Dictionaries and collections

One 、 Additions, deletions and modifications to the dictionary

1. Add and change
Dictionaries [ key ] = value - When the key exists, it is to modify the corresponding value of the corresponding key ; When the key does not exist, the key value pair is added
Dictionaries .setdefault( key , value ) - Add key value pair ( When the key exists, the original value will not be modified )

cat = {
    'name': ' Beautiful ', 'age': 2, 'color': ' white '}
print(cat)      # {'name': ' Beautiful ', 'age': 2, 'color': ' white '}

Add method 1

cat['breed'] = ' Blue cat '
print(cat)		# {'name': ' Beautiful ', 'age': 2, 'color': ' white ', 'breed': ' Blue cat '}
cat['aihao'] = 'sheep'
print(cat)		# {'name': ' Beautiful ', 'age': 2, 'color': ' white ', 'breed': ' Blue cat ', 'aihao': 'sheep'}

Add method 2

cat.setdefault('weight', 8)
print(cat)		# {'name': ' Beautiful ', 'age': 2, 'color': ' white ', 'weight': 8}

modify

cat['age'] = 3
print(cat)		# {'name': ' Beautiful ', 'age': 3, 'color': ' white '}

 practice ： stay students Add the key value pair corresponding to the score to the students who have no score in , The score value is zero 
students = [
    {
    'name': 'stu1', 'tel': '1234', 'score': 89},
    {
    'name': 'stu2', 'tel': '465', 'score': 80},
    {
    'name': 'stu3', 'tel': '678'},
    {
    'name': 'stu3', 'score': 78},
    {
    'name': 'stu4', 'tel': '234'}
]
for new_students in students:
    new_students.setdefault('score', 0)
print(students)		# [{'name': 'stu1', 'tel': '1234', 'score': 89}, {'name': 'stu2', 'tel': '465', 'score': 80}, {'name': 'stu3', 'tel': '678', 'score': 0}, {'name': 'stu3', 'score': 78}, {'name': 'stu4', 'tel': '234', 'score': 0}]

2. Delete - Delete key value pair
   del Dictionaries 【 key 】 - Delete the key value pair corresponding to the specified key （ If the key does not exist, an error will be reported ）
   Dictionaries .pop（ key ） - Take out the value corresponding to the specified key in the dictionary

print(cat)		# {'name': ' Beautiful ', 'age': 3, 'color': ' white ', 'weight': 8}
del cat['color']
print(cat)		# {'name': ' Beautiful ', 'age': 3, 'color': ' white ', 'weight': 8}


print(cat)		# {'name': ' Beautiful ', 'age': 3, 'color': ' white ', 'weight': 8}
del cat['weight']
print(cat)		# {'name': ' Beautiful ', 'age': 3, 'color': ' white ', 'weight': 8}

del_value = cat.pop('name')
print(del_value)

Two 、 Dictionary related operations and functions

1. The relevant operation
   Dictionary not supported +、*, Nor does it support size comparison operators , Only support ==、!=
   in and not in - Dictionary in and not in The judgment is whether the key exists
   key in Dictionaries
   key not in Dictionaries

dic1 = {
    'a': 10, 'b': 20, 'c': 30}
print(10 in dic1)           # False
print('a' in dic1)          # True

2. Correlation function
len
dict( data ) - 1) The data itself is a sequence
2) The elements in the sequence must be a small sequence with only two elements
3) The first element of a small sequence must be immutable data

data = [('a', 10), ('b', 20)]
print(dict(data))       # {'a': 10, 'b': 20}

data2 = ['ab', range(2), [10, 20]]
print(dict(data2))      # {'a': 'b', 0: 1, 10: 20}

‘’’
Dictionary into list / Tuples
‘’’

dic2 = {
    'a': 10, 'b': 20, 'c': 30}
print(list(dic2))       # ['a', 'b', 'c']

