Setting numpy slice in lambda function

Question

I want to create a lambda function that takes two numpy arrays and sets a slice of the first to the second and returns the newly set numpy array.

Considering you can't assign things in lambda functions is there a way to do something similar to this?

The context of this is that I want to set the centre of a zeros array to another array in a single line, and the only solution I could come up with is to use reduce and lambda functions.

I.e. I'm thinking about the condensation of this (where b is given):

a = numpy.zeros( numpy.array(b.shape) + 2)
a[1:-1,1:-1] = b

Into a single line. Is this possible? This is just an exercise in oneliners. I have the code doing what I want it to do, I'm just wondering about this for the fun of it :).

Answer 1

This is ugly; you should not use it. But it is oneline lambda as you've asked:

f = lambda b, a=None, s=slice(1,-1): f(b, numpy.zeros(numpy.array(b.shape) + 2))\
                      if a is None else (a.__setitem__([s]*a.ndim, b), a)[1]

What is __setitem__?

obj.__setitem__(index, value) is equivalent to obj[index] = value in this case. Example:

class A:
    def __setitem__(self, index, value):
       print 'index=%s, value=%s' % (index, value)

a = A()
a[1, 2] = 3

It prints:

index=(1, 2), value=3

Why does __setitem__() return None?

There is a general convention in Python that methods such as list.extend(), list.append() that modify an object in-place should return None. There are exceptions e.g., list.pop().

Y Combinator in Python

Here's blog post On writing Python one-liners which shows how write nameless recursive functions using lambdas (the link is suggested by @Peter Hansen).