Python列表、元祖、字典的常用方法
#######################################灰魔法: list类中提供的方法 ####################################### # li = [11, 22, 33, 22, 44] # 参数 # 1. 原来值最后追加 # 对象.方法(..) # li对象调用append方法 # li.append(5) # li.append("alex") # li.append([1234,2323]) # print(li) # 2 清空列表 # li.clear() # print(li) # 3 拷贝,浅拷贝 # v = li.copy() # print(v) # 4. 计算元素出现的次数 # v = li.count(22) # print(v) # 5. 扩展原列表,参数:可迭代对象 # li = [11, 22, 33, 22, 44] # li.append([9898,"不得了"]) # [11, 22, 33, 22, 44, [9898, '不得了']] # li.extend([9898,"不得了"]) # for i in [9898,"不得了"]: # li.append(i) # [11, 22, 33, 22, 44, 9898, '不得了'] # # li.extend("不得了") # print(li) # 6. 根据值获取当前值索引位置(左边优先) # li = [11, 22, 33, 22, 44] # v= li.index(22) # print(v) # 7. 在指定索引位置插入元素 # li = [11, 22, 33, 22, 44] # li.insert(0,99) # print(li) # 8、 删除某个值(1.指定索引;2. 默认最后一个),并获取删除的值 # li = [11, 22, 33, 22, 44] # v = li.pop() # print(li) # print(v) # li = [11, 22, 33, 22, 44] # v = li.pop(1) # print(li) # print(v) # 9. 删除列表中的指定值,左边优先 # li = [11, 22, 33, 22, 44] # li.remove(22) # print(li) # PS: pop remove del li[0] del li[7:9] clear # 10 将当前列表进行翻转 # li = [11, 22, 33, 22, 44] # li.reverse() # print(li) # 11 列表的排序 # li = [11,44, 22, 33, 22] # li.sort() # li.sort(reverse=True) # print(li) ### 欠 # cmp # key # sorted ####################################### 深灰魔法 ####################################### # 1. 列表格式 # 2. 列表中可以嵌套任何类型 # 中括号括起来 # ,分割每个元素 # 列表中的元素可以是 数字,字符串,列表,布尔值..所有的都能放进去 # “集合”,内部放置任何东西 """ # 3. # 索引取值 print(li[3]) # 4 切片,切片结果也是列表 print(li[3:-1]) # 5 for循环 # while循环 for item in li: print(item) """ # 列表元素,可以被修改 # li = [1, 12, 9, "age", ["石振文", ["19", 10], "庞麦郎"], "alex", True] ############## 6 索引 # 修改 # li[1] = 120 # print(li) # li[1] = [11,22,33,44] # print(li) # 删除,第一种方式 # del li[1] # print(li) ############## 7 切片 # 修改 # li[1:3] = [120,90] # print(li) # 删除 # del li[2:6] # print(li) # 8 in 操作 # li = [1, 12, 9, "age", ["石振文", ["19", 10], "庞麦郎"], "alex", True] # v1 = "石振文" in li # print(v1) # v2 = "age" in li # print(v2) ###### 列表中的元素, # 9 操作 # li = [1, 12, 9, "age", ["石振文", ["19", 10], "庞麦郎"], "alex", True] # li[4][1][0] # [1] # li = [1, 12, 9, "age", ["石振文", ["19", 10], "庞麦郎"], "alex", True] # s = "pouaskdfauspdfiajsdkfj" # s = 123 # a = "123" # int(a) # a = 123 # str(a) # 10 转换 # 字符串转换列表 li = list("asdfasdfasdf"), 内部使用for循环 # s = "pouaskdfauspdfiajsdkfj" # new_li = list(s) # print(new_li) # 列表转换成字符串, # 需要自己写for循环一个一个处理: 既有数字又有字符串 # li = [11,22,33,"123","alex"] # # r = str(li) # '[11,22,33,"123","alex"]' # # print(r) # s = "" # for i in li: # s = s + str(i) # print(s) # 直接使用字符串join方法:列表中的元素只有字符串 # li = ["123","alex"] # v = "".join(li) # print(v) ### 补充:字符串创建后,不可修改 # v = "alex" # v = v.replace('l','el') # print(v) # li = [11,22,33,44] # li[0] # li[0] = 999 # s = "alex" # li[0] # s[0] = "E" # li = [11,22,33,44] # print(li) # print(li) # print(li) # print(li) # print(li) # print(li) # print(li) # print(li) # 列表,有序;元素可以被修改 # 列表 # list # li = [111,22,33,44] #################################################################################################### # 元组,元素不可被修改,不能被增加或者删除 # tuple # tu = (11,22,33,44) # tu.count(22),获取指定元素在元组中出现的次数 # tu.index(22) ####################################### 深灰魔法 ####################################### # 1. 