Check For Perfect Squares In A List

I want to identify numbers which are perfect squares in a list of numbers, for example: a = [3, 4, 8, 16, 21, 58, 144] # return [4, 16, 144]

Solution 1:

Once approach is to build a predicate (a function returning true or false) and apply it with filter():

>>> defis_perfect_square(n):
        returnround(n ** 0.5) ** 2 == n

>>> list(filter(is_perfect_square, [3, 4, 8, 16, 21, 58, 144]))
[4, 16, 144]

Or for those who prefer list comprehensions over filter():

>>> [x for x in [3, 4, 8, 16, 21, 58, 144] if is_perfect_square(x)]
[4, 16, 144]

The perfect square test works by taking the square root of a number and rounding it to the nearest integer, re-squaring it and comparing it to a the original number. The square root step can suffer a little round-off error, but the re-squaring of the rounded integer will be exact. This should be somewhat robust except for very large inputs.

Solution 2:

You can combine math.sqrt() and is_integer() to filter the list, like the following:

import math

a = [3, 4, 8, 16, 21, 58, 144]
print [x for x in a ifmath.sqrt(x).is_integer()]

Solution 3:

You can try:

>>>square_list = []>>>a = [3, 4, 8, 16, 21, 58, 144]>>>defperfect_square(number):
    return (number**0.5) % int(number**0.5) == 0

>>>for data in a:
    if perfect_square(data):
        square_list.append(data)


>>>print square_list
[4, 16, 144]

Solution 4:

Using this test for perfect squares:

defis_square(n):
    for b inrange(n):
        n -= (2 * b + 1)
        if n <= 0:
            returnnot n

... you can find those that appear in a with a simple list comprehension:

>>> [n for n in a if is_square(n)][4, 16, 144]