python - Squaring all elements in a list

Question

I am told to

Write a function, square(a), that takes an array, a, of numbers and returns an array containing each of the values of a squared.

At first, I had

def square(a):
    for i in a: print i**2

But this does not work since I'm printing, and not returning like I was asked. So I tried

    def square(a):
        for i in a: return i**2

But this only squares the last number of my array. How can I get it to square the whole list?

Answer 1

You could use a list comprehension:

def square(list):
    return [i ** 2 for i in list]

Or you could map it:

def square(list):
    return map(lambda x: x ** 2, list)

Or you could use a generator. It won't return a list, but you can still iterate through it, and since you don't have to allocate an entire new list, it is possibly more space-efficient than the other options:

def square(list):
    for i in list:
        yield i ** 2

Or you can do the boring old for-loop, though this is not as idiomatic as some Python programmers would prefer:

def square(list):
    ret = []
    for i in list:
        ret.append(i ** 2)
    return ret