Higher-order function from old Python version to Python 3

Question

Does anyone know how to translate this HOF from previous version of Python to Python 3?

apply_each = lambda fns, args=[]: map(apply, fns, [args]*len(fns))

This comes from the book titled: Text Processing in Python by David Mertz. I'm not able to use the apply function in Python 3 since it has been deprecated. I've tried using func(*args, **kwargs) instead of apply(func, args, kwargs), but I get this TypeError: 'float' object is not iterable. I've also found those notes which show what the function is supposed to return: https://wiki.python.org/moin/TextProcessingInPython

here's my try at the code:

apply_each = lambda fns, args=[]: map(fns, *args)
args = [5.4, 6.3, 6.3]
print(list(apply_each(approx, args)))

juanpa.arrivillaga · Accepted Answer · 2019-02-03 19:27:26Z

1

You can define your own .apply:

>>> def apply(f, args):
...     return f(*args)
...
>>> apply_each = lambda fns, args=[]: map(apply, fns, [args]*len(fns))

Or all in one go:

apply_each = lambda fns, args=[]: map(lambda f, args: f(*args), fns, [args]*len(fns))

But this is poor style, the only advantage of a lambda is that it is anonymous, if you are going to assign it to a name, that defeats the purpose. This is better style:

def apply(f, args):
    return f(*args)

def apply_each(fns, args=()):
    return map(apply, fns, [args]*len(fns))

So:

>>> import operator as op
>>> list(apply_each([op.add, op.sub, op.mul, op.truediv], args=[1,2]))
[3, -1, 2, 0.5]

edited Feb 3, 2019 at 19:27

answered Feb 3, 2019 at 19:19

juanpa.arrivillaga

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