day06数据类型(中)

1. 列表

列表(list), 是一个有序可变的容器,在里面可以存放多个不同类型的元素。

1.1 定义

user_list =  ["苍","有","大"]
number_list = [98,88,666,12,-1]
data_list = [1,True,"Alex","宝强","贾乃亮"]
user_list = []
user_list.append("铁锤")
user_list.append(123)
user_list.append(True)
print(user_list) # ["铁锤",123,True]

不可变类型:字符串、布尔、整型(已最小,内部数据无法进行修改)

可变类型:列表(内部数据元素可以修改)

1.2 独有功能

Python中为所有的列表类型的数据提供了一批独有的功能。

在开始学习列表的独有功能之前,先来做一个字符串和列表的对比:

  • 字符串,不可变,即:创建好之后内部就无法修改。【独有功能都是新创建一份数据】
name = "alex"
data = name.upper()
print(name)
print(data)
  • 列表,可变,即:创建好之后内部元素可以修改。【独有功能基本上都是直接操作列表内部,不会创建新的一份数据】
user_list = ["车子","妹子"]
user_list.append("嫂子")

print(user_list) # ["车子","妹子","嫂子"

列表中的常见独有功能如下:

  1. 追加,在原列表中尾部追加值。
data_list = []

v1 = input("请输入姓名")
data_list.append(v1)

v2 = input("请输入姓名")
data_list.append(v2)

print(data_list) # ["alex","eric"]
# 案例1
user_list = []

while True:
    user = input("请输入用户名(Q退出):")
    if user == "Q":
        break
    user_list.append(user)
    
print(user_list)
welcome = "欢迎使用NB游戏".center(30, '*')
print(welcome)

user_count = 0
while True:
    count = input("请输入游戏人数:")
    if count.isdecimal():
        user_count = int(count)
        break
    else:
        print("输入格式错误,人数必须是数字。")


message = "{}人参加游戏NB游戏。".format(user_count)
print(message)


user_name_list = []

for i in range(1, user_count + 1):
    tips = "请输入玩家姓名({}/{}):".format(i, user_count)
    name = input(tips)
    user_name_list.append(name)

print(user_name_list)
  1. 批量追加,将一个列表中的元素逐一添加另外一个列表。
tools = ["搬砖","菜刀","榔头"]
tools.extend( [11,22,33] ) # weapon中的值逐一追加到tools中
print(tools) # ["搬砖","菜刀","榔头",11,22,33]
tools = ["搬砖","菜刀","榔头"]
weapon = ["AK47","M6"]
#tools.extend(weapon) # weapon中的值逐一追加到tools中
#print(tools) # ["搬砖","菜刀","榔头","AK47","M6"]

weapon.extend(tools)
print(tools) # ["搬砖","菜刀","榔头"]
print(weapon) # ["AK47","M6","搬砖","菜刀","榔头"]
# 等价于(扩展)
weapon = ["AK47","M6"]
for item in weapon:
    print(item)

# 输出:
#  AK47
#  M6
tools = ["搬砖","菜刀","榔头"]
weapon = ["AK47","M6"]
for item in weapon:
    tools.append(item)  
print(tools) # ["搬砖","菜刀","榔头","AK47","M6"]
  1. 插入,在原列表的指定索引位置插入值
user_list = ["苍","有",""]
user_list.insert(0,"马蓉")
user_list.insert(2,"李小璐")
print(user_list)
# 案例
name_list = []
while True:
    name = input("请输入购买火车票用户姓名(Q/q退出):")
    if name.upper() == "Q":
        break
    if name.startswith("刁"):
        name_list.insert(0, name)
    else:
        name_list.append(name)
print(name_list)
  1. 在原列表中根据值删除(从左到右找到第一个删除)【慎用,里面没有会报错】
user_list = ["王宝强","陈羽凡","Alex","贾乃亮","Alex"]
user_list.remove("Alex")
print(user_list)


user_list = ["王宝强","陈羽凡","Alex","贾乃亮","Alex"]
if "Alex" in user_list:
	user_list.remove("Alex")
print(user_list)


user_list = ["王宝强","陈羽凡","Alex","贾乃亮","Alex"]
while True:
    if "Alex" in user_list:
        user_list.remove("Alex")
	else:
        break
print(user_list)
# 案例:自动抽奖程序
import random  # random.choice()参数为列表

data_list = ["iphone12", "二手充气女友", "大保健一次", "泰国5日游", "避孕套"]

while data_list:
    name = input("自动抽奖程序,请输入自己的姓名:")

