How to pass arguments to a function inside a dictionary?

Question

As switch case is not available in Python, I made use of dictionary for this to perform arithmetic operation:

a = 10
b = 5

def add(a, b):
    return a + b 

def sub(a, b):
    return a - b

def mul(a, b):
    return a * b

def div(a, b):
    return a / b

def default():
    return "not a valid option"

opt = {
   1: add,
   2: sub,
   3: mul,
   4: div
}

op = 2

def getdetails(op):
    return opt.get(op, default)()

getdetails(op)

How do I pass my a and b value to my function when I enter option which hits the dictionary to call particular function?

"switch case is not available in Python" Yes, it is, since version 3.10, see structural pattern matching. — matszwecja, Jan 20 '22 at 14:56
https://docs.python.org/3/whatsnew/3.10.html#simple-pattern-match-to-a-literal see the release notes of python 3.10 — 3dSpatialUser, Jan 20 '22 at 14:58

Tomerikoo · Answer 1 · 2022-01-20T16:29:34.240

Since Python 3.10, there actually is a "switch-case" feature, called pattern matching:

a = 10
b = 5

op = 2

match op:
    case 1:
        print(a + b)
    case 2:
        print(a - b)
    case 3:
        print(a * b)
    case 4:
        print(a / b)
    case _:
        print("not a valid option")

The last case, _, is a wildcard.

For earlier versions you can easily "emulate" something similar to a switch-case:

a = 10
b = 5

op = 2

def case(value):
    return op == value

if case(1):
    print(a + b)
elif case(2):
    print(a - b)
elif case(3):
    print(a * b)
elif case(4):
    print(a / b)
else:
    print("not a valid option")

Or, for keeping with your approach, I would do some small changes to your code:

No need to implement all those functions - there is the operator module for that.
I wouldn't even bother with a default function - as others mentioned, the different signature will cause you headaches ahead.
Instead, I would go for forgiveness instead of permission and assume the op exists, and if there is a KeyError - we know it is the default.
Finally, your actual mistake is that you simply didn't pass the arguments - opt.get(op, default)() should simply be opt.get(op, default)(a, b), apart from the problem mentioned above with default.

from operator import add, sub, mul, truediv

a = 10
b = 5

opt = {
   1: add,
   2: sub,
   3: mul,
   4: truediv
}

op = 2

try:
    print(opt[op](a, b))
except KeyError:
    print("not a valid option")

Danil Perestoronin · Answer 2 · 2022-01-20T15:29:57.777

0

Tou should change:

def default():
    return "not a valid option"

def getdetails(op):
    return opt.get(op,default)()

getdetails(op)

To

def default(*_):
    return "not a valid option"

def getdetails(op):
    return opt.get(op, default)

res = getdetails(op)(a,b)

And it will work.

Full code example:

a = 10
b = 5


def add(a, b):
    return a + b


def sub(a, b):
    return a - b


def mul(a, b):
    return a * b


def div(a, b):
    return a / b


def default(*_):
    return "not a valid option"


opt = {
    1: add,
    2: sub,
    3: mul,
    4: div
}


def getdetails(op):
    return opt.get(op, default)


op = 2

res = getdetails(op)(a,b)

print(res)

edited Jan 20 '22 at 15:29

answered Jan 20 '22 at 15:15

Danil Perestoronin

1,063
1
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9

can you explain me what is it `*_` – Jan 20 '22 at 15:56
@coreke2042 `*_` - for ignoring arguments of the function `default` (because you don't use it anyway). `*` - used to pass a variable number of arguments to a function, `_` - for ignoring them More info can be found here https://stackoverflow.com/questions/13726155/is-an-acceptable-way-to-ignore-arguments-in-python – Danil Perestoronin Jan 20 '22 at 16:09

3dSpatialUser · Answer 3 · 2022-01-20T16:15:44.270

Your code looks ok to me, but I made a change in your default function, so that there is no error when op is not available in your dictionary. Also add the function call:

a=10
b=5

def add(a,b):
    return a+b 

def sub(a,b):
    return a-b

def mul(a,b):
    return a*b

def div(a,b):
    return a/b

def default(*args): # a, b are input, to prevent an positional arguments error add *args.
    return "not a valid option"

opt={
   1: add,
   2: sub,
   3: mul,
   4: div
}

op=2
# Call the function:
print(opt.get(op, default)(a,b))
>>> 5
print( opt.get(None, default)(a,b) )
>>> 'not a valid option'

EDIT:

Why the *args part in the function? *args is one method to capture variable for a function which are not defiend:

def test(*args):
    for arg in args:
        print(arg, end=' ') # end=' ' adds an space instead of a new line after print.

test(1, 2, 5)
>>> 1, 2, 3

What happens when a function is called without *args?:

def test():
    print('TEST')

test(1, 2)
>>> TypeError: test() takes 0 positional arguments but 2 were given

You are expecting a function from your dictionary and you are going to add the arguments a and b. Therefore you have to deal with them in each function in the dictionary.

What you also could do is:

def default(a,b):
    # Just ignore a,b
    return 'not valid'

That will also work, but if you want to add more arguments for other functions in the future the advantage of *args is taht it can take as much arguments as you want, there is no maximum.

can you explain in detail about *args why it is used in default () — , Jan 20 '22 at 16:10

Mad Physicist · Answer 4 · 2022-01-20T15:08:55.760

Considering that default is a special case, I would handle it manually instead of with get:

# instead of baking your own, use stdlib
from operator import add, sub, mul, div, mod

opt = {
   '+': add,
   '-': sub,
   '*': mul,
   '/': div,
   '%': mod,
}

def getresult(op, a, b):
    try:
        return opt[op](a, b)
    except KeyError:
        return default()

getresult('*', 10, 5)

If you do want to use get, you have to normalize the signature of default to accept the same arguments as the other options:

def default(*args, **kwargs):
    return 'Invalid option'

def getresult(op, a, b):
    return opt.get(op)(a, b)

How to pass arguments to a function inside a dictionary?

4 Answers4