list of integer square roots python

Question

Chances are this is a duplicate question but I've looked through the duplicate questions and I don't see an answer for python.

I'm trying to create a simple program where I can hard code a list of integers and filter out all the numbers that have integer square roots. Here's what I have so far:

# Python Program to display the number of integer square roots in a list
import math

# Change this list for testing
terms = [1,2,3,4,5,6,7,9,10,11,12,13,16,24,36]

all_roots = []  #list to hold all square roots from terms list

for i in terms:
    all_roots.append(math.sqrt(i))

#integer_roots = list(filter(lambda x: [????], all_roots)) #not sure what I should put in the "[????]" part

print(all_roots) #prints integer and floating square roots
print(integer_roots) #supposed to print only square roots that are integers

To clarify, print(all_roots) currently displays:

[1.0, 1.4142135623730951, 1.7320508075688772, 2.0, 2.23606797749979, 2.449489742783178, 2.6457513110645907, 3.0, 3.1622776601683795, 3.3166247903554, 3.4641016151377544, 3.605551275463989, 4.0, 4.898979485566356, 6.0]

But I want print(integer_roots) to display

[1, 2, 3, 4, 6]

Answer 1

import math
terms = [1,2,3,4,5,6,7,9,10,11,12,13,16,24,36]

integer_roots = [] 

for i in terms:
    tmp = math.sqrt(i)
    if (tmp.is_integer()):
         integer_roots.append(int(tmp))

print(integer_roots)

Answer 2

A list comprehension would help with a generator expression inside to prevent duplicate function calls:

>>> [int(i) for i in (math.sqrt(i) for i in terms) if i.is_integer()]
[1, 2, 3, 4, 6]

Or if you want a purely functional approach, since you suggest using filter() in your post:

>>> list(map(int, filter(float.is_integer, map(math.sqrt, terms))))
[1, 2, 3, 4, 6]

Of course this can be written out with a regular for loop, which is less performant but totally acceptable, particularly given the OP seems new to Python:

lst = []
for i in terms:
    s = math.sqrt(i)
    if s.is_integer():
         lst.append(int(s))

Answer 3

import math arr = list(map(math.isqrt ,[1,2,3,4])) print(arr)