How to create an independent copy of a function in python?
Question:
Is it possible in python to create an un-linked copy of a function? For example, if I have
a = lambda(x): x
b = lambda(x): a(x)+1
I want b(x) to always return x+1, regardless if a(x) is modified not. Currently, if I do
a = lambda(x): x
b = lambda(x): a(x)+1
print a(1.),b(1.)
a = lambda(x): x*0
print a(1.),b(1.)
the output is
1. 2.
0. 1.
Instead of being
1. 2.
0. 2.
as I would like to. Any idea on how to implement this? It seems that using deepcopy does not help for functions. Also keep in mind that a(x) is created externally and I can’t change its definition. I’ve also looked into using this method, but it did not help.
Answers:
You could define b like this:
b = lambda x, a=a: a(x)+1
This makes a a parameter of b, and therefore a local variable. You default it to the value of a in the current environment, so b will hold onto that value. You don’t need to copy a, just keep its current value, so that if a new value is created, you have the one you wanted.
That said, this sounds like something unusual happening, and if you tell us more about what’s going on, there’s likely a better answer.
I might need to know a little more about your constraints before I can give a satisfactory answer. Why couldn’t you do something like
a = lambda(x): x
c = a
b = lambda(x): c(x)+1
Then no matter what happens to a, b will stay the same. This works because of the somewhat unusual way that assignment works in Python. When you do c = a, the name c is linked to the object that a links to. You can use id() to look at that object’s id and see that they are the same; c and a point to the same object.
c = a
id(a)
>>> 4410483968
id(c)
>>> 4410483968
Then, when you redefine a with a = lambda x: x*0, you’re actually doing two things in one line. First, lambda x: x*0 creates a new function object, then the assignment causes the name a to be linked to the new object. As you can see, a is now pointing to a different object:
a = lambda x: x*0
id(a)
>>>> 4717817680
But if you look at id(c), it still points to the old object!
id(c)
>>>> 4410483968
This is because when we redefined a, we merely created a new object and linked a to it. c remains linked to the old one.
So if you redefine a as you do in the question, you get the output you want:
print a(1.),b(1.)
>>>> 0.0,2.0
Is it possible in python to create an un-linked copy of a function? For example, if I have
a = lambda(x): x
b = lambda(x): a(x)+1
I want b(x) to always return x+1, regardless if a(x) is modified not. Currently, if I do
a = lambda(x): x
b = lambda(x): a(x)+1
print a(1.),b(1.)
a = lambda(x): x*0
print a(1.),b(1.)
the output is
1. 2.
0. 1.
Instead of being
1. 2.
0. 2.
as I would like to. Any idea on how to implement this? It seems that using deepcopy does not help for functions. Also keep in mind that a(x) is created externally and I can’t change its definition. I’ve also looked into using this method, but it did not help.
You could define b like this:
b = lambda x, a=a: a(x)+1
This makes a a parameter of b, and therefore a local variable. You default it to the value of a in the current environment, so b will hold onto that value. You don’t need to copy a, just keep its current value, so that if a new value is created, you have the one you wanted.
That said, this sounds like something unusual happening, and if you tell us more about what’s going on, there’s likely a better answer.
I might need to know a little more about your constraints before I can give a satisfactory answer. Why couldn’t you do something like
a = lambda(x): x
c = a
b = lambda(x): c(x)+1
Then no matter what happens to a, b will stay the same. This works because of the somewhat unusual way that assignment works in Python. When you do c = a, the name c is linked to the object that a links to. You can use id() to look at that object’s id and see that they are the same; c and a point to the same object.
c = a
id(a)
>>> 4410483968
id(c)
>>> 4410483968
Then, when you redefine a with a = lambda x: x*0, you’re actually doing two things in one line. First, lambda x: x*0 creates a new function object, then the assignment causes the name a to be linked to the new object. As you can see, a is now pointing to a different object:
a = lambda x: x*0
id(a)
>>>> 4717817680
But if you look at id(c), it still points to the old object!
id(c)
>>>> 4410483968
This is because when we redefined a, we merely created a new object and linked a to it. c remains linked to the old one.
So if you redefine a as you do in the question, you get the output you want:
print a(1.),b(1.)
>>>> 0.0,2.0