    # 随机从data_list抽取一个值出来
    value = random.choice(data_list) # "二手充气女友"
    print( "恭喜{},抽中{}.".format(name, value) )
    
    data_list.remove(value) # "二手充气女友"
  1. 在原列表中根据索引踢出某个元素(根据索引位置删除)
user_list = ["王宝强","陈羽凡","Alex","贾乃亮","Alex"]
#               0       1      2      3       4
user_list.pop(1)
print(user_list) #  ["王宝强","Alex","贾乃亮","Alex"]

user_list.pop()
print(user_list) # ["王宝强","Alex","贾乃亮"]

item = user_list.pop(1)
print(item) # "Alex"
print(user_list) # ["王宝强","贾乃亮"]
# 案例:排队买火车票

# ["alex","李杰","eric","武沛齐","老妖","钢蛋"]
user_queue = []

while True:
    name = input("北京~上海火车票,购买请输入姓名排队(Q退出):")
    if name == "Q":
        break
    user_queue.append(name)

ticket_count = 3
for i in range(ticket_count):
    username = user_queue.pop(0)
    message = "恭喜{},购买火车票成功。".format(username)
    print(message)

# user_queue = ["武沛齐","老妖","钢蛋"]
faild_user = "、".join(user_queue) # "武沛齐、老妖、钢蛋"
faild_message = "非常抱歉,票已售完,以下几位用户请选择其他出行方式,名单:{}。".format(faild_user)
print(faild_message)
  1. 清空原列表
user_list = ["王宝强","陈羽凡","Alex","贾乃亮","Alex"]
user_list.clear()
print(user_list) # []
  1. 根据值获取索引(从左到右找到第一个删除)【慎用,找不到报错】
user_list = ["王宝强","陈羽凡","Alex","贾乃亮","Alex"]
#               0       1      2       3      4
if "Alex" in user_list:
	index = user_list.index("Alex")
	print(index) # 2
else:
    print("不存在")
  1. 列表元素排序
# 数字排序
num_list = [11, 22, 4, 5, 11, 99, 88]
print(num_list)
num_list.sort()  # 让num_list从小到大排序
num_list.sort(reverse=True)  # # 让num_list从大到小排序
print(num_list)


# 字符串排序
user_list = ["王宝强", "Ab陈羽凡", "Alex", "贾乃亮", "贾乃", "1"]
#       [29579, 23453, 24378]
#       [65, 98, 38472, 32701, 20961]
#       [65, 108, 101, 120]
#       [49]
print(user_list)
"""
sort的排序原理
    [ "x x x" ," x x x x x " ]
"""
user_list.sort()
print(user_list)

注意:排序时内部元素无法进行比较时,程序会报错(尽量数据类型统一)

  1. 反转列表
user_list = ["王宝强","陈羽凡","Alex","贾乃亮","Alex"]
user_list.reverse()

print(user_list) # ['Alex', '贾乃亮', 'Alex', '陈羽凡', '王宝强']

1.3 公共功能

  1. 相加,两个列表相加获取生成一个新的列表。
data = ["赵四","刘能"] + ["宋晓峰","范德彪"]
print(data) # ["赵四","刘能","宋晓峰","范德彪"]

v1 = ["赵四","刘能"]
v2 = ["宋晓峰","范德彪"]
v3 = v1 + v2
print(v3) # ["赵四","刘能","宋晓峰","范德彪"]
  1. 相乘,列表*整型 将列表中的元素再创建N份并生成一个新的列表
data = ["赵四","刘能"] * 2
print(data) # ["赵四","刘能","赵四","刘能"]

