ios 在swift中将双精度值四舍五入到x个小数位数

Question

提问by Leighton

Can anyone tell me how to round a double value to x number of decimal places in Swift?

谁能告诉我如何在 Swift 中将 double 值四舍五入到 x 位小数？

I have:

我有：

var totalWorkTimeInHours = (totalWorkTime/60/60)

With totalWorkTimebeing an NSTimeInterval (double) in second.

与totalWorkTime作为一个NSTimeInterval（双）在第二。

totalWorkTimeInHourswill give me the hours, but it gives me the amount of time in such a long precise number e.g. 1.543240952039......

totalWorkTimeInHours会给我小时数，但它给了我这么长的精确数字的时间量，例如 1.543240952039 ......

How do I round this down to, say, 1.543 when I print totalWorkTimeInHours?

打印时如何将其舍入到 1.543 totalWorkTimeInHours？

Answer 1

采纳答案by zisoft

You can use Swift's roundfunction to accomplish this.

您可以使用 Swift 的round函数来完成此操作。

To round a Doublewith 3 digits precision, first multiply it by 1000, round it and divide the rounded result by 1000:

要Double以 3 位精度舍入 a ，首先将其乘以 1000，将其舍入并将舍入后的结果除以 1000：

let x = 1.23556789
let y = Double(round(1000*x)/1000)
print(y)  // 1.236

Other than any kind of printf(...)or String(format: ...)solutions, the result of this operation is still of type Double.

除了任何一种printf(...)或String(format: ...)解决方案，此操作的结果仍然是类型Double。

EDIT:
Regarding the comments that it sometimes does not work, please read this:

编辑：
关于它有时不起作用的评论，请阅读：

What Every Programmer Should Know About Floating-Point Arithmetic

每个程序员都应该了解的关于浮点运算的知识

Answer 2

回答by Sebastian

Extension for Swift 2

Swift 2 的扩展

A more general solution is the following extension, which works with Swift 2 & iOS 9:

更通用的解决方案是以下扩展，它适用于 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
    }
}

Extension for Swift 3

Swift 3 的扩展

In Swift 3 roundis replaced by rounded:

在 Swift 3 中round被替换为rounded：

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
    }
}

Example which returns Double rounded to 4 decimal places:

返回四舍五入到小数点后 4 位的 Double 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

Answer 3

回答by vacawama

How do I round this down to, say, 1.543 when I printtotalWorkTimeInHours?

打印时如何将其舍入到 1.543 totalWorkTimeInHours？

To round totalWorkTimeInHoursto 3 digits for printing, use the Stringconstructor which takes a formatstring:

要四舍五入totalWorkTimeInHours到 3 位数进行打印，请使用String带有format字符串的构造函数：

print(String(format: "%.3f", totalWorkTimeInHours))

Answer 4

回答by Imanou Petit

With Swift 5, according to your needs, you can choose one of the 9 following stylesin order to have a rounded result from a Double.

使用 Swift 5，您可以根据需要选择以下 9 种样式之一，以便从Double.

#1. Using `FloatingPointrounded()`method

#1. 使用`FloatingPointrounded()`方法

In the simplest case, you may use the Doublerounded()method.

在最简单的情况下，您可以使用该Doublerounded()方法。

let roundedValue1 = (0.6844 * 1000).rounded() / 1000
let roundedValue2 = (0.6849 * 1000).rounded() / 1000
print(roundedValue1) // returns 0.684
print(roundedValue2) // returns 0.685

#2. Using `FloatingPointrounded(_:)`method

#2. 使用`FloatingPointrounded(_:)`方法

let roundedValue1 = (0.6844 * 1000).rounded(.toNearestOrEven) / 1000
let roundedValue2 = (0.6849 * 1000).rounded(.toNearestOrEven) / 1000
print(roundedValue1) // returns 0.684
print(roundedValue2) // returns 0.685

#3. Using Darwin `round`function

#3. 使用达尔文`round`函数

Foundation offers a roundfunction via Darwin.

Foundationround通过 Darwin提供了一个函数。

import Foundation

let roundedValue1 = round(0.6844 * 1000) / 1000
let roundedValue2 = round(0.6849 * 1000) / 1000
print(roundedValue1) // returns 0.684
print(roundedValue2) // returns 0.685

#4. Using a `Double`extension custom method built with Darwin `round`and `pow`functions

#4. 使用由`Double`Darwin`round`和`pow`函数构建的扩展自定义方法

If you want to repeat the previous operation many times, refactoring your code can be a good idea.

如果您想多次重复之前的操作，重构您的代码可能是一个好主意。

import Foundation

extension Double {
    func roundToDecimal(_ fractionDigits: Int) -> Double {
        let multiplier = pow(10, Double(fractionDigits))
        return Darwin.round(self * multiplier) / multiplier
    }
}

let roundedValue1 = 0.6844.roundToDecimal(3)
let roundedValue2 = 0.6849.roundToDecimal(3)
print(roundedValue1) // returns 0.684
print(roundedValue2) // returns 0.685

#5. Using `NSDecimalNumberrounding(accordingToBehavior:)`method

#5. 使用`NSDecimalNumberrounding(accordingToBehavior:)`方法

If needed, NSDecimalNumberoffers a verbose but powerful solution for rounding decimal numbers.

