zoukankan      html  css  js  c++  java
  • dict 方法总结整理

    #!/usr/bin/env python
    __author__ = "lrtao2010"
    
    #Python 3.7.0 字典常用方法
    
    #字典的key是唯一的,且不能被修改,value可以是任意值
    #字典是无序的
    
    # info = {'name':'tom','age':'10'}
    #
    # info1 = {
    #     'name':'tom',
    #     'age':'10'
    # }
    
    # info = {'name':'tom','age':'10'}
    # v1 = info['name']
    # v2 = info['age']
    #
    # print(v1,v2)
    # tom 10
    
    # info = {
    #     'name':[
    #         11,
    #         22,
    #         33,
    #         {
    #             'here':'ok'
    #         }
    #
    #     ]
    # }
    #
    # v = info['name'][3]['here']
    # print(v)
    # ok
    
    #clear(self) 清空字典
    # info = {"name":"tom"}
    # info.clear()
    # print(info)
    
    #copy(self) 浅复制
    # info1 = {"name":['tom']}
    # info2 = info1.copy()
    # print(info2)
    # info1["name"][0] = 'lilei'
    # print(info1)
    # print(info2)
    # {'name': ['tom']}
    # {'name': ['lilei']}
    # {'name': ['lilei']}
    
    #深复制
    # import copy
    # info1 = {"name":['tom']}
    # info2 = copy.deepcopy(info1)
    # print(info2)
    # info1["name"][0] = 'lilei'
    # print(info1)
    # print(info2)
    # {'name': ['tom']}
    # {'name': ['lilei']}
    # {'name': ['tom']}
    
    #fromkeys(*args, **kwargs)使用给定的键建立字典,默认对应值为None
    # keys_in = ('name','age')
    # keys_out1 = dict.fromkeys(keys_in)
    # keys_out2 = dict.fromkeys(keys_in,10)
    # print(keys_out1)
    # print(keys_out2)
    # {'name': None, 'age': None}
    # {'name': 10, 'age': 10}
    
    #get(self, *args, **kwargs) 返回指定键的值,如果键不存在,返回默认值
    # info1 = {"name":['tom']}
    # print(info1.get('name','No'))
    # print(info1.get('name1','No'))
    # ['tom']
    # No
    
    # v = info1['name1'] #键不存在会报异常
    # print(v)
    #     v = info1['name1']
    # KeyError: 'name1'
    
    #items(self) 遍历键值对
    # info1 = {"name":['tom']}
    # v = info1.items()
    # print(v)
    # dict_items([('name', ['tom'])])
    
    #keys(self) 遍历键
    # info1 = {"name":['tom']}
    # v1 = info1.keys()
    # print(v1)
    # dict_keys(['name'])
    
    #values(self) 遍历值
    # info1 = {"name":['tom']}
    # v2 = info1.values()
    # print(v2)
    # dict_values([['tom']])
    
    # info1 = {"name":['tom'],'age':'18','city':'BJ'}
    #
    # for k,v in info1.items():
    #     print("items()",k,v)
    # for k1 in info1.keys():
    #     print('keys()',k1)
    # for v1 in info1.values():
    #     print("values()",v1)
    # items() name ['tom']
    # items() age 18
    # items() city BJ
    # keys() name
    # keys() age
    # keys() city
    # values() ['tom']
    # values() 18
    # values() BJ
    
    
    #pop(self, k, d=None)
    # 删除指定的键,并返回值,要是键不存在,并且没有给定d值,报异常,给定d值,返回d值
    # info1 = {"name":['tom'],'age':'18'}
    # v = info1.pop('name')
    # print(v)
    # print(info1)
    # ['tom']
    # {'age': '18'}
    
    # info1 = {"name":['tom'],'age':'18'}
    # v = info1.pop('name1')
    # print(v)
    # print(info1)
    #     v = info1.pop('name1')
    # KeyError: 'name1'
    
    # info1 = {"name":['tom'],'age':'18'}
    # v = info1.pop('name1','Not find')
    # print(v)
    # print(info1)
    # Not find
    # {'name': ['tom'], 'age': '18'}
    
    #popitem(self) 随机删除一个,并返回键值对
    
    # info1 = {"name":['tom'],'age':'18','city':'BJ'}
    # v = info1.popitem()
    # print(v)
    # print(info1)
    #
    # ('city', 'BJ')
    # {'name': ['tom'], 'age': '18'}
    
    
    
    #setdefault(self, *args, **kwargs)
    #向字典中添加键值,如果添加的键在字典中不存在,直接加入键值,默认值为None
    # 键存在,则返回对应的值
    # info1 = {"name":['tom']}
    # v1 = info1.setdefault('name','lei')
    # v2 = info1.setdefault('city')
    # v3 = info1.setdefault('age','18')
    # print(v1)
    # print(v2)
    # print(v3)
    # print(info1)
    # ['tom']
    # None
    # 18
    # {'name': ['tom'], 'city': None, 'age': '18'}
    
    
    #update(self, E=None, **F) 将字典2中的键值对更新到字典1中,要是有重复的键,将被覆盖
    # info1 = {'name': ['tom'],'age': '18'}
    # info2 = {'name': ['lilei'], 'city': 'BJ'}
    # info1.update(info2)
    # print(info1)
    # print(info2)
    # {'name': ['lilei'], 'age': '18', 'city': 'BJ'}
    # {'name': ['lilei'], 'city': 'BJ'}
    
    # info1 = {'name': ['tom'],'age': '18'}
    # info1.update(name=123,city='SH')
    # print(info1)
    # {'name': 123, 'age': '18', 'city': 'SH'}
    #
    
    
    ##################################################################
    
    #删除k,v
    # info = {'name':'tom','age':'10'}
    # del info['age']
    # print(info)
    # {'name': 'tom'}
  • 相关阅读:
    C#磁吸屏幕窗体类库
    准备
    我写的诗
    How to turn off a laptop keyboard
    How to tell which commit a tag points to in Git?
    Why should I care about lightweight vs. annotated tags?
    How to get rid of “would clobber existing tag”
    Facebook, Google and Twitter threaten to leave Hong Kong over privacy law changes
    The need for legislative reform on secrecy orders
    Can a foreign key be NULL and/or duplicate?
  • 原文地址:https://www.cnblogs.com/lrtao2010/p/9297032.html
Copyright © 2011-2022 走看看