《零基礎入門學習Python》（20）--函式：內嵌函式和閉包
阿新 • • 發佈：2018-11-09
前言
之前我們講了函式的一些相關知識，瞭解了什麼是全域性變數什麼是區域性變數，接下來我們看看怎麼將區域性變數變為全域性變數。還要將將內嵌函式的相關概念
知識點
- global關鍵字
使用global關鍵字，可以修改全域性變數：
>>> 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 >>> ---------------------
- 內部函式
函式內部新建立另一個函式
>>> def fun1():
print('fun1()正在被呼叫...')
def fun2():
print('fun2()正在被呼叫...')
fun2()
>>> fun1()
fun1()正在被呼叫...
fun2()正在被呼叫...
- 閉包(Closure)
- 如果在一個內部函式裡，對外部作用域（但不是在全域性作用域的）變數進行引用，那麼內部函式就會認為是閉包
>>> def FunX(x): def FunY(y): return x*y return FunY >>> i =FunX(8) >>> type(i) <class 'function'> >>> i(5) 40 >>> FunX(8)()#錯誤舉例 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關鍵字
用於宣告區域性變數
>>> 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：其存放在棧空間，如果修改，則會在區域性函式作用域中遮蔽外部變數，就會出現引用了沒有定義的變數值的錯誤。
解決辦法：在python3之前沒有直接的解決辦法，只能間接通過的容器來解決，因為容器型別不存放在棧裡邊，所以變數不會被遮蔽。
第一種解決方法：（因為佇列沒有存放在棧中）
>>> def Fun1():
x = [5]
def Fun2():
x[0] *= x[0]
return x[0]
return Fun2()
>>> Fun1()
25第二種解決方法：python3之後可以使用nonlocal關鍵字來解決了，在使用函式外部變數的時候在使用前先用關鍵字nonlocal宣告一下：
>>> def Fun1():
x = 5
def Fun2():
nonlocal x
x *= x
return x
return Fun2()
>>> Fun1()
25
課後作業
測試題
- 如果希望在函式中修改全域性變數的值，應該使用什麼關鍵字？
global 關鍵字
例如：
>>> count = 5
>>> def MyFun():
global count
count = 10
print(count)
>>> MyFun()
10
>>> count
10
- 在巢狀的函式中，如果希望在內部函式修改外部函式的變數，應該使用什麼關鍵字？
#nolocal關鍵字
#例如：
>>> def Fun1():
x = 5
def Fun2():
nonlocal x
x *= x
return x
return Fun2()
>>> Fun1()
25
- Python的函式可以巢狀，但要注意訪問的作用域問題，請問以下程式碼存在什麼問題呢？
def outside():
print('I am outside!')
def inside():
print('I am inside!')
inside()使用巢狀函式要注意一點就是作用域問題，inside()函式是內嵌在outside()函式中的，所以除了outside()可以呼叫，在外邊或者別的函式體裡都無法對其進行呼叫的。
正確的呼叫應該是(注意空格的位置)：
def outside():
print('I am outside!')
def inside():
print('I am inside!')
