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Home/ Questions/Q 8125301
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Editorial Team
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Editorial Team
Asked: June 6, 20262026-06-06T06:46:16+00:00 2026-06-06T06:46:16+00:00

I have a list areas :: [Double]. Now I want to filter this list

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I have a list areas :: [Double]. Now I want to filter this list for those which are actually integral values. I want to do something like this for my predicate:

isInteger :: Double -> Bool
isInteger x = abs (fromIntegral (floor x) - x) < delta
  where delta = 0.00001

However, I would guess there is a better way to do this. Is there a Haskell idiom for checking if a real value is an integer?

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  1. Editorial Team
    Editorial Team
    2026-06-06T06:46:17+00:00Added an answer on June 6, 2026 at 6:46 am

    This looks fine and idiomatic to me, though you probably want to use round rather than floor. You could also consider using approxRational and checking that the denominator of the result is 1:

    isInteger x = denominator (approxRational x 0.00001) == 1
    
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