下面是我以前如何将浮点数截断到两位小数

NSLog(@" %.02f %.02f %.02f", r, g, b);

我检查了文档和电子书，但无法弄清楚。谢谢！

根据David的回应，到目前为止，我最好的解决方案是：

import Foundation

extension Int {
    func format(f: String) -> String {
        return String(format: "%\(f)d", self)
    }
}

extension Double {
    func format(f: String) -> String {
        return String(format: "%\(f)f", self)
    }
}

let someInt = 4, someIntFormat = "03"
println("The integer number \(someInt) formatted with \"\(someIntFormat)\" looks like \(someInt.format(someIntFormat))")
// The integer number 4 formatted with "03" looks like 004

let someDouble = 3.14159265359, someDoubleFormat = ".3"
println("The floating point number \(someDouble) formatted with \"\(someDoubleFormat)\" looks like \(someDouble.format(someDoubleFormat))")
// The floating point number 3.14159265359 formatted with ".3" looks like 3.142

我认为这是最类似于Swift的解决方案，将格式化操作直接与数据类型相关联。可能是某处有一个内置的格式化操作库，或者它可能很快就会发布。请记住，该语言仍处于测试阶段。

一个简单的方法是：

import Foundation // required for String(format: _, _)

print(String(format: "hex string: %X", 123456))
print(String(format: "a float number: %.5f", 1.0321))

我发现String.localizedStringWithFormat工作得很好：

例：

let value: Float = 0.33333
let unit: String = "mph"

yourUILabel.text = String.localizedStringWithFormat("%.2f %@", value, unit)

这是不需要复杂解决方案的非常快速和简单的方法。

let duration = String(format: "%.01f", 3.32323242)
// result = 3.3

这里的大多数答案都是有效的。但是，如果您经常格式化数字，请考虑扩展Float类以添加一个返回格式化字符串的方法。请参见下面的示例代码。通过使用数字格式化程序和扩展名，这一目标得以实现。

extension Float {
    func string(fractionDigits:Int) -> String {
        let formatter = NSNumberFormatter()
        formatter.minimumFractionDigits = fractionDigits
        formatter.maximumFractionDigits = fractionDigits
        return formatter.stringFromNumber(self) ?? "\(self)"
    }
}

let myVelocity:Float = 12.32982342034

println("The velocity is \(myVelocity.string(2))")
println("The velocity is \(myVelocity.string(1))")

控制台显示：

The velocity is 12.33
The velocity is 12.3

SWIFT 3.1更新

extension Float {
    func string(fractionDigits:Int) -> String {
        let formatter = NumberFormatter()
        formatter.minimumFractionDigits = fractionDigits
        formatter.maximumFractionDigits = fractionDigits
        return formatter.string(from: NSNumber(value: self)) ?? "\(self)"
    }
}

您还不能通过字符串插值来做到这一点。最好的选择仍然是NSString格式：

println(NSString(format:"%.2f", sqrt(2.0)))

从python推断，似乎合理的语法可能是：

@infix func % (value:Double, format:String) -> String {
    return NSString(format:format, value)
}

然后，您可以将它们用作：

M_PI % "%5.3f"                // "3.142"

您可以为所有数字类型定义类似的运算符，但不幸的是，我还没有找到使用泛型的方法。

Swift 5更新

从至少Swift 5开始，它String直接支持format:初始化程序，因此无需使用NSString并且@infix不再需要该属性，这意味着上面的示例应写为：

println(String(format:"%.2f", sqrt(2.0)))

func %(value:Double, format:String) -> String {
    return String(format:format, value)
}

Double.pi % "%5.3f"         // "3.142"

为什么要这么复杂？您可以改用以下方法：

import UIKit

let PI = 3.14159265359

round( PI ) // 3.0 rounded to the nearest decimal
round( PI * 100 ) / 100 //3.14 rounded to the nearest hundredth
round( PI * 1000 ) / 1000 // 3.142 rounded to the nearest thousandth

