Created
March 3, 2017 09:44
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iOS solution to convert large numbers to smaller format. See : http://stackoverflow.com/a/35504720/1661338
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extension Int { | |
func formatUsingAbbrevation () -> String { | |
let numFormatter = NSNumberFormatter() | |
typealias Abbrevation = (threshold:Double, divisor:Double, suffix:String) | |
let abbreviations:[Abbrevation] = [(0, 1, ""), | |
(1000.0, 1000.0, "K"), | |
(100_000.0, 1_000_000.0, "M"), | |
(100_000_000.0, 1_000_000_000.0, "B")] | |
// you can add more ! | |
let startValue = Double (abs(self)) | |
let abbreviation:Abbrevation = { | |
var prevAbbreviation = abbreviations[0] | |
for tmpAbbreviation in abbreviations { | |
if (startValue < tmpAbbreviation.threshold) { | |
break | |
} | |
prevAbbreviation = tmpAbbreviation | |
} | |
return prevAbbreviation | |
} () | |
let value = Double(self) / abbreviation.divisor | |
numFormatter.positiveSuffix = abbreviation.suffix | |
numFormatter.negativeSuffix = abbreviation.suffix | |
numFormatter.allowsFloats = true | |
numFormatter.minimumIntegerDigits = 1 | |
numFormatter.minimumFractionDigits = 0 | |
numFormatter.maximumFractionDigits = 1 | |
return numFormatter.stringFromNumber(NSNumber (double:value))! | |
} | |
} | |
let testValue:[Int] = [598, -999, 1000, -1284, 9940, 9980, 39900, 99880, 399880, 999898, 999999, 1456384, 12383474] | |
testValue.forEach() { | |
print ("Value : \($0) -> \($0.formatUsingAbbrevation ())") | |
} | |
///// | |
// Result : | |
// Value : 598 -> 598 | |
// Value : -999 -> -999 | |
// Value : 1000 -> 1K | |
// Value : -1284 -> -1.3K | |
// Value : 9940 -> 9.9K | |
// Value : 9980 -> 10K | |
// Value : 39900 -> 39.9K | |
// Value : 99880 -> 99.9K | |
// Value : 399880 -> 0.4M | |
// Value : 999898 -> 1M | |
// Value : 999999 -> 1M | |
// Value : 1456384 -> 1.5M | |
// Value : 12383474 -> 12.4M | |
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I don't find the license for this extension, is a MIT one?
I appreciate this work anyway. Thank you :)