Mapping multiple functions to a single value

Question

I am working on some python exercises. I understand there are more efficient ways of solving this problem, however, I am trying to get a better understanding of what the question is asking.

Q: Use a map and lambda to pass multiple functions to a single value.

It is important to note here that I have not assigned a variable name as list. Here is what I have tried:

import math

mult_funcs = [math.sqrt, math.log]

test_func = list(map(lambda x: x, mult_funcs))
test_func(2)

---------------------------------------------------------------------------
TypeError                                 Traceback (most recent call last)
<ipython-input-89-6561142477ba> in <module>
      1 test_func = list(map(lambda x: x, func_lst))
----> 2 test_func(2)

TypeError: 'list' object is not callable

Investigating SO led me to this solution: Apply multiple functions to the same argument in functional Python

Now, I have the code below which works. However, I am curious how to modify the lambda function so the value can change from 2 to any other number. In other words, I'd like to be able to do a regular function call (e.g., test_func(4)) instead of setting the lambda equal to a variable.

arg = 2
test_func = list(map(lambda x: x(arg), mult_funcs))

In the first variant you create a list (what the list(...) call does) and assign it to test_func. In the next line you try to call this list. In the second variant you can just replace the arg in the second line by 4 or anything else. First line can then be omitted. — Michael Butscher
– Michael Butscher, Commented Jan 12, 2020 at 1:57

Mark · Accepted Answer · 2020-01-12 01:56:57Z

2

If you want to call test_func it needs to be a function. So if you need to start with:

test_func = lambda x: ...

With that you can use map within that function and that map will need another lambda. Something like:

import math

mult_funcs = [math.sqrt, math.log]

test_func = lambda x: list(map(lambda f: f(x), mult_funcs))
test_func(2)
# [1.4142135623730951, 0.6931471805599453]

Which will capture the passed in number in a closure and apply the functions to it in the second lambda.

answered Jan 12, 2020 at 1:56

Mark

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2 Comments

Kelly Bundy Over a year ago

Hmm, why use lambda for test_func rather than def?

Mark Over a year ago

@HeapOverflow Because that’s the assignment. But I agree, a def makes sense.

Austin · Accepted Answer · 2020-01-12 02:04:44Z

1

Your first code snippet did not work because test_func at the end of lambda operation is a list and you can't do a function-call like thing on your list.

The most pythonic way would be to iterate over than using lambda:

import math

mult_funcs = [math.sqrt, math.log]

arg = 2
test_func = [x(arg) for x in mult_funcs]
# or: test_func = list(map(lambda x: x(arg), mult_funcs))

print(test_func)

In fact, the variable test_func is not a function here, but you get the idea.

edited Jan 12, 2020 at 2:04

answered Jan 12, 2020 at 1:59

Austin

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1 Comment

tlk27 Over a year ago

That makes sense and I appreciate the alternative approach to solving this problem. In the spirit of the exercise question I was required to use map and lambda.

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