字典2

Posted 2020-10-22 l-pxiaotiancai

 1 v=dict.fromkeys(["k1",12356,"494949"],123)
 2 {‘494949‘: 123, 12356: 123, ‘k1‘: 123}
 3 
 4 v=dict.fromkeys(["k1",12356,"494949"],123)
 5 v1=v.get("k0",1223)（另一种获得字典中值得方法）
 6 dict
 7 
 8 v=dict.fromkeys(["k1",12356,"494949"],123)
 9 v1=v.setdefault("k0",100) 设置一个值，如果存在则取出已经存在的值，如果不存再添加进去
10 dict
11 
12 v=dict.fromkeys(["k1",12,"494949"],123)
13 v.update({"k1":15,"k3":45})
14 v.update(k1=15,k3=45)此种写法也是可以的
15 {‘k1‘: 15, 12: 123, ‘k3‘: 45, ‘494949‘: 123}