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

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

Answer 2

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

so how would i do the square of numbers from 1-20 using the above function

Answer 3

One more map solution:

def square(a):
    return map(pow, a, [2]*len(a))

Answer 4

from array import *

array('i', [5, 6, 4])
print(*list(map(pow, vals, [2] * len(vals))), sep='\n')

Answer 5

Use a list comprehension (this is the way to go in pure Python):

>>> l = [1, 2, 3, 4]
>>> [i**2 for i in l]
[1, 4, 9, 16]

Or numpy (a well-established module):

>>> numpy.array([1, 2, 3, 4])**2
array([ 1,  4,  9, 16])

In numpy, math operations on arrays are, by default, executed element-wise. That's why you can **2 an entire array there.

Other possible solutions would be map-based, but in this case I'd really go for the list comprehension. It's Pythonic :) and a map-based solution that requires lambdas is slower than LC.

Answer 6

import numpy as np
a = [2 ,3, 4]
np.square(a)