下面是我以前如何将一个浮点数截断到小数点后两位
NSLog(@" %.02f %.02f %.02f", r, g, b);
我查了文档和电子书,但还没找到答案。谢谢!
下面是我以前如何将一个浮点数截断到小数点后两位
NSLog(@" %.02f %.02f %.02f", r, g, b);
我查了文档和电子书,但还没找到答案。谢谢!
当前回答
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
其他回答
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 ""
}
}
这里有一个“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)")
延伸的力量
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: 速度是12.3
这里大多数答案都是有效的。但是,如果要经常格式化数字,可以考虑扩展Float类,添加一个返回格式化字符串的方法。参见下面的示例代码。这一个通过使用数字格式化器和扩展实现了相同的目标。
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)"
}
}
我目前为止最好的解决方案,以下是David的回答:
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上。很可能在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。请记住,该语言仍处于测试阶段。