元组及内置方法

# 1. 直接定义获得
test_tuple = ("zcy",)
test_tuple1 = ("zcy",20,["play","eat"])

# 2.利用tuple()方法转换列表
test_list = [1,2,3]
test_tuple = tuple(test_list)

# 1.元祖索引,这两个列表里的值代表各自位置索引
test_tuple1 = (0, 1, 2, 3, 4) # 从左是从0开始数
test_tuple2 = (-5,-4,-3,-2,-1) # 从右是-1开始数

# 2.取值
# 2.1 可以利用索引取值。
test_tuple = ("zcy",20,["play","eat"])

# 取出"zcy"
test_tuple[0]

# 取出20
test_tuple[-2]

# 取出"play"
test_tuple[2][0]

# for k in d.keys():
#     print(k)
#
# for k in d:
#     print(k)

# for v in d.values():
#     print(v)

# for k,v in d.items():
#     print(k,v)

# print(list(d.keys()))
# print(list(d.values()))
# print(list(d.items()))

# 队列：LIFO,先进先出（上楼的那种自动扶梯）
l = []
#入栈操作
l.append(‘first‘)
l.append(‘second‘)
l.append(‘third‘)
print(l)
#出队操作
print(l.pop(0))
print(l.pop(0))
print(l.pop(0))
# [‘first‘, ‘second‘, ‘third‘]
# first
# second
# third

# 堆栈: LIFO,后进先出（在行李箱里面放衣服）
l = []
#入栈操作
l.append(‘first‘)
l.append(‘second‘)
l.append(‘third‘)
print(l)
#出队操作
print(l.pop())
print(l.pop())
print(l.pop())
# [‘first‘, ‘second‘, ‘third‘]
# third
# second
# first

# d={‘k1‘:111}
#1、d.clear()

#2、d.update()
# d.update({‘k2‘:222,‘k3‘:333,‘k1‘:111111111111111})
# print(d)

#3、d.get() ：根据key取值，容错性好
# print(d[‘k2‘])  # key不存在则报错