这是一种非常快速和简单的方法，不需要复杂的解决方案。

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

使用以下方法

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

println(output)

为什么要把它弄得这么复杂?你可以用这个代替:

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

看它在游乐场工作。

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

Vincent Guerci的ruby / python %操作符，为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]

Swift2示例:iOS设备的屏幕宽度格式化浮点数去除小数

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

那么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))")

打印:112 465 848 348 508.46