v1 = ["赵四","刘能"]
v2 = v1 * 2
print(v1) # ["赵四","刘能"]
print(v2) # ["赵四","刘能","赵四","刘能"]
  1. 运算符in包含
    由于列表内部是由多个元素组成,可以通过in来判断元素是否在列表中。
user_list = ["狗子","二蛋","沙雕","alex"] 
result = "alex" in user_list
# result = "alex" not in user_list
print(result) #  True

if "alex" in user_list:
    print("在,把他删除")
    user_list.remove("alex")
else:
    print("不在")
user_list = ["狗子","二蛋","沙雕","alex"] 
if "alex" in user_list:
    print("在,把他删除")
    user_list.remove("alex")
else:
    print("不在")
# 案例
user_list = ["狗子","二蛋","沙雕","alex"] 
if "alex" in user_list:
    print("在,把他删除")
    user_list.remove("alex")
else:
    print("不在")
# 案例
user_list = ["王宝强","陈羽凡","Alex","贾乃亮","Alex"]
if "Alex" in user_list:
	index = user_list.index("Alex")
	user_list.pop(index)
# 案例:敏感词替换
text = input("请输入文本内容:") # 按时打发第三方科技爱普生豆腐啊;了深刻的房价破阿偶打飞机
forbidden_list = ["草","欧美","日韩"]
for item in forbidden_list:
    text = text.replace(item,"**")
print(text)

注意:列表检查元素是否存在时,是采用逐一比较的方式,效率会比较低。

  1. 获取长度
user_list = ["范德彪","刘华强",'尼古拉斯赵四']
print( len(user_list) )
  1. 索引
# 读
user_list = ["范德彪","刘华强",'尼古拉斯赵四']
print( user_list[0] )
print( user_list[2] )
print( user_list[3] ) # 报错
# 改
user_list = ["范德彪","刘华强",'尼古拉斯赵四']
user_list[0] = "武沛齐"
print(user_list) # ["武沛齐","刘华强",'尼古拉斯赵四']
# 删
user_list = ["范德彪","刘华强",'尼古拉斯赵四']
del user_list[1]

user_list.remove("刘华强")
ele = user_list.pop(1)

注意:超出索引范围会报错。
提示:由于字符串是不可变类型,所以他只有索引读的功能,而列表可以进行 读、改、删

  1. 切片,多个元素的操作(很少用)
# 读
user_list = ["范德彪","刘华强",'尼古拉斯赵四']

print( user_list[0:2] ) # ["范德彪","刘华强"]
print( user_list[1:] )  # ["刘华强",'尼古拉斯赵四']
print( user_list[:-1] ) # ["范德彪","刘华强"]
# 改
user_list = ["范德彪", "刘华强", '尼古拉斯赵四']
user_list[0:2] = [11, 22, 33, 44]
print(user_list) # 输出 [11, 22, 33, 44, '尼古拉斯赵四']

user_list = ["范德彪", "刘华强", '尼古拉斯赵四']
user_list[2:] = [11, 22, 33, 44]
print(user_list) # 输出 ['范德彪', '刘华强', 11, 22, 33, 44]

user_list = ["范德彪", "刘华强", '尼古拉斯赵四']
user_list[3:] = [11, 22, 33, 44]
print(user_list) # 输出 ['范德彪', '刘华强', '尼古拉斯赵四', 11, 22, 33, 44]


user_list = ["范德彪", "刘华强", '尼古拉斯赵四']
user_list[10000:] = [11, 22, 33, 44]
print(user_list) # 输出 ['范德彪', '刘华强', '尼古拉斯赵四', 11, 22, 33, 44]


