I’m writing this code for the problem #18 in Codeabbey. I need to calculate the square root of an array of numbers [150, 0, 5, 1 10, 3]
I have to divide this array in three arrays (x,n) [[150, 0], [5, 1], [10 3]]
where x: is the number I want to calculate the square root of and n: is the number of times I have to try the formula r = (r + x / r) / 2 to get the result which is r, r is starting with 1. There’s no problem with that, the problem is when I have to append the result because if r is 3.0
I have to append it as an integer: 3
but if r is 3.964
I have to append it as a floating.
def squareRoot(): rawArr = [int(n) for n in input().split()] arr = [rawArr[i:i+2] for i in range(0,len(rawArr)-1,2)] results = [] for a in arr: x,n = a r = 1.0 for i in range(0,n): r = (r + x / r) / 2 if r.is_integer: results.append(str(int(r))) else: results.append(str(round(r,3))) return " ".join(results)
The input is:
150 0 5 1 10 3
and the output is:
'1 3 3'
This is what I get if I don’t use is_integer():
'1 3.0 3.196xxxxx'
What the output should be:
1 3 3.196
I can’t see where the problem is.
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Answer
is_integer
is a method you run on float
type. You forgot to invoke it, so it evaluates to True
as it returns a builtin which is something (not nothing).
Just replace
if r.is_integer:
with
if r.is_integer():