列表字典组合

# coding=utf-8

stus = []

stu1 = {"no":"111","name":"张三","age":23,"nation":"汉","birthday":"1998-01-01","father":{"name":"张建国","company":"中国移动","job":"客户经理"},"mother":{"name":"吴燕玲","company":"师大附中","job":"语文老师"}}

stus.append(stu1)

print(stus)

name = stus[-1]["name"]

print(name)

stu2 = {"no":"112","name":"李四","age":23,"nation":"汉","birthday":"1998-01-01","father":{"name":"李建国","company":"中国联通","job":"客户经理"},"mother":{"name":"张丽丽","company":"人民医院","job":"外科医生"}}

stus.append(stu2)

print(stus)

name = stus[-1]["name"]

print(name)

fatherName = stus[-1]["father"]["name"]

print(fatherName)

列表字典组合

# coding=utf-8

stus = []

stu1 = {"no":"111","name":"张三","age":23,"nation":"汉","birthday":"1998-01-01","father":{"name":"张建国","company":"中国移动","job":"客户经理"},"mother":{"name":"吴燕玲","company":"师大附中","job":"语文老师"}}

stus.append(stu1)

print(stus)

name = stus[-1]["name"]

print(name)

stu2 = {"no":"112","name":"李四","age":23,"nation":"汉","birthday":"1998-01-01","father":{"name":"李建国","company":"中国联通","job":"客户经理"},"mother":{"name":"张丽丽","company":"人民医院","job":"外科医生"}}

stus.append(stu2)

print(stus)

name = stus[-1]["name"]

print(name)

fatherName = stus[-1]["father"]["name"]

print(fatherName)

列表字典组合

# coding=utf-8

stus = []

stu1 = {"no":"111","name":"张三","age":23,"nation":"汉","birthday":"1998-01-01","father":{"name":"张建国","company":"中国移动","job":"客户经理"},"mother":{"name":"吴燕玲","company":"师大附中","job":"语文老师"}}

stus.append(stu1)

print(stus)

name = stus[-1]["name"]

print(name)

stu2 = {"no":"112","name":"李四","age":23,"nation":"汉","birthday":"1998-01-01","father":{"name":"李建国","company":"中国联通","job":"客户经理"},"mother":{"name":"张丽丽","company":"人民医院","job":"外科医生"}}

stus.append(stu2)

print(stus)

name = stus[-1]["name"]

print(name)

fatherName = stus[-1]["father"]["name"]

print(fatherName)

#键值对存储key不能重复，value可以重复

# coding=utf-8

lisa = {"name":"Lisa","age":23}

print(lisa)

print(lisa["age"])

lisa["age"] = 26

lisa["height"] = 166

print(lisa)

lisa.pop("age")

print(lisa)

左闭右开[0:2]取第一个和第二个

反向索引最后一个下标是-1

# coding=utf-8

books = ["红楼梦","三国演义","水浒传","西游记","聊斋志异"]

print(books[0:4])

print(books[0:-1])

# coding=utf-8

books = ["红楼梦","三国演义","水浒传","西游记","聊斋志异"]

print(books[0])

int取整

%取余

# coding=utf-8

n=154

a=int(n/100)

b=int(n%100/10)

c=n%10

print(a)

print(b)

print(c)

re=a*a*a+b*b*b+c*c*c

if n==re:

print(n,end="")

print("是水仙花数")

else:

print(n,end="")

print("不是水仙花数")

154除以100取余54,54除以10等于5.4取整为5

# coding=utf-8

score = 30

if score >=60:

print("考试通过！")

if score >= 90:

print("名列前茅,重点表扬")

else:

print("继续努力,潜力巨大")

else:

print("考试未通过，还需努力")

注意缩进

整型（整数）不需要加双引号，加了双引号就是字符串

双引号一定标识字符串

# coding=utf-8

la = "德语"

if la == "德语" :

print("Ich liebe dich")

elif la == "韩语" :

print("사랑해")

elif la == "英语" :

print("I love you")

else :

print("我爱你")

由于是用记事本打开编辑的所以对编码有要求最终保存为utf-8 文件开头对应utf-8   猜测是这个原因，也许我用个正经编辑器就不会这样

注意 （） not and or的优先级

# coding=gbk

r1 = 1 != 1 or 8 < 2 and 6  > 3

print(r1)

r2 = 1 != 1 or not 8 < 1 and 6 > 3

print(r2)

r3 = 1 != 1 or not (8 < 1 and 6 > 3)

print(r3)

x = 700

re = (1000 - x) * 30

print(re)

if(re >= 1 and re <= 10000) :

print("结果落在区间内")

else :

print("结果未落在区间内")

# coding=gbk

x=1600

re = (1000-x)*30

print(re)

if re >10000 :

print("结果大于10000")

else :

print("结果小于10000")

True False  首字母是大写的

# coding=gbk

if 8 > 6 :

print("判断条件成立")

else :

print("判断条件不成立")

yntaxError: Non-ASCII character '\xe5' in file index.py on line 7, but no encoding declared; see http://python.org/dev/p

开头加# -*- coding: UTF-8 -*-

不好使就加#coding=gbk  总有一款适合你