姓名:刘浩然。 2020年大目标Python

day12 Python元祖

前戏

#元祖:元素不可被改变,不能白增加或者删除
#tuple
#tu = (11,22,33,44)
#tu.count(22),获取指定元素在元祖中出现的次数
#tu.index(22),获取元素的下标

• 不能增加或者删除元素

1、书写格式

tu = (111,"alex",(11,22),[(33,44)],True,33,44,)
print(tu)

结果:
(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

tu = (111,"alex",(11,22),[(33,44)],True,33,44,)
print(tu[1])

结果:
alex

3、切片

tu = (111,"alex",(11,22),[(33,44)],True,33,44,)
print(tu[1:3])

结果:
('alex', (11, 22))
#>=1   <3

4、可以被for循环,可迭代对象

tu = (111,"alex",(11,22),[(33,44)],True,33,44,)
for i in tu:
    print(i)

结果:
111
alex
(11, 22)
[(33, 44)]
True
33
44

5、转换

s = "charon"
li = ["charon","pluto"]
tu = ("pluto","charon",)
v = tuple(s)
print(v)
v1 = tuple(li)
print(v1)
v2 = list(tu)
print(v2)
v3 = "_".join(tu)
print(v3)
li.extend((11,22,33,))
print(li)
v4 = li.pop()
print(v4)


结果:
('c', 'h', 'a', 'r', 'o', 'n')
('charon', 'pluto')
['pluto', 'charon']
pluto_charon
['charon', 'pluto', 11, 22, 33]
33

6、元祖的一级元素不可修改,删除,增加

tu = (111,"alex",(11,22),[(33,44)],True,33,44,)
v = tu[3][0]
print(v)
tu[3][0] = 2
print(tu)


结果:
(33, 44)
(111, 'alex', (11, 22), [2], True, 33, 44)
#元祖不可以更改,但是元祖里面的列表是可以更改的

  

posted @ 2019-01-31 15:01  pluto2charon  阅读(150)  评论(0编辑  收藏  举报