如何使两个目录条目始终引用相同的浮点值

发布于 2024-09-08 20:47:41 字数 358 浏览 3 评论 0原文

考虑一下：

>>> foo = {}
>>> foo[1] = 1.0
>>> foo[2] = foo[1]
>>> foo
{1: 0.0, 2: 0.0}
>>> foo[1] += 1.0
{1: 1.0, 2: 0.0}

这就是发生的事情。但是，我想要的是最后一行内容为：

 {1: 1.0, 2: 1.0}

意味着两者都引用相同的值，即使该值发生变化。我知道上面的代码是这样工作的，因为数字在 Python 中是不可变的。有没有比创建自定义类来存储值更容易的方法？

原文

Consider this:

>>> foo = {}
>>> foo[1] = 1.0
>>> foo[2] = foo[1]
>>> foo
{1: 0.0, 2: 0.0}
>>> foo[1] += 1.0
{1: 1.0, 2: 0.0}

This is what happens. However, what I want would be that the last line reads:

 {1: 1.0, 2: 1.0}

Meaning that both refer to the same value, even when that value changes. I know that the above works the way it does because numbers are immutable in Python. Is there any way easier than creating a custom class to store the value?

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柠檬色的秋千 2024-09-15 20:47:41

只有可变对象才有可能，因此您必须用一些可变对象包装不可变值。事实上，任何可变对象都可以，例如内置列表：

>>> n = [0]
>>> d = { 1 : n, 2 : n }
>>> d
{1: [0], 2: [0]}
>>> d[1][0] = 3
>>> d
{1: [3], 2: [3]}

但是创建自己的类或对象有什么困难呢？

>>> n = type( "number", ( object, ), { "val" : 0, "__repr__" : lambda self: str(self.val) } )()
>>> d = { 1 : n, 2 : n }
>>> d
{1: 0, 2: 0}
>>> d[1].val = 9
>>> d
{1: 9, 2: 9}

工作得一样好；）

It is possible only with mutable objects, so you have to wrap your immutable value with some mutable object. In fact any mutable object will do, for example built-in list:

>>> n = [0]
>>> d = { 1 : n, 2 : n }
>>> d
{1: [0], 2: [0]}
>>> d[1][0] = 3
>>> d
{1: [3], 2: [3]}

but what's hard in creating your own class or object?

>>> n = type( "number", ( object, ), { "val" : 0, "__repr__" : lambda self: str(self.val) } )()
>>> d = { 1 : n, 2 : n }
>>> d
{1: 0, 2: 0}
>>> d[1].val = 9
>>> d
{1: 9, 2: 9}

Works just as fine ;)

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溇涏 2024-09-15 20:47:41

在Python中拥有一种指针的更简单的方法是将你的值打包到一个列表中。

>>> foo = {}
>>> foo[1] = [1.0]
>>> foo[2] = foo[1]

>>> foo
{1: [1.0], 2: [1.0]}

>>> foo[1][0]+=100 # note the [0] to write in the list

>>> foo
{1: [101.0], 2: [101.0]}

有效！

The easier way to have a kind of pointer in python is pack you value in a list.

>>> foo = {}
>>> foo[1] = [1.0]
>>> foo[2] = foo[1]

>>> foo
{1: [1.0], 2: [1.0]}

>>> foo[1][0]+=100 # note the [0] to write in the list

>>> foo
{1: [101.0], 2: [101.0]}

Works !

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~没有更多了~