python collection of hash dictionary

python types of data may be hash, not hash

什么是hash呢？

　 　Hash typically translated as "hash", also transliterated as a "hash", that is, the input of any length, through a hash algorithm, a hash value is converted into a fixed-length output, output.
　This conversion is a compression map, i.e., the space hash value is typically much smaller than the input space, different inputs may hash to the same output, it is impossible from the hash value
is uniquely determined input value. A message is simply an arbitrary length is compressed to a function of a fixed length message digest.

Objects may be used include: number, string, a tuple can not be used is a list of the hash function, the dictionary.

Characteristic hash: the hash value calculation is based on the feature value calculated, which requires the hash value must be fixed, so the hash value must be immutable

The variable type of data is not the hash, such as List, dictionaries, set: addresses with different values, different values ​​for the same address

a = [1,2,3]
b = [1,2,3]
print(id(a),id(b))  #2110426618696 2110426618824
a.append(4)
print(id(a))    #2110426618696

Numerical, alphabetical, strings, numbers, immutable tuples: the same values ​​for the same address, different address values ​​different

#改变a的值后 内存地址也改变了（id）

Determine how the variable immutable?

Summary: to see a change of value is not the same id, id like a variable, you can not hash, changed values, id change, was immutable, it can hash

dictionary:

Also known as key data dict dict = {key: value}

Storing the data dictionary lookup faster

python3.6 version of the above default dictionary is ordered, how to write on how to print out, but do not go out and people say, when the dictionary is unordered

Dictionary key: immutable data type of the key hash can not, the variable data is not a variable data hash immutable data can be hashed

Dictionary values: free to

dic = {'01':'邓新','2':'情哥哥','14':'小冯','98':'啦'}
print(dic['01'])    # 通过键去查找
print(dic['2']) # 通过键去查找

increase:

dic[11] = '小米'
print(dic)
dic.setdefault(99,'华为') # 99键 华为  值
#1.先去字典中通过99这个键去查有没有值
#2.如果是值 就不添加
#3.如果是None  就添加     返回的是添加的值

delete:

del dic #删除整个字典
del dic[键]  #通过键删除
print()

print(dic.pop(99))  # pop 通过指点键删除   pop也有返回值 返回的是被删除的值

dic.clear() #清空
print(dic)

dic.popitem()   #随机删除   python3.6 删除最后一个    python3.5   随机删除
print(dic)

change:

dic1 = {'33':'jj','44':'TT'}
dic2 = {'33':'gg','66':'pp'}
dic1.update({'33':'gg','66':'yy'})  #更新     被更新的内容如果在     要更新的内容中那值就会覆盖
#两个字典中，都没有一样的就合并
dic1.update(dic2)

dic2['33'] = '啦' #字典的增    是字典中没有的时候才叫增，如果字典中键存在的时候才叫改

check:

dic1 = {'33':'jj','44':'TT'}
print(dic1['33'])  #通过键查找  如果键不存在就报错
print(dic1.get('33'))  #通过键查找   键不存在不报错  返回None
print(dic1.setdefault('33'))   #不存在返回None

Other operations:

dic1 = {'33':'jj','44':'TT'}
for i in dic1:
    print(i)

for i in dic1:
    print(dic1.get(i))

for i in dic1.keys:
    print(dic1.get(i))
    
 print(dic1.keys())     #高仿列表 dic—_keys(['33','66'])    不能使用下标

print(dic1.values())    #高仿列表 dic—_values(['jj','TT'])  不能使用下标

#取出键值
print(dic1.items())  # 取出键值 每部分用元组包起来

Deconstruction:

a,b,c = '4','5','6'   ['4','5','6']  ('4','5','6')  {'1':8,'2':9,'3':0}     True,False,True
print(a,b,c)

for k,v in dic1.items():
    print(k,b)
    
#面试题
a = 10
b = 20
a,b = b,a
print(a,b)

supplement:

dic2 = {'33':'gg','66':'pp'}
ret1 = dic2.get('99':'啦啦啦')
ret2 = dic2.pop('99':'没有')
print(ret1,ret2)

#坑：
dic3 = dict.formkeys([1,2,3,4,5],[1,2,3])   #第一个是键 要是可迭代的对象
#第二个是值，值是无所谓    是因为创建的新字典的值用的都是一个   值是一个可变数据类型
print(dic3)

dic3[3] = '啦啦啦'

Nesting dictionary:

dic = {'汪峰':{'1':'la','2':'啦'},'陈冠希':{'张柏芝':'喜剧之王','阿娇':'千机变'}}
dic1 = dic['陈冠希']
print(dic1['张柏芝'])

set:

set collection of stored data deduplication natural disorder can not use the photo

se = {1,2,3,4,5,1,1,3,4,2,45,3,2}
print(se)

#面试题
lst = [1,2,3,41,12,1,1,12,2]
print(list(set(lst)))

for i in {1,2,3,4}:
    print(i)

increase:

s = {"刘嘉玲", '关之琳', "王祖贤"}

s.add("郑裕玲") # 重复的内容不会被添加到set集合中
print(s)

s = {"刘嘉玲", '关之琳', "王祖贤"}
s.update("麻花藤") # 迭代更新
print(s)
s.update(["张曼⽟", "李若彤","李若彤"])
print(s)

delete:

s = {"刘嘉玲", '关之琳', "王祖贤","张曼⽟", "李若彤"}
item = s.pop() # 随机弹出⼀个.
print(s)
print(item)
s.remove("关之琳") # 直接删除元素
# s.remove("⻢⻁疼") # 不存在这个元素. 删除会报错
print(s)
s.clear() # 清空set集合.需要注意的是set集合如果是空的. 打印出来是set() 因为要和
dict区分的.
print(s) # set()

check:

# set是⼀个可迭代对象. 所以可以进⾏for循环
for el in s:
 print(el)

Common operations:

s1 = {"刘能", "赵四", "⽪⻓⼭"}
s2 = {"刘科⻓", "冯乡⻓", "⽪⻓⼭"}
# 交集
# 两个集合中的共有元素
print(s1 & s2) # {'⽪⻓⼭'}
print(s1.intersection(s2)) # {'⽪⻓⼭'}
# 并集
print(s1 | s2) # {'刘科⻓', '冯乡⻓', '赵四', '⽪⻓⼭', '刘能'}
print(s1.union(s2)) # {'刘科⻓', '冯乡⻓', '赵四', '⽪⻓⼭', '刘能'}
# 差集
print(s1 - s2) # {'赵四', '刘能'} 得到第⼀个中单独存在的
print(s1.difference(s2)) # {'赵四', '刘能'}
# 反交集
print(s1 ^ s2) # 两个集合中单独存在的数据 {'冯乡⻓', '刘能', '刘科⻓', '赵四'}
print(s1.symmetric_difference(s2)) # {'冯乡⻓', '刘能', '刘科⻓', '赵四'}
s1 = {"刘能", "赵四"}
s2 = {"刘能", "赵四", "⽪⻓⼭"}
# ⼦集
print(s1 < s2) # set1是set2的⼦集吗? True
print(s1.issubset(s2))
# 超集
print(s1 > s2) # set1是set2的超集吗? False
print(s1.issuperset(s2))

This collection set can be made ⽣ chest upwards changed. is not the hash. frozenset we can use the data to store. frozenset is immutable. ⼀ is a type of data may be hashed

`s ``=` `frozenset``([``"赵本⼭"``, ``"刘能"``, ``"⽪⻓⼭"``, ``"⻓跪"``])``dic ``=` `{s:``'123'``} ``# 可以正常使⽤了``print``(dic)`