‘’’
3. Dictionary related methods
Dictionaries .clear() - Empty dictionary
Dictionaries .copy - Copy the original dictionary to produce an identical new dictionary
Dictionaries .update( Sequence ) - Add all the elements in the sequence to the dictionary ( If it exists, it will cover ). The sequence must be a dictionary or a sequence that can be converted into a dictionary
items、keys、values
Dictionaries .keys() - Get all dictionary keys , Return a sequence （ Not a dictionary or tuple ）
Dictionaries .value() - Get all the values of the dictionary , Return a sequence （ Not a dictionary or tuple ）
Dictionaries .items() - Get all the key value pairs in the dictionary , Return a sequence （ Not a dictionary or tuple ）
‘’’

data3 = {
    'd': 100, 'e': 200}

dic2.update(data3)
print(dic2)             # {'a': 10, 'b': 20, 'c': 30, 'd': 100, 'e': 200}

print(dic2.keys())      # dict_keys(['a', 'b', 'c', 'd', 'e'])
print(dic2.values())    # dict_values([10, 20, 30, 100, 200])
print(dic2.items())     # dict_items([('a', 10), ('b', 20), ('c', 30), ('d', 100), ('e', 200)])

‘’’
4. Dictionary derivation
{ Key expression ： The expression of the value for Variable in Sequence }
{ Key expression ： The expression of the value for Variable in Sequence if Conditional statements }
‘’’

 practice ： Exchange the keys and values of a dictionary through the derivation of the dictionary 
'''{'a': 10, 'b': 201} --> {10:'a', 20, 'b'}'''
new_dic2= {
    dic2[key]: key for key in dic2}
print(new_dic2)			#  Beautiful 

3、 ... and 、 aggregate

1. What is a collection （set）
Collections are containers ; take {} As a sign of the container , Multiple elements are separated by commas ：{ Elements 1, Elements 2, Elements 3,…}
The set is variable ; Sets are unordered ;
Elements ： Immutable data 、 The element is the only ( It has the function of automatic weight removal )

1)  Empty set 
set1 = set()
print(type(set1), len(set1))        # <class 'set'> 0

2)  Sets are unordered 
print({
    1, 2, 3} == {
    3, 1, 2})

3)  Elements must be immutable data 
set2 = {
    1, 'abc', (2, 3)}
print(set2)

# set3 = {1, 'abc', [2, 3]}
# print(set3) #  Report errors ！

# 4)  The element is the only 
set4 = {
    1, 2, 3, 1, 1, 3}
print(set4)            # {1, 2, 3}

2. Addition, deletion, modification and query of set elements

1. check - Traverse
   for Variable in aggregate ：
   The loop body

nums = {
    23, 90, 89, 78}
for x in nums:
    print(x)

2. increase
   ‘’’
   aggregate .add( Elements ) - Add the specified element to the collection
   aggregate .update( Sequence ) - Add all the elements in the sequence to the collection
   ‘’’

nums.add(45)
print(nums)         # {45, 78, 23, 89, 90}

nums.update('abc')
print(nums)         # {'a', 'b', 45, 78, 23, 89, 90, 'c'}

3. Delete
   ‘’’
   aggregate .remove( Elements ) - Deletes the specified element , If the element does not exist, an error will be reported
   aggregate .discard( Elements ) - Deletes the specified element , If the element does not exist, no error will be reported
   ‘’’

nums.remove(89)
print(nums)     # {'a', 'b', 45, 78, 'c', 23, 90}

nums.discard('a')
print(nums)     # {45, 78, 'c', 23, 90, 'b'}

# nums.remove(100) #  Report errors 
nums.discard(100)

4. Change - A collection cannot directly modify the value of an element , If you have to change it, delete the element you want to change , Add new elements

nums.discard('b')
nums.update('B')
print(nums)

