small python list of dictionaries practice

1. Use code verification "name" is in the dictionary keys?

   info = { 'name': 'Gang egg', 'hobby': 'hammer', 'age': '18', ... 100} key-value pairs

   info = {'name':'王刚蛋','hobby':'铁锤','age':'18'}
   info_lis = list(info.keys())
   if 'name' in info_lis:
       print("'name'在字典的键中。")
   else:
       print('没有此键')

2. Use code verification "alex" if the value of the dictionary?

   info = {'name':'王刚蛋','hobby':'铁锤','age':'18',...100个键值对}

   info = {'name':'王刚蛋','hobby':'铁锤','age':'18',1:'alex'}
   info_lis = list(info.values())
   if 'alex' in info_lis:
       print("'alex'在字典的值中。")
   else:
       print('没有此值。')

3. There follows

   v1 = {'武沛齐','李杰','太白','景女神'}
   v2 = {'李杰','景女神}

   • Please get v1 and v2 and the intersection of output

   • Please get v1 and v2 union and output

   • Please give v1 and v2 and the output differential current

   • Please give v2 and v1 and difference output

     v1 = {'武沛齐','李杰','太白','景女神'}
     v2 = {'李杰','景女神'}
     
     # - 请得到 v1 和 v2 的交集并输出
     print(v1 & v2)
     # - 请得到 v1 和 v2 的并集并输出
     print(v1 | v2)
     # - 请得到 v1 和 v2 的 差集并输出
     print(v1 - v2)
     # - 请得到 v2 和 v1 的 差集并输出
     print(v2 - v1)

4. Cycle prompts the user, and inputs the content added to the list (if the input is stopped cycle n or N)

   lis = []
   while 1:
       msg = input('请输入：（N/n退出）').strip()
       if msg.upper() == 'N':break
       else:
           lis.append(msg)
           print('*'*30)
       print(lis)
       print('*'*30)

5. Cycle prompts the user, and inputs the contents to the collection (n if the input is stopped or N cycle)

   set1 = set()
   while 1:
       msg = input('请输入：（N/n退出）').strip()
       if msg.upper() == 'N':break
       else:
           set1.add(msg)
           print('*'*30)
       print(set1)
       print('*'*30)

6. Write code to achieve

   v1 = {'alex','武sir','肖大'}
   v2 = []
   
   # 循环提示用户输入，如果输入值在v1中存在，则追加到v2中，如果v1中不存在，则添加到v1中。（如果输入N或n则停止循环）

   v1 = {'alex','武sir','肖大'}
   v2 = []
   while 1:
       print('*'*30)
       msg = input('请输入：（N/n退出）').strip()
       if msg.upper() == 'N':break
       elif msg in v1:
           v2.append(msg)
       else:
           v1.add(msg)
       print('*'*30)
       print(v1)
       print(v2)

7. Using a loop to print out at the results:

* 
** 
*** 
**** 
***** 

for i in range(1,6):
    print('*'*i)

**** 
***
** 
*

for i in range(-5,0):
    print('*'* -i)

* 
*** 
***** 
******* 
*********

for i in range(1,10,2):
    print('*'* i)

1. Knock seven game from a start count. 7 or 7 encounter multiple (17, 27 does not contain this number) to be. In the table programming to complete knock knock ⼀ VII. ⼀ given arbitrary number n from a start count the number n to end. each number on the list, a multiple of 7 or 7 appears in the number of process (17, 27 does not contain this number). Add an then to the list a 'bang'

For example, input START 10. lst = [1, 2, 3, 4, 5, 6, 'bang', 8, 9, 10]

lis = []
while 1:
    n = input('请输入任意一个数字：（Q/q退出）').strip()
    if n.upper() == 'Q':break
    for i in range(1,int(n)+1):
        if i % 7 == 0:
            lis.append('咣')
        else:
            lis.append(i)
        print(f'当前数字为：{i},结果为：{lis}')

    print('*'* 60)
    print(f'输入的数字是：{n}')
    print(f'结果为：{lis}')