python_元组(tuple)

#tuple#元组不可以修改,不能对其进行增加或删除操作#元组是有序的
#定义
tu_1 = () #定义一个空元组
tu_2 = (1,2,'alex',[3,4],(5,6,7),True,{'name':"liming"})
print(tu_1,tu_2)
print('定义结束!!!')
#索引取值
def suoyin():
vi_one = tu_2[3] #取出[3,4]
vi_two = tu_2[2][1] #取出alex中的l
vi_three = tu_2[4] #取出(5,6,7)
vi_four = tu_2[5] #取出True
print(vi_one)
print(vi_two )
print(vi_three )
print(vi_four )
suoyin()
print('索引取值结束!!!')
#切片取值
def qiepian():
vi_five = tu_2[2:4] #不包含4
vi_six = tu_2[3:] #输出3之后的值,包含3的值
print(vi_five) #输出(‘alex’,[3,4])
print(vi_six) #输出[3,4],(5,6,7),True,{'name':'liming'}
qiepian()
print('切片取值结束!!!')

#可以进行for循环
for i in tu_2:
print(i,end=' ') #end=' '表示不换行输出,用空格隔开
print('for循环结束!!!')
#tuple()
# 将列表转换为元组
li = [1,2,3,4]
print(tuple(li))

#字符串转换为元组
st = 'ascll'
print(tuple(st))

#将元组转换为列表
tu_3 = (1,2,3)
print(list(tu_3))

#将元组转换为字符串
s = ''
tu_4 = (2,3,4,'a')
for i in tu_4:
s = s + str(i)
print(s)

#将元组转换为字符串,元素全部为字符串
tu_5 = ('a','b','c')
v = ''.join(tu_5)
print(v)

#计算元组中元素出现的次数
tu_6 = (1,2,2,3,4,5,5,5,6,7)
v_1 = tu_6.count(1) #统计1在元组中的个数
v_2 = tu_6.count(5) #统计5在元组中的个数
print(v_1)
print(v_2)
print('统计元素出现的次数结束!!!')
#查找元素在元组中的索引值
tu_7 = (1,2,5,6,7,4,3,7,'dfs')
v_3 = tu_7.index(4) #输出4的索引值5
#v_4 = tu_7.index(0) #当查找的元素不在元组中会报错
v_5 = tu_7.index('dfs') #输出’dfs'的索引值8
v_6 = tu_7.index(7) #输出7的索引值4,索引次数从左边开始计算
print(v_3)
print(v_5)
print(v_6)
print('查找元素在元组中的索引值结束!!!')
#对元组中的二级元素进行修改
tu = (111,"alex",(11,22),[(33,44)],True,4,5)
tu[3][0] = 567 #通过这种赋值方式对元组二级进行修改,不能对一级元素进行修改
#tu[1] = 'df' #报错
print(tu)

原文地址:https://www.cnblogs.com/shadowfolk/p/14286740.html