del
不仅更易于理解,而且似乎比pop()稍微快一些:
$ python -m timeit -s "d = {'f':1,'foo':2,'bar':3}" "for k in d.keys():" " if k.startswith('f'):" " del d[k]"
1000000 loops, best of 3: 0.733 usec per loop
$ python -m timeit -s "d = {'f':1,'foo':2,'bar':3}" "for k in d.keys():" " if k.startswith('f'):" " d.pop(k)"
1000000 loops, best of 3: 0.742 usec per loop
编辑:感谢Alex Martelli提供了如何进行此基准测试的指导。希望我没有犯任何错误。
首先测量复制所需的时间:
$ python -m timeit -s "d = {'f':1,'foo':2,'bar':3}" "d1 = d.copy()"
1000000 loops, best of 3: 0.278 usec per loop
复制字典的基准测试:
$ python -m timeit -s "d = {'f':1,'foo':2,'bar':3}" "d1 = d.copy()" "for k in d1.keys():" " if k.startswith('f'):" " del d1[k]"
100000 loops, best of 3: 1.95 usec per loop
$ python -m timeit -s "d = {'f':1,'foo':2,'bar':3}" "d1 = d.copy()" "for k in d1.keys():" " if k.startswith('f'):" " d1.pop(k)"
100000 loops, best of 3: 2.15 usec per loop
减去复制成本后,我们得到pop()
的时间为1.872微秒,del
的时间为1.672微秒。