Python: Can I partially apply reduce with an initializer?

Question

The initializer comes after the iterable. This causes problems for partial application. Consider these (trivial) examples:

In [1]: from functools import reduce, partial

In [2]: f = partial(reduce, lambda a,b: a+b, 100)

In [3]: f([1,2,3])
---------------------------------------------------------------------------
TypeError                                 Traceback (most recent call last)
<ipython-input-3-816cce84b257> in <module>()
----> 1 f([1,2,3])

TypeError: reduce() arg 2 must support iteration

In [4]: f = partial(reduce, lambda a,b: a+b, initializer=100)

In [5]: f([1,2,3])
---------------------------------------------------------------------------
TypeError                                 Traceback (most recent call last)
<ipython-input-5-816cce84b257> in <module>()
----> 1 f([1,2,3])

TypeError: reduce() takes no keyword arguments

Is there some trick to get around this?

The problem seems to be the commas _inside_ your lambda. Try parenthesizing the whole thing. — jhpratt, Apr 09 '18 at 02:15
Ah, ok. I don't use `reduce` a ton, so I didn't remember the order off the top of my head. — jhpratt, Apr 09 '18 at 02:23
this was just a trivial example of the pattern. That wouldn't work in all situations. — Solomon Bothwell, Apr 09 '18 at 02:35
So do you have an actual example, where the only option is to pass an initializer to reduce? — Sohaib Farooqi, Apr 09 '18 at 02:40
Well, in Python the main case is if its possible the list is empty — Solomon Bothwell, Apr 09 '18 at 03:05

score 1 · Accepted Answer · answered Apr 09 '18 at 02:22

1

Just change the order of arguments:

>>> f = partial(lambda func, init, iterable: reduce(func, iterable, init), 
                lambda a,b: a+b, 100)
>>> f([1,2,3])
106

answered Apr 09 '18 at 02:22

Marat

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Cool solution!! – Solomon Bothwell Apr 09 '18 at 02:22

score 0 · Answer 2 · answered Apr 09 '18 at 02:24

Marat gave a great solution. For fun, I wrote foldl with the arguments flipped. I'm curious how this would compare performance-wise to using a lambda to flip the arguments like Marat did.

def foldl(f: Callable, acc: Any, xs: Iterable) -> Any:
    if not xs:
        return acc
    return foldl(f, f(acc, xs[0]), xs[1:])

Python: Can I partially apply reduce with an initializer?

2 Answers2