Dynamically generating a function tree in Python.

Question

I want to do is something like this

def mult(x, y):
     return x * y

def add(x, y):
     return x + y

treeFunction = mult(2, add(x, y))

#outputs 10 (2 * ( 2 + 3 ))
print treeFunction(2, 3)

basically, is there a way during run-time to change add()'s return value by multiplying it by some value.

Chris Eberle · Accepted Answer · 2011-09-11 02:25:08Z

7

I think what you're asking for is a lambda function:

treeFunction = lambda x,y: mult(2, add(x, y))

Then just use it as:

treeFunction(2, 3)

EDIT

Have a read about closures in python. Definitely worth your time to learn.

edited Sep 11, 2011 at 2:25

answered Sep 11, 2011 at 1:47

Chris Eberle

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5 Comments

David Z Over a year ago

There are a number of complex and general ways to do this, but I agree that for this particular example, the easiest solution is a lambda function.

Chris Eberle Over a year ago

The other way I interpreted this question was as asking for an interpreter for some domain-specific language. This code was easier to write.

JBernardo Over a year ago

is it hard to write def treeFunction(x,y): return mult(2, add(x,y)) ? giving a name to a lambda is plain wrong

Femaref Over a year ago

No, it isn't. In this case, yes, there is no difference, but what if it is in a function, and you want to assign the variable differing calls depending on a condition? Using anonymous functions isn't new. I doesn't clutter the namespace and allows you to generate function calls on the fly.

John Over a year ago

After sleeping on this and re-reading it all you have answered my question, but it didn't really help with my problem =P. My mistake. Thanks for the link to the closure stuff, very interesting.

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