前言
之前我们讲了函数的一些相关知识，了解了什么是全局变量什么是局部变量，接下来我们看看怎么将局部变量变为全局变量。还要将将内嵌函数的相关概念
知识点
- global关键字
使用global关键字，可以修改全局变量：
>>> count = 5
>>> def Myfun():
count = 10
print(10)
>>> Myfun()
10
>>> print(count)
5
--------------------------------
>>> def Myfun():
global count
count = 10
print(10)
>>> Myfun()
10
>>> print(count)
10
>>>
---------------------
- 内部函数
函数内部新创建另一个函数
>>> def fun1():
print('fun1()正在被调用...')
def fun2():
print('fun2()正在被调用...')
fun2()
>>> fun1()
fun1()正在被调用...
fun2()正在被调用...
- 闭包(Closure)
- 如果在一个内部函数里，对外部作用域（但不是在全局作用域的）变量进行引用，那么内部函数就会认为是闭包
>>> def FunX(x):
def FunY(y):
return x*y
return FunY
>>> i =FunX(8)
>>> type(i)
<class 'function'>
>>> i(5)
40
>>> FunX(8)()#错误举例
Traceback (most recent call last):
File "<pyshell#17>", line 1, in <module>
FunX(8)()
TypeError: FunY() missing 1 required positional argument: 'y'
>>> FunX(8)(5)
40
>>> FunY(5)
Traceback (most recent call last):
File "<pyshell#19>", line 1, in <module>
FunY(5)
NameError: name 'FunY' is not defined
>>> - nonlocal关键字
用于声明局部变量
>>> def Fun1():
x = 5
def Fun2():
x *= x
return x
return Fun2()
>>> Fun1()
Traceback (most recent call last):
File "<pyshell#51>", line 1, in <module>
Fun1()
File "<pyshell#50>", line 6, in Fun1
return Fun2()
File "<pyshell#50>", line 4, in Fun2
x *= x
UnboundLocalError: local variable 'x' referenced before assignment
原因：其存放在栈空间，如果修改，则会在局部函数作用域中屏蔽外部变量，就会出现引用了没有定义的变量值的错误。
解决办法：在python3之前没有直接的解决办法，只能间接通过的容器来解决，因为容器类型不存放在栈里边，所以变量不会被屏蔽。
第一种解决方法：（因为队列没有存放在栈中）
>>> def Fun1():
x = [5]
def Fun2():
x[0] *= x[0]
return x[0]
return Fun2()
>>> Fun1()
25第二种解决方法：python3之后可以使用nonlocal关键字来解决了，在使用函数外部变量的时候在使用前先用关键字nonlocal声明一下：
>>> def Fun1():
x = 5
def Fun2():
nonlocal x
x *= x
return x
return Fun2()
>>> Fun1()
25
课后作业
测试题
- 如果希望在函数中修改全局变量的值，应该使用什么关键字？
global 关键字
例如：
>>> count = 5
>>> def MyFun():
global count
count = 10
print(count)
>>> MyFun()
10
>>> count
10
- 在嵌套的函数中，如果希望在内部函数修改外部函数的变量，应该使用什么关键字？
#nolocal关键字
#例如：
>>> def Fun1():
x = 5
def Fun2():
nonlocal x
x *= x
return x
return Fun2()
>>> Fun1()
25
- Python的函数可以嵌套，但要注意访问的作用域问题，请问以下代码存在什么问题呢？
def outside():
print('I am outside!')
def inside():
print('I am inside!')
inside()使用嵌套函数要注意一点就是作用域问题，inside()函数是内嵌在outside()函数中的，所以除了outside()可以调用，在外边或者别的函数体里都无法对其进行调用的。
正确的调用应该是(注意空格的位置)：
def outside():
print('I am outside!')
def inside():
print('I am inside!')
