Square a list (or array?) of numbers in Python

Question

I'm coming from a MATLAB background and this simple operation so far seems to be extremely complicated to implement in Python, according to the answers in other stacks. Typically, most answers use a for loop.

The best I've seen so far is

import numpy
start_list = [5, 3, 1, 2, 4]
b = list(numpy.array(start_list)**2)

Is there a simpler way?

Answer 1

_{Note: Since we already have duplicates for the vanilla Python, list comprehensions and map and that I haven't found a duplicate to square a 1D numpy array, I thought I'd keep my original answer that uses numpy}

numpy.square()

If you're coming from MATLAB to Python, you are definitely correct to try to use numpy for this. With numpy you can use numpy.square() which returns the element-wise square of the input:

>>> import numpy as np
>>> start_list = [5, 3, 1, 2, 4]
>>> np.square(start_list)
array([25,  9,  1,  4, 16])

numpy.power()

There is also a more generic numpy.power()

>>> np.power(start_list, 2)
array([25,  9,  1,  4, 16])