user_list = ["范德彪", "刘华强", '尼古拉斯赵四']
user_list[-10000:1] = [11, 22, 33, 44]
print(user_list) # 输出 [11, 22, 33, 44, '刘华强', '尼古拉斯赵四']
# 删
user_list = ["范德彪", "刘华强", '尼古拉斯赵四']
del user_list[1:]
print(user_list) # 输出 ['范德彪']
  1. 步长
user_list = ["范德彪","刘华强",'尼古拉斯赵四',"宋小宝","刘能"]
#              0        1        2          3       4
print( user_list[1:4:2] )  # ["刘华强", "宋小宝"]
print( user_list[0::2] )   # ["范德彪", '尼古拉斯赵四', "刘能"]
print( user_list[1::2] )   # ["刘华强","宋小宝"]
print( user_list[4:1:-1] ) # ["刘能", "宋小宝", '尼古拉斯赵四']
# 案例:实现列表的翻转
user_list = ["范德彪","刘华强",'尼古拉斯赵四',"宋小宝","刘能"]
new_data = user_list[::-1]
print(new_data)


data_list = ["范德彪","刘华强",'尼古拉斯赵四',"宋小宝","刘能"]
data_list.reverse()
print(data_list)

# 给你一个字符串请实现字符串的翻转?
name = "刘小伟"
name[::-1]
  1. for循环
user_list = ["范德彪","刘华强",'尼古拉斯赵四',"宋小宝","刘能"]
for item in user_list:
	print(item)
user_list = ["范德彪","刘华强",'尼古拉斯赵四',"宋小宝","刘能"]

for index in range( len(user_list) ):
    item = user_index[index]
    print(item)

切记,循环的过程中对数据进行删除会踩坑【面试题】

# 错误方式, 有坑,结果不是你想要的。
# 千万不要在循环的过程中,边循环获取列表的数据,边删除列表的数据
user_list = ["刘德华", "范德彪", "刘华强", '刘尼古拉斯赵四', "宋小宝", "刘能"]
for item in user_list:  # 0 1 2 3
    if item.startswith("刘"):
        user_list.remove(item)
        
print(user_list)  # ['范德彪', '刘尼古拉斯赵四', '宋小宝']
# 正确方式,倒着删除。
user_list = ["刘的话", "范德彪", "刘华强", '刘尼古拉斯赵四', "宋小宝", "刘能"]
for index in range(len(user_list) - 1, -1, -1):
  # 5 4 3 2 1 0
    item = user_list[index]
    if item.startswith("刘"):
        user_list.remove(item)
print(user_list)

1.4 转换

  • int、bool无法转换成列表
  • st
name = "刘小伟"

data = list(name)  # ["刘","小","伟"]
print(data)

超前

v1 = (11,22,33,44) # 元组
vv1 = list(v1)     # 列表 [11,22,33,44]

v2 = {"alex","eric","dsb"} # 集合
v2 = list(v2) # 列表 ["alex","eric","dsb"]

1.5 其他

1.5.1 嵌套

列表属于容器,内部可以存放各种数据,所以它也支持列表的嵌套,如:

data = [ "谢广坤",["海燕","赵本山"],True,[11,22,[999,123],33,44],"宋小宝" ]

对于嵌套的值,可以根据之前学习的索引知识点来进行学习,例如:

data = [ "谢广坤",["海燕","赵本山"],True,[11,22,33,44],"宋小宝" ]

print( data[0] ) # "谢广坤"
print( data[1] ) # ["海燕","赵本山"]
print( data[0][2] ) # "坤"
print( data[1][-1] ) # "赵本山"

data.append(666)
print(data) # [ "谢广坤",["海燕","赵本山"],True,[11,22,33,44],"宋小宝",666]

data[1].append("谢大脚")
print(data) # [ "谢广坤",["海燕","赵本山","谢大脚"],True,[11,22,33,44],"宋小宝",666 ]


del data[-2]
print(data) # [ "谢广坤",["海燕","赵本山","谢大脚"],True,[11,22,33,44],666 ]


data[-2][1] = "alex"
print(data) # [ "谢广坤",["海燕","赵本山","谢大脚"],True,[11,"alex",33,44],666 ]


data[1][0:2] = [999,666]
print(data) # [ "谢广坤",[999,666,"谢大脚"],True,[11,"alex",33,44],666 ]
# 创建用户列表
#    用户列表应该长: [ ["alex","123"],["eric","666"] ]

