Implementing fuction that takes in array and multiplies each element by given int [duplicate]

Answer 1

This will give you the desired result:

def multAll(A, k):
    return [i * k for i in A]

It is using the list comprehension pattern that allows you to edit each element of the list.

Answer 2

Since you're calling the multAll function without expecting a returning value, you should update the list A in-place by updating its item values through indices:

def multAll(A, k):
    for i in range(len(A)):
        A[i] *= k

Answer 3

When return i happens for the first time in your function, the function stops and returns the current i.

def multAll(A, k):
    return_value = []
    for i in A:
        i = i*k
        return_value.append(i)
    return return_value

Like this, a complete list return_value is created, and that list is returned.

Answer 4

As written, your function should not return the given list; it should return the value of A[0] * k, the scalar 50.

Let's walk through the code for a moment, as you should have done with a print statement before posting.

for i in A:
    # i is a local variable; it takes on the value of A[0], or 5
    i = i*k
    # i is now 50
    return i
    # You return 50 to the calling program -- which ignores the return value.

At this point, the function is done, gone, exited, and will execute no more until you call it again.

To get the list (you have a list, not an **array*) altered, you can change each element in place. Don't leave the function until you've processed all of the elements.

for i in range(len(A)):
    A[i] *= k

return

Answer 5

Numbers are not the objects, so In loop every time you get only the next value of a list. And to that value, you do that operation. You should iterate over indexes, and assign a value to a specific place in the list. You could use the range function to do that.

Another problem is that you return value after calculating the first value in the list, so you should remove one indention.