Question

我无法找到关于这个非常简单的程序中发生的事情的简明信息：

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!!!

这个简单测试的输出是：

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]]

我显然不明白python如何处理每个案例。任何人都可以提供一个链接，以便我可以阅读它，或者对正在发生的事情做一个简短的解释吗？

非常感谢。

Answer 1

这是a fantastic visualization tool for python code。

在其中运行您的代码，一切都应该在一分钟内变得清晰。

Answer 2

这里有两件重要的事情：

变量只是指向对象的标签
列表在python中是可变的，而整数则不是。

当您发现a和b都被修改时，那是因为它们都指向同一个对象。您可以id(a)和id(b)进行确认。

关于示例，请注意a[:]将创建一个新对象，该对象是a的副本。但是，它将是浅拷贝，而不是深拷贝。这解释了为什么在示例4中，您仍然可以同时修改a和b。它们指向不同的列表对象，但是一个元素是由它们共享的另一个列表。

Answer 3

案例1：名称b正在反弹。

案例2：a和b绑定到同一个对象。

案例3：a的浅表副本绑定到b。列表不同，但列表中的对象是相同的。

案例4：a的浅表副本绑定到b，然后其中一个对象发生变异。

重新绑定不会发生变异，并且变异不会重新绑定。

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'

Python变量处理，我不明白

4 个答案: