Why doesn't functools.partial return a real function (and how to create one that does)?
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Published on 2012-11-20T22:47:15Z
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2012/11/20
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So I was playing around with currying functions in Python and one of the things that I noticed was that functools.partial returns a partial object rather than an actual function. One of the things that annoyed me about this was that if I did something along the lines of:
five = partial(len, 'hello')
five('something')
then we get
TypeError: len() takes exactly 1 argument (2 given)
but what I want to happen is
TypeError: five() takes no arguments (1 given)
Is there a clean way to make it work like this? I wrote a workaround, but it's too hacky for my taste (doesn't work yet for functions with varargs):
def mypartial(f, *args):
argcount = f.func_code.co_argcount - len(args)
params = ''.join('a' + str(i) + ',' for i in xrange(argcount))
code = '''
def func(f, args):
def %s(%s):
return f(*(args+(%s)))
return %s
''' % (f.func_name, params, params, f.func_name)
exec code in locals()
return func(f, args)
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