# user_list = [["alex","123"],["eric","666"],]
# user_list.append(["alex","123"])
# user_list.append(["eric","666"])


user_list = []
while True:
    user = input("请输入用户名:")
    pwd = input("请输入密码:")

    data = []
    data.append(user)
    data.append(pwd)
    
    user_list.append(data)
user_list = []
while True:
    user = input("请输入用户名(Q退出):")
    if user == "Q":
        break
    pwd = input("请输入密码:")
    data = [user,pwd]
    user_list.append(data)

print(user_list)

1.6阶段作业

  1. 写代码,有如下列表,按照要求实现每个功能
li = ["alex", "WuSir", "ritian", "barry", "刘小伟"]
  • 计算列表的长度并输出
length = len(li)
print(f'列表的长度{length}')
  • 列表中追加元素"seven",并输出添加后的列表
li.append('seven')
print(li)  # ['alex', 'WuSir', 'ritian', 'barry', '刘小伟', 'seven']
  • 请在列表的第1个索引位置插入元素"Tony",并输出添加后的列表
li.insert(1, 'Tony')
print(li) # ['alex', 'Tony', 'WuSir', 'ritian', 'barry', '刘小伟']
  • 请修改列表第2个索引位置的元素为"Kelly",并输出修改后的列表
li[2] = 'Kelly'
print(li) # ['alex', 'WuSir', 'Kelly', 'barry', '刘小伟']
  • 请将列表的第3个位置的值改成 “妖怪”,并输出修改后的列表
li[3] = '妖怪'
print(li) #['alex', 'WuSir', 'ritian', '妖怪', '刘小伟']
  • 请将列表 data=[1,"a",3,4,"heart"] 的每一个元素追加到列表 li 中,并输出添加后的列表
for item in data:
  li.append(item)
  
print(li)  # ['alex', 'WuSir', 'ritian', 'barry', '刘小伟', 1, 'a', 3, 4, 'heart']

li.extend(data)
  • 请将字符串 s = "qwert"的每一个元素到列表 li 中。
for i in list(s):
  li.append(i)
 
print(li) # ['alex', 'WuSir', 'ritian', 'barry', '刘小伟', 'q', 'w', 'e', 'r', 't']

li.extend(s) # extend本质就是遍历列表,然后追加
  • 请删除列表中的元素"barry",并输出添加后的列表
li.remove('barry')
print(li) # ['alex', 'WuSir', 'ritian', '刘小伟']
  • 请删除列表中的第2个元素,并输出删除元素后的列表
del li[1]
li.pop(1)
print(li)  # ['alex', 'ritian', 'barry', '刘小伟']
  • 请删除列表中的第2至第4个元素,并输出删除元素后的列表
del li[2:4] 
li[2:4] = ''
print(li) # ['alex', '刘小伟']

方法二 ,倒序删除
for i in range(3, 0, -1):
  li.pop(i)

print(li)
  1. 写代码,有如下列表,利用切片实现每一个功能
li = [1, 3, 2, "a", 4, "b", 5,"c"]
  • 通过对li列表的切片形成新的列表 [1,3,2]
  • 通过对li列表的切片形成新的列表 [“a”,4,“b”]
  • 通过对li列表的切片形成新的列表 [1,2,4,5]
  • 通过对li列表的切片形成新的列表 [3,“a”,“b”]
  • 通过对li列表的切片形成新的列表 [3,“a”,“b”,“c”]
  • 通过对li列表的切片形成新的列表 [“c”]
  • 通过对li列表的切片形成新的列表 [“b”,“a”,3]
print(li[0:3])
print(li[3:6])
print(li[::2])
print(li[1:-1:2])
print(li[1::2])
print(li[-1:-2:-1])
print(li[-3::-2])
  1. 写代码,有如下列表,按照要求实现每一个功能。
lis = [2, 3, "k", ["qwe", 20, ["k1", ["tt", 3, "1"]], 89], "ab", "adv"]
  • 将列表lis中的第2个索引位置的值变成大写,并打印列表。
  • 将列表中的数字3变成字符串"100"
  • 将列表中的字符串"tt"变成数字 101
  • 在 "qwe"前面插入字符串:“火车头”
lis[2] = lis[2].upper()
print(lis)
lis[1] = 100
lis[3][2][1][1] =100
print(lis)
lis[3][2][1][0] = 101
print(lis)
lis[3].insert(0, '火车头') 
print(lis)
  1. 请用代码实现循环输出元素和值:users = [“刘小伟”, “景女神”, “肖大侠”] ,如
users = ["刘小伟", "景女神", "肖大侠"]
for user in users:
  print(f'{users.index(user)} {user}')
 
