为了方便使用，我创建了一个扩展:

extension Double {
    var threeDigits: Double {
        return (self * 1000).rounded(.toNearestOrEven) / 1000
    }
    
    var twoDigits: Double {
        return (self * 100).rounded(.toNearestOrEven) / 100
    }
    
    var oneDigit: Double {
        return (self * 10).rounded(.toNearestOrEven) / 10
    }
}

var myDouble = 0.12345
print(myDouble.threeDigits)
print(myDouble.twoDigits)
print(myDouble.oneDigit)

打印结果如下:

0.123
0.12
0.1

感谢其他答案的启发!

对于许多应用程序，需要精确的小数位数来舍入。

对于其他一些，您没有这样的限制，并希望“压缩”输出大小，但仍然希望避免将数字转换为字符串(反之亦然)，就像导出具有数百万个数字的JSON一样。

在这种情况下，你可以使用'trick'来四舍五入显数(尾数)，而不是整数。在这种情况下，你会得到最后的小数点 大多数数字仍然最多保留3位小数点后(如果需要的话)，而有些数字会稍微多一些。

这是你在使用十进制数时所期望的结果:

5.2472 5.2516 5.2556 5.26 5.264

而不是:

5.24731462499949 5.251488374999099 5.25566283399894 5.259839374999501 5.264012208999702

let value = 5.24731462499949

print(value)
// 5.24731462499949

let valueSignificandRounded = round((1000 * 10) * value.significand) / (1000 * 10)

let valueRounded = CGFloat(sign: v.sign, exponent: v.exponent, significand: valueSignificandRounded)

print(valueRounded)
// 5.2472

Lots of example are using maths, the problem is floats are approximations of real number, there is no way to express 0.1 (1/10) exactly as a float just as there is no exact way to express ⅓ exactly using decimal points, so you need to ask your self exactly what your are trying to achieve, if you just want to display them leave them as they are in code, trying to round them is going to justify give you less accurate result as you are throwing away precious, round ⅓ in decimal notation to 1 decimal place is not going to give you a number closer to ⅓, us NumberFormate to round it, if you have something like a viewModel class it can be used to return a string representation to your models numbers. NumberFormaters give you lots of control on how numbers are formatted and the number of decimal places you want.

你可以使用Swift的round函数来实现这一点。

要对Double进行3位精度的四舍五入，首先将其乘以1000，四舍五入，然后将四舍五入结果除以1000:

let x = 1.23556789
let y = Double(round(1000 * x) / 1000)
print(y) /// 1.236

与任何类型的printf(…)或String(format:…)解决方案不同，此操作的结果仍然是Double类型。

编辑: 关于它有时不能工作的评论，请阅读这篇:每个程序员都应该知道的关于浮点算术的事情

在Swift 3.0和Xcode 8.0中:

extension Double {
    func roundTo(places: Int) -> Double {
        let divisor = pow(10.0, Double(places))
        return (self * divisor).rounded() / divisor
    }
}

像这样使用这个扩展:

let doubleValue = 3.567
let roundedValue = doubleValue.roundTo(places: 2)
print(roundedValue) // prints 3.56

为了方便使用，我创建了一个扩展:

extension Double {
    var threeDigits: Double {
        return (self * 1000).rounded(.toNearestOrEven) / 1000
    }
    
    var twoDigits: Double {
        return (self * 100).rounded(.toNearestOrEven) / 100
    }
    
    var oneDigit: Double {
        return (self * 10).rounded(.toNearestOrEven) / 10
    }
}

var myDouble = 0.12345
print(myDouble.threeDigits)
print(myDouble.twoDigits)
print(myDouble.oneDigit)

打印结果如下:

0.123
0.12
0.1

感谢其他答案的启发!