Transforming an array: take the square root of each perfect square, else square the number

Question

SquareOrSquareRoot should get an array of integers and return a new array. If the number at index i is a "square" number the returned array at index i should have its square root. If the original number is not a "square" number then the returned array in index i should be the number squared.

Coming from Python (where this could be done in a single line using list comprehension or using map), I find it very odd that this is probably one of the shortest ways to achieve that in Go (please prove me wrong).

package main

import (
    "fmt"
    "math"
)

func SquareOrSquareRoot(arr []int) []int{
    arr_to_return := make([]int, len(arr))
    for index, value := range arr {
        val_sqrt := math.Sqrt(float64(value))
        if val_sqrt == math.Trunc(val_sqrt) {
            arr_to_return[index] = int(val_sqrt)
        } else {
            arr_to_return[index] = value * value
        }
    }

    return arr_to_return
}

func main() {
    arr := []int{100, 101, 5, 5, 1, 1}
    fmt.Println(SquareOrSquareRoot(arr))
    // [10 10201 25 25 1 1]
}

Answer 1

Naming conventions aside (short names in camelCase are typically used in Go), this is the right way of doing what you want. The authors of Go have a different view of what is more readable & maintainable than the authors of Python :-)