Map list of functions to list of arguments in python

Question

I have a simple example as so:

import numpy as np
func_dict1 = {0: np.sin, 1: np.cos, 2: np.tan}
out = map(func_dict1.get, np.array([0, 2, 0]))

Here I am picking out three functions by their dictionary keys. Now I want to pass unique arguments to each function like so:

[f(x) for f,x in zip(out, [3,1,2])]

which renders the output:

[0.1411200080598672, 1.557407724654902, 0.9092974268256817]

But how can I do this with map?

I thought this would work, but it does not:

map(out, [3,1,2])

Where am I going wrong? And is there any benefit to using map over list comprehension? My prior is that it is faster but I confess to not being an expert on the subject.

smp55 · Accepted Answer · 2021-06-25 15:43:48Z

3

map is designed to take a single function and apply it to every item in an iterable. You are applying a different function to different items. I think the list comprehension is an elegant way of doing it.

answered Jun 25, 2021 at 15:43

smp55

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Jörn Hees · Accepted Answer · 2021-06-25 16:24:01Z

WARNING: you probably don't want to use map and this answer might confuse you more than it helps ;).

However, as you asked how you can make map do this and as it's python, let's take the challenge: one way to achieve what you want is by wrapping your out in an object that is callable (so behaves like a function) and on each call also advances to the next function. For example like this:

# yours
import numpy as np
func_dict1 = {0: np.sin, 1: np.cos, 2: np.tan}
out = map(func_dict1.get, np.array([0, 2, 0]))

# extend like this
class FuncIterCaller:
    def __init__(self, funcs):
        self.funcs = funcs

    def __call__(self, *args, **kwds):
        return next(self.funcs)(*args, **kwds)

res = map(FuncIterCaller(out), [3,1,2])

# to see what's inside:
print(list(res))

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