# 输出
0 刘小伟
1 景女神
2 肖大侠
  1. 请用代码实现循环输出元素和值:users = [“武沛齐”,“景女神”,“肖大侠”] ,如:
users = ["刘小伟", "景女神", "肖大侠"]
for user in users:
  print(f'{users.index(user)+1} {user}')
  
# 输出
1 武沛齐
2 景女神
3 肖大侠
  1. 写代码实现以下功能
  • 如有变量 goods = [‘汽车’,‘飞机’,‘火箭’] 提示用户可供选择的商品:
goods = ['汽车','飞机','火箭']
for good in goods:
  print(f'{goods.index(good)},{good}')

# 输出
0,汽车
1,飞机
2,火箭
  • 用户输入索引后,将指定商品的内容拼接打印,如:用户输入0,则打印 您选择的商品是汽车。
goods = ['汽车','飞机','火箭']
for good in goods:
  print(f'{goods.index(good)},{good}')

try:

    choice = input('请输入商品编号:(0-2)')

    print(f'你选择的商品是:{goods[int(choice)]}')
except:
    print('输入错误!')
  1. 利用for循环和range 找出 0 ~ 50 以内能被3整除的数,并追加到一个列表。
data_lst = []
for n in range(51):
  if n % 3 == 0:
    data_lst.append(n)
print(data_lst)
  1. 利用for循环和range 找出 0 ~ 50 以内能被3整除的数,并插入到列表的第0个索引位置,最终结果如下:
data_lst = [48,45,42...]
data_lst = []
for n in range(51):
  if n % 3 == 0:
    data_lst.insert(0,n)
print(data_lst)
  1. 查找列表li中的元素,移除每个元素的空格,并找出以"a"开头,并添加到一个新列表中,最后循环打印这个新列表。
li = ["alexC", "AbC ", "egon", " riTiAn", "WuSir", "  aqc"]
li = ["alexC", "AbC ", "egon", " riTiAn", "WuSir", "  aqc"]
new_lst = []

for i in li:
    if i.strip().startswith('a'):
        new_lst.append(i.strip())
print(new_lst)
  1. 将以下车牌中所有 的车牌搞到一个列表中,并输出京牌车辆的数量。
data = ["京1231", "冀8899", "京166631", "晋989"]
data = ["京1231", "冀8899", "京166631", "晋989"]
data_京 = []
for item in data:
    if item.startswith('京'):
        data_京.append(item)

print(data_京)

# 方法2:
data = ["京1231", "冀8899", "京166631", "晋989"]
data_京 = []
for item in data:
    if not item.startswith('京'):
      continue
    data_京.append(item)
count = len(data_京)
print(f'京牌的车号数据为:{count}')

2. 元组(tuple)

列表(list), 是一个有序可变的容器,在里面可以存放多个不同类型的元素。

元组(tuple),是一个有序不可变的容器,在里面可以存放多个不同类型的元素。

如何体现不可变呢?

记住一句话:《“我儿子永远不能换成是别人,但我儿子可以长大”》

2.1 定义

v1 = (11,22,33)
v2 = ("李杰","Alex")
v3 = (True,123,"Alex",[11,22,33,44])

# 建议:议在元组的最后多加一个逗v3 = ("李杰","Alex",)
d1 = (1)  # 1
d2 = (1,) # (1,)

d3 = (1,2)
d4 = (1,2)

**注意:**建议在元组的最后多加一个逗号,用于标识他是一个元组。

# 面试题
1. 比较值 v1 = (1) 和 v2 = 1 和 v3 = (1,) 有什么区别?
v1, v2相同, v3是元组
2. 比较值 v1 = ( (1),(2),(3) ) 和 v2 = ( (1,) , (2,) , (3,),) 有什么区别?
v1元组内的元素是整型数字,v2元组内嵌套元组
              v3 = (1,2,3)

