7. Set set of Python data type

Collection set of data types
Set does not allow duplicate unordered collections and cannot be evaluated by subscript because unordered
1. Create
Create an empty collection
S ={} The default type is a dictionary, so it is not an empty collection, the empty collection is as follows
s = Set ()
s = {11,12,12,34,23} #字典是有键值对, the collection has no
s = set () #括号内可以接收可以迭代的元素, str list tuple dict
2. Conversion
s = ([11,123,1234,12345])
L = "123"
L = [1,2,3,4]
L = (1,2,3,4)
s = Set (L)
3. Methods
#集合中元素无序不重复
St = {11,22,33,44,55,66}
#集合中添加元素

St.add (+) print (ST)

A = {11,22,33,55,66}b = {55,66,77,11}

#diference方法

#a. Difference (b), a differs from the elements of B, and a remains the same

Reta = a.difference (b) print (a) print (Reta)

#b. Difference (a), B is different from element A, B remains unchanged

Print (b) Retb = B.difference (a) print (RETB)

#difference_update方法, find the element a differs from B and re-overwrite a

A.difference_update (b) print (a)

#找出b不同于a的元素并覆盖b

B.difference_update (a) print (b)

#找出a集合不同于b结合的的元素, the B collection is different from the elements of a set, and can be assigned to a new collection

Retab = a.symmetric_difference (b) retba =  b.symmetric_difference (a) print (retab) print (RETBA)

#找出a集合不同于b结合的的元素, the B collection differs from the elements of a set and updates the found results to a combined

A.symmetric_difference_update (b) print (a)

#找出b集合不同于a结合的的元素, a sets the elements that are different from the B collection and updates the found results to the B combination

B.symmetric_difference_update (a) print (b)

#discard移除集合中的元素, if the element is removed, there is no error

Print (a) a.discard (one) print (a) A.discard (456) print (a)

#remove移除集合中的元素, if the element is removed, there is no error

Print (a) a.remove (a) print (a)

#111元素不存在与a中, the following will be an error

A.remove (111) A.remove ()

#intersaction取两个集合的交集, and can be assigned to a new collection

Retinab = a.intersection (b) print (RETINAB)

#a. Intersection_update (b) takes the intersection of two sets and adds the resulting collection to the collection A

A.intersection_update (b) print (a)

#并集union
#取a集合和b集合的并集

RETUAB = a.union (b) print (RETUAB)

#update将b集合更新追加到a集合中

A.update (b) print (a)

#a. Isdisjoint (b)

T = a.isdisjoint (b) print (t)

#a. Issubset (b) A is not a sub-sequence of b

Print (A.issubset (b))

#a. Issuperset (b) is not the parent sequence of B

Print (A.issuperset (b))

#a. Pop () to remove an element, you can assign the removed value to a variable

Ta = A.pop () print (TA)

#练习: Look for differences and update the old dictionary
Updates that are present in the #将old_dict, new_dict, old_dict deletions that do not exist in New_dict
#new_dict在old_dict中不存在的, append to old.

Old_dict = {    "#1": {' hostname ': ' C1 ', ' Cpu_count ': 2, ' mem_capicity ': +}, '    #2 ': {' hostname ': ' C1 ', ' Cpu_count ': 2 , ' mem_capicity ': "    #3": {' hostname ': ' C1 ', ' Cpu_count ': 2, ' mem_capicity ':}}new_dict = {    "#1": {' hostname ': ' C1 ', ' Cpu_count ': 2, ' mem_capicity ': + ', '    #3 ': {' hostname ': ' C1 ', ' Cpu_count ': 2, ' mem_capicity ': +}, '    #4 ' : {' hostname ': ' C2 ', ' Cpu_count ': 2, ' mem_capicity ':}}k1 = Old_dict.keys () K2 = New_dict.keys () K1_set = set (k1) K2_set = Set (k2) K1_d_k2 = k1_set.difference (k2_set) print (K1_D_K2) for I1 in K1_d_k2:    old_dict.pop (i1) print (old_dict) k2_d_ K1 = k2_set.difference (k1_set) print (K2_D_K1) for i2 in K2_D_K1:    old_dict[i2] = New_dict[i2]print (old_dict) k1_inter _K2 = k1_set.intersection (k2_set) print (K1_INTER_K2) for i3 in K1_INTER_K2:    old_dict[i3] = New_dict[i3]print (old_ dict) print (old_dict==new_dict)

7. Set set of Python data type