Question

我在Python程序中遇到过一种行为，我已经能够压缩成一个较小的程序。我很确定我已经遇到它并解决了它和过去，但不记得是怎么回事。

考虑以下Python程序：

import random

class B:
    def __init__(self):
        self.attr=random.randrange(0,10)

class A:
    def __init__(self):
        self.b=B()

list=[]
print "First pass"
for i in range(0,10):
    a=A()
    print "Random number in b: ",a.b.attr
    list.insert(0,a)

print "Second pass"
for l in list:
    print "Random number in b: ",a.b.attr

它通常输出如下内容：

First pass
Random number in b:  9
Random number in b:  7
Random number in b:  1
Random number in b:  5
Random number in b:  9
Random number in b:  7
Random number in b:  0
Random number in b:  6
Random number in b:  2
Random number in b:  7

Second pass
Random number in b:  7
Random number in b:  7
Random number in b:  7
Random number in b:  7
Random number in b:  7
Random number in b:  7
Random number in b:  7
Random number in b:  7
Random number in b:  7
Random number in b:  7

我们如何解决这个问题？我知道它与Python如何复制对象等有关，但有：

import copy
list.insert(0,copy.deepcopy(a))

不能像人们期望的那样解决它。

Answer 1

也许不是

print "Second pass"
for l in list:
    print "Random number in b: ",a.b.attr

使用

print "Second pass"
for l in list:
    print "Random number in b: ", l.b.attr

将引用对象的对象插入到列表中

1 个答案: