[0, 2, 4, 6, 8]
-:func:`enumerate(iter) <enumerate>` counts off the elements in the iterable,
-returning 2-tuples containing the count and each element. ::
+:func:`enumerate(iter, start=0) <enumerate>` counts off the elements in the
+iterable returning 2-tuples containing the count (from *start*) and
+each element. ::
>>> for item in enumerate(['subject', 'verb', 'object']):
... print(item)
Creating new iterators
----------------------
-:func:`itertools.count(n) <itertools.count>` returns an infinite stream of
-integers, increasing by 1 each time. You can optionally supply the starting
-number, which defaults to 0::
+:func:`itertools.count(start, step) <itertools.count>` returns an infinite
+stream of evenly spaced values. You can optionally supply the starting number,
+which defaults to 0, and the interval between numbers, which defaults to 1::
itertools.count() =>
0, 1, 2, 3, 4, 5, 6, 7, 8, 9, ...
itertools.count(10) =>
10, 11, 12, 13, 14, 15, 16, 17, 18, 19, ...
+ itertools.count(10, 5) =>
+ 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, ...
:func:`itertools.cycle(iter) <itertools.cycle>` saves a copy of the contents of
a provided iterable and returns a new iterator that returns its elements from
for i in [1,2,3]:
product *= i
-A related function is `itertools.accumulate(iterable, func=operator.add) <itertools.accumulate`.
-It performs the same calculation, but instead of returning only the
-final result, :func:`accumulate` returns an iterator that also yields
-each partial result::
+A related function is :func:`itertools.accumulate(iterable, func=operator.add)
+<itertools.accumulate>`. It performs the same calculation, but instead of
+returning only the final result, :func:`accumulate` returns an iterator that
+also yields each partial result::
itertools.accumulate([1,2,3,4,5]) =>
1, 3, 6, 10, 15
Documentation for the :mod:`itertools` module.
+Documentation for the :mod:`functools` module.
+
Documentation for the :mod:`operator` module.
:pep:`289`: "Generator Expressions"