看到它在Playground中有效。

PS：解决方案来自：http : //rrike.sh/xcode/rounding-various-decimal-places-swift/

import Foundation

extension CGFloat {
    var string1: String {
        return String(format: "%.1f", self)
    }
    var string2: String {
        return String(format: "%.2f", self)
    }
}

用法

let offset = CGPoint(1.23, 4.56)
print("offset: \(offset.x.string1) x \(offset.y.string1)")
// offset: 1.2 x 4.6

一个更优雅，更通用的解决方案是重写ruby / python %运算符：

// Updated for beta 5
func %(format:String, args:[CVarArgType]) -> String {
    return NSString(format:format, arguments:getVaList(args))
}

"Hello %@, This is pi : %.2f" % ["World", M_PI]

斯威夫特4

let string = String(format: "%.2f", locale: Locale.current, arguments: 15.123)

您仍然可以像在Objective-C中一样在Swift中使用NSLog，而无需使用@符号。

NSLog("%.02f %.02f %.02f", r, g, b)

编辑：与Swift合作一段时间以来，我还想添加此变化

    var r=1.2
    var g=1.3
    var b=1.4
    NSLog("\(r) \(g) \(b)")

输出：

2014-12-07 21:00:42.128 MyApp[1626:60b] 1.2 1.3 1.4

细节

• Xcode 9.3，Swift 4.1
• Xcode 10.2.1（10E1001），Swift 5

解决方案1

func rounded（）->双精度

(5.2).rounded()
// 5.0
(5.5).rounded()
// 6.0
(-5.2).rounded()
// -5.0
(-5.5).rounded()
// -6.0

func rounded（_规则：FloatingPointRoundingRule）-> Double

let x = 6.5

// Equivalent to the C 'round' function:
print(x.rounded(.toNearestOrAwayFromZero))
// Prints "7.0"

// Equivalent to the C 'trunc' function:
print(x.rounded(.towardZero))
// Prints "6.0"

// Equivalent to the C 'ceil' function:
print(x.rounded(.up))
// Prints "7.0"

// Equivalent to the C 'floor' function:
print(x.rounded(.down))
// Prints "6.0"

变异func round（）

var x = 5.2
x.round()
// x == 5.0
var y = 5.5
y.round()
// y == 6.0
var z = -5.5
z.round()
// z == -6.0

变异函数回合（_规则：FloatingPointRoundingRule）

// Equivalent to the C 'round' function:
var w = 6.5
w.round(.toNearestOrAwayFromZero)
// w == 7.0

// Equivalent to the C 'trunc' function:
var x = 6.5
x.round(.towardZero)
// x == 6.0

// Equivalent to the C 'ceil' function:
var y = 6.5
y.round(.up)
// y == 7.0

// Equivalent to the C 'floor' function:
var z = 6.5
z.round(.down)
// z == 6.0

解决方案2

extension Numeric {

    private func _precision(number: NSNumber, formatter: NumberFormatter) -> Self? {
        if  let formatedNumString = formatter.string(from: number),
            let formatedNum = formatter.number(from: formatedNumString) {
                return formatedNum as? Self
        }
        return nil
    }

    private func toNSNumber() -> NSNumber? {
        if let num = self as? NSNumber { return num }
        guard let string = self as? String, let double = Double(string) else { return nil }
        return NSNumber(value: double)
    }

    func precision(_ minimumFractionDigits: Int,
                   roundingMode: NumberFormatter.RoundingMode = NumberFormatter.RoundingMode.halfUp) -> Self? {
        guard let number = toNSNumber() else { return nil }
        let formatter = NumberFormatter()
        formatter.minimumFractionDigits = minimumFractionDigits
        formatter.roundingMode = roundingMode
        return _precision(number: number, formatter: formatter)
    }

    func precision(with numberFormatter: NumberFormatter) -> String? {
        guard let number = toNSNumber() else { return nil }
        return numberFormatter.string(from: number)
    }
}

用法

_ = 123.44.precision(2)
_ = 123.44.precision(3, roundingMode: .up)

let numberFormatter = NumberFormatter()
numberFormatter.minimumFractionDigits = 1
numberFormatter.groupingSeparator = " "
let num = 222.3333
_ = num.precision(2)

