Function with lists/scalars as arguments

Question

If I define a function as:

def somefunc(a, b, c):
    out = [ (a+b)/c for a, b, c in zip(a, b, c) ]
    return out

all the arguments must be lists of the same length, otherwise I would get the output truncated. Example: somefunc([1, 2], [3, 4], [5, 5]) would give me [0.8, 1.2].

It would be really convenient to be able to input a single number instead of a list when it doesn't change, like: somefunc([1, 2], [3, 4], 5). Of course any of the arguments can be reduced to a single value instead of a list. This can be achieved using numpy with this function definition:

from numpy import array

def somefunc(a, b, c):

    a=array(a)
    b=array(b)
    c=array(c)

    out = (a+b)/c
    return out

Is there a way to do the same without numpy (and without converting all the arguments one by one)?

No, zip expects iterable arguments; if your function takes non-iterable arguments, you will have to convert them. — jonrsharpe
– jonrsharpe, Commented Jul 23, 2014 at 15:55
Note: If you want NumPy features, the easiest way is going to be to use NumPy. Each feature can of course be implemented without NumPy, but if that was easy, NumPy wouldn't be here. — user2357112
– user2357112, Commented Jul 25, 2014 at 8:26

Hans Then · Accepted Answer · 2014-07-23 16:24:07Z

2

Sure, you can make an infinite list repeating any argument that is not a list in itself.

from itertools import repeat

def broadcast(o):
    return o if hasattr(o, '__iter__') else repeat(o)

def somefunc(a, b, c):
    out = [ (a+b)/c for a, b, c in zip(broadcast(a), broadcast(b), broadcast(c)) ]
    return out

>>> somefunc([1, 2], [3, 4], 5)
[0.8, 1.2]
>>> somefunc([1, 2], 5, [3, 4])
[2.0, 1.75]
>>>

edited Jul 23, 2014 at 16:24

answered Jul 23, 2014 at 16:00

Hans Then

11.4k3 gold badges36 silver badges52 bronze badges

Sign up to request clarification or add additional context in comments.

3 Comments

gaggio Over a year ago

Ok, but it doesn't work if the scalar argument is not the last: somefunc([1, 2], 3, [5, 5])

Hans Then Over a year ago

I think you commented the wrong answer. rroowwllaanndd's solution does not work if either of the first two arguments are scalars. Mine does.

gaggio Over a year ago

ooops you are right, but now the other has been edited and it works as well. thanks

shx2 · Accepted Answer · 2014-07-25 08:20:49Z

1

Numpy is awesome, and is perfect for vectorizing computaions like the one you posted. I'd recommend you do use it.

Also, you don't need to convert all the arguments to arrays, only one, and it is better to use asarray, which would avoid creating a new copy when one isn't necessary.

def somefunc(a, b, c):
    return (asarray(a) + b) / c

edited Jul 25, 2014 at 8:20

answered Jul 23, 2014 at 19:14

shx2

64.8k17 gold badges139 silver badges166 bronze badges

Collectives™ on Stack Overflow

Function with lists/scalars as arguments

2 Answers 2

3 Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

3 Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related