inside()
outside()- 请问为什么代码A没有报错，但代码B却报错了？应该如何修改？
代码A：
>>> def outside():
var = 5
def inside():
var = 3
print(var)
inside()
>>> outside()
3
代码B
>>> def outside():
var = 5
def inside():
print(var)
var = 3
inside()
>>> outside()
Traceback (most recent call last):
File "<pyshell#81>", line 1, in <module>
outside()
File "<pyshell#80>", line 6, in outside
inside()
File "<pyshell#80>", line 4, in inside
print(var)
UnboundLocalError: local variable 'var' referenced before assignment
报错原因分析：
说的是变量var没有被定义就拿来使用，肯定错啦！
这里outside()函数里有一个var变量，但要注意的是，内嵌函数inside()也有一个同名的变量，Python为了保护变量的作用域，故将outside()的var变量屏蔽起来，因此此时是无法访问到外层的var变量的。
应该修改为：
>>> def outside():
var = 5
def inside():
nonlocal var
print(var)
var = 3
inside()
>>> outside()
5
- 请问如何访问funIn()呢？
>>> def funOut():
def funIn():
print('宾果！你成功访问到我啦！')
return funIn()
-------------直接调用----------------
>>> funOut()
宾果！你成功访问到我啦！
>>> def funOut():
def funIn():
print('宾果！你成功访问到我啦！')
return funIn
>>> funOut
<function funOut at 0x000000A7EC476F28>
>>> funOut()
<function funOut.<locals>.funIn at 0x000000A7EC476E18>
>>> funOut()()
宾果！你成功访问到我啦！
- 以下是“闭包”的一个例子，请你目测下回打印什么内容？
>>> def funX():
x = 5
def funY():
nonlocal x
x += 1
return x
return funY
>>> funX()()
6
>>> funX()()
6
>>> funX()()
6
>>> a = funX()
>>> print(a())
6
>>> print(a())
7
>>> print(a())
8
#有些鱼油可能会比较疑惑，这怎么跟全局变量一样了？局部变量x不是应该在每次调用的时候都重新初始化了吗？
#其实大家仔细看就明白了，当a = funX() 时，只要a变量没有被重新赋值，funX()就没有被释放，也就是说局部变量x就没有被重新初始化。
所以当全局变量不适用时，可考虑用闭包更稳定和安全，你还可以参考
游戏中的角色移动闭包在实际开发中的作用
动动手
请用已学过的知识编写程序，统计下边这个长字符串各个字符出现的次数，并找到小甲鱼送给大家的一句话。
string1 ='''%%$@_$^__#)^)&!_+]!*@&^}@[@%]()%+$&[(_@%+%$*^@$^!+]!&_#)_*}{}}!}_]$[%}@[{_@#_^{*
@##&{#&{&)*%(]{{([*}@[@&]+!!*{)!}{%+{))])[!^})+)$]#{*+^((@^@}$[**$&^{$!@#$%)!@(&
+^!{%_$&@^!}$_${)$_#)!({@!)(^}!*^&!$%_&&}&_#&@{)]{+)%*{&*%*&@%$+]!*__(#!*){%&@++
!_)^$&&%#+)}!@!)&^}**#!_$([$!$}#*^}$+&#[{*{}{((#$]{[$[$$()_#}!@}^@_&%^*!){*^^_$^
]@}#%[%!^[^_})+@&}{@*!(@$%$^)}[_!}(*}#}#___}!](@_{{(*#%!%%+*)^+#%}$+_]#}%!**#!^_
省略大部分。
'''
list1 = []
for each in string1:
if each not in list1:
if each == '\\n':
print('\\n',string1.count(each))
else:
print(each,string1.count(each))
list1.append(each)
运行结果：
================= RESTART: C:\Users\ThinkPad\Desktop\test.py =================
% 26
$ 30
@ 28
_ 30
^ 27
# 24
) 26
& 25
! 33
+ 19
] 14
* 28
} 30
[ 15
( 18
{ 27
6
省 1
略 1
大 1
部 1
分 1
。 1- 请用已学过的知识编写程序，找出小甲鱼藏在下边这个长字符串中的密码，密码的埋藏点符合以下规律：
1)每位密码为单个小写字母
2)每位密码的左右两边有且均只有三个大写字母str1 = '''ACFlCTLIQlAIVMTqHFkswqbDDHtpgcWaXSSglUYKE
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qKWGc'''
countA = 0 # 统计前边的大写字母
countB = 0 # 统计小写字母
countC = 0 # 统计后边的大写字母
length = len(str1)
for i in range(length):
if str1[i] == '\n':
continue
"""
|如果str1[i]是大写字母：
|-- 如果已经出现小写字母：
|-- -- 统计后边的大写字母
|-- 如果未出现小写字母：
|-- -- 清空后边大写字母的统计
|-- -- 统计前边的大写字母
"""
if str1[i].isupper():
if countB:
countC += 1
else:
countC = 0
countA += 1
"""
|如果str1[i]是小写字母：
|-- 如果小写字母前边不是三个大写字母（不符合条件）：
|-- -- 清空所有记录，重新统计
|-- 如果小写字母前边是三个大写字母（符合条件）：
|-- -- 如果已经存在小写字母：
|-- -- -- 清空所有记录，重新统计（出现两个小写字母）
|-- -- 如果该小写字母是唯一的：
|-- -- -- countB记录出现小写字母，准备开始统计countC
"""
if str1[i].islower():
if countA != 3:
countA = 0
countB = 0
countC = 0
else:
if countB:
countA = 0
countB = 0
countC = 0
else:
countB = 1
countC = 0
target = i
"""
|如果前边和后边都是三个大写字母：
|-- 如果后边第四个字母也是大写字母（不符合条件）：
|-- -- 清空记录B和C，重新统计
|-- 如果后边仅有三个大写字母（符合所有条件）：
|-- -- 打印结果，并清空所有记录，进入下一轮统计
"""
if countA == 3 and countC == 3:
if i+1 != length and str1[i+1].isupper():
countB = 0
countC = 0
else:
print(str1[target], end='')
countA = 3
countB = 0
countC = 0
运行结果：
================= RESTART: C:\Users\ThinkPad\Desktop\test.py =================
ilovefishc