Python list pop（）比list [1：]慢得多

Question

I recently wrote a quick and dirty BFS implementation, to find diamonds in a directed graph. The BFS loop looked like this:

while toVisit:
    y = toVisit.pop()
    if y in visited: return "Found diamond"
    visited.add(y)
    toVisit.extend(G[y])

( G is the graph - a dictionary from node names to the lists of their neighbors)

Then comes the interesting part: I thought that list.pop() is probably too slow, so I ran a profiler to compare the speed of this implementation with deque.pop - and got a bit of an improvement. Then I compared it with y = toVisit[0]; toVisit = toVisit[1:] y = toVisit[0]; toVisit = toVisit[1:] , and to my surprise, the last implementation is actually the fastest one.

Does this make any sense? Is there any performance reason to ever use list.pop() instead of the apparently much faster two-liner?

Answer 1

You have measured wrong. With cPython 2.7 on x64, I get the following results:

$ python -m timeit 'l = list(range(10000))' 'while l: l = l[1:]'
10 loops, best of 3: 365 msec per loop
$ python -m timeit 'l = list(range(10000))' 'while l: l.pop()'
1000 loops, best of 3: 1.82 msec per loop
$ python -m timeit 'import collections' \
         'l = collections.deque(list(range(10000)))' 'while l: l.pop()'
1000 loops, best of 3: 1.67 msec per loop

Answer 2

使用生成器来提高性能

python -m timeit 'import itertools' 'l=iter(xrange(10000))' 'while next(l, None): l,a = itertools.tee(l)'

1000000 loops, best of 3: 0.986 usec per loop