python 3 列表元组字典集合

#list
empty_list = []
numlist = [1,5,3,4,2]
mixlist = ['name',100,200]
list_in_list = [1,2,[3,4,5]]
list_use_fuction = list("hello word")
print(list_use_fuction)

#list拆分
hero = ['dd',24,173]
name,age,height = hero
print(name)

#list索引
print(hero[1])
print(list_in_list[2][2])
print([1,2,3,4,5][-1])#list可以自身索引不需名字

#切片
print(list_use_fuction[::2])#取偶数位的字母

#函数
print(len(list_use_fuction))
print(max(numlist))
print(sum(numlist))
print(sorted(numlist))
i = reversed(numlist)
next(i)#reversed返回的是一个迭代器，需要用next返回

#list改变
numlist[0] = 100;
print(numlist)
numlist[1:3] = "hello"
print(numlist)#[100, 'h', 'e', 'l', 'l', 'o', 4, 2]
numlist[:] = [6,7,2,5,1,8,9,0,6,6]
print(numlist)

#list方法
print(numlist.index(7))#检索
print(numlist.count(6))
numlist.append(666)#增加元素
print(numlist)
numlist.insert(0,100)
print(numlist)
numlist.remove(6)#删除元素
print(numlist)
first = numlist.pop(0)#删除并返回
del numlist[1:]#删除多个元素
print(numlist)

#列表推导式
#【表达式 for 变量 in 列表 if 条件】
#求1000以内3和5的倍数之和
print(sum([i for i in range(0,1000) if i%3 == 0 or i%5 == 0 ]))
#使用生成器表达式，不是真正创建列表，而是返回一个生成器，可以用迭代的方法获得生成器的值，最大的优点就是节省内存
#具体就是将上面的[]换成()
print(sum((i for i in range(0,1000) if i%3 == 0 or i%5 == 0 )))

#copy and deepcopy
import copy
a = [1,2,3]
b = copy.copy(a)
c = copy.deepcopy(a)
a.remove(1)
print(a)
print(b)
print(c)

#元组
t1 = tuple()
t2 = ()
t3 = (1,2,3)
t4 = tuple("hello")
print(t4)

#zip
t5 = [1,2,3,4,5]
t6 = '123'
for i in zip(t4,t5,t6):#接收多个序列，每个序列顺序选取一个值放入一个元组
    print(i)

#dict
d = dict()
hero = {'name':'dd','age':24,'weight':75}
print(hero)
print(hero['name'])
#functions
print(hero.keys())
print(hero.values())
for k,v in hero.items():
    print(k + ' ===> ' + str(v))

#set
null = set()
aset = set("aaaab")
bset = set('abc')
print(aset)
print(aset.intersection(bset))#交集，并集，差集
print(aset.union(bset))
print(bset.difference(aset))
print(aset.symmetric_difference(bset))#对称差，保留两个集合中不同的元素
#issubset,isuperset判断子集和超集
#function
bset.add('d')
bset.remove('a')
print(bset)
bset.clear()