Python Constructing Multi level Menu Function with Dictionary
- 2021-07-13 05:52:35
- OfStack
Related knowledge points:
#key-value
# The dictionary is unordered because it has no subscript and passes through key Looking for
info={
'stu01':"liuhaolai",
'stu02':"wangshulin"
}
print(info['stu01'])
info['stu03']=' Liu **'# If this does not exist key Is added directly
info['stu04']=' Wang ##'
print(info)
#del
del info['stu03']
print(info)
#pop
info.pop('stu04')
print(info)
print(info.get('stu03'))# I'm not sure if there is any time in the dictionary
print('stu01' in info)# Judge whether there is
print(info.values())# Print value
print(info.keys())# Print key
# Nesting of dictionaries
place={
' Liaoning ':{
' Dalian ':[' It's beautiful ',' Very clean '],
' Shenyang ':[' Very large ',' Some mess '],
' Fuxin ':[' Very small ',' Very comfortable ']
},
' Jiangsu ':{
' Nanjing ':[' Provincial capitals ',' Extremely low cost performance '],
' Suzhou ':[' Combination of ancient and modern times ',' Small bridge flowing water '],
' Zhenjiang ':[' Some mess ','1 Be like ']
}
}
print("------")
print(place)
place[' Jiangsu '][' Zhenjiang '][1]='1 So-so '
print(place)
print(place.values())# Print value
print(place.keys())# Print key
place.setdefault(' Hebei ',{' Tangshan :[0,1]'})# Check, add if you don't, and get Different
print(place)
print(place.setdefault(' Jiangsu ',{' Suzhou :[0,1]'}))# Check, add if you don't have it
info={
'stu01':"liuhaolai",
'stu02':"wangshulin",
'stu03':'lucky',
'stu04':'pangpang'
}
new={
'stu01':'david',
'stu02':'tree',
'stu05':'lala'
}
new.setdefault('stu03','david')# Check, add if you don't have it
print(new)
info.update(new)# Update existing overrides, create none
print(info)
print(info.items())# Turn dictionary into list
c=dict.fromkeys([6,7,8],[1,{"name":'number'}])# Initialization 1 A new dictionary
print(c)
# Loop of dictionary
info2={
'stu01':"liuhaolai",
'stu02':"wangshulin",
'stu03':'lucky',
'stu04':'pangpang'
}
print(info2.get('stu01'))
print(info2.setdefault('stu01'))
for i in info2:# Recommend! ! ! ! ! !
print(i,info[i])
print("------")
for k,v in info2.items():# Effect 1 Sample, but inefficient
print(k,v)--------------------------------------------------------------------------------
Text:
Requirements:
