Reputation: 2084
I want to round a Double to the nearest non zero number that follows the decimal.
For example:
x = 0.002341
rounded = 0.002
x = 0.000048123
rounded = 0.00005
For cases where the base number is > 0, it should perform as such
x = 1.000234
rounded = 1.0002
I know I can use Double(round(1000*x)/1000)
if I know the number of digits, but I want it to work for any number. Is there a swift function that does this?
Upvotes: 4
Views: 1727
Reputation: 93181
You haven't specified what happened when x = 1.002341
for examples. Base on your limited cases you can use this:
func myRound (var number : Double) -> Double {
if number == 0 {
return number
}
let integerPart = floor(number)
var base = 1.0
number -= integerPart
while abs(number) < 1 {
number *= 10.0
base *= 10.0
}
return integerPart + round(number) / base
}
Upvotes: 1
Reputation: 726809
You can have a little fun with logarithms to solve this:
func roundFirst(x:Double) -> Double {
if x == 0 {
return x;
}
let mul : Double = pow(10, floor(log10(abs(x))))
return round(x/mul)*mul
}
The non-fractional part of log10(abs(x))
gives you a positive or negative power of ten of the inverse of the number which you use as your multiplier. floor
drops the fraction, and pow(10,...)
gives you a multiplier to use in your rounding trick.
I tried this in the playground with a few numbers. Here is what I've got:
println(roundFirst(0.002341)) // 0.002
println(roundFirst(0.000048123)) // 5e-05
println(roundFirst(0.0)) // 0.0
println(roundFirst(2.6)) // 3.0
println(roundFirst(123.0)) // 100
println(roundFirst(-0.002341)) // -0.002
println(roundFirst(-0.000048123)) // -5e-05
println(roundFirst(-2.6)) // -3.0
println(roundFirst(-123.0)) // -100
Upvotes: 3