I was writing an answer to this question when noticed that my simple implementation didn't produce correct results. While hunting down the bug, I noticed the following:
In [1]: import itertools
In [2]: gen = itertools.cycle((0,1,2))
In [3]: zip(gen, range(3))
Out[3]: [(0, 0), (1, 1), (2, 2)]
In [4]: zip(gen, range(3))
Out[4]: [(1, 0), (2, 1), (0, 2)]
For whatever reason, gen
's next()
method is called one additioinal time.
To illustrate this, I used the following:
class loudCycle(itertools.cycle):
def next(self):
n = super(loudCycle, self).next()
print n
return n
In [6]: gen = loudCycle((0,1,2))
In [7]: zip(gen, range(3))
0
1
2
0
Out[7]: [(0, 0), (1, 1), (2, 2)]
The zip() function is not a generator function, it just returns an iterators.
Definition and Usage. The zip() function returns a zip object, which is an iterator of tuples where the first item in each passed iterator is paired together, and then the second item in each passed iterator are paired together etc.
Python zip() The zip() function takes iterables (can be zero or more), aggregates them in a tuple, and returns it.
It can take zero or more arguments and it returns us an iterator to the tuple.
This happens because zip
evaluates iterators from left to right, meaning that, after three steps, it calls next()
on gen
and only then on iter(range(3))
(or something like that) and encounters a StopIteration
. To get around this, use the shorter (finite) iterable as the left-most argument:
In [8]: zip(range(3), gen)
0
1
2
Out[8]: [(0, 0), (1, 1), (2, 2)]
Your self-answer is exactly right, and presents a very good solution -- if one of the arguments to zip
is always shorter than the other. However, in situations where you don't know which will be shorter, you might find islice
useful. islice
also provides an easy workaround if you want the first item in your tuples to be from your generator. In your case, you could do this:
>>> import itertools
>>> gen = itertools.cycle(('a', 'b', 'c'))
>>> seq = range(3)
>>> zip(itertools.islice(gen, len(seq)), seq)
[('a', 0), ('b', 1), ('c', 2)]
>>> zip(itertools.islice(gen, len(seq)), seq)
[('a', 0), ('b', 1), ('c', 2)]
Your answer is probably better in this case -- it's certainly simpler -- but I thought I'd add this as a supplement.
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