CODEWITHSUNDEEP
pythonlambdapython-internals
sashab
sashab

Reputation: 1564

Lambda-function comparison in python

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?

Upvotes: 4

Views: 4052

Answers (1)

Ignacio Vazquez-Abrams
Ignacio Vazquez-Abrams

Reputation: 799310

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

Upvotes: 6

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