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1 #比较时间差,判断执行有时。 2 3 import time 4 def panduan(x): 5 if x%5==0 and x%7==0: 6 return True 7 else: 8 return False 9 start_time=time.time() 10 print reduce((lambda x,y:x+y),filter(panduan,range(1001))) 11 end_time=time.time() 12 print end_time-start_time 13 14 sum=0 15 start_time1=time.time() 16 for i in range(1001): 17 if i%5==0 and i%7==0: 18 sum=sum+i 19 print sum 20 end_time1=time.time() 21 print end_time1-start_time1 22 23 #执行结果 24 #14210 25 #0.00799989700317 26 #14210 27 #0.00500011444092
两种方法执行1000以内即被5又被7整除的数的和,比较执行所需时间
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原文地址:http://www.cnblogs.com/tingfenglin/p/4434876.html