A question regarding the Lambda function in Python

Question

Why in the following code, the output is 22?

In my understanding, we have a function that needs two arguments, but it has been defined with only one! However, the first time we use it in mydoubler = myfunc(2), it assigns the argument(2) to variable n, but the second time we use it in print(mydoubler(11), it uses the argument(11) to set the value of the variable a! Why is that? Does Lambda work like a recursive function?

   def myfunc(n):
        return lambda a : a * n

   mydoubler = myfunc(2)

   print(mydoubler(11))

DonnumS · Accepted Answer · 2020-06-24 15:10:25Z

1

Basically what happens is this:

mydoubler = myfunc(2) is actually the same as writing mydoubler = lambda a : a * 2

The reason for this is that myfunc(2) returns lambda a : a * 2

So now mydoubler = lambda a : a * 2

Then when mydoubler(11) is called, it simply returns 11 * 2

answered Jun 24, 2020 at 15:10

DonnumS

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Stephane Desnault · Accepted Answer · 2020-06-24 15:17:31Z

1

You're returning a lambda, which is a one-liner function, NOT a number. The code below does the EXACT SAME thing, but is maybe a bit clearer as to its purpose:

def multiplier_factory(constant_factor):
    # Define our new function
    def multiplier(factor):
        result = constant_factor * factor
        return result
    # Return the FUNCTION, not a number
    return multiplier

doubler = multiplier_factory(2)
tripler = multiplier_factory(3)

print (doubler(1))  # prints 2
print (tripler(1))  # prints 3

print (doubler('a'))  # prints 'aa'
print (tripler('a'))  # prints 'aaa'

answered Jun 24, 2020 at 15:17

Stephane Desnault

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Victor Corte · Accepted Answer · 2020-06-24 15:03:24Z

0

lambda a: a * n is the same of:

def somefunction(a):
    return a * n

When you called myfunc(2) you dynamically created a function that is the same of:

def somefunction(a):
    return a * 2

answered Jun 24, 2020 at 15:03

Victor Corte

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Bram Dekker · Accepted Answer · 2020-06-24 15:06:56Z

0

myfunc returns a function. So mydoubler is a function and is described by lamda a : a * 2. Then you call that function with the argument 22 and so naturally 11 * 2 = 22 is printed. Lambda functions are not per se recursive, they are just a shorter way of writing a simple function. In your case you can also write:

def myfunc(n):
    def multiplier(a):
        return a * n

    return multiplier

answered Jun 24, 2020 at 15:06

Bram Dekker

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