lambda: confused about parameter transmission

Question

I come from the SQL world, and I am a newbie to this. I found on w3shools.com similar code:

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

mydoubler = myfunc(2, 6)
print(mydoubler(15))

I cannot figure out how the a parameter, which I asssume is "15" in the call, is made available to the lambda. I searched and I could not find a description of the logical and programmatic paradigm.

Answer 1

lambda function is an anonymous function, this mean that a lambda expression is a function by itself but is inline. here, you created a function that will return a lambda expression that calculates the products of a and n and m so the returned value will be that. we know that when we return a function, we can store what we returned in a variable, in our case, we returned a lambda function so you need to enter the a value. if it is not clear, ask me

Answer 2

Lambda function is a function, but without a name. The only difference is, ho definition looks like. Your code snipped could be translated to:

def myfunc(n, m):

    def some_func(a):
        return a * n * m

    return some_func

mydoubler = myfunc(2, 6)
print(mydoubler(15))

And the result would be the same. From the official Python docs Lambda Expressions:

Like nested function definitions, lambda functions can reference variables from the containing scope

Because of that, values of n and m are available for the function. Value of a is passed when you are executing mydoubler.

Answer 3

this is a detailed answer:

first of all, you created the function myfunc that requires 2 arguments "n" and "m". then you will return a lambda function that takes an argument "a" and returns the product of "a","n" and "m". then you declared a variable mydoubler whose value is the returned value of myfunc which is that lambda. imagine that mydoubler = lambda a: a * 2 * 6. and finally you printed the returned value of mydoubler which is a function and the argument a is 15. if you didnt understand the lambda format for mydoubler, here is it but as a normal function: def mydoubler(a): return a * 2 * 6 dont forget, 2 is the value of n and 6 is the value of m