Python学习------第十天
数据容器-----元组
定义格式，特点，相关操作
元组一旦定义，就无法修改
元组内只有一个数据，后面必须加逗号
"""
#元组
(1,"hello",True)
#定义元组
t1 = (1,"hello")
t2 = ()
t3 = tuple()
print(f"t1的类型是{type(t1)}")
#单个元素后面需要加逗号
t4 = ("hello",)
print(f"t4的类型是{type(t4)}")
#元素的嵌套
t5 = ((1,2,3),(4,5,6))
print(f"t5的类型是{type(t5),}，内容是{t5}")
#下标索引去取内容
element = t5[1][1]
print(element)
#index查找方法
t6 = ("heima","heima","hij","sda")
index = t6.index("hij")
print(f"在元组t6中查找hij，的下标是：{index}")
#元组的操作：count统计方法
num = t6.count("heima")
print(f"在元组t6中，heima的数量是{num}")
t8 = ("hins","heima","heima","hij","sda")
num1 = len(t8)
print(f"t8元组中，元素的个数为{num1}")
#while循环
#for 循环遍历
t1 = (1,2,3,4,5,6)
for element in t1:
print(f"t1中的元素分别为{element}")
t1 = (1,2,3,4,5,6)
index = 0
while index<len(t1):
print(f"t1中的元素分别为{t1[index]}")
index += 1
#元组不支持修改元素
t1 = (1,2,3)
t1[0]=4
print(t1)
"""
#元组内的列表内容可以修改
t2 = (1,2,3,[4,5,6,7])
t2[3][0]=8
print(t2)
t1 = ("周杰伦",11,["football","music"])
index = t1.index("周杰伦")
print(f"周杰伦年龄所在的下标位置是{index}")
name = t1[0]
print(f"该学生的姓名为{name}")
t1[2][0]=()
print(t1)
t1[2][0]="coding"
print(t1)
2.掌握字符串的常见操作
#字符串的替换
#得到的是一个新字符串而并非将原有字符串修改
my_str = "itheima and itcast"
my_str2 = my_str.replace("and","beautiful")
print(my_str2)
print(my_str)
"""
#元组
(1,"hello",True)
#定义元组
t1 = (1,"hello")
t2 = ()
t3 = tuple()
print(f"t1的类型是{type(t1)}")
#单个元素后面需要加逗号
t4 = ("hello",)
print(f"t4的类型是{type(t4)}")
#元素的嵌套
t5 = ((1,2,3),(4,5,6))
print(f"t5的类型是{type(t5),}，内容是{t5}")
#下标索引去取内容
element = t5[1][1]
print(element)
#index查找方法
t6 = ("heima","heima","hij","sda")
index = t6.index("hij")
print(f"在元组t6中查找hij，的下标是：{index}")
#元组的操作：count统计方法
num = t6.count("heima")
print(f"在元组t6中，heima的数量是{num}")
t8 = ("hins","heima","heima","hij","sda")
num1 = len(t8)
print(f"t8元组中，元素的个数为{num1}")
#while循环
#for 循环遍历
t1 = (1,2,3,4,5,6)
for element in t1:
print(f"t1中的元素分别为{element}")
t1 = (1,2,3,4,5,6)
index = 0
while index<len(t1):
print(f"t1中的元素分别为{t1[index]}")
index += 1
#元组不支持修改元素
t1 = (1,2,3)
t1[0]=4
print(t1)
#元组内的列表内容可以修改
t2 = (1,2,3,[4,5,6,7])
t2[3][0]=8
print(t2)
#练习
t1 = ("周杰伦",11,["football","music"])
index = t1.index("周杰伦")
print(f"周杰伦年龄所在的下标位置是{index}")
name = t1[0]
print(f"该学生的姓名为{name}")
t1[2][0]=()
print(t1)
t1[2][0]="coding"
print(t1)
my_str = "itheima and itcast"
value = my_str[0]
print(value)
value2 = my_str[-1]
print(value2)
#字符串不支持修改
my_str[0]="1"
print(my_str)
#字符串的index方法
my_str = "itheima and itcast"
value = my_str.index("and")
print(f"在字符串中查找and，其起始下标是{value}")
#字符串的替换
#得到的是一个新字符串而并非将原有字符串修改
my_str = "itheima and itcast"
my_str2 = my_str.replace("and","beautiful")
print(my_str2)
print(my_str)
#字符串的切分
my_str = "itheima and itcast"
mystr2 = my_str.split()
print(my_str)
print(mystr2,f"类型是{type(mystr2)}")
#字符串的归整操作(去前后空格)
#不传入参数，去除收尾空格
my_str = " itheima and itcast "
newstr = my_str.strip()
print(my_str)
print(newstr)
#去除收尾指定元素
my_str = "2314itheima and itcast231"
newstr = my_str.strip("2314")
print(newstr)
#统计元素出现次数
count = my_str.count("i")
print(count)
#统计长度
my_str = "2314itheima and itcast231"
length = len(my_str)
print(length)
#字符串的遍历
#while
mystr = "lili is a good boy"
index = 0
while index <len(mystr):
print(f"该字符串的元素为{mystr[index]}")
index += 1
#for循环
mystr = "lili is a good boy"
for element in mystr:
print(f"该字符串的元素为{element}")
"""
mystr = "itheima itcast boxuegu"
count = mystr.count("it")
print(f"字符串中it一共有{count}个")
newstr = mystr.replace(" ","|")
print(newstr)
new2 = newstr.split("|")
print(new2)
原文地址：https://blog.csdn.net/weixin_48671989/article/details/143914704
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