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 ha

Answer 1

Use numpy.

import numpy as np
b = list(np.array(a)**2)

Answer 2

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

Answer 3

you can do

square_list =[i**2 for i in start_list]

which returns

[25, 9, 1, 4, 16]  

or, if the list already has values

square_list.extend([i**2 for i in start_list])  

which results in a list that looks like:

[25, 9, 1, 4, 16]  

Note: you don't want to do

square_list.append([i**2 for i in start_list])

as it literally adds a list to the original list, such as:

[_original_, _list_, _data_, [25, 9, 1, 4, 16]]

Answer 4

numbers = []
for i in range(10):numbers.append(i)
result = map(lambda x: x * x * x,numbers)
print(list(result))