Anonymous function managing function's outputs, without lambda

Question

Say I have a function that returns a tuple of outputs:

def f():
    return 'zero', 1

But I have to supply in-line a new anonymous function that returns only e.g. the zeroth element of that returned tuple. This is an easy one-liner with a lambda:

lambda : f()[0]  # returns 'zero'

Is there a way to do the same thing using functools or similar?

Given that lambda was allegedly 'planned to [be removed] from Python 3, as one of "Python's glitches"', I'm assuming there's a way to do this without lambda, but I haven't been able to figure it out.

I know that functools.partial can pretty much be a substitute for lambda when managing a function's input arguments. But managing a function's outputs? Haven't figured it out.

chepner · Accepted Answer · 2018-01-03 14:36:46Z

2

You are perhaps looking for operator.itemgetter:

from operator import itemgetter

get_zeroth = itemgetter(0)
assert get_zeroth(f()) == 'zero'

answered Jan 3, 2018 at 14:36

chepner

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score 1 · Accepted Answer · 2018-01-03 15:11:27Z

1

You can try behaviour to produce a poor-man's conditional expression

print(f()[f()[0]==' '])

Since [f()[0]!=' '] produces a boolean , As you said it produce a tuple so [f()[0]==' '] will always be wrong so it will give 0 which is index no which you want.

edited Jan 3, 2018 at 15:11

answered Jan 3, 2018 at 15:05

user9158931

Comments

fuglede · Accepted Answer · 2018-01-25 14:16:52Z

0

The current answers appear to be producing the value itself -- which you seem to be content with achieving through f()[0] -- as opposed to a callable producing the value. Is lazy evaluation (as you have with your lambda example) a requirement? If not, the problem is no longer that of post-composing f with a projection but rather that of defining a constant function, and I'll nominate

itertools.repeat(f()[0]).__next__

edited Jan 25, 2018 at 14:16

answered Jan 24, 2018 at 21:01

fuglede

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