5.c元组类型内置方法

元组类型内置方法

元组是不可变的列表,即元组的值不可更改,因此元组一般只用于只存不取的需求。也因此元组可以被列表取代掉,所以元组相比较列表使用的很少。元组相比较列表的优点为:列表的值修改后,列表的结构将会发生改变,而元组只需要存储,因此列表在某种程度上而言需要占用更多的内存。

优先掌握(*****)

  1. 索引取值
  2. 切片(顾头不顾尾,步长)
  3. 长度len
  4. 成员运算in和not in
  5. 循环
  6. count
  7. index

1.索引取值

# tuple之索引取值
name_tuple = ('nick', 'jason', 'tank', 'sean')
# name_tuple[0] = 'nick handsom'  # 报错

print(f"name_tuple[0]: {name_tuple[0]}")
name_tuple[0]: nick

2.切片(顾头不顾尾,步长)

# tuple之切片
name_tuple = ('nick', 'jason', 'tank', 'sean')

print(f"name_tuple[1:3:2]: {name_tuple[1:3:2]}")
name_tuple[1:3:2]: ('jason',)

3.长度len

# tuple之长度
name_tuple = ('nick', 'jason', 'tank', 'sean')

print(f"name_tuple: {name_tuple}")
name_tuple: ('nick', 'jason', 'tank', 'sean')

4.成员运算in和not in

# tuple之成员运算
name_tuple = ('nick', 'jason', 'tank', 'sean')

print(f"'nick' in name_tuple: {'nick' in name_tuple}")
'nick' in name_tuple: True

5.循环

# tuple之循环
name_tuple = ('nick', 'jason', 'tank', 'sean')

for name in name_tuple:
    print(name)
nick
jason
tank
sean

6.count()

# tuple之count()
name_tuple = ('nick', 'jason', 'tank', 'sean')

print(f"name_tuple.count('nick'): {name_tuple.count('nick')}")
name_tuple.count('nick'): 1

7.index()

# tuple之index()
name_tuple = ('nick', 'jason', 'tank', 'sean')

print(f"name_tuple.index('nick'): {name_tuple.index('nick')}")
name_tuple.index('nick'): 0
posted @ 2019-05-08 15:23  yellowcloud  阅读(367)  评论(0编辑  收藏  举报
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