完整样本

func option1<T: Numeric>(value: T, numerFormatter: NumberFormatter? = nil) {
    print("Type: \(type(of: value))")
    print("Original Value: \(value)")
    let value1 = value.precision(2)
    print("value1 = \(value1 != nil ? "\(value1!)" : "nil")")
    let value2 = value.precision(5)
    print("value2 = \(value2 != nil ? "\(value2!)" : "nil")")
    if let value1 = value1, let value2 = value2 {
        print("value1 + value2 = \(value1 + value2)")
    }
    print("")
}

func option2<T: Numeric>(value: T, numberFormatter: NumberFormatter) {
    print("Type: \(type(of: value))")
    print("Original Value: \(value)")
    let value1 = value.precision(with: numberFormatter)
    print("formated value = \(value1 != nil ? "\(value1!)" : "nil")\n")
}

func test(with double: Double) {
    print("===========================\nTest with: \(double)\n")
    let float = Float(double)
    let float32 = Float32(double)
    let float64 = Float64(double)
    let float80 = Float80(double)
    let cgfloat = CGFloat(double)

    // Exapmle 1
    print("-- Option1\n")
    option1(value: double)
    option1(value: float)
    option1(value: float32)
    option1(value: float64)
    option1(value: float80)
    option1(value: cgfloat)

    // Exapmle 2

    let numberFormatter = NumberFormatter()
    numberFormatter.formatterBehavior = .behavior10_4
    numberFormatter.minimumIntegerDigits = 1
    numberFormatter.minimumFractionDigits = 4
    numberFormatter.maximumFractionDigits = 9
    numberFormatter.usesGroupingSeparator = true
    numberFormatter.groupingSeparator = " "
    numberFormatter.groupingSize = 3

    print("-- Option 2\n")
    option2(value: double, numberFormatter: numberFormatter)
    option2(value: float, numberFormatter: numberFormatter)
    option2(value: float32, numberFormatter: numberFormatter)
    option2(value: float64, numberFormatter: numberFormatter)
    option2(value: float80, numberFormatter: numberFormatter)
    option2(value: cgfloat, numberFormatter: numberFormatter)
}

test(with: 123.22)
test(with: 1234567890987654321.0987654321)

输出量

===========================
Test with: 123.22

-- Option1

Type: Double
Original Value: 123.22
value1 = 123.22
value2 = 123.22
value1 + value2 = 246.44

Type: Float
Original Value: 123.22
value1 = nil
value2 = nil

Type: Float
Original Value: 123.22
value1 = nil
value2 = nil

Type: Double
Original Value: 123.22
value1 = 123.22
value2 = 123.22
value1 + value2 = 246.44

Type: Float80
Original Value: 123.21999999999999886
value1 = nil
value2 = nil

Type: CGFloat
Original Value: 123.22
value1 = 123.22
value2 = 123.22
value1 + value2 = 246.44

-- Option 2

Type: Double
Original Value: 123.22
formatted value = 123.2200

Type: Float
Original Value: 123.22
formatted value = 123.220001221

Type: Float
Original Value: 123.22
formatted value = 123.220001221

Type: Double
Original Value: 123.22
formatted value = 123.2200

Type: Float80
Original Value: 123.21999999999999886
formatted value = nil

Type: CGFloat
Original Value: 123.22
formatted value = 123.2200

===========================
Test with: 1.2345678909876544e+18

-- Option1

Type: Double
Original Value: 1.2345678909876544e+18
value1 = 1.23456789098765e+18
value2 = 1.23456789098765e+18
value1 + value2 = 2.4691357819753e+18

Type: Float
Original Value: 1.234568e+18
value1 = nil
value2 = nil

Type: Float
Original Value: 1.234568e+18
value1 = nil
value2 = nil

Type: Double
Original Value: 1.2345678909876544e+18
value1 = 1.23456789098765e+18
value2 = 1.23456789098765e+18
value1 + value2 = 2.4691357819753e+18

Type: Float80
Original Value: 1234567890987654400.0
value1 = nil
value2 = nil

Type: CGFloat
Original Value: 1.2345678909876544e+18
value1 = 1.23456789098765e+18
value2 = 1.23456789098765e+18
value1 + value2 = 2.4691357819753e+18