书写格式 # tu = (111,"alex",(11,22),[(33,44)],True,33,44,) # 一般写元组的时候,推荐在最后加入 , # 元素不可被修改,不能被增加或者删除 # 2. 索引 # v = tu[0] # print(v) # 3. 切片 # v = tu[0:2] # print(v) # 4. 可以被for循环,可迭代对象 # for item in tu: # print(item) # 5. 转换 # s = "asdfasdf0" # li = ["asdf","asdfasdf"] # tu = ("asdf","asdf") # # v = tuple(s) # print(v) # v = tuple(li) # print(v) # v = list(tu) # print(v) # v = "_".join(tu) # print(v) # li = ["asdf","asdfasdf"] # li.extend((11,22,33,)) # print(li) # 6.元组的一级元素不可修改/删除/增加 # tu = (111,"alex",(11,22),[(33,44)],True,33,44,) # # 元组,有序。 # # v = tu[3][0][0] # # print(v) # # v=tu[3] # # print(v) # tu[3][0] = 567 # print(tu) #################################################################################################### # 字典 # dict # dict # dic = { # "k1": 'v1', # "k2": 'v2' # } # 1 根据序列,创建字典,并指定统一的值 # v = dict.fromkeys(["k1",123,"999"],123) # print(v) # 2 根据Key获取值,key不存在时,可以指定默认值(None) # v = dic['k11111'] # print(v) # v = dic.get('k1',111111) # print(v) # 3 删除并获取值 # dic = { # "k1": 'v1', # "k2": 'v2' # } # v = dic.pop('k1',90) # print(dic,v) # k,v = dic.popitem() # print(dic,k,v) # 4 设置值, # 已存在,不设置,获取当前key对应的值 # 不存在,设置,获取当前key对应的值 # dic = { # "k1": 'v1', # "k2": 'v2' # } # v = dic.setdefault('k1111','123') # print(dic,v) # 5 更新 # dic = { # "k1": 'v1', # "k2": 'v2' # } # dic.update({'k1': '111111','k3': 123}) # print(dic) # dic.update(k1=123,k3=345,k5="asdf") # print(dic) # 6 keys() 7 values() 8 items() get update ########## # 1、基本机构 # info = { # "k1": "v1", # 键值对 # "k2": "v2" # } #### 2 字典的value可以是任何值 # info = { # "k1": 18, # "k2": True, # "k3": [ # 11, # [], # (), # 22, # 33, # { # 'kk1': 'vv1', # 'kk2': 'vv2', # 'kk3': (11,22), # } # ], # "k4": (11,22,33,44) # } # print(info) #### 3 布尔值(1,0)、列表、字典不能作为字典的key # info ={ # 1: 'asdf', # "k1": 'asdf', # True: "123", # # [11,22]: 123 # (11,22): 123, # # {'k1':'v1'}: 123 # # } # print(info) # 4 字典无序 # info = { # "k1": 18, # "k2": True, # "k3": [ # 11, # [], # (), # 22, # 33, # { # 'kk1': 'vv1', # 'kk2': 'vv2', # 'kk3': (11,22), # } # ], # "k4": (11,22,33,44) # } # print(info) # 5、索引方式找到指定元素 # info = { # "k1": 18, # 2: True, # "k3": [ # 11, # [], # (), # 22, # 33, # { # 'kk1': 'vv1', # 'kk2': 'vv2', # 'kk3': (11,22), # } # ], # "k4": (11,22,33,44) # } # # v = info['k1'] # # print(v) # # v = info[2] # # print(v) # v = info['k3'][5]['kk3'][0] # print(v) # 6 字典支持 del 删除 # info = { # "k1": 18, # 2: True, # "k3": [ # 11, # [], # (), # 22, # 33, # { # 'kk1': 'vv1', # 'kk2': 'vv2', # 'kk3': (11,22), # } # ], # "k4": (11,22,33,44) # } # del info['k1'] # # del info['k3'][5]['kk1'] # print(info) # 7 for循环 # dict # info = { # "k1": 18, # 2: True, # "k3": [ # 11, # [], # (), # 22, # 33, # { # 'kk1': 'vv1', # 'kk2': 'vv2', # 'kk3': (11,22), # } # ], # "k4": (11,22,33,44) # } # for item in info: # print(item) # # for item in info.keys(): # print(item) # for item in info.values(): # print(item) # for item in info.keys(): # print(item,info[item]) # for k,v in info.items(): # print(k,v) # True 1 False 0 # info ={ # "k1": 'asdf', # True: "123", # # [11,22]: 123 # (11,22): 123, # # {'k1':' v1'}: 123 # # } # print(info)