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)