另一种方法是:
[fr.index(x) for x in sorted(fr, reverse=True)[:3]]
当我们比较它们的速度时...
import heapq
fr = [8, 4, 1, 1, 12]
def method_one():
for i in xrange(10000):
res = [fr.index(x) for x in sorted(fr, reverse=True)[:3]]
def method_two():
for i in xrange(10000):
heapq.nlargest(3, xrange(len(fr)), key=fr.__getitem__)
if __name__ == '__main__':
import timeit
print timeit.repeat(stmt='method_one()',
setup='from __main__ import method_one',
number=100)
print timeit.repeat(stmt='method_two()',
setup='from __main__ import method_two',
number=100)
我们获得:
[1.1253619194030762, 1.1268768310546875, 1.128382921218872]
[2.5129621028900146, 2.529547929763794, 2.492828130722046]
0
在哪里? - thefourtheye0
。 - KBN