Dictionary

2023/03/26
李慕庭

函數

\(f(x) = y\)
一個\(x\)只能對到一個\(y\)
\(x\) 不重複

Dictionary

\(D(key) = value\)
一個\(key\)只能對到一個\(value\)
\(key\)不重複

dictionary 中每個元素都由key & value 組成
key只能是不變的資料型態 :
- string, value, tuple..
value可以是任何的資料型態 :
- string, value, list, another dictionary…

建立dictionary

 score = {"Fireball": 100, "Cookie": 60, "Chocolate":20}
 
print(type(score))
print(score)

 cookies = dict(Chocolate=1, Mocha=2)
print(cookies)

取得dictionary中的特定元素

傳入 \(key\) 值
和list, string tuple 傳入位置index不同

 score = {"Fireball":100, "Cookie":60, "Chocolate":20}
 
print(score["Fireball"])
print(score["Cookie"])
print(score["Chocolate"])

會有什麼問題？

key error

 score = {"Fireball":100, "Cookie":60, "Chocolate":20}
 
print(score["Mocha"])

How to solve key error - if

 score = {"Fireball":100, "Cookie":60, "Chocolate":20}
 
if "Mocha" in score:
    print(score["Mocha"])

How to solve key error - get()

如果找不到這個key值就會回傳None
也可以自訂找不到回傳的值

 score = {"Fireball":100, "Cookie":60, "Chocolate":20}
 
print(score.get("Cookie"))   # 60
print(score.get("Mocha"))    # None
print(score.get("Mocha", 0)) # 0

遍歷所有dictionary中的元素

在list中

 rank = [1, 3, 4, 6, 8]
 
for i in range(5):
    print(rank[i])
    
for i in rank:
    print(i)

在dictionary中如果照做？

會只取到key值
因此底下的程式碼只會輸出

Fireball
Cookie
Chocolate

 score = {"Fireball":100, "Cookie":60, "Chocolate":20}
 
for k in score:
    print(k)

用item來遍歷!

 score = {"Fireball":100, "Cookie":60, "Chocolate":20}
 
for k in score.items():
    print(k)

Output

回傳的元素型態是tuple
包含了一個key跟一個value

('Fireball', 100)
('Cookie', 60)
('Chocolate', 20)

tuple

tuple 的特性

不可變動 – 因此做為dictionary的\(key\)
所秏儲存空間較少

建立tuple

 score = 1, 2, 3
 
print(score)

如果想建立只有一個元素的tuple

 score1 = 1
print(score1)
 
score2 = 1,
print(score2)

也可以直接加上小括號來設tuple

 score = (1, 2, 3)
print(score)

甚至是空的tuple

 nothing = ()
 
print(nothing)
print(type(nothing))

將其他資料型態轉換為tuple (string)

 name = "Fireball"
tuple_name = tuple(name)
 
print(type(tuple_name))
print(tuple_name)

將其他資料型態轉換為tuple (list)

 score = [1, 2, 3]
tuple_score = tuple(score)
 
print(type(tuple_score))
print(tuple_score)

新增元素至dictionary

 score = {"Fireball":100, "Cookie":60}
print(score)
 
score["Chocolate"] = 20
print(score)

更改dictionray中key對應到的值

 score = {"Fireball":100, "Cookie":60, "Chocolate":20}
print(score)
 
score["Fireball"] = 0
print(score)

刪除dictionary中的特定元素

 score = {"Fireball":100, "Cookie":60, "Chocolate":20}
print(score)
 
del score["Fireball"]
print(score)

如果刪到不存在的key值

 score = {"Fireball":100, "Cookie":60, "Chocolate":20}
 
del score["Mocha"]

也可以一次刪掉所有的元素

 score = {"Fireball":100, "Cookie":60, "Chocolate":20}
 
score.clear()

跟list哪裡不一樣

存取元素不是依照順序/index，而是key值
不像list能按照value來刪除

其他用法

輸出dictionary中元素數量

 score = {"Fireball":100, "Cookie":60, "Chocolate":20}
 
print(len(score))

輸出所有的key值

 score = {"Fireball":100, "Cookie":60, "Chocolate":20}
 
print(score.keys())

輸出所有的value值

 score = {"Fireball":100, "Cookie":60, "Chocolate":20}
 
print(score.values())

輸出所有的元素

 score = {"Fireball":100, "Cookie":60, "Chocolate":20}
 
print(score.items())

所有元素按key排序輸出

 score = {"Fireball":100, "Mocha":60, "Chocolate":20}
 
print(sorted(score.items()))
 
for k in sorted(score.items()):
    print(k)

所有元素按value排序輸出

 score = {"Fireball":100, "Mocha":60, "Chocolate":20}
 
print(sorted(score.items(), key=lambda x:x[1]))
 
for k in sorted(score.items(), key=lambda x:x[1]):
    print(k)

課後練習題

NEOJ 3035 - 有好心的程式碼提示

Dictionary 2023/03/26 李慕庭

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