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
:
Using String(format:): Typecast Double to String with %.3f format specifier and then back to Double Double(String(format: "%.3f", 10.123546789))! Or extend Double to handle N-Decimal places: extension Double { func rounded(toDecimalPlaces n: Int) -> Double { return Double(String(format: "%.\(n)f", self))! } } By calculation multiply with 10^3, round it and then divide by 10^3... (1000 * 10.123546789).rounded()/1000 Or extend Double to handle N-Decimal places: extension Double { func rounded(toDecimalPlaces n: Int) -> Double { let multiplier = pow(10, Double(n)) return (multiplier * self).rounded()/multiplier } }
其他回答
:
Using String(format:): Typecast Double to String with %.3f format specifier and then back to Double Double(String(format: "%.3f", 10.123546789))! Or extend Double to handle N-Decimal places: extension Double { func rounded(toDecimalPlaces n: Int) -> Double { return Double(String(format: "%.\(n)f", self))! } } By calculation multiply with 10^3, round it and then divide by 10^3... (1000 * 10.123546789).rounded()/1000 Or extend Double to handle N-Decimal places: extension Double { func rounded(toDecimalPlaces n: Int) -> Double { let multiplier = pow(10, Double(n)) return (multiplier * self).rounded()/multiplier } }
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)
二重值舍入到小数的x位 不。小数点后的数
var x = 1.5657676754
var y = (x*10000).rounded()/10000
print(y) // 1.5658
var x = 1.5657676754
var y = (x*100).rounded()/100
print(y) // 1.57
var x = 1.5657676754
var y = (x*10).rounded()/10
print(y) // 1.6
这是一个完整的代码
Swift 3.0/4.0/5.0, Xcode 9.0 GM/9.2及以上
let doubleValue : Double = 123.32565254455
self.lblValue.text = String(format:"%.f", doubleValue)
print(self.lblValue.text)
输出- 123
let doubleValue : Double = 123.32565254455
self.lblValue_1.text = String(format:"%.1f", doubleValue)
print(self.lblValue_1.text)
输出- 123.3
let doubleValue : Double = 123.32565254455
self.lblValue_2.text = String(format:"%.2f", doubleValue)
print(self.lblValue_2.text)
输出- 123.33
let doubleValue : Double = 123.32565254455
self.lblValue_3.text = String(format:"%.3f", doubleValue)
print(self.lblValue_3.text)
输出- 123.326
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
}