-- Option 2

Type: Double
Original Value: 1.2345678909876544e+18
formatted value = 1 234 567 890 987 650 000.0000

Type: Float
Original Value: 1.234568e+18
formatted value = 1 234 567 939 550 610 000.0000

Type: Float
Original Value: 1.234568e+18
formatted value = 1 234 567 939 550 610 000.0000

Type: Double
Original Value: 1.2345678909876544e+18
formatted value = 1 234 567 890 987 650 000.0000

Type: Float80
Original Value: 1234567890987654400.0
formatted value = nil

Type: CGFloat
Original Value: 1.2345678909876544e+18
formatted value = 1 234 567 890 987 650 000.0000

extension Double {
  func formatWithDecimalPlaces(decimalPlaces: Int) -> Double {
     let formattedString = NSString(format: "%.\(decimalPlaces)f", self) as String
     return Double(formattedString)!
     }
 }

 1.3333.formatWithDecimalPlaces(2)

到目前为止，获得最多投票的答案取决于NSString方法，并且要求您已导入Foundation。

但是，完成此操作后，您仍然可以访问NSLog。

因此，如果您要问如何继续在Swift中使用NSLog，我认为问题的答案很简单：

import Foundation

//It will more help, by specify how much decimal Point you want.
let decimalPoint = 2
let floatAmount = 1.10001
let amountValue = String(format: "%0.*f", decimalPoint, floatAmount)

这是一个“纯粹的”快速解决方案

 var d = 1.234567
operator infix ~> {}
@infix func ~> (left: Double, right: Int) -> String {
    if right == 0 {
        return "\(Int(left))"
    }
    var k = 1.0
    for i in 1..right+1 {
        k = 10.0 * k
    }
    let n = Double(Int(left*k)) / Double(k)
    return "\(n)"
}
println("\(d~>2)")
println("\(d~>1)")
println("\(d~>0)")

扩展的力量

extension Double {
    var asNumber:String {
        if self >= 0 {
            var formatter = NSNumberFormatter()
            formatter.numberStyle = .NoStyle
            formatter.percentSymbol = ""
            formatter.maximumFractionDigits = 1
            return "\(formatter.stringFromNumber(self)!)"
        }
        return ""
    }
}

let velocity:Float = 12.32982342034

println("The velocity is \(velocity.toNumber)")

输出：速度为12.3

少打字的方式：

func fprint(format: String, _ args: CVarArgType...) {
    print(NSString(format: format, arguments: getVaList(args)))
}

您也可以通过这种方式创建运算符

operator infix <- {}

func <- (format: String, args:[CVarArg]) -> String {
    return String(format: format, arguments: args)
}

let str = "%d %.1f" <- [1453, 1.123]

同样舍入：

extension Float
{
    func format(f: String) -> String
    {
        return NSString(format: "%\(f)f", self)
    }
    mutating func roundTo(f: String)
    {
        self = NSString(format: "%\(f)f", self).floatValue
    }
}

extension Double
{
    func format(f: String) -> String
    {
        return NSString(format: "%\(f)f", self)
    }
    mutating func roundTo(f: String)
    {
        self = NSString(format: "%\(f)f", self).doubleValue
    }
}

x = 0.90695652173913
x.roundTo(".2")
println(x) //0.91

使用以下方法

let output = String.localizedStringWithFormat(" %.02f %.02f %.02f", r, g, b)

println(output)

Vincent Guerci的ruby / python％运算符的版本，已针对Swift 2.1更新：

func %(format:String, args:[CVarArgType]) -> String {
  return String(format:format, arguments:args)
}

"Hello %@, This is pi : %.2f" % ["World", M_PI]

上面有很多不错的答案，但是有时候模式比“％.3f”这样的傻瓜更合适。这是我在Swift 3中使用NumberFormatter的方法。

extension Double {
  func format(_ pattern: String) -> String {
    let formatter = NumberFormatter()
    formatter.format = pattern
    return formatter.string(from: NSNumber(value: self))!
  }    
}

let n1 = 0.350, n2 = 0.355
print(n1.format("0.00#")) // 0.35
print(n2.format("0.00#")) // 0.355

