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

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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

转载于:https://www.cnblogs.com/wanbin/p/9514690.html