测试效率 timeit cProfile
timeit使用
def f1(lIn): l1 = sorted(lIn) # O(nlogn) C语言的 l2 = [i for i in l1 if i<0.5] # O(n) return [i*i for i in l2] # O(k) def f2(lIn): l1 = [i for i in lIn if i<0.5] # O(n) l2 = sorted(l1) # O(klogk) C语言的 return [i*i for i in l2] # O(k) def f3(lIn): l1 = [i*i for i in lIn] # O(n) l2 = sorted(l1) # O(nlogn) C语言的 return [i for i in l2 if i<(0.5*0.5)] # O(n) import timeit import random l = [random.random() for i in range(100000)] t1 = timeit.Timer("f1(l)","from __main__ import random,l,f1") t2 = timeit.Timer("f2(l)","from __main__ import random,l,f2") t3 = timeit.Timer("f3(l)","from __main__ import random,l,f3") print(t1.timeit(number=50)) # 3.0630057093816947 print(t2.timeit(number=50)) # 1.6970076176407773 print(t3.timeit(number=50)) # 3.30195772223185
或使用Ipython
def f1(lIn): l1 = sorted(lIn) # O(nlogn) C语言的 l2 = [i for i in l1 if i<0.5] # O(n) return [i*i for i in l2] # O(k) def f2(lIn): l1 = [i for i in lIn if i<0.5] # O(n) l2 = sorted(l1) # O(klogk) C语言的 return [i*i for i in l2] # O(k) def f3(lIn): l1 = [i*i for i in lIn] # O(n) l2 = sorted(l1) # O(nlogn) C语言的 return [i for i in l2 if i<(0.5*0.5)] # O(n) import random lIn = [random.random() for i in range(100000)] %timeit f1(lIn) 75.8 ms ± 917 µs per loop (mean ± std. dev. of 7 runs, 10 loops each) %timeit f2(lIn) 39.3 ms ± 560 µs per loop (mean ± std. dev. of 7 runs, 10 loops each) %timeit f3(lIn) 80 ms ± 1.09 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
使用cProfile
def f1(lIn): l1 = sorted(lIn) # O(nlogn) C语言的 l2 = [i for i in l1 if i<0.5] # O(n) return [i*i for i in l2] # O(k) def f2(lIn): l1 = [i for i in lIn if i<0.5] # O(n) l2 = sorted(l1) # O(klogk) C语言的 return [i*i for i in l2] # O(k) def f3(lIn): l1 = [i*i for i in lIn] # O(n) l2 = sorted(l1) # O(nlogn) C语言的 return [i for i in l2 if i<(0.5*0.5)] # O(n) import cProfile import random lIn = [random.random() for i in range(100000)] cProfile.run('f1(lIn)') cProfile.run('f2(lIn)') cProfile.run('f3(lIn)') """ 7 function calls in 0.125 seconds Ordered by: standard name ncalls tottime percall cumtime percall filename:lineno(function) 1 0.000 0.000 0.120 0.120 11.py:1(f1) 1 0.020 0.020 0.020 0.020 11.py:3(<listcomp>) 1 0.009 0.009 0.009 0.009 11.py:4(<listcomp>) 1 0.005 0.005 0.125 0.125 <string>:1(<module>) 1 0.000 0.000 0.125 0.125 {built-in method builtins.exec} 1 0.091 0.091 0.091 0.091 {built-in method builtins.sorted} 1 0.000 0.000 0.000 0.000 {method 'disable' of '_lsprof.Profiler' objects} 7 function calls in 0.044 seconds Ordered by: standard name ncalls tottime percall cumtime percall filename:lineno(function) 1 0.000 0.000 0.041 0.041 11.py:6(f2) 1 0.007 0.007 0.007 0.007 11.py:7(<listcomp>) 1 0.006 0.006 0.006 0.006 11.py:9(<listcomp>) 1 0.003 0.003 0.044 0.044 <string>:1(<module>) 1 0.000 0.000 0.044 0.044 {built-in method builtins.exec} 1 0.028 0.028 0.028 0.028 {built-in method builtins.sorted} 1 0.000 0.000 0.000 0.000 {method 'disable' of '_lsprof.Profiler' objects} 7 function calls in 0.068 seconds Ordered by: standard name ncalls tottime percall cumtime percall filename:lineno(function) 1 0.000 0.000 0.065 0.065 11.py:11(f3) 1 0.006 0.006 0.006 0.006 11.py:12(<listcomp>) 1 0.010 0.010 0.010 0.010 11.py:14(<listcomp>) 1 0.002 0.002 0.068 0.068 <string>:1(<module>) 1 0.000 0.000 0.068 0.068 {built-in method builtins.exec} 1 0.049 0.049 0.049 0.049 {built-in method builtins.sorted} 1 0.000 0.000 0.000 0.000 {method 'disable' of '_lsprof.Profiler' objects} """