2.2 独有功能

2.3 公共功能

  1. 相加,两个列表相加获取生成一个新的列表。
data = ("赵四","刘能") + ("宋晓峰","范德彪")
print(data) # ("赵四","刘能","宋晓峰","范德彪")

v1 = ("赵四","刘能")
v2 = ("宋晓峰","范德彪")
v3 = v1 + v2
print(v3) # ("赵四","刘能","宋晓峰","范德彪")
  1. 相乘,列表*整型 将列表中的元素再创建N份并生成一个新的列表。
data = ("赵四","刘能") * 2
print(data) # ("赵四","刘能","赵四","刘能")

v1 = ("赵四","刘能")
v2 = v1 * 2
print(v1) # ("赵四","刘能")
print(v2) # ("赵四","刘能","赵四","刘能")
  1. 获取长度
user_list = ("范德彪","刘华强",'尼古拉斯赵四',)
print( len(user_list) )
  1. 索引
user_list = ("范德彪","刘华强",'尼古拉斯赵四',)
print( user_list[0] )
print( user_list[2] )
print( user_list[3] )  # 报错
  1. 切片
user_list = ("范德彪","刘华强",'尼古拉斯赵四',)
print( user_list[0:2] )
print( user_list[1:] )
print( user_list[:-1] )
  1. 步长
# 字符串 & 元组
user_list = ("范德彪","刘华强",'尼古拉斯赵四',"宋小宝","刘能")
data = user_list[::-1]

# 列表
user_list = ["范德彪","刘华强",'尼古拉斯赵四',"宋小宝","刘能"]
data = user_list[::-1]

user_list.reverse()
print(user_list)
print( user_list[1:4:2] )  # ["刘华强","宋小宝"]
print( user_list[0::2] )  # ["范德彪",'尼古拉斯赵四',"刘能"]
print( user_list[1::2] ) # ["刘华强","宋小宝"]
print( user_list[4:1:-1] ) # ["刘能", "宋小宝", '尼古拉斯赵四']
  1. for 循环
user_list = ("范德彪","刘华强",'尼古拉斯赵四',"宋小宝","刘能")
for item in user_list:
	print(item)
user_list = ("范德彪","刘华强",'尼古拉斯赵四',"宋小宝","刘能")
for item in user_list:
 if item == '刘华强':
	 continue
 print(name) #"范德彪",'尼古拉斯赵四',"宋小宝","刘能"

目前:只有 str、list、tuple 可以被for循环。 “xxx” [11,22,33] (111,22,33)

# len + range + for + 索引
user_list = ("范德彪","刘华强",'尼古拉斯赵四',"宋小宝","刘能")
for index in range(len(user_list)):
    item = user_list[index]
    print(item)

2.4 转换

其他类型转换为元组,使用tuple(其他类型),目前只有字符串和列表可以转换为元组。

data = tuple(其他)

# str / list
name = "刘小伟"
data = tuple(name)
print(data) # 输出 ("刘","小","伟")
name = ["刘小伟",18,"pythonav"]
data = tuple(name)
print(data) # 输出 ("刘小伟",18,"pythonav")

2.5 嵌套

其他类型转换为元组,使用tuple(其他类型),目前只有字符串和列表可以转换为元组。

tu = ( '今天姐姐不在家', '姐夫和小姨子在客厅聊天', ('姐夫问小姨子税后多少钱','小姨子低声说道说和姐夫还提钱') )
tu1 = tu[0]
tu2 = tu[1]
tu3 = tu[2][0]
tu4 = tu[2][1]
tu5 = tu[2][1][3]

print(tu1) # 今天姐姐不在家
print(tu2) # 姐夫和小姨子在客厅聊天
print(tu3) # 姐夫问小姨子税后多少钱
print(tu4) # 小姨子低声说道说和姐夫还提钱
print(tu5) # 低

