谁能告诉我如何在Swift中舍入一个双数值到x位小数点后数位?
我有:
var totalWorkTimeInHours = (totalWorkTime/60/60)
totalWorkTime是一个NSTimeInterval (double),单位为秒。
totalWorkTimeInHours会给我小时数,但它给我的时间量是如此长的精确数字,例如1.543240952039......
当我打印totalWorkTimeInHours时,我如何将其四舍五入到1.543 ?
谁能告诉我如何在Swift中舍入一个双数值到x位小数点后数位?
我有:
var totalWorkTimeInHours = (totalWorkTime/60/60)
totalWorkTime是一个NSTimeInterval (double),单位为秒。
totalWorkTimeInHours会给我小时数,但它给我的时间量是如此长的精确数字,例如1.543240952039......
当我打印totalWorkTimeInHours时,我如何将其四舍五入到1.543 ?
当前回答
小数点后特定数字的代码为:
var a = 1.543240952039
var roundedString = String(format: "%.3f", a)
这里是%。3f告诉swift将这个数字四舍五入到小数点后3位。如果你想要双倍的数字,你可以使用这个代码:
//字符串为Double
var rounded弦= Double(字符串)(格式:%)。3f”,b型)
其他回答
这个解决方案对我很有效。XCode 13.3.1 & Swift 5
extension Double {
func rounded(decimalPoint: Int) -> Double {
let power = pow(10, Double(decimalPoint))
return (self * power).rounded() / power
}
}
测试:
print(-87.7183123123.rounded(decimalPoint: 3))
print(-87.7188123123.rounded(decimalPoint: 3))
print(-87.7128123123.rounded(decimalPoint: 3))
结果:
-87.718
-87.719
-87.713
Swift 2扩展
一个更通用的解决方案是以下扩展,适用于Swift 2和iOS 9:
extension Double {
/// Rounds the double to decimal places value
func roundToPlaces(places:Int) -> Double {
let divisor = pow(10.0, Double(places))
return round(self * divisor) / divisor
}
}
Swift 3扩展
在Swift 3中,round被圆润取代:
extension Double {
/// Rounds the double to decimal places value
func rounded(toPlaces places:Int) -> Double {
let divisor = pow(10.0, Double(places))
return (self * divisor).rounded() / divisor
}
}
返回Double四舍五入到小数点后4位的示例:
let x = Double(0.123456789).roundToPlaces(4) // x becomes 0.1235 under Swift 2
let x = Double(0.123456789).rounded(toPlaces: 4) // Swift 3 version
基于Yogi的回答,这里有一个Swift函数来完成这项工作:
func roundToPlaces(value:Double, places:Int) -> Double {
let divisor = pow(10.0, Double(places))
return round(value * divisor) / divisor
}
如果你想四舍五入Double值,你可能想要使用Swift Decimal,这样你就不会在尝试用这些四舍五入值计算时引入任何可能出现的错误。如果使用Decimal,则它可以准确地表示舍入浮点值的十进制值。
所以你可以这样做:
extension Double {
/// Convert `Double` to `Decimal`, rounding it to `scale` decimal places.
///
/// - Parameters:
/// - scale: How many decimal places to round to. Defaults to `0`.
/// - mode: The preferred rounding mode. Defaults to `.plain`.
/// - Returns: The rounded `Decimal` value.
func roundedDecimal(to scale: Int = 0, mode: NSDecimalNumber.RoundingMode = .plain) -> Decimal {
var decimalValue = Decimal(self)
var result = Decimal()
NSDecimalRound(&result, &decimalValue, scale, mode)
return result
}
}
然后,你可以像这样得到四舍五入的十进制值:
let foo = 427.3000000002
let value = foo.roundedDecimal(to: 2) // results in 427.30
如果你想用指定的小数位数来显示它(以及为用户的当前语言环境本地化字符串),你可以使用NumberFormatter:
let formatter = NumberFormatter()
formatter.maximumFractionDigits = 2
formatter.minimumFractionDigits = 2
if let string = formatter.string(for: value) {
print(string)
}
斯威夫特5
使用字符串方法
var yourDouble = 3.12345
//to round this to 2 decimal spaces i could turn it into string
let roundingString = String(format: "%.2f", myDouble)
let roundedDouble = Double(roundingString) //and than back to double
// result is 3.12
但是使用扩展更容易被接受
extension Double {
func round(to decimalPlaces: Int) -> Double {
let precisionNumber = pow(10,Double(decimalPlaces))
var n = self // self is a current value of the Double that you will round
n = n * precisionNumber
n.round()
n = n / precisionNumber
return n
}
}
然后你可以使用:
yourDouble.round(to:2)