In this Kata, you're to complete the function `sum_square_even_root_odd`.

You will be given a list of numbers, `nums`, as the only argument to the function. Take each number in the list and *square* it if it is even, or *square root* the number if it is odd. Take this new list and find the *sum*, rounding to two decimal places.

The list will never be empty and will only contain values that are greater than or equal to zero.

Good luck!

Source: codewars | Difficulty: EASY
Tags: Fundamentals

Starter code: def sum_square_even_root_odd(nums):
	

Solutions:
def sum_square_even_root_odd(nums):
	return round(sum((n ** 2 if n % 2 == 0 else n ** 0.5 for n in nums)), 2)


sum_square_even_root_odd = lambda l: round(sum((x ** [2, 0.5][x & 1] for x in l)), 2)


def sum_square_even_root_odd(nums):
	return round(sum((a ** (2, 0.5)[a % 2] for a in nums)), 2)


import math

def sum_square_even_root_odd(nums):
	return float('%.2f' % sum([x * x if x % 2 == 0 else math.sqrt(x) for x in nums]))


def sum_square_even_root_odd(nums):
	return round(sum((n ** ((2 - n % 2) / (1 + n % 2)) for n in nums)), 2)


def sum_square_even_root_odd(nums):
	return round(sum((i ** [2, 0.5][i & 1] for i in nums)), 2)