零基础入门学习Python（20）--函数：内嵌函数和闭包

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
<function FunX.<locals>.FunY at 0x000000A7EC476840>
>>> i(5)
40
>>> FunX(8)(5)
40
>>> FunY(5)
Traceback (most recent call last):
  File "<pyshell#36>", 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。

>>> def Fun1():
        x = [5]
        def Fun2():
            x[0] *= x[0]
            return x[0]
        return Fun2()
>>> Fun1()
25

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：

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 =================
% 6104
$ 6046
@ 6157
_ 6112
^ 6030
# 6115
) 6186
& 6043
! 6079
+ 6066
] 6152
* 6034
} 6105
[ 6108
( 6154
{ 6046
\n 1219
G 1
O 2
D 1
L 1
U 1
C 1
K 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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bVIhEAxNwOUDQCOCDIBHsQlFxCbHErKiBsvQNjuIGnIzZiwoVTdUZnucSaKdHrxJeGQfTawFJefGJFFa
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UEEFhWRoTKGRwFmPpZPOsalsrNOlkTdXqGKoWlWkqJvYYrtLiTfhdItMbbhdumSYgCcQVUARnYFWkCsC
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LvAxlgliymfbjTwDSqNTLcAEIerzpUShNKkuehciEAYeGJOFFcOvurJFUyfZhQqzfaneiBcrPhaphSCB
StQiJvtoPhHcLUbriDUIfifVFnNtJfGlmcbQJSXsZfVjpbqnkMllUmHzLvEuSarSlNlNHsWvYlfhjMEA
epYTjoRGZeAAtURFeDfDxTYmOmONuQQBdcdncFGjhHmKlwqmUWoXuIXIxaaXnNThgPGtIlynrUIPLQTG
xXhattDrfBGbZRveKbgjzxJLdYREQlMeLtcIEUoyJocdAfUbymxuFLVjGkOQniiPParqoyQYfDYAQTkM
WLicLxpEFkBbwlKrTyYilKTtYkpVGxtOjYmcBDOrwFhFiGutmpTyTarUbVUeSevBTdaPDpjRkaEmLJMg
WsMhSGfIcBChcqrRKgKpjvGnFipjswgjetRtniMagakbCXAjpzWTtMlgZGCJwGyglpcLebrKWhgwJfWV
qGifWNEpCtjuejHoyVCdIxzMYGnfoslgTNAJdtVBWDVoGLzHSAVBTnhNIvAOExQNiJOIPPiHkdaRbfaP
ixDDoCDOOeAqvQJFxLWDICfGmufyxmaMshbvcrtjqqVtffZTnbtCOQfzRMGwOQEKaAmSWjnYdNgvdkmd
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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