如果需要，NSDecimalNumber提供详细但功能强大的四舍五入十进制数解决方案。

import Foundation

let scale: Int16 = 3

let behavior = NSDecimalNumberHandler(roundingMode: .plain, scale: scale, raiseOnExactness: false, raiseOnOverflow: false, raiseOnUnderflow: false, raiseOnDivideByZero: true)

let roundedValue1 = NSDecimalNumber(value: 0.6844).rounding(accordingToBehavior: behavior)
let roundedValue2 = NSDecimalNumber(value: 0.6849).rounding(accordingToBehavior: behavior)

print(roundedValue1) // returns 0.684
print(roundedValue2) // returns 0.685

#6. Using `NSDecimalRound(_:_:_:_:)`function

#6. 使用`NSDecimalRound(_:_:_:_:)`功能

import Foundation

let scale = 3

var value1 = Decimal(0.6844)
var value2 = Decimal(0.6849)

var roundedValue1 = Decimal()
var roundedValue2 = Decimal()

NSDecimalRound(&roundedValue1, &value1, scale, NSDecimalNumber.RoundingMode.plain)
NSDecimalRound(&roundedValue2, &value2, scale, NSDecimalNumber.RoundingMode.plain)

print(roundedValue1) // returns 0.684
print(roundedValue2) // returns 0.685

#7. Using `NSStringinit(format:arguments:)`initializer

#7. 使用`NSStringinit(format:arguments:)`初始化程序

If you want to return a NSStringfrom your rounding operation, using NSStringinitializer is a simple but efficient solution.

如果您想NSString从舍入操作中返回 a ，使用NSString初始化程序是一个简单但有效的解决方案。

import Foundation

let roundedValue1 = NSString(format: "%.3f", 0.6844)
let roundedValue2 = NSString(format: "%.3f", 0.6849)
print(roundedValue1) // prints 0.684
print(roundedValue2) // prints 0.685

#8. Using `Stringinit(format:_:)`initializer

#8. 使用`Stringinit(format:_:)`初始化程序

Swift's Stringtype is bridged with Foundation's NSStringclass. Therefore, you can use the following code in order to return a Stringfrom your rounding operation:

Swift 的String类型与 Foundation 的NSString类相连。因此，您可以使用以下代码String从舍入操作中返回 a ：

import Foundation

let roundedValue1 = String(format: "%.3f", 0.6844)
let roundedValue2 = String(format: "%.3f", 0.6849)
print(roundedValue1) // prints 0.684
print(roundedValue2) // prints 0.685

#9. Using `NumberFormatter`

#9. 使用`NumberFormatter`

If you expect to get a String?from your rounding operation, NumberFormatteroffers a highly customizable solution.

如果您希望String?从您的舍入操作中获得一个，则NumberFormatter提供高度可定制的解决方案。

import Foundation

let formatter = NumberFormatter()
formatter.numberStyle = NumberFormatter.Style.decimal
formatter.roundingMode = NumberFormatter.RoundingMode.halfUp
formatter.maximumFractionDigits = 3

let roundedValue1 = formatter.string(from: 0.6844)
let roundedValue2 = formatter.string(from: 0.6849)
print(String(describing: roundedValue1)) // prints Optional("0.684")
print(String(describing: roundedValue2)) // prints Optional("0.685")

Answer 5

回答by Claudio Guirunas

In Swift 2.0 and Xcode 7.2:

在 Swift 2.0 和 Xcode 7.2 中：

let pi: Double = 3.14159265358979
String(format:"%.2f", pi)

Example:

例子：

Answer 6

回答by Krunal Patel

This is a fully worked code

这是一个完整的代码

Swift 3.0/4.0/5.0 , Xcode 9.0 GM/9.2 and above

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)

output - 123

输出 - 123

let doubleValue : Double = 123.32565254455
self.lblValue_1.text = String(format:"%.1f", doubleValue)
print(self.lblValue_1.text)

output - 123.3

输出 - 123.3

let doubleValue : Double = 123.32565254455
self.lblValue_2.text = String(format:"%.2f", doubleValue)
print(self.lblValue_2.text)

output - 123.33

输出 - 123.33

let doubleValue : Double = 123.32565254455
self.lblValue_3.text = String(format:"%.3f", doubleValue)
print(self.lblValue_3.text)

output - 123.326

输出 - 123.326

Answer 7

回答by Ash

Building on Yogi's answer, here's a Swift function that does the job:

基于 Yogi 的回答，这里有一个 Swift 函数可以完成这项工作：

func roundToPlaces(value:Double, places:Int) -> Double {
    let divisor = pow(10.0, Double(places))
    return round(value * divisor) / divisor
}

Answer 8

回答by jaiswal Rajan

In Swift 3.0 and Xcode 8.0:

在 Swift 3.0 和 Xcode 8.0 中：

extension Double {
    func roundTo(places: Int) -> Double {
        let divisor = pow(10.0, Double(places))
        return (self * divisor).rounded() / divisor
    }
}

