From 591176e544dfeec4586e239d5d07c6420a650f2c Mon Sep 17 00:00:00 2001 From: Ezio Melotti Date: Mon, 4 Aug 2014 17:01:16 +0300 Subject: [PATCH] #18588: update the timeit examples to be consistent. --- Doc/library/timeit.rst | 19 ++++++++++--------- 1 file changed, 10 insertions(+), 9 deletions(-) diff --git a/Doc/library/timeit.rst b/Doc/library/timeit.rst index 824a8a31605..19b5e4e7db9 100644 --- a/Doc/library/timeit.rst +++ b/Doc/library/timeit.rst @@ -28,22 +28,23 @@ can be used to compare three different expressions: .. code-block:: sh - $ python -m timeit '"-".join(str(n) for n in range(100))' - 10000 loops, best of 3: 40.3 usec per loop - $ python -m timeit '"-".join([str(n) for n in range(100)])' - 10000 loops, best of 3: 33.4 usec per loop - $ python -m timeit '"-".join(map(str, range(100)))' - 10000 loops, best of 3: 25.2 usec per loop + $ python3 -m timeit '"-".join(str(n) for n in range(100))' + 10000 loops, best of 3: 30.2 usec per loop + $ python3 -m timeit '"-".join([str(n) for n in range(100)])' + 10000 loops, best of 3: 27.5 usec per loop + $ python3 -m timeit '"-".join(map(str, range(100)))' + 10000 loops, best of 3: 23.2 usec per loop This can be achieved from the :ref:`python-interface` with:: >>> import timeit >>> timeit.timeit('"-".join(str(n) for n in range(100))', number=10000) - 0.8187260627746582 + 0.3018611848820001 >>> timeit.timeit('"-".join([str(n) for n in range(100)])', number=10000) - 0.7288308143615723 + 0.2727368790656328 >>> timeit.timeit('"-".join(map(str, range(100)))', number=10000) - 0.5858950614929199 + 0.23702679807320237 + Note however that :mod:`timeit` will automatically determine the number of repetitions only when the command-line interface is used. In the