print the code which defined a lambda function [duplicate]

Question

I'd like to be able to print the definition code of a lambda function.

Example if I define this function through the lambda syntax:

>>>myfunction = lambda x: x==2 >>>print_code(myfunction) 

I'd like to get this output:

x==2 

Answer 1

As long as you save your code to a source file you can retrieve the source code of an object using the inspect module.

example: open editor type:

myfunction = lambda x: x==2 

save as lamtest.py

open shell type python to get to interactive python type the following:

>>>from lamtest import myfunc >>>import inspect >>>inspect.getsource(myfunc) 

the result:

'myfunc = lambda x: x==2\n' 

Answer 2

It will only work for mathematical based operations, but you might look at SymPy's Lambda() object. It was designed exactly for this purpose:

>>> from sympy import * >>> x = Symbol('x') >>> l = Lambda(x, x**2) >>> l Lambda(_x, _x**2) >>> l(3) 9 

It even supports pretty printing:

>>> pprint(l)  ⎛    2⎞ Λ⎝x, x ⎠ 

To do your equals example, us the SymPy Eq() object:

>>> l1 = Lambda(x, Eq(x, 2)) >>> l1 Lambda(_x, _x == 2) >>> l1(2) True 

It supports partial argument expansion:

>>> y = Symbol('y') >>> l2 = Lambda((x, y), x*y + x) >>> l2(1) Lambda(_y, 1 + _y) >>> l2(1, 2) 3 

And of course, you get the advantage of getting all of SymPy's computer algebra:

>>> l3 = Lambda(x, sin(x*pi/3)) >>> pprint(l3(1))   ⎽⎽⎽ ╲╱ 3  ─────   2   

By the way, if this sounds like a shameless plug, it's because it is. I am one of the developers of SymPy.