z=y=x=10
print(x,y,z)
m=10
n=20
m,n=n,m # #交叉赋值
print(m,n)
nums = [11,22,33,44,55]
a,b,c,d,e = nuns # nums包含多个值,就好比一个压缩包,解压赋值因此得名
print(a,b,c,d,e)
salaries=[111,222,333,444,555]
x,y,z*_=salaries=[111,222,333,444,555] #取前三个值
print(x,y,z)
#*—会将没有对应关系的值存成列表,*—只能帮我们取两头的值,无法取中间的值
print(not True)
print(True and 10>3)
print(True and 10>3 and 10 and 0)
print(3>2 or o)
print("egon" in "hello egon") #判断一个字符串是否在一个大字符串中间
print("e" in "hello egon") #判断一个字符串是否在一个大字符串中间
print("egon" notin "hello egon")
x=10
print(id(x))
x=11
print(id(x))
#id不同
x=3.1
print(id(x))
x=3.2
print(id(x))
#id不同
x="abc"
print(id(x))
x="bbb"
print(id(x))
#id不同
l=['aaa','bbb','ccc']
print(id(l))
l[0]='AAA'
print(id(l))
#id相同
dic={'k1':'111','k2',:'222'}
print(id(dic))
dic[ki]=333
print(id(dic))
#id相同
原文:https://www.cnblogs.com/hm666/p/12431736.html