Accessing variables defined in external function from internal function

Question

Suppose I have the following function that returns a function:

def make_square_matrix_creator(dim):

    mat = np.zeros([dim, dim])

    def square_matrix_creator(value):
        mat += value
        return mat
    return square_matrix_creator

Now, this code doesn't work, because the internal function can't access mat.

f = make_square_matrix_creator(4)

f(3)

UnboundLocalError: local variable 'mat' referenced before assignment

I know there are a few ways to get around this; I can make mat global:

def make_square_matrix_creator(dim):

    global mat
    mat = np.zeros([dim, dim])

    def square_matrix_creator(value):
        global mat
        mat += value
        return mat
    return square_matrix_creator

It works, but this has all of the problems associated with making global objects within functions

I can pass mat as a default argument to the internal function;

def make_square_matrix_creator(dim):

    mat = np.zeros([dim, dim])

    def square_matrix_creator(value, mat=mat):
        mat += value
        return mat
    return square_matrix_creator

But when I try this out in my real-world example, I run into problems with mutable defaults. Are there other options for giving an internal function access to objects created in its parent function?

If you are using Python 3.x, you can look into [nonlocal](https://docs.python.org/3/reference/simple_stmts.html#grammar-token-nonlocal_stmt) statement. — Anand S Kumar, Oct 13 '15 at 16:50
http://eli.thegreenplace.net/2011/05/15/understanding-unboundlocalerror-in-python/ — Cong Ma, Oct 13 '15 at 17:16

score 0 · Answer 1 · answered Oct 13 '15 at 16:58

For Python 2.7, you can use [:] to indicate in-place mutation in the nested function. Example -

def make_square_matrix_creator(dim):
    mat = np.zeros([dim, dim])
    def square_matrix_creator(value):
        mat[:] = mat + value
        return mat
    return square_matrix_creator

I tested this in Python 3.4 (Sadly, I don't have a Python 2.7 with numpy to test it out , I tested in Python 2.7 with normal lists and it works for normal lists) . Demo -

In [50]: def make_square_matrix_creator(dim):
   ....:         mat = np.zeros([dim, dim])
   ....:         def square_matrix_creator(value):
   ....:                 mat[:] = mat + value
   ....:                 return mat
   ....:         return square_matrix_creator
   ....:

In [53]: f = make_square_matrix_creator(4)

In [54]: f(3)
Out[54]:
array([[ 3.,  3.,  3.,  3.],
       [ 3.,  3.,  3.,  3.],
       [ 3.,  3.,  3.,  3.],
       [ 3.,  3.,  3.,  3.]])

Cong Ma · Answer 2 · 2015-10-13T17:21:44.417

You can try creating an alias.

# This class is a minimally working shim for the numpy object
# because I don't want to import the full numpy package.
class K(object):
    def __init__(self, other):
        self.v = other

    def __iadd__(self, other):
        self.v += other

def outer(x):
    mat = K(x)
    def inner(y):
        alias = mat   # Create alias to mutable object.  alias is now modifiable
        alias += 1    # Modify alias to work around Python 2.x limitation.
        return mat
    return inner

>>> f = outer(5)
>>> res = f(1)
>>> res
<__main__.K at 0x102acb1d0>
>>> res.v
6
>>> f(1)
<__main__.K at 0x102acb1d0>
>>> res.v
7

Accessing variables defined in external function from internal function

2 Answers2