inside()
outside()- 請問為什麼程式碼A沒有報錯，但程式碼B卻報錯了？應該如何修改？
程式碼A：
>>> def outside():
var = 5
def inside():
var = 3
print(var)
inside()
>>> outside()
3
程式碼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
報錯原因分析：
說的是變數var沒有被定義就拿來使用，肯定錯啦！
這裡outside()函式裡有一個var變數，但要注意的是，內嵌函式inside()也有一個同名的變數，Python為了保護變數的作用域，故將outside()的var變數遮蔽起來，因此此時是無法訪問到外層的var變數的。
應該修改為：
>>> def outside():
var = 5
def inside():
nonlocal var
print(var)
var = 3
inside()
>>> outside()
5
- 請問如何訪問funIn()呢？
>>> def funOut():
def funIn():
print('賓果！你成功訪問到我啦！')
return funIn()
-------------直接呼叫----------------
>>> funOut()
賓果！你成功訪問到我啦！
>>> def funOut():
def funIn():
print('賓果！你成功訪問到我啦！')
return funIn
>>> funOut
<function funOut at 0x000000A7EC476F28>
>>> funOut()
<function funOut.<locals>.funIn at 0x000000A7EC476E18>
>>> funOut()()
賓果！你成功訪問到我啦！
- 以下是“閉包”的一個例子，請你目測下回列印什麼內容？
>>> 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
#有些魚油可能會比較疑惑，這怎麼跟全域性變數一樣了？區域性變數x不是應該在每次呼叫的時候都重新初始化了嗎？
#其實大家仔細看就明白了，當a = funX() 時，只要a變數沒有被重新賦值，funX()就沒有被釋放，也就是說區域性變數x就沒有被重新初始化。
所以當全域性變數不適用時，可考慮用閉包更穩定和安全，你還可以參考
遊戲中的角色移動閉包在實際開發中的作用
動動手
請用已學過的知識編寫程式，統計下邊這個長字串各個字元出現的次數，並找到小甲魚送給大家的一句話。
string1 ='''%%[email protected]_$^__#)^)&!_+]!*@&^}@[@%]()%+$&[([email protected]%+%$*^@$^!+]!&_#)_*}{}}!}_]$[%}@[{[email protected]#_^{*
@##&{#&{&)*%(]{{([*}@[@&]+!!*{)!}{%+{))])[!^})+)$]#{*+^((@^@}$[**$&^{[email protected]#$%)[email protected](&
+^!{%_$&@^!}$_${)$_#)!({@!)(^}!*^&!$%_&&}&_#&@{)]{+)%*{&*%*&@%$+]!*__(#!*){%&@++
!_)^$&&%#+)}[email protected]!)&^}**#!_$([$!$}#*^}$+&#[{*{}{((#$]{[$[$$()_#}[email protected]}^@_&%^*!){*^^_$^
]@}#%[%!^[^_})[email protected]&}{@*!(@$%$^)}[_!}(*}#}#___}!](@_{{(*#%!%%+*)^+#%}$+_]#}%!**#!^_
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)
執行結果：
================= 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- 請用已學過的知識編寫程式，找出小甲魚藏在下邊這個長字串中的密碼，密碼的埋藏點符合以下規律：
1)每位密碼為單個小寫字母
2)每位密碼的左右兩邊有且均只有三個大寫字母str1 = '''ACFlCTLIQlAIVMTqHFkswqbDDHtpgcWaXSSglUYKE
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qKWGc'''
countA = 0 # 統計前邊的大寫字母
countB = 0 # 統計小寫字母
countC = 0 # 統計後邊的大寫字母
length = len(str1)
for i in range(length):
if str1[i] == '\n':
continue
"""
|如果str1[i]是大寫字母：
|-- 如果已經出現小寫字母：
|-- -- 統計後邊的大寫字母
|-- 如果未出現小寫字母：
|-- -- 清空後邊大寫字母的統計
|-- -- 統計前邊的大寫字母
"""
if str1[i].isupper():
if countB:
countC += 1
else:
countC = 0
countA += 1
"""
|如果str1[i]是小寫字母：
|-- 如果小寫字母前邊不是三個大寫字母（不符合條件）：
|-- -- 清空所有記錄，重新統計
|-- 如果小寫字母前邊是三個大寫字母（符合條件）：
|-- -- 如果已經存在小寫字母：
|-- -- -- 清空所有記錄，重新統計（出現兩個小寫字母）
|-- -- 如果該小寫字母是唯一的：
|-- -- -- countB記錄出現小寫字母，準備開始統計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
"""
|如果前邊和後邊都是三個大寫字母：
|-- 如果後邊第四個字母也是大寫字母（不符合條件）：
|-- -- 清空記錄B和C，重新統計
|-- 如果後邊僅有三個大寫字母（符合所有條件）：
|-- -- 列印結果，並清空所有記錄，進入下一輪統計
"""
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
執行結果：
================= RESTART: C:\Users\ThinkPad\Desktop\test.py =================
ilovefishc