问题
In defining variable of a list object, for example:
x = [1,2,0.2,3,4]
y = x
x.sort()
I would expect that y is still equal to [1, 2, 0.2, 3, 4], but it does not. The value of y changed as x changed. To counter this, I found that using y = x.copy() can preserve the value in the first line.
On the other hand, another example :
x = 5
y = x
x = 4
from this the value of y is still 5, it does not change as x change.
My question : is this due to the design in list's class, or there is another explanation? I found the dynamic change also happen when using x.append(value). Any insight is appreciated. Regards, Arief
回答1:
Every variable is just a pointer to an Python object, if you have two variables pointing to the same object then you'll see the changes in each of them (and .sort works in-place, if you want a new list you should use x = sorted(x)). However if you re-assign a variable then it will point to a different object.
I included some images to better visualize what's happening (not high-quality but I hope it conveys the message).
x = [1,2,0.2,3,4]
y = x
If you copy (it's a shallow copy so the list-contents still refer to the same items!):
x = [1,2,0.2,3,4]
y = x.copy()
Your second case is just the same:
x = 5
y = x
But then you re-assign the variable x (so it points to another object thereafter):
x = 4
回答2:
The problem is, y and x are just references to the class list.
When you do something like:
y=x
You are coping the reference of the class and not creating another one.
When using copy you are doing a shallow copy that are creating a new class, copying all elements again to this new object.
Python manual presents a explanation and others operators used to actually copy a full class.
来源：https://stackoverflow.com/questions/44401453/reference-list-object-by-a-variable-is-different-than-for-some-other-objects