在这里，我希望始终显示2个小数，但仅在不为零的情况下才显示第三个小数。

Swift 4 Xcode 10更新

extension Double {
    var asNumber:String {
        if self >= 0 {
            let formatter = NumberFormatter()
            formatter.numberStyle = .none
            formatter.percentSymbol = ""
            formatter.maximumFractionDigits = 2
            return "\(formatter.string(from: NSNumber(value: self)) ?? "")"
        }
        return ""
    }
}

关于Double和CGFloat类型的扩展呢？

extension Double {

   func formatted(_ decimalPlaces: Int?) -> String {
      let theDecimalPlaces : Int
      if decimalPlaces != nil {
         theDecimalPlaces = decimalPlaces!
      }
      else {
         theDecimalPlaces = 2
      }
      let theNumberFormatter = NumberFormatter()
      theNumberFormatter.formatterBehavior = .behavior10_4
      theNumberFormatter.minimumIntegerDigits = 1
      theNumberFormatter.minimumFractionDigits = 1
      theNumberFormatter.maximumFractionDigits = theDecimalPlaces
      theNumberFormatter.usesGroupingSeparator = true
      theNumberFormatter.groupingSeparator = " "
      theNumberFormatter.groupingSize = 3

      if let theResult = theNumberFormatter.string(from: NSNumber(value:self)) {
         return theResult
      }
      else {
         return "\(self)"
      }
   }
}

用法：

let aNumber: Double = 112465848348508.458758344
Swift.print("The number: \(aNumber.formatted(2))")

印刷品：112465848348348.46

@infix func ^(left:Double, right: Int) -> NSNumber {
    let nf = NSNumberFormatter()
    nf.maximumSignificantDigits = Int(right)
    return  nf.numberFromString(nf.stringFromNumber(left))
}


let r = 0.52264
let g = 0.22643
let b = 0.94837

println("this is a color: \(r^3) \(g^3) \(b^3)")

// this is a color: 0.523 0.226 0.948

我不知道小数点后两位，但是这是打印小数点后零位的浮点数的方法，因此我想可以是2位，3位...（注意：您必须将CGFloat转换为Double才能通过转换为String（format :)，否则它将看到零值）

func logRect(r: CGRect, _ title: String = "") {
    println(String(format: "[ (%.0f, %.0f), (%.0f, %.0f) ] %@",
        Double(r.origin.x), Double(r.origin.y), Double(r.size.width), Double(r.size.height), title))
}

Swift2示例：格式化Float的iOS设备的屏幕宽度，删除了小数

print(NSString(format: "Screen width = %.0f pixels", CGRectGetWidth(self.view.frame)))

@克里斯汀·迪特里希（Christian Dietrich）：

代替：

var k = 1.0
    for i in 1...right+1 {
        k = 10.0 * k
    }
let n = Double(Int(left*k)) / Double(k)
return "\(n)"

也可能是：

let k = pow(10.0, Double(right))
let n = Double(Int(left*k)) / k
return "\(n)"

[更正：] 抱歉造成混淆*-当然，这适用于Doubles。我认为，最实用的方法（如果您希望数字四舍五入而不是被截断）将是这样的：

infix operator ~> {}
func ~> (left: Double, right: Int) -> Double {
    if right <= 0 {
        return round(left)
    }
    let k = pow(10.0, Double(right))
    return round(left*k) / k
}

仅对于Float，只需将Double替换为Float，将pow替换为powf，将round替换为roundf。
更新：我发现使用返回类型Double而不是String最为实用。对于String输出，它的工作原理相同，即：

println("Pi is roughly \(3.1415926 ~> 3)")

打印：Pi大约为3.142，
因此您可以以相同的方式将其用于字符串（甚至可以编写：println（d〜> 2）），但是除此之外，您还可以使用它直接舍入值，即：

d = Double(slider.value) ~> 2

或任何您需要的...

采用

CGFloat 

要么

Float.roundTo(places:2)