Python變量處理，我不明白

Question

我無法找到關於這個非常簡單的程序中發生的事情的簡明信息：

print 'case 1'
# a and b stay different
a = [1,2,3]
b = a
b = [4,5,6]

print 'a =',a
print 'b =',b

print
print 'case 2'
# a and b becomes equal
a = [1,2,3]
b = a
b[0] = 4 
b[1] = 5 
b[2] = 6 

print 'a =',a
print 'b =',b
print

print 'case 3'
# a and b stay different now
a = [1,2,3]
b = a[:]
b[0] = 4 
b[1] = 5 
b[2] = 6 
print 'a =',a
print 'b =',b
print

print 'case 4'
# now the funny thing
a=[1,2,[3]]
b=a[:]
b[0]    = 4 
b[1]    = 5 
b[2][0] = 6 # this modifies b and a!!!

這個簡單測試的輸出是：

case 1
a = [1, 2, 3]
b = [4, 5, 6]

case 2
a = [4, 5, 6]
b = [4, 5, 6]

case 3
a = [1, 2, 3]
b = [4, 5, 6]

case 4
a = [1, 2, [6]]
b = [4, 5, [6]]

我顯然不明白python如何處理每個案例。 任何人都可以提供一個鏈接，以便我可以閱讀它，或者對正在發生的事情做一個簡短的解釋嗎？

非常感謝。

Answer 1

這是一個奇妙的python代碼可視化工具 。

在其中運行您的代碼，一切都應該在一分鍾內變得清晰。

Answer 2

這里有兩件重要的事情：

• 變量只是指向對象的標簽
• 列表在python中是可變的，而整數則不是。

當你發現a和b都被修改了，那就是因為它們都指向同一個對象。 您可以執行id(a)和id(b)來確認這一點。

關於例子，請注意a[:]將創建一個新的對象，它是一個副本a 。 但是，它將是淺拷貝，而不是深拷貝。 這解釋了為什么在示例4中，您仍然可以修改a和b 。 它們指向不同的列表對象，但是一個元素是由它們共享的另一個列表。

Answer 3

案例1：名稱b反彈。

情況2： a和b綁定到同一個對象。

案例3：的淺拷貝a必然b 。 列表不同，但列表中的對象是相同的。

情況4：的淺拷貝a被綁定到b ，然后將對象中的一個被突變。

重新綁定不會發生變異，並且變異不會重新綁定。

Answer 4

print 'case 1'
# a and b stay different
a = [1,2,3]
b = a              #At this point 'b' and 'a' are the same, 
                   #just names given to the list
b = [4,5,6]        #At this point you assign the name 'b' to a different list

print 'a =',a
print 'b =',b

print
print 'case 2'
# a and b becomes equal
a = [1,2,3]        #At this point 'b' and 'a' are the same, 
                   #just names given to the list
b = a
b[0] = 4           #From here you modify the list, since both 'a' and 'b' 
                   #reference the same list, you will see the change in 'a'
b[1] = 5 
b[2] = 6 


print 'case 3'
# a and b stay different now
a = [1,2,3]
b = a[:]              #At this point you COPY the elements from 'a' into a new 
                      #list that is referenced by 'b'
b[0] = 4              #From here you modify 'b' but this has no connection to 'a'
b[1] = 5 
b[2] = 6 
print 'a =',a
print 'b =',b
print

print 'case 4'
# now the funny thing
a=[1,2,[3]]
b=a[:]           #At this point you COPY the elements from 'a' into a new 
                 #list that is referenced by 'b'
b[0]    = 4      #Same as before 
b[1]    = 5 
b[2][0] = 6 # this modifies b and a!!!    #Here what happens is that 'b[2]' holds 
                #the same list as 'a[2]'. Since you only modify the element of that 
                #list that will be visible in 'a', try to see it as cases 1/2 just  
                #'recursively'. If you do b[2] = 0, that won't change 'a'