练习题1:判断是否可以实现,如果可以请写代码实现。

li = ["alex", [11,22,(88,99,100,),33],  "LiuSir",  ("ritian", "barry",),  "wenzhou"]
#        0               1                 2                3                4

# 1.请将 "LiuSir" 修改成 "刘小伟"
li[2] = "刘小伟"
index = li.index("Liusir")
li[index] = "刘小伟"

# 2.请将 ("ritian", "barry",) 修改为 ['日天','日地']
li[3] = ['日天','日地']

# 3.请将 88 修改为 87
li[1][2][0] = 87 # (报错,)

# 4.请将 "wenzhou" 删除,然后再在列表第0个索引位置插入 "周周"
# li.remove("wenzhou")
# del li[-1]
li.insert(0,"周周")

练习题2:记住一句话:《“我儿子永远不能换成是别人,但我儿子可以长大”》

data = ("123",666,[11,22,33], ("alex","李杰",[999,666,(5,6,7)]) )

# 1.将 “123” 替换成 9   报错

# 2.将 [11,22,33] 换成 "刘小伟"    报错

# 3.将 11 换成 99
data[2][0] = 99
print(data)  # ("123",666,[99,22,33], ("alex","李杰",[999,666,(5,6,7)]) )

# 4.在列表 [11,22,33] 追加一个44
data[2].append(44)
print(data) # ("123",666,[11,22,33,44], ("alex","李杰",[999,666,(5,6,7)]) )

练习题3:动态的创建用户并添加到用户列表中。

# 创建用户 5个
# user_list = [] # 用户信息
user_list = [ ("alex","132"),("admin","123"),("eric","123") ]

while True:
    user = input("请输入用户名:")
    if user == "Q":
        break
    pwd = input("请输入密码:")
    item = (user,pwd,) # ,代表元组
    user_list.append(item)
    
# 实现:用户登录案例
print("登录程序")
username = input("请输入用户名:")
password = input("请输入密码:")

is_success = False

for item in user_list:
    # item = ("alex","132")   ("admin","123")    ("eric","123")
    if username == item[0] and password == item[1]:
        is_success = True
        break

if is_success:
    print("登录成功")
else:
    print("登录失败")

总结

  1. 概述
  • 列表,以后写程序会用的非常多,要多些多练。
  • 元组,以后写程序用的不是很多,主要以了解其特殊和用法为主。
  1. 列表和元组的区别。
  2. 可变类型和不可变类型。
  3. 列表独有功能 & 公共功能(不用特地去记,多做题目去用,以后每天都会有相关的练习题)。
  4. 列表和元组等数据的嵌套
  5. 元组中 (1) 和 (1,) 的区别。
  6. 元组的元素不能被替换,但元组的元素如果是可变类型,可变类型内部是可以修改的。

作业

  1. 以下哪些数据类型转换为布尔值为False
1
""   # False
-19
[]   # False
[11,22]
(1)
(1,2,3)
()   # False
  1. 运算符操作
v1 = [] or "alex"  # "alex"
v2 = [11,22] and (1,2,)  # (1, 2,)
  1. 比较:a = [1,2,3]b = [(1),(2),(3) ] 以及 c = [(1,),(2,),(3,) ] 的区别?
a 是列表,元素为 1, 2, 3, b是列表,元素为1, 2, 3 c是列表, 元素为(1,)(2,)(3,)元组
  1. 将字符串text = "liuxiaowei|alex|eric"根据 | 分割为列表,然后列表转换为元组类型。
text = "liuxiaowei|alex|eric"
new_lst = text.split('|')
print(tuple(new_lst))
  1. 根据如下规则创建一副扑克牌(排除大小王)。
# 花色列表
color_list = ["红桃","黑桃","方片","梅花"]

# 牌值
num_list = []
for num in range(1,14):
    num_list.append(num)
    
result = []
# 请根据以上的花色和牌值创建一副扑克牌(排除大小王)
# 最终result的结果格式为: [ ("红桃",1), ("红桃",2) ... ]
for color in color_list:
    for num in num_list:
        pk = (color, num,)
        result.append(pk)

print(result)