Python对象混淆:a=b，修改b和a变化！ [英] Python objects confusion: a=b, modify b and a changes!

问题描述

直到今晚我才知道 Python.做这样的事情的正确方法是什么?这是我的代码:

a = ["一", "二", "三"]b = a # 这里我想要一个完整的副本，当 b 改变时，对 a 绝对没有影响b.append["四"]print a # a 现在有四"

基本上我想知道，而不是 b = a 步骤，我将如何正确制作列表或字典的副本，以便在 b 更改时a 不会随之改变吗?

解决方案

您遇到的是引用的概念.Python 中的所有对象都有一个引用，当你为 a 和 b 分配一到两个名称时，这会导致 a 和 b 指向 same 对象.

<预><代码>>>>a = 范围(3)>>>b = a # 相同的对象>>>b.append(3)>>>a, b # 相同的内容([0, 1, 2, 3], [0, 1, 2, 3])

使用列表，您可以使用 b = a[ 创建一个新列表 b，它是另一个 a 的副本:].

<预><代码>>>>a = 范围(3)>>>b = a[:] # 使 b 成为 a 的新副本>>>b.append(3)>>>a, b # a 保持不变([0, 1, 2], [0, 1, 2, 3])

对于任何对象的更通用的解决方案，请使用复制模块.浅拷贝将复制您正在复制的对象中存储的引用，而深拷贝将递归地创建所有对象的新副本.

<预><代码>>>>a = [范围(2)，范围(3)]>>>b = copy.copy(a) # a 的浅拷贝，等价于 a[:]>>>b[0] = 范围(4)>>>a, b # 设置 b 的元素使 a 保持不变([[0, 1], [0, 1, 2]], [[0, 1, 2, 3], [0, 1, 2]])>>>b[1].append(3)>>>a, b # 修改 b 的元素会修改 a 中的元素([[0, 1], [0, 1, 2, 3]], [[0, 1, 2, 3], [0, 1, 2, 3]])>>>a = [范围(2)，范围(3)]>>>b = copy.deepcopy(a) # a 的深度递归副本>>>b[1].append(3)>>>a, b # 修改 b 中的任何内容都会保持 a 不变([[0, 1], [0, 1, 2]], [[0, 1], [0, 1, 2, 3]])

I thought i knew Python until tonight. What is the correct way to do something like this? Here's my code:

a = ["one", "two", "three"]
b = a  # here I want a complete copy that when b is changed, has absolutely no effect on a
b.append["four"]
print a  # a now has "four" in it

Basically i want to know, instead of the b = a step, how would I correctly make a copy of a list or dictionary so that when b is changed a does not change along with it?

解决方案

What you are experiencing is the concept of references. All objects in Python have a reference and when you assign one to two names a and b, this results in both a and b pointing to the same object.

>>> a = range(3)
>>> b = a                     # same object
>>> b.append(3)
>>> a, b                      # same contents
([0, 1, 2, 3], [0, 1, 2, 3])

With lists, you can make create a new list b that is a copy of another a using b = a[:].

>>> a = range(3)
>>> b = a[:]                  # make b a new copy of a
>>> b.append(3)
>>> a, b                      # a is left unchanged
([0, 1, 2], [0, 1, 2, 3])

For a more general solution for any object, use the copy module. A shallow copy will copy the references stored within the object you're copying, whereas a deep copy will recursively make new copies of all objects.

>>> a = [range(2), range(3)]
>>> b = copy.copy(a)          # shallow copy of a, equivalent to a[:]
>>> b[0] = range(4)
>>> a, b                      # setting an element of b leaves a unchanged
([[0, 1], [0, 1, 2]], [[0, 1, 2, 3], [0, 1, 2]])
>>> b[1].append(3)
>>> a, b                      # modifying an element of b modifies the element in a
([[0, 1], [0, 1, 2, 3]], [[0, 1, 2, 3], [0, 1, 2, 3]])

>>> a = [range(2), range(3)]
>>> b = copy.deepcopy(a)      # deep recursive copy of a
>>> b[1].append(3)
>>> a, b                      # modifying anything in b leaves a unchanged
([[0, 1], [0, 1, 2]], [[0, 1], [0, 1, 2, 3]])

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