Swift - 如果小数等于 0,如何从浮点数中删除小数?

Swift - How to remove a decimal from a float if the decimal is equal to 0?

我显示的距离只有一位小数,我想删除这个小数以防它等于 0(例如:1200.0Km),我怎么能在 swift 中做到这一点? 我这样显示这个数字:

let distanceFloat: Float = (currentUser.distance! as NSString).floatValue
distanceLabel.text = String(format: "%.1f", distanceFloat) + "Km"

Swift 3/4:

var distanceFloat1: Float = 5.0
var distanceFloat2: Float = 5.540
var distanceFloat3: Float = 5.03

extension Float {
    var clean: String {
       return self.truncatingRemainder(dividingBy: 1) == 0 ? String(format: "%.0f", self) : String(self)
    }
}

print("Value \(distanceFloat1.clean)") // 5
print("Value \(distanceFloat2.clean)") // 5.54
print("Value \(distanceFloat3.clean)") // 5.03

Swift 2(原始答案)

let distanceFloat: Float = (currentUser.distance! as NSString).floatValue
distanceLabel.text = String(format: distanceFloat == floor(distanceFloat) ? “%.0f" : "%.1f", distanceFloat) + "Km"

或作为扩展:

extension Float {
    var clean: String {
        return self % 1 == 0 ? String(format: "%.0f", self) : String(self)
    }
}

NSNumberFormatter 是你的朋友

let distanceFloat: Float = (currentUser.distance! as NSString).floatValue
let numberFormatter = NSNumberFormatter()
numberFormatter.positiveFormat = "###0.##"
let distance = numberFormatter.stringFromNumber(NSNumber(float: distanceFloat))!
distanceLabel.text = distance + " Km"

使用 NSNumberFormatter:

let formatter = NumberFormatter()
formatter.minimumFractionDigits = 0
formatter.maximumFractionDigits = 2

// Avoid not getting a zero on numbers lower than 1
// Eg: .5, .67, etc...
formatter.numberStyle = .decimal

let nums = [3.0, 5.1, 7.21, 9.311, 600.0, 0.5677, 0.6988]

for num in nums {
    print(formatter.string(from: num as NSNumber) ?? "n/a")
}

Returns:

3

5.1

7.21

9.31

600

0.57

0.7

这是完整的代码。

let numberA: Float = 123.456
let numberB: Float = 789.000

func displayNumber(number: Float) {
    if number - Float(Int(number)) == 0 {
        println("\(Int(number))")
    } else {
        println("\(number)")
    }
}

displayNumber(numberA) // console output: 123.456
displayNumber(numberB) // console output: 789

下面是最重要的一行。

func displayNumber(number: Float) {
  1. Int(number).
  2. 去除浮点数的小数位
  3. Returns 剥离后的数字回浮点运算与Float(Int(number)).
  4. number - Float(Int(number))
  5. 的十进制数
  6. if number - Float(Int(number)) == 0
  7. 检查十进制数是否为空

if和else语句中的内容不需要解释。

extension 是强大的方法。

分机:

Swift 2 的代码(不是 Swift 3 或更高版本):

extension Float {
    var cleanValue: String {
        return self % 1 == 0 ? String(format: "%.0f", self) : String(self)
    }
}

用法:

var sampleValue: Float = 3.234
print(sampleValue.cleanValue)

3.234

sampleValue = 3.0
print(sampleValue.cleanValue)

3

sampleValue = 3
print(sampleValue.cleanValue)

3


示例游乐场文件是 here.

也许stringByReplacingOccurrencesOfString可以帮到你:)

let aFloat: Float = 1.000

let aString: String = String(format: "%.1f", aFloat) // "1.0"

let wantedString: String = aString.stringByReplacingOccurrencesOfString(".0", withString: "") // "1"

您可以使用已经提到的扩展程序,但此解决方案更短一些:

extension Float {
    var shortValue: String {
        return String(format: "%g", self)
    }
}

用法示例:

var sample: Float = 3.234
print(sample.shortValue)

更新 swift 3 的已接受答案:

extension Float {
    var cleanValue: String {
        return self.truncatingRemainder(dividingBy: 1) == 0 ? String(format: "%.0f", self) : String(self)
    }
}

用法只是:

let someValue: Float = 3.0

print(someValue.cleanValue) //prints 3

这也可能有帮助。

extension Float {
    func cleanValue() -> String {
        let intValue = Int(self)
        if self == 0 {return "0"}
        if self / Float (intValue) == 1 { return "\(intValue)" }
        return "\(self)"
    }
}

用法:

let number:Float = 45.23230000
number.cleanValue()

要将其格式化为字符串,请遵循此模式

let aFloat: Float = 1.123

let aString: String = String(format: "%.0f", aFloat) // "1"
let aString: String = String(format: "%.1f", aFloat) // "1.1"
let aString: String = String(format: "%.2f", aFloat) // "1.12"
let aString: String = String(format: "%.3f", aFloat) // "1.123"

要将其转换为 Int,请遵循此模式

let aInt: Int = Int(aFloat) // "1"

当您使用 String(format: 初始化器时,Swift 将根据以下数字自动根据需要四舍五入最后一位。

Swift 4 试试这个。

    extension CGFloat{
        var cleanValue: String{
            //return String(format: 1 == floor(self) ? "%.0f" : "%.2f", self)
            return self.truncatingRemainder(dividingBy: 1) == 0 ? String(format: "%.0f", self) : String(format: "%.2f", self)//
        }
    }

//使用方法 - 如果在(.)点后输入超过两个字符,它会自动裁剪最后一个字符,只显示点后两个字符。

let strValue = "32.12"
print(\(CGFloat(strValue).cleanValue)

格式化为最大小数位数,没有尾随零

当需要自定义输出精度时,这种情况很好。 这个解决方案似乎与 NumberFormatter + NSNumber 大致一样快,但一个好处可能是我们在这里避免了 NSObject。

extension Double {
    func string(maximumFractionDigits: Int = 2) -> String {
        let s = String(format: "%.\(maximumFractionDigits)f", self)
        var offset = -maximumFractionDigits - 1
        for i in stride(from: 0, to: -maximumFractionDigits, by: -1) {
            if s[s.index(s.endIndex, offsetBy: i - 1)] != "0" {
                offset = i
                break
            }
        }
        return String(s[..<s.index(s.endIndex, offsetBy: offset)])
    }
}

(也适用于 extension Float,但不适用于 macOS 专用类型 Float80

用法:myNumericValue.string(maximumFractionDigits: 2)myNumericValue.string()

maximumFractionDigits: 2 的输出:

1.0 → "1"
0.12 → "0.12"
0.012 → "0.01"
0.0012 → "0"
0.00012 → "0"

简单:

Int(floor(myFloatValue))

Swift 5 对于 Double 它与@Frankie 对 float

的回答相同
var dec: Double = 1.0
dec.clean // 1

用于扩展

extension Double {
    var clean: String {
       return self.truncatingRemainder(dividingBy: 1) == 0 ? String(format: "%.0f", self) : String(self)
    }

}

Swift 5.5 轻松搞定

只需使用新的 formatted() api 和默认的 FloatingPointFormatStyle:

let values: [Double] = [1.0, 4.5, 100.0, 7]
for value in values {
    print(value.formatted(FloatingPointFormatStyle()))
}
// prints "1, 4.5, 100, 7"