in range(20)))) [[0, 1, 4], [9, 16, 25], [36, 49, 64], [81, 100, 121], [144, 169, 196], [225, 256, 289], [324, 361]]  >>> [''.join(s) for s in split_every(6, 'Hello world')] ['Hello ', 'world']  >>> list(split_every(100, [])) []