两个list 求交集效率对比

__author__ = 'daitr'
#--coding:utf-8--
import datetime
#方法一:
#a=[2,3,4,5]
#b=[2,5,8]
#tmp = [val for val in a if val in b]
#print tmp
##[2, 5]

#方法二
#print list(set(a).intersection(set(b)))
#print list(set(a).union(set(b)))
#print list(set(b).difference(set(a))) # b中有而a中没有的

a,b= [],[]
for i in range(1,60000):
    a.append(i)

for j in range(40000,100000):
    b.append(j)
print a[-1]
print b[-1]

time1 = datetime.datetime.now()
#print time1

tmp = [val for val in a if val in b]
#print tmp

time2 = datetime.datetime.now()
#print time2
d1 = time2-time1
print d1

list(set(a).intersection(set(b)))
#print temp2

time3 = datetime.datetime.now()
#print time3
d2 = time3-time2
print d2

 

posted on 2013-07-24 12:10  呆头龙  阅读(1096)  评论(0编辑  收藏  举报

点击这里给我发消息