Python 字典（dict）操作基础

曾经终败给现在 2021-12-16 08:34 595阅读 0赞

# Python 字典与结构化数据 #

### 1.字典基本结构 ###

#### 1.1 创建存储生日的字典 ####

birthday = {
         'Alice':'Ar 1','Bob':'July 7'}

#### 1.2 查询生日的小程序 ####

while True:
        print('Enter a name(blank to quit):')
        name = input()
        if name == '':
            break
    
        if name in birthday:
            print(birthday[name] + 'is the birthday of ' + name)
        else:
            print("I don't know the birthday of " + name)
            print("Tell me plz")
            newbday = input()
            birthday[name] = newbday
            print('Birthday database updated')

Enter a name(blank to quit):
    Alice
    Ar 1is the birthday of Alice
    Enter a name(blank to quit):
    Bob
    July 7is the birthday of Bob
    Enter a name(blank to quit):
    Chen
    I don't know the birthday of Chen
    Tell me plz
    July 22th
    Birthday database updated
    Enter a name(blank to quit):
    Chen
    July 22this the birthday of Chen
    Enter a name(blank to quit):
    
    I don't know the birthday of  
    Tell me plz
    
    Birthday database updated
    Enter a name(blank to quit):

birthday

{'Alice': 'Ar 1', 'Bob': 'July 7', 'Chen': 'July 22th'}

#字典可以修改value,那么，怎么修改key
    birthday['Alice'] = 'Apr 1st'
    birthday['Bob'] = 'July 7th'
    birthday

{'Alice': 'Apr 1st', 'Bob': 'July 7th', 'Chen': 'July 22th'}

### 2. 字典的方法 ###

#### 2.1 **字典的 values() , keys() ,items()** ####

for v in birthday.values():
        print (v)
    for v in birthday.keys():
        print(v)
    for v in birthday.items():
        print(v)

Apr 1st
    July 7th
    July 22th
    Alice
    Bob
    Chen
    ('Alice', 'Apr 1st')
    ('Bob', 'July 7th')
    ('Chen', 'July 22th')

'Alice' in birthday
    'Apr 1st' in birthday.values()
    'Apr 1st' in birthday.keys()

False

#### 2.2 **get方法** ####

有对应键，返回键对应值；没有对应键存在，返回备用值

print('I think Rachel was born on '+ birthday.get('Rachel','a rainy day'))

I think Rachel was born on a rainy day

#### 2.3 **setdafault() 方法** ####

为键赋予默认取值，返回 键的当前值；

setdefault(‘keys’,’default’)

# 字符统计程序
    message = 'In the preceding example, the kernel is chosen \ because it represents an expansion of polynomials and can conveniently compute \ high-dimensional inner products. In this example the kernel is chosen \ because of its functional form in the representation '
    
    count = {}
    for char in message:
        count.setdefault(char,0)
        count[char] = count[char] + 1
    
    print (count)

{'I': 2, 'n': 26, ' ': 37, 't': 14, 'h': 9, 'e': 32, 'p': 9, 'r': 10, 'c': 10, 'd': 4, 'i': 16, 'g': 2, 'x': 3, 'a': 12, 'm': 6, 'l': 9, ',': 1, 'k': 2, 's': 15, 'o': 14, 'b': 2, 'u': 5, 'f': 4, 'y': 2, 'v': 1, '-': 1, '.': 1}

#### 2.4 **漂亮打印 pprint 模块** ####

import pprint
    pprint.pprint(count)
    
    # 等价于 print(pprint.pformat(count))

{' ': 37,
     ',': 1,
     '-': 1,
     '.': 1,
     'I': 2,
     'a': 12,
     'b': 2,
     'c': 10,
     'd': 4,
     'e': 32,
     'f': 4,
     'g': 2,
     'h': 9,
     'i': 16,
     'k': 2,
     'l': 9,
     'm': 6,
     'n': 26,
     'o': 14,
     'p': 9,
     'r': 10,
     's': 15,
     't': 14,
     'u': 5,
     'v': 1,
     'x': 3,
     'y': 2}

### 3.小练习 ###

#### 3.1 **井字棋** ####

theBoard = {
         'top-l':'' , 'top-m':'' , 'top-r':'',
                'mid-l':'' , 'mid-m':'' , 'mid-r':'',
                'low-l':'' , 'low-m':'' , 'low-r':''}
    def printBoard(board):
        print(board['top-l'] + ' | ' + board['top-m'] + '| '+board['top-r'])
        print('-+-+-')
        print(board['mid-l'] + ' | ' + board['mid-m'] + '| '+board['mid-r'])
        print('-+-+-')
        print(board['low-l'] + ' | ' + board['low-m'] + '| '+board['low-r'])
        print('-+-+-')
    
    def gameStart():
        printBoard(theBoard)
        i = 0
        while True:
            chess = ['O','X']
    
            print('where do you wanna place')
            position = input()
            if theBoard[position]:
                print('already full')
                print('Another choice plz')
                position = input()
    
            theBoard[position] = chess[i]
            printBoard(theBoard)
            i = 1 - i
            #判断棋盘是否已满，笨办法
            k = 0
            for value in theBoard.values():
                if value:
                    k = k+1
    
            if k >=9 :
                print('All full,the end')
                break

#### 3.2 **嵌套的字典和列表** ####

*  对复杂事物进行建模，字典列表中可能包含其他字典列表；
 *  列表适合包含一组有序的值，字典适合包含关联的键和值

#示例程序：统计野餐事物总数
    
    allGuests = {
         'Ali':{
         'apple':16,'bread':12},
                 'Bob':{
         'ham':1,'coffee':32,'apple':3},
                'Carol':{
         'bread':23,'coffee':2}}
    
    def totalBrought(guests,item):
        numBrought = 0
        for k,v in guests.items():                   #字典中item的循环 k,v
            numBrought = numBrought + v.get(item,0)
        return numBrought
    
    print('Number of things being brought:')
    print('- apples :'+ str(totalBrought(allGuests,'apple')))
    print('- bread ：' + str(totalBrought(allGuests,'bread')))
    print('- coffee ：'+ str(totalBrought(allGuests,'coffee')))

Number of things being brought:
    - apples :19
    - bread ：35
    - coffee ：34

#### 3.3 **好玩游戏的物品清单** ####

stuff = {
         'rope':5,'torch':2,'gold':100,'sword':1}
    
    def inventoryDisplay(stuff):
        numTotal = 0
        print('Inventory:')
        for k,v in stuff.items():
            print( k +':'+ str(v))
            numTotal = numTotal + v
        print ('total:'+ str(numTotal))
    
    #屠龙战利品 以列表形式存储
    dragonLoot = ['rope','rope','torch','gold','gold','gold','diamond']
    
    #列表到字典的函数
    def addToInventory(loot,stuff):
        countLoot = {}
        for k in loot:
            countLoot.setdefault(k,0)
            countLoot[k] = countLoot[k] + 1
        for l in countLoot.keys():
            stuff.setdefault(l,0)
            stuff[l] = stuff[l] + countLoot[l]
        return stuff
    
    
    inventoryDisplay(stuff)
    print(addToInventory(dragonLoot,stuff))

Inventory:
    rope:5
    torch:2
    gold:100
    sword:1
    total:108
    {'rope': 7, 'torch': 3, 'gold': 103, 'sword': 1, 'diamond': 1}

> 参考文献  
> 《Python编程快速上手–让繁琐工作自动化》

转载于:https://www.cnblogs.com/blogofnickchen/p/7221634.html