Swift 4:将数字格式化为友好的K.

Question

目前正在开发一个简单的功能,对我来说做得很好..

例如:如果我有1000,它将打印1.0K,或1,000,000它将是1M,一切正常,直到这里,

如果我想将1,000,000,000转为1B怎么办？

我尝试了以下 - >

func formatPoints(from: Int) -> String {
     let number = Double(from)
     let thousand = number / 1000
     let million = number / 1000000
     let billion = number / 1000000000

     if million >= 1.0 { 
     return "\(round(million*10)/10)M" 
} else if thousand >= 1.0 { 
     return "\(round(thousand*10)/10)K"
} else if billion >= 1.0 {
        return ("\(round(billion*10/10))B")
} else {
    return "\(Int(number))"}
}

print(formatPoints(from: 1000000000))

但它返回1000.0M,而不是1B

谢谢!

Answer 1

我个人不喜欢在格式化这样的数字时舍入,我更喜欢截断.在接受的答案中,1,515四舍五入是为了给你2k什么时候应该得到的1.5k.因此,如果您希望截断舍入,请考虑此方法:

extension Double {
    func truncate(places: Int) -> Double {
        return Double(floor(pow(10.0, Double(places)) * self)/pow(10.0, Double(places)))
    }
}

func formatNumber(_ n: Int) -> String {

    let num = abs(Double(n))
    let sign = (n < 0) ? "-" : ""

    switch num {

    case 1_000_000_000...:
        var formatted = num / 1_000_000_000
        formatted = formatted.truncate(places: 1)
        return "\(sign)\(formatted)B"

    case 1_000_000...:
        var formatted = num / 1_000_000
        formatted = formatted.truncate(places: 1)
        return "\(sign)\(formatted)M"

    case 1_000...:
        var formatted = num / 1_000
        formatted = formatted.truncate(places: 1)
        return "\(sign)\(formatted)K"

    case 0...:
        return "\(n)"

    default:
        return "\(sign)\(n)"

    }

}

对于需要进一步截断的特定情况,您可以非常轻松地微调此方法,例如100k代替100.5k或1M替代1.1M.此方法也处理否定.

print(formatNumber(1515)) // 1.5K
print(formatNumber(999999)) // 999.9K
print(formatNumber(1000999)) // 1.0M

Answer 2

以下if-else语句逻辑向您展示了什么是先发生的，什么是最后发生的：

import Foundation

func formatPoints(from: Int) -> String {

    let number = Double(from)
    let thousand = number / 1000
    let million = number / 1000000
    let billion = number / 1000000000
    
    if billion >= 1.0 {
        return "\(round(billion*10)/10)B"
    } else if million >= 1.0 {
        return "\(round(million*10)/10)M"
    } else if thousand >= 1.0 {
        return ("\(round(thousand*10/10))K")
    } else {
        return "\(Int(number))"
    }
}

print(formatPoints(from: 1000))             /*  1.0 K  */
print(formatPoints(from: 1000000))          /*  1.0 M  */
print(formatPoints(from: 1000000000))       /*  1.0 B  */

亿必须先走。