用法

Question

以下是我之前将浮点数截断为两位小数

的方法

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

我检查了文档和电子书，但一直无法弄清楚。谢谢！

Answer 1

一个简单的方法是：

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

Answer 2

到目前为止我的最佳解决方案，来自David's response：

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Swift的解决方案，将格式化操作直接绑定到数据类型。很可能某个地方有一个内置的格式化操作库，或者它很快就会发布。请注意，该语言仍处于测试阶段。

Answer 3

我发现String.localizedStringWithFormat工作得很好：

示例：

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

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

Answer 4

这是一种非常快和简单的方式，他们不需要复杂的解决方案。

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

Answer 5

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类以添加返回格式化字符串的方法。请参阅下面的示例代码这个通过使用数字格式化器和扩展来实现相同的目标。

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)"
    }
}

Answer 6

你不能用（或）字符串插值来做。你最好的选择仍然是NSString格式：

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

从python中推断，似乎合理的语法可能是：

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

然后允许您将它们用作：

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

您可以为所有数字类型定义类似的运算符，遗憾的是我还没有找到使用泛型的方法。

Answer 7

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

Answer 8

为什么要这么复杂？您可以改为使用它：

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

在Playground看到它的工作。

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

Answer 9

更优雅和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]

Answer 10

Swift 4

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

Answer 11

详细

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

解决方案1

func rounded() -> Double

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

func rounded(_ rule: 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"

mutating 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

mutating func round(_ rule: 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

===========================
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

Answer 12

你仍然可以在没有@符号的情况下在Objective-C中使用Swift中的NSLog。

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

编辑：在使用Swift之后一段时间我想添加此变体

    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

Answer 13

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

 1.3333.formatWithDecimalPlaces(2)

Answer 14

到目前为止获得最多投票的答案依赖于NSString方法，并且要求您已经导入了基金会。

完成此操作后，仍然可以访问NSLog。

所以我认为这个问题的答案，如果你问如何继续在Swift中使用NSLog，那就简单了：

import Foundation

Answer 15

这里是＆＃34;纯粹的＆＃34;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)")

Answer 16

//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)

Answer 17

减少打字方式：

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

Answer 18

扩展权力

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

Answer 19

Double和CGFloat类型的扩展名如何：

some hash

用法：

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)"
      }
   }
}

打印：112 465 848 348 508.46

Answer 20

使用以下方法

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

println(output)

Answer 21

上面有很多好的答案，但有时一种模式比“％.3f”类型的gobbledygook更合适。这是我在Swift 3中使用NumberFormatter的看法。

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

这里我想要总是显示2位小数，但只有在它不为零时才是第三位。

Answer 22

Vincent Guerci的ruby / python％运算符版本，针对Swift 2.1进行了更新：

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

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

Answer 23

还有四舍五入：

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

Answer 24

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]

Answer 25

Swift2示例：iOS设备的屏幕宽度格式化Float删除小数

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

Answer 26

我不知道两个小数位，但是这里是你如何打印零小数位的浮点数，所以我想这可以是2个地方，3个地方......（注意：你必须将CGFloat转换为Double以传递给String（格式:)，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))
}

Answer 27

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 ""
    }
}

Answer 28

@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

Answer 29

iOS 15+ 之后推荐使用此解决方案：

2.31234.formatted(.number.precision(.fractionalLength(1)))

Answer 30

@Christian Dietrich：

而不是：

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)"

[校正：] 抱歉混淆* - 当然这适用于双打。我认为，最实用的（如果你想要数字被舍入，而不是被截断），它就会是这样的：

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
}

仅限Float，只需将Double替换为Float，使用powf替换pow并使用roundf替换。
更新：我发现使用返回类型Double而不是String是最实际的。它对于String输出也是一样的，即：

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

打印：Pi约为3.142
所以你可以用相同的方式使用它（你甚至可以写：println（d~> 2）），但另外你也可以用它来直接舍入值，即：

d = Double(slider.value) ~> 2

或任何你需要的......

Answer 31

使用

CGFloat

或

Float.roundTo(places:2)

Swift中的精确字符串格式说明符

31 个答案:

SWIFT 3.1更新

用法

详细

解决方案1

解决方案2

用法

完整样本

输出

Swift中的精确字符串格式说明符

31 个答案:

SWIFT 3.1更新

用法

详细

解决方案1 ​​

解决方案2

用法

完整样本

输出

解决方案1