python入門學習篇七
阿新 • • 發佈:2021-12-30
列表的內建方法
# l1 = [11, 22, 33, 44, 55, 66] # l1.reverse() # print(l1) # l1 = [99, 11, 22, 33, 44, 55, 66] # l1.sort(reverse=True) # 預設是升序排列 # print(l1) # 列表比較 # l1 = [999, 888] # l2 = [111, 222, 333, 444, 555] # print(l1 > l2) # 列表切片 l = [1, 2, 3, 4, 5, 6, 7, 8, 9] # print(l[1]) # print(l[1:5])# print(l[1:]) # 冒號右邊不寫,代表從開始位置一直切到末尾 # print(l[:5]) # 冒號左邊不寫,代表從頭開始一直切斷索引指定位置 # print(l[1:8:2]) # # print(l[-1]) # -1位置取得就是末尾資料 # print(l[-8:-1:2]) # print(l[::-1]) # 冒號左右兩邊都不寫,代表全都要 # s = 'helloworld' # print(s[1:]) # print(s[:6])
字典的內建方法
# 1. 定義字典 d = {'username': 'ly', 'age': 12} # 2. 第二種方式dict# d1 = dict(name='ly', age=18, gender='male') # print(d1) # 瞭解 # info = dict([['name', 'tony'], ('age', 18)]) # print(info) # dic = { # 'name': 'xxx', # 'age': 18, # 'hobbies': ['play game', 'basketball'] # } # 1. 取值 # print(dic['name1111']) # print(dic['hobbies']) # 2. 第二種方式, 掌握 # print(dic.get('name1111', 666)) # None# print(dic.get('name', 666)) # None # print(dic.get('hobbies')) # 3. 修改值 # dic['name'] = 'ly' # k值 存在,直接進行修改操作 # dic['pwd'] = '123456' # k值不存在,會往字典中新增一個k:v # print(dic) # l = [1, 2, 3] # 0-2 # # l[4] = 666 # print(l[4]) # 4. 求長度 # l = [1, 2, 3, 4, 5, 6] # print(len(l)) # print(len(dic)) # 5. 成員運算 # print('name' in dic) # print('name' not in dic) # 6. 刪除 # 第一種方式 # del dic['name'] # del dic['hobbies'] # print(dic) # 第二種方式 # dic.pop('name') # dic.pop('hobbies') # print(dic) dic = { 'name': 'xxx', 'age': 18, 'hobbies': ['play game', 'basketball'] } # 7. 字典三劍客 # print(dic.keys()) # dict_keys(['name', 'age', 'hobbies']) => 列表 # print(dic.values()) # dict_values(['xxx', 18, ['play game', 'basketball']]) => 列表 # print(dic.items()) # dict_items([('name', 'xxx'), ('age', 18), ('hobbies', ['play game', 'basketball'])]) # 8. 迴圈字典 # for i in dic: # print(i) # print(dic[i]) # print(dic.get(i)) # [('name', 'xxx'), ('age', 18), ('hobbies', ['play game', 'basketball'])] # k,v = ('name', 'xxx') # for k, v in dic.items(): # print(k, v) for i in dic.keys(): print(i) for j in dic.values(): print(j) 字典需要了解的方法 dic = { 'name': 'xxx', 'age': 18, 'hobbies': ['play game', 'basketball'] } # print(dic.popitem()) # print(dic) # dic1 = { # 'name':'aaa' # } # dic1.update({'name': 'ly', 'pwd': 123}) # print(dic1) # dic1 = dict.fromkeys(['k1', 'k2', 'k3'], []) # print(dic1) # dic1['k1'] = [] # dic1['k1'].append(666) # # dic1['k1'].append(777) # # dic1['k1'].append(888) # print(dic1) # setdefault # print(dic.setdefault('name1111', 666)) # print(dic)
元組的內建方法
# 1.型別轉換 關鍵字:tuple tuple(111) # 不行 tuple(1.11) # 不行 tuple('helloworld') # 行 ... # 支援for迴圈的資料型別都可以轉為元組 # 第一道筆試題: t1 = (111) t2 = (1.22) t3 = ('helloworld') t4 = ('a', 'b') t5 = ('c', ) '''當元組中只有一個元素的時候,也要加逗號''' print(type(t1)) # <class 'int'> print(type(t2)) # <class 'float'> print(type(t3)) # <class 'str'> print(type(t4)) # <class 'tuple'> print(type(t5)) # <class 'tuple'> # 求長度 len(tuple1) # 第二道筆試題 t1 = (111, 222, [444, 555, 666]) # print(t1) # print(t1[2]) t1[2].append(777) print(t1)
集合的內建方法
而集合型別既沒有索引也沒有key與值對應,所以無法取得單個的值,而且對於集合來說,主要用於去重與關係元素,根本沒有取出單個指定值這種需 d = {} # 預設是空字典 s = set() # 這才是定義空集合 # 第二道題:去重,並且保留原來的順序 ll = [11, 22, 22, 22, 33, 33, 44, 44, 55, 66, 77, 77, 88] # ss = set(ll) # l1 = list(ss) # print(l1) new_list = [] for i in ll: if i not in new_list: new_list.append(i) print(new_list)
集合的運算關係
friends1 = {"zero", "kevin", "jason", "egon"} friends2 = {"Jy", "ricky", "jason", "egon"} # 求合集,並集 print(friends1 | friends2) # 求交集 print(friends1 & friends2) # 求friends1差集 print(friends1 - friends2) # 求friends2差集 print(friends2 - friends1) # 求對稱差集 print(friends1 ^ friends2) # 求父集,子集 print(friends1 > friends2) print(friends1 < friends2)
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