3. Mathematical set operation ：&( intersection )、|( Combine )、-( Difference set )、^( Symmetric difference set )、>/<( True subset )、>=/<=( A subset of )

nums1 = {
    1, 2, 3, 4, 5, 6, 7}
nums2 = {
    5, 6, 7, 8, 9}

1. aggregate 1 & aggregate 2 – Get the common elements of two collections ( Get is in the set 1 It's gathering again 2 The elements inside )

print(nums1 & nums2)        # {5, 6, 7}

2. aggregate 1 | aggregate 2 – Get all the elements of the two sets

print(nums1 | nums2)        # {1, 2, 3, 4, 5, 6, 7, 8, 9}

3. aggregate 1 - aggregate 2 – Get collection 1 In addition to being included in the set 2 Parts other than middle

print(nums1 - nums2)        # {1, 2, 3, 4}
print(nums2 - nums1)        # {8, 9}

4. aggregate 1 ^ aggregate 2 – Merge two sets , Remove the middle common part

print(nums1 - nums2)        # {1, 2, 3, 4}

5. A subset of （ It could be equal ） And true subsets （ It's really smaller than it ）
   aggregate 1 > aggregate 2 --> Judgment set 2 Is it a collection 1 The proper subset of

print({
    10, 20, 30, 40} > {
    1, 2})        # False
print({
    10, 20, 30, 40} > {
    10, 30})      # True
print({
    10, 20, 30, 40} > {
    10, 20, 30, 40})     # False
print({
    10, 20, 30, 40} >= {
    10, 20, 30, 40})     # True

Dictionary and set work

1. Define a list , Save in list 6 Student information ( Student information includes : full name 、 Age 、 achievement ( Single subject )、 Telephone 、 Gender ( male 、 Woman 、 Unknown ) )

   nums = [{
         'name': 'student1', 'age': 18, 'math': 90, 'tel': 9832642, ' Gender ': ' male '},
           {
         'name': 'student3', 'age': 19, 'math': 98, 'tel': 1367687, ' Gender ': ' male '},
           {
         'name': 'student4', 'age': 15, 'math': 56, 'tel': 8736588, ' Gender ': ' Woman '},
           {
         'name': 'student5', 'age': 17, 'math': 98, 'tel': 1382363, ' Gender ': ' Woman '},
           {
         'name': 'student6', 'age': 19, 'math': 56, 'tel': 7486447, ' Gender ': ' male '},
           {
         'name': 'student7', 'age': 16, 'math': 46, 'tel': 5346535, ' Gender ': ' Unknown '}
   ]
   

   1. Count the number of failed students

      a = nums
      b = 0
      for x in a[2:]:
          s = x['math']
          if s < 60:
              b += 1
      print(b)
      

   2. Print the name of the failed student and the corresponding grade

      a = nums
      for x in a[2:]:
          if x['math'] < 60:
              print(x['name'], x['math'])
      

   3. Print the mobile phone tail number is 8 The name of the student

for stu in list1:
    if stu['tel'][-1] == '8':
        print(' The tail number of the mobile phone is 8：', stu['name'])

4. Print the highest score and the corresponding student's name

   b = nums[0]['math']
   for x in nums[1:]:
       a = x['math']
       if a > b:
           b = a
           print(' The highest ：', b)
   for x in nums:
       if x['math'] == b:
           print(x['name'])
   

5. Delete all students of Unknown Gender

new_list1 = [stu for stu in list1 if stu[' Gender '] != ' Unknown ']
print(new_list1)

6. Sort the list by student grade from large to small ( Struggle for a moment , If you can't, just give up )

list1.sort(key=lambda item: item['math'], reverse=True)
print(list1)

2. Use three sets to represent the names of selected students in three disciplines ( A student can take multiple courses at the same time )

   1. How many students are there in total
   2. Find the number of people who chose only the first subject and the corresponding name
   3. Find the number of students who have chosen only one subject and the corresponding name
   4. Find the number of students who have chosen only two subjects and their corresponding names
   5. Find the number of students who have chosen three courses and their corresponding names