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  • day5:字典dict

    1, 判断是不是列表

    li = ['lis3a', 'mary', 'lucy', 'hh', 'kk', 'gg', 'mm', 'oo', 'vv']
    if type(li) == list:
        print("list")

    2,字典的key必须是不可变数据类型,也就是bool,int,str,tuple,value可以是任意的数据类型,有点二分查找去查询,存储大量的键值对,特点是无序的,3.5以前版本

    3,字典的增加与修改方式,没有的添加,有的覆盖

    dict1 = {"name":"lisa","age":"22","hobby":"piano"}
    dict1["height"]="165" #没有的添加进去
    dict1["name"]="mary" #有的覆盖,就相当于修改了
    print(dict1)
    
    运行结果:
    {'name': 'mary', 'age': '22', 'hobby': 'piano', 'height': '165'}

    4,添加的另外一种方式setdefault

    dict1 = {"name":"lisa","age":"22","hobby":"piano"}
    
    dict1.setdefault("hometown") # 不设置value默认为None
    print(dict1)
    
    dict1.setdefault("name")  # 本来有的不会更新
    print(dict1)
    
    dict1.setdefault("gender","female") # 设置了value
    print(dict1)
    
    dict1.setdefault(5,True) # 支持很多数据类型,详见上一条
    print(dict1)

    5,删除,pop函数,必须制定key,不指定会报错,制定成value也会报错,制定的不存在的key 也会崩溃

    dict1 = {"name":"lisa","age":"22","hobby":"piano"}
    print(dict1.pop())
    
    运行结果:
    Traceback (most recent call last):
      File "/Users/guolixiao/PycharmProjects/lisa's_practise/boys/5.1_dict.py", line 28, in <module>
        print(dict1.pop())
    TypeError: pop expected at least 1 arguments, got 0
    
    dict1 = {"name":"lisa","age":"22","hobby":"piano"}
    print(dict1.pop(“name”))
    
    运行结果:
    lisa
    dict1 = {"name":"lisa","age":"22","hobby":"piano"}
    print(dict1.pop("gender"))
    
    运行结果:
    Traceback (most recent call last):
      File "/Users/guolixiao/PycharmProjects/lisa's_practise/boys/5.1_dict.py", line 29, in <module>
        print(dict1.pop("gender"))
    KeyError: 'gender'
    dict1 = {"name":"lisa","age":"22","hobby":"piano"}
    print(dict1.pop("gender",None))
    # 这个可以专门用来表示没有这个key时候的返回值,或者提示“没有此键”,等也用来防止程序崩溃
    运行结果:
    None

    6,删除的另外一种popitem ,不需要参数,3.5以前随机删除,3.5以后是最后一个

    7,del函数可以删除摸一个,或者删除整个dict,这个和列表类似

    8,清空字典,clear函数,这个也和字典类似

    9,改,update函数

    dict1 = {"name":"lisa","age":"22","hobby":"piano"}
    dict2 ={'name':"mary","gender":"female"}
    dict2.update(dict1)
    print(dict1)
    print(dict2)
    #用dict1来更新dict2,没有的添加,有的更新value
    
    运行结果:
    {'name': 'lisa', 'age': '22', 'hobby': 'piano'}
    {'name': 'lisa', 'gender': 'female', 'age': '22', 'hobby': 'piano'}

    10,查,区分,value,key,item

    dict1 = {"name":"lisa","age":"22","hobby":"piano"}
    print(dict1.keys())
    print(dict1.values())
    print(dict1.items())
    
    运行结果:
    dict_keys(['name', 'age', 'hobby'])
    dict_values(['lisa', '22', 'piano'])
    dict_items([('name', 'lisa'), ('age', '22'), ('hobby', 'piano')])
    dict1 = {"name":"lisa","age":"22","hobby":"piano"} 
    for i in dict1:  # 默认输出的是key和下一句效果是一样的
        print(i)
    
    for i in dict1.keys():
        print(i)
    
    运行结果:
    name
    age
    hobby
    name
    age
    hobby

    11,穿插点字典以外的知识点:

    a,b,c =1,2,3
    print(a,b,c)
    a,b = b,a
    print(a,b,c)
    
    运行结果:
    1 2 3
    2 1 3
    a,b = [1,2]
    print(a,b)
    a,b =[1,2],[3,4]
    print(a,b)
    # 永远按照最小的单位进行拆包
    运行结果:
    1 2
    [1, 2] [3, 4]

    12,iterm打印出来的是元祖,可以进行拆包赋值给k,v

    dict1 = {"name":"lisa","age":"22","hobby":"piano"}
    for i in dict1.values():
        print(i)
    
    for i in dict1.items():
        print(i)
    
    for k,v in dict1.items():
        print(k,v)
    
    运行结果:
    lisa
    22
    piano
    ('name', 'lisa')
    ('age', '22')
    ('hobby', 'piano')
    name lisa
    age 22
    hobby piano

    13,查,两种,首选get函数,因为不存在是不会崩溃

    v1 = dict1.get("name")
    print(v1)
    
    v1 = dict1.get("name1")
    print(v1)
    
    v1 =dict1.get("name1","没有此键")
    print(v1)
    
    运行结果:
    lisa
    None
    没有此键
    dict1 = {"name":"lisa","age":"22","hobby":"piano"}
    v1 =dict1["name"]
    print(v1)
    
    运行结果:
    lisa
    
    
    dict1 = {"name":"lisa","age":"22","hobby":"piano"}
    v1 =dict1["name1"]
    print(v1)
    
    运行结果:
    Traceback (most recent call last):
      File "/Users/guolixiao/PycharmProjects/lisa's_practise/boys/5.1_dict.py", line 82, in <module>
        v1 =dict1["name1"]
    KeyError: 'name1'

    14,一道练习题

    str1 ="hiohph65nooihoh89nlohiolh76nklnl4"
    # 找出里面有多少个数字,65算一个,89算一个
    for i in str1:
        if i.isdigit():
            break
        else:
            str1 = str1.replace(i,"  ") #用空格替换非数字,这里一定要把新生成的字符串赋值回去
    
    l = str1.split()
    print(len(l))
    
    运行结果:
    4
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  • 原文地址:https://www.cnblogs.com/lisa-blog/p/10023271.html
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