(一)字典(無序,能夠修改,鍵不能夠重複,值能夠重複)app
每對鍵值組合稱爲項,字典跟序列不一樣,序列講究順序,字典講究映射,不講順序。three
注意:字典的鍵必須是獨一無二,而值能夠多種數據類型,可是鍵取自不可改變的數據類型(如字符串、數值型或元組)。鍵值惟一,值不惟一。ip
1.字典的建立(至少五中方法)rem
a = dict(one=1,two=2,three=3,four="肆")字符串
b = {"one":1,"two":"貳","three":3,"four":4}get
c = dict(zip(["one","two","three","four","five"],["壹","貳","叄","肆","伍"]))it
d = dict([("two",2),("one",1),("three",3),("four,4")]) 或d = dict((("two",2),("one",1),("three",3),("four,4")))date
e = dict({"three":3,"two":"貳","one":1})遍歷
g = dict.fromkeys(seq[,val]) 建立一個新字典,以序列seq中元素,作字典,val作值數據類型
2.字典的訪問
建立一個空子典 dict1 = {}
dict2 = dict.fromkeys(range(32),"贊")
print(dict2.keys()) #輸出全部鍵組成的列表
print(dict2.values()) #輸出全部組成的列表
print(dict2.items()) #輸出全部的鍵值對組成的列表
3.獲取內容
dict3 = {}
dict4 = dict3.fromkeys(range(5),"贊")
print(dict4.get(4)) #輸出0~4都是贊,而其餘值都是None
print(dict4.get(10),"木有") #給不存在的鍵賦值默認值
4.清空字典clear()刪除字典內全部元素
dict5 = {"name":"張三","sex":"男","age":18}
print(dict5.clear())
5.刪除字典
dict6 = {"name":"李麗","sex":"女","age":18}
del dict6
dict6.pop(name) #經過指定鍵來鍵對應的值,同時顯示刪除鍵所對應的值
dict6.popitem() #隨機刪除一項鍵值對
6.拷貝字典,返回一個字典的淺複製
dict6 = {"name":"李麗","sex":"女","age":18}
dict7 = dict6.copy()
print(dict7)
7.更新字典
dict8 = {"米奇":"老鼠","湯姆":"貓","小白":"豬"}
dict8.update(小白="狗")
注意:當一個字典中有重複的鍵所對應的值時,後面的健值對會替換掉前面的健值對
8.統計字典的長度
dict8 = {"米奇":"老鼠","湯姆":"貓","小白":"豬"}
len(dict8)
str(dict8) 以字符串形式輸出字典
print("value:",dict8.__contains__("米奇")) 判斷鍵是否在字典內,若是在返回True,不在返回False
9.遍歷嵌套字典
def dict9_list(d,l):
for x in d.keys():
if type(d[x]) == dict:
dict9_list(di[x],l)
else:
l.append(d[x])
d = {1:"a",2:"b",3:{4:"c",5:"d",6:{7:"e"}},8:"f"}
l = []
dict9_list(d,l)
print(l)
(二)集合,你是個人惟一,個人心中只有你(無序,不能夠修改,不能夠重複)(能夠用做去重元素)
1.集合建立
set1 ={1,2,3,4,5,6}
set2 = set([1,2,3,4,5,6])
set1 == set2
2.訪問集合
set3 = {1,2,3,4,5,6,3,7,8,9,9}
for each in set3:
print(each,end=" ")
能夠使用 in和not in 來判斷一個元素是否在集合中已經存在
>>>2 in set3 #True >>>2 not in set3 #False
使用add()方法能夠給集合添加元素,使用remove()方法刪除集合中已知的元素
>>>set3.add(11)
>>>set3
{1,2,3,4,5,6,7,8,9,11}
>>>set3.remove(5)
>>>set3
{1, 2, 3, 4, 6, 7, 8, 9}
3.不可變集合(冰凍集合frozenset())
>>>set4 = frozenset({1,2,3,4,5})
>>>set4.add(6)
報錯:由於冰凍集合沒有add屬性