Level 1.3 menu
2. You can choose to enter each submenu in turn
3. Press b to return step by step, and press q to exit
# Author : David Liu
# Be Happy!
place={
' Liaoning 1':{
' Dalian ':{
' Ganjingzi district ':[' Dalian University of Technology '],
' Shahekou District ':[' Digital Plaza ']
},
},
' Liaoning 2':{
' Shenyang ': {
' Huanggu District ': [' Liaoning University '],
' Tiexi District ': [' Grove ']
},
},
' Liaoning 3':{
' Fuxin ': {
' Haizhou District ':[' To be fat '],
' Xihe district ':[' Xihe District People's government ']
}
}
}
exit_flag=True
while exit_flag:
for i in place:
print(i)
choice=input(" Select Enter, press q Quit >>")
if choice in place:
while exit_flag:
for i2 in place[choice]:
print('\t',i2)
choice2=input(" Select Enter, press b Back, press q Quit >>")
if choice2 in place[choice]:
while exit_flag:
for i3 in place[choice][choice2]:
print('\t\t', i3)
choice3 = input(" Select Enter, press b Back, press q Quit >>")
if choice3 in place[choice][choice2]:
for i4 in place[choice][choice2][choice3]:
print('\t\t\t',i4)
choice4 = input(" Finally 1 Layer, press b Back, press q Quit >>")
if choice4=='b':
pass# Useless, occupying
elif choice4=='q':
exit_flag = False
if choice3 == 'b':
break
elif choice3 == 'q':
exit_flag = False
if choice2 == 'b':
break
elif choice2 == 'q':
exit_flag = False
elif choice == 'q':
exit_flag = False
print('\n'," Has quit! ".center(50,'-'))
ps: Let's take a look at the Python dictionary implementation level 3 menu
################################################
# Task Name: 3级菜单 #
# Description：打印省、市、县3级菜单 #
# 可返回上1级 #
# 可随时退出程序 #
#----------------------------------------------#
# Author：Oliver Lee #
################################################
zone = {
'山东' : {
'青岛' : ['4方','黄岛','崂山','李沧','城阳'],
'济南' : ['历城','槐荫','高新','长青','章丘'],
'烟台' : ['龙口','莱山','牟平','蓬莱','招远']
},
'江苏' : {
'苏州' : ['沧浪','相城','平江','吴中','昆山'],
'南京' : ['白下','秦淮','浦口','栖霞','江宁'],
'无锡' : ['崇安','南长','北塘','锡山','江阴']
},
'浙江' : {
'杭州' : ['西湖','江干','下城','上城','滨江'],
'宁波' : ['海曙','江东','江北','镇海','余姚'],
'温州' : ['鹿城','龙湾','乐清','瑞安','永嘉']
},
'安徽' : {
'合肥' : ['蜀山','庐阳','包河','经开','新站'],
'芜湖' : ['镜湖','鸠江','无为','3山','南陵'],
'蚌埠' : ['蚌山','龙子湖','淮上','怀远','固镇']
},
'广东' : {
'深圳' : ['罗湖','福田','南山','宝安','布吉'],
'广州' : ['天河','珠海','越秀','白云','黄埔'],
'东莞' : ['莞城','长安','虎门','万江','大朗']
}
}
province_list = list(zone.keys()) #省列表
# flag = False
# flag1 = False
while True:
print(" 省 ".center(50,'*'))
for i in province_list:
print(province_list.index(i)+1,i) #打印省列表
pro_id = input("请输入省编号,或输入q(quit)退出：") #省ID
if pro_id.isdigit():
pro_id = int(pro_id)
if pro_id > 0 and pro_id <= len(province_list):
pro_name = province_list[pro_id-1] #根据省ID获取省名称
city_list = list(zone[pro_name].keys()) #根据省名称获取对应的值，从新字典中获取key，即市列表
while True:
print(" 市 ".center(50,'*'))
for v in city_list:
print(city_list.index(v)+1,v) #打印市列表
city_id = input("请输入市编号,或输入b(back)返回上级菜单，或输入q(quit)退出：")
if city_id.isdigit():
city_id = int(city_id)
if city_id > 0 and city_id <= len(city_list):
city_name = city_list[city_id-1] #根据市ID获取市名称
town_list = zone[pro_name][city_name] #根据省名称获取对应的值，从新字典中获取值，即县列表
while True:
print(" 县 ".center(50,'*'))
for j in town_list:
print(town_list.index(j)+1,j)
back_or_quit = input("输入b(back)返回上级菜单，或输入q(quit)退出：")
if back_or_quit == 'b':
break #终止此层while循环，跳转到上1层While。
elif back_or_quit == 'q':
# flag1 = True
# break #根据标志位结束程序。
exit()
else:
print("输入非法！")
else:
print("编号%d不存在。"%city_id)
elif city_id == 'b':
break
elif city_id == 'q':
# flag = True
# break
exit()
else:
print("输入非法!")
# if flag1:
# break
else:
print("编号%d不存在。"%pro_id)
elif pro_id == 'q':
break
else:
print("输入非法!")
# if flag or flag1:
# breakSummarize