python-Collection, Dictionary

Set and dictionary set (variable, unordered, non-repeating) initialization

#set()->new empty set ibjecta=set()print(a)                        #输出:set()#set(iterable)->new set objectprint(set(range(5)))            #输出:{0, 1, 2, 3, 4}print(set(list(range(5))))      #输出:{0, 1, 2, 3, 4}s={}                            #1print(s)print(type(s))                  #输出:<class ‘dict‘>s1={(1,2),3,‘a‘}print(s1)                       #输出:{(1, 2), ‘a‘, 3}s2={[1],(1,),1}                 #2print(s2)                       #TypeError: unhashable type: ‘list‘

1 When there is no value in {}, the system will be judged as a dictionary; 2set elements must be hash, that is, immutable;

Note: Set can iterate, but there is no index because he is unordered;

Set operation

#set(elem)增加s={1,}s.add(2)print(s)                #输出:{1,2}s.add(1)print(s)                #输出:{1,2}#1#update(*other)#合并其他元素到set中，other必须是可迭代对象s1={‘da‘,31,‘dd‘,234,‘f‘,25,}s2={312,"2",31,"312ds"}print(s1.update(s2))            #输出:Noneprint(s1)                       #输出:{‘f‘, ‘da‘, ‘dd‘, 234, ‘312ds‘, ‘2‘, 312, 25, 31}

1 If an element exists, nothing is done

#remove(elem)#从set中移除一个元素，不存在抛异常a={‘f‘, ‘da‘, ‘dd‘, 234, ‘312ds‘, ‘2‘, 312, 25, 31}print(a.remove(‘f‘))            #输出:Noneprint(a)                        #输出:{‘2‘, 234, ‘da‘, ‘312ds‘, 312, 25, ‘dd‘, 31}#discard(elem)#从set中移除一个元素，不存在什么也不做b={234, ‘312ds‘, ‘2‘, 312, 25, 31}print(b.discard(234))               #输出:Noneprint(b)                            #输出:{‘2‘, 312, 25, ‘312ds‘, 31}#1#pop()->item#移除并返回任意元素c={312, 25, 31,1789}print(c.pop())print(c)#clear()#移除所有元素d={312, 25, 31,1789}print(d.clear())print(d)

1 in contrast to remove and discard, priority should be given to using discard his error hints are milder;

Note: Set for non-linear unordered, all nature has not changed this said, can delete one, add one to simulate change

Dictionary (variable, unordered, can not be duplicated) initialization

#空字典d1={}d2=dict()print(d1)print(d2)#使用key=value来初始化d3=dict(a=12,b=3)print(d3)                       #输出:{‘a‘: 12, ‘b‘: 3}#以下几种方法，必须是二元可迭代结构才能d4=dict(((1,‘a‘),(2,‘b‘)))print(d4)                       #输出:{1: ‘a‘, 2: ‘b‘}d5=dict(([1,‘1‘],[2,‘2‘]))print(d5)                       #输出:{1: ‘1‘, 2: ‘2‘}d6=dict([[1,‘1‘],[2,‘2‘]])print(d6)                       #输出:{1: ‘1‘, 2: ‘2‘}#dict(mapping,**kwarg)使用一个字典构建领一个字典d7=dict(d6,a=2)print(d7)#最简单的定义方法d8={‘a‘:1,‘q‘:3}print(d8)#类方法dict.fromkeys(iterable,value),用来统一赋值d=dict.fromkeys(range(5))print(d)d=dict.fromkeys(range(5),10)print(d)

Access to dictionary elements

#d[key]，不存在抛异常d={1: ‘1‘, 2: ‘2‘, ‘a‘: 2}print(d[‘a‘])#get(key,[,default])  返回key对应的值value，后面可是设置缺省值，不存在返回Noned={1: ‘1‘, 2: ‘2‘, ‘a‘: 2}print(d.get(2,"exit"))    #输出:2 print(d.get(23,"no exit"))   #输出:no exitprint(d)#setdefault(key,[,default])  返回key对应的值value，后面可是设置缺省值，不存在返回Noneprint(d.setdefault(23,"no exit"))print(d)                       #输出:{1: ‘1‘, 2: ‘2‘, ‘a‘: 2, 23: ‘no exit‘}#1

1: From the above it seems that if you see the dictionary there is no content and to join, then you need to use SetDefault

Additions and modifications to dictionaries

#d[key]=value，如果key不存在那么是添加，如果存在是修改d={1: ‘1‘, 2: ‘2‘, ‘a‘: 2}d["wuyanzu"]="123"print(d)                #输出:{1: ‘1‘, 2: ‘2‘, ‘a‘: 2, ‘wuyanzu‘: ‘123‘}#update([other])->None#使用另一个字典的kv来更新本字典，key不存在就添加，存在就覆盖d1=dict(apple=‘a‘,pear=‘p‘)print(d1.update(d))      #输出:Noneprint(d1)                #输出:{‘apple‘: ‘a‘, ‘pear‘: ‘p‘, 1: ‘1‘, 2: ‘2‘, ‘a‘: 2, ‘wuyanzu‘: ‘123‘}

Deletion of dictionaries

a={1: ‘1‘, 2: ‘2‘, ‘a‘: 2, ‘wuyanzu‘: ‘123‘}#pop(key,[,default])#key存在print(a.pop(1))                 #输出:1，返回的是keyprint(a)                        #输出:{2: ‘2‘, ‘a‘: 2, ‘wuyanzu‘: ‘123‘}#key不存在，default存在a={1: ‘1‘, 2: ‘2‘, ‘a‘: 2, ‘wuyanzu‘: ‘123‘}print(a.pop(3,"no exist"))       #输出:no exist#key不存在，default不存在a={1: ‘1‘, 2: ‘2‘, ‘a‘: 2, ‘wuyanzu‘: ‘123‘}print(a.pop(3))                  #抛异常，keyerrora={1: ‘1‘, 2: ‘2‘, ‘a‘: 2, ‘wuyanzu‘: ‘123‘}#popitem移除并返回任意一个键值对，直到空字典时，抛出异常print(a.popitem())print(a.clear())print(a)#输出(‘wuyanzu‘, ‘123‘)None{}d={‘a‘:1,‘b‘:‘b‘,‘c‘:[1,3,5]}#deldel d[‘c‘]print(d)#输出:{‘a‘: 1, ‘b‘: ‘b‘}

Del Delete is to reduce the reference to the object, not to delete the contents of the reference in memory;

Dictionary traversal

d={‘a‘:1,‘b‘:‘b‘,‘c‘:[1,3,5]}#key遍历for k in d.keys():    print(k)#输出abcfor k in d:    print(d[k])         #默认也是对key进行遍历;#value遍历for v in d.values():    print(v)#输出1b[1, 3, 5]#k,v都遍历for k,v in d.items():    print(k,v)#输出a 1b bc [1, 3, 5]

The dictionary key and set requirements are consistent, must be hashable (immutable type);

python-Collection, Dictionary