标签:lambda ict print zip int col 方法 lis item
方法一: list1 = [‘k1‘,‘k2‘,‘k3‘] list2 = [‘v1‘,‘v2‘,‘v3‘] dic = dict(map(lambda x,y:[x,y],list1,list2)) >>> print(dic) {‘k3‘: ‘v3‘, ‘k2‘: ‘v2‘, ‘k1‘: ‘v1‘} 方法二: >>> dict(zip(list1,list2)) {‘k3‘: ‘v3‘, ‘k2‘: ‘v2‘, ‘k1‘: ‘v1‘} >>> l1=[1,2,3,4,5,6] >>> l2=[4,5,6,7,8,9] >>> >>> >>> >>> {k:v for k,v in zip(l1,l2)} {1: 4, 2: 5, 3: 6, 4: 7, 5: 8, 6: 9} >>> >>> >>> >>> x = {1: 4, 2: 5, 3: 6, 4: 7, 5: 8, 6: 9} >>> {v:k for k,v in x.items()} #反过来 将字典中的v和k调换 {4: 1, 5: 2, 6: 3, 7: 4, 8: 5, 9: 6}
标签:lambda ict print zip int col 方法 lis item
原文地址:https://www.cnblogs.com/brf-test/p/12728115.html