Use this extension like so:

像这样使用这个扩展：

let doubleValue = 3.567
let roundedValue = doubleValue.roundTo(places: 2)
print(roundedValue) // prints 3.56

Answer 9

回答by Ramazan Karami

The code for specific digits after decimals is:

小数点后特定数字的代码为：

var a = 1.543240952039
var roundedString = String(format: "%.3f", a)

Here the %.3f tells the swift to make this number rounded to 3 decimal places.and if you want double number, you may use this code:

这里 %.3f 告诉 swift 将这个数字四舍五入到小数点后 3 位。如果你想要双数，你可以使用这个代码：

// String to Double

// 字符串转双

var roundedString = Double(String(format: "%.3f", b))

Answer 10

回答by Naishta

Swift 4, Xcode 10

斯威夫特 4，Xcode 10

yourLabel.text =  String(format:"%.2f", yourDecimalValue)

ios 在swift中将双精度值四舍五入到x个小数位数

提问by Leighton

采纳答案by zisoft

回答by Sebastian

Extension for Swift 2

Swift 2 的扩展

Extension for Swift 3

Swift 3 的扩展

回答by vacawama

回答by Imanou Petit

#1. Using `FloatingPointrounded()`method

#1. 使用`FloatingPointrounded()`方法

#2. Using `FloatingPointrounded(_:)`method

#2. 使用`FloatingPointrounded(_:)`方法

#3. Using Darwin `round`function

#3. 使用达尔文`round`函数

#4. Using a `Double`extension custom method built with Darwin `round`and `pow`functions

#4. 使用由`Double`Darwin`round`和`pow`函数构建的扩展自定义方法

#5. Using `NSDecimalNumberrounding(accordingToBehavior:)`method

#5. 使用`NSDecimalNumberrounding(accordingToBehavior:)`方法

#6. Using `NSDecimalRound(_:_:_:_:)`function

#6. 使用`NSDecimalRound(_:_:_:_:)`功能

#7. Using `NSStringinit(format:arguments:)`initializer

#7. 使用`NSStringinit(format:arguments:)`初始化程序

#8. Using `Stringinit(format:_:)`initializer

#8. 使用`Stringinit(format:_:)`初始化程序

#9. Using `NumberFormatter`

#9. 使用`NumberFormatter`

回答by Claudio Guirunas

回答by Krunal Patel

回答by Ash

回答by jaiswal Rajan

回答by Ramazan Karami

回答by Naishta

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最近更新

标签

ios 在swift中将双精度值四舍五入到x个小数位数

提问by Leighton

采纳答案by zisoft

回答by Sebastian

Extension for Swift 2

Swift 2 的扩展

Extension for Swift 3

Swift 3 的扩展

回答by vacawama

回答by Imanou Petit

#1. Using FloatingPointrounded()method

#1. 使用FloatingPointrounded()方法

#2. Using FloatingPointrounded(_:)method

#2. 使用FloatingPointrounded(_:)方法

#3. Using Darwin roundfunction

#3. 使用达尔文round函数

#4. Using a Doubleextension custom method built with Darwin roundand powfunctions

#4. 使用由DoubleDarwinround和pow函数构建的扩展自定义方法

#5. Using NSDecimalNumberrounding(accordingToBehavior:)method

#5. 使用NSDecimalNumberrounding(accordingToBehavior:)方法

#6. Using NSDecimalRound(_:_:_:_:)function

#6. 使用NSDecimalRound(_:_:_:_:)功能

#7. Using NSStringinit(format:arguments:)initializer

#7. 使用NSStringinit(format:arguments:)初始化程序

#8. Using Stringinit(format:_:)initializer

#8. 使用Stringinit(format:_:)初始化程序

#9. Using NumberFormatter

#9. 使用NumberFormatter

回答by Claudio Guirunas

回答by Krunal Patel

回答by Ash

回答by jaiswal Rajan

回答by Ramazan Karami

回答by Naishta

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标签

#1. Using `FloatingPointrounded()`method

#1. 使用`FloatingPointrounded()`方法

#2. Using `FloatingPointrounded(_:)`method

#2. 使用`FloatingPointrounded(_:)`方法

#3. Using Darwin `round`function

#3. 使用达尔文`round`函数

#4. Using a `Double`extension custom method built with Darwin `round`and `pow`functions

#4. 使用由`Double`Darwin`round`和`pow`函数构建的扩展自定义方法

#5. Using `NSDecimalNumberrounding(accordingToBehavior:)`method

#5. 使用`NSDecimalNumberrounding(accordingToBehavior:)`方法

#6. Using `NSDecimalRound(_:_:_:_:)`function

#6. 使用`NSDecimalRound(_:_:_:_:)`功能

#7. Using `NSStringinit(format:arguments:)`initializer

#7. 使用`NSStringinit(format:arguments:)`初始化程序

#8. Using `Stringinit(format:_:)`initializer

#8. 使用`Stringinit(format:_:)`初始化程序

#9. Using `NumberFormatter`

#9. 使用`NumberFormatter`