python学习Day2

学习简述：三元运算，列表，元组，字符串，字典
三元运算其实就是为了缩减代码而产生，可以把三句话用一句来表示。a,b,c = 1,2,4
d = a if b>c else c
意思就是如果b>c是正确的，那么d = a,否则d = c,相当于：
a,b,c = 1,2,4
if b>c:
d = a
else:
d = c下面我们好好说一说列表：
首先列表用" [] "符号表示，我们来说说列表涉及的切片，增，删，插入，改等各种方法
#切片
names = ['aa','bb','cc','dd']
print (names[1:3])

print (names[-3:-1])#默认是从左往右
#增
names = ['aa','bb','cc','dd']
names.append("ee")
print (names)

#插入
names = ['aa','bb','cc','dd']
names.insert(1,"ee") #1:你想插入的位置，'ee':你想要加入的内容
print (names)

#删
names = ['aa','bb','cc','dd']
names.remove('cc')#
del names[0] #
names.pop(0) #这三者等价
print (names)

#改
names = ['aa','bb','cc','dd']
names[0] = 'niuniu'
print (names)

#索引
names = ['aa','bb','cc','dd','aa']
print (names.index('bb')) #索引找的这个东西必须是在列表里的

#合并
names = ['aa','bb','cc','dd','aa']
names2 = [1,2,3,4]
names.extend(names2)
print(names)说说还有些价值的浅copy
names = ['aa','bb','cc','dd','aa']
p1 = names.copy()
p2 = names.copy()
names[0] = '11'
print (names)
print (p1)
print (p2)输出结果你会发现，只有names的'aa'变为了'11',而p1,p2中的未改变，因为copy是浅copy，若再往深改进，你就会看到变化
names = ['aa','bb',['age',12],'dd','aa']
p1 = names.copy()
p2 = names.copy()
names[2][0] = 'salary'
print (names)
print (p1)
print (p2)

#结果
['aa', 'bb', ['salary', 12], 'dd', 'aa']
['aa', 'bb', ['salary', 12], 'dd', 'aa']
['aa', 'bb', ['salary', 12], 'dd', 'aa']

这并不代表没用，比如如果夫妻两人相创建一个联合账号，丈夫加入取走500，那么妻子等于也少了500
person = ['names',['salary',10000]]
p1 = person[:]
p2 = person[:]
p1[0] = 'husband'
p2[0] = 'wife'
p1[1][1] = '500'
print (p1)
print (p2)
#结果
['husband', ['salary', '500']]
['wife', ['salary', '500']]元组其实就相当于一个小的列表，但是注意，元组是只读的，只能切片和索引。字符串
特性：不可修改   
可以切片name = 'niu'
name.capitalize() #首字母大写
name.casefold() #大写全部变小写
name.center(50,"-") #输出 '---------------------niu----------------------'
name.count('niu') #统计niu出现次数
name.encode() #将字符串编码成bytes格式
name.endswith("niu") #判断字符串是否以niu结尾
"niu\tZ".expandtabs(10) #输出'niu Z'， 将\t转换成多长的空格
name.find('A') #查找A,找到返回其索引， 找不到返回-1 字典

特性：字典是无序的，但key是唯一的，所以天生去重
#改
score = {
'studen1':90,
'student2':78,
'student3':88,
'student4':56
}
score['student2'] = 100
print (score)
# >>>{'studen1': 90, 'student2': 100, 'student3': 88, 'student4': 56}

#增
score['student5'] = 23
print (score)
# >>>{'studen1': 90, 'student2': 100, 'student3': 88, 'student4': 56, 'student5': 23}

#删
del score['student2']
score.pop('student2')
score.popitem() #随机删
print (score)

#查
print(score['student4'])
print(score.get('student4'))

#其他
print(score.values())
print(score.keys())

#循环
for i in score:
print (i,score[i])