Lambda-function comparison in python

Question

In python you cannot directly compare functions created by lambda expressions:

>>> (lambda x: x+2) == (lambda x: x+2)
False

I made a routine to hash the disassembly.

import sys
import dis
import hashlib
import contextlib


def get_lambda_hash(l, hasher=lambda x: hashlib.sha256(x).hexdigest()):
    @contextlib.contextmanager
    def capture():
        from cStringIO import StringIO
        oldout, olderr = sys.stdout, sys.stderr
        try:
            out=[StringIO(), StringIO()]
            sys.stdout, sys.stderr = out
            yield out
        finally:
            sys.stdout, sys.stderr = oldout, olderr
            out[0] = out[0].getvalue()
            out[1] = out[1].getvalue()

    with capture() as out:
        dis.dis(l)

    return hasher(out[0])

The usage is:

>>>> get_lambda_hash(lambda x: x+2) == get_lambda_hash(lambda x: x+1)
False

>>>> get_lambda_hash(lambda x: x+2) == get_lambda_hash(lambda x: x+2)
True

Is there any more elegant solution for this problem?

If you need comparisons, perhaps a lambda isn't the tool for the job? — jonrsharpe
– jonrsharpe, Commented Feb 20, 2015 at 17:03
Should the expressions be syntactical or semantical equal? The latter can't be done at all. — willeM_ Van Onsem
– willeM_ Van Onsem, Commented Feb 20, 2015 at 17:10
@jonrsharpe consider you have a library to read long files with rows and columns into memory. You can specify columns you want to select, and also you can specify lambdas to do some trickery when selecting e.g. select sum of two columns. Now you want to do cache where key is which columns(and lambdas!) you select. — sashab
– sashab, Commented Feb 20, 2015 at 17:10
Yes, but it's not clear why that has led you to this requirement, which seems to indicate potential for broader improvements in design. This is likely an XY problem; see e.g. meta.stackexchange.com/q/66377/248731 — jonrsharpe
– jonrsharpe, Commented Feb 20, 2015 at 17:11

Ignacio Vazquez-Abrams · Accepted Answer · 2015-02-20 18:08:02Z

6

If you insist on performing this insane bit of insanity, compare the bytecode and constants of each.

>>> import operator
>>> coco = operator.attrgetter('co_code', 'co_consts')
>>> coco((lambda x: x+2).__code__) == coco((lambda x: x+2).__code__)
True
>>> coco((lambda x: x+2).__code__) == coco((lambda x: x+1).__code__)
False
>>> def foo(y):
...   return y + 2
... 
>>> coco((lambda x: x+2).__code__) == coco(foo.__code__)
True

answered Feb 20, 2015 at 18:08

Ignacio Vazquez-Abrams

804k160 gold badges1.4k silver badges1.4k bronze badges

Sign up to request clarification or add additional context in comments.

2 Comments

zch Over a year ago

This might be not good enough - for example for def f(x): return lambda y: x+y it would consider f(1) and f(2) to be the same function.

sashab Over a year ago

Also not good if data used in lambda changed, like if in lambda x: some_dict[x] dictionary some_dict changed.

Collectives™ on Stack Overflow

Lambda-function comparison in python

1 Answer 1

2 Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

2 Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related