我有一个非常长的十进制数(比如说
17.9384693864596069567
),我想将小数部分截断到几位小数(输出结果应该是 17.9384
)。我不想对数字四舍五入为17.9385
。如何实现这个功能?你可以通过将其作为Double
的扩展来进一步整理它:
extension Double {
func truncate(places : Int)-> Double {
return Double(floor(pow(10.0, Double(places)) * self)/pow(10.0, Double(places)))
}
}
你可以像这样使用它:
var num = 1.23456789
// return the number truncated to 2 places
print(num.truncate(places: 2))
// return the number truncated to 6 places
print(num.truncate(places: 6))
0.23456789.truncate(2)
,你会得到0.23000000000000001
。 - Jeroen BakkerString(format: "%.0f", ratio*100)
在这里,0代表您想要允许的小数位数,本例中为零。 Ratio是一种像0.5556633这样的双精度浮点数。
希望这有所帮助。我已经弄清楚了。
只需将数字向下取整,使用一些花哨的技巧即可。
let x = 1.23556789
let y = Double(floor(10000*x)/10000) // leaves on first four decimal places
let z = Double(floor(1000*x)/1000) // leaves on first three decimal places
print(y) // 1.2355
print(z) // 1.235
所以,将数字乘以1,0的个数为你想要的小数位数,向下取整,再除以你所乘的数字。就这样。
extension Double {
/// Rounds the double to decimal places value
func roundToPlaces(_ places:Int) -> Double {
let divisor = pow(10.0, Double(places))
return (self * divisor).rounded() / divisor
}
func cutOffDecimalsAfter(_ places:Int) -> Double {
let divisor = pow(10.0, Double(places))
return (self*divisor).rounded(.towardZero) / divisor
}
}
let a:Double = 1.228923598
print(a.roundToPlaces(2)) // 1.23
print(a.cutOffDecimalsAfter(2)) // 1.22
let number = 17.9384693864596069567;
let merichle = Float(String(format: "%.1f", (number * 10000)).dropLast(2))!/10000
//merichle = 17.9384
Double(String(format: "%.2f", b))
- Merichle在Swift 5中,也可以通过为Decimal创建扩展来进行截断到小数。
extension Decimal {
func truncation(by digit: Int) -> Decimal {
var initialDecimal = self
var roundedDecimal = Decimal()
NSDecimalRound(&roundedDecimal, &initialDecimal, digit, .plain)
return roundedDecimal
}
使用十进制值的用例
value = Decimal(2.56430).truncation(by:2)
值为 2.560000(截断后)
.plain
会导致 NSDecimalRound 向上或向下舍入。将其替换为 .down
,以始终向下舍入,这将使其具有截断行为。 - stef将此代码复制到您的应用程序中...
import Foundation
func truncateDigitsAfterDecimal(number: Double, afterDecimalDigits: Int) -> Double {
if afterDecimalDigits < 1 || afterDecimalDigits > 512 {return 0.0}
return Double(String(format: "%.\(afterDecimalDigits)f", number))!
}
然后你可以像这样调用这个函数:
truncateDigitsAfterDecimal(number: 45.123456789, afterDecimalDigits: 3)
将会产生以下结果:
45.123
SwiftUI: 如果您想在视图中格式化输出,但不涉及计算,那么SwiftUI提供了一种方便的方式来使用C格式说明符作为Text()函数的一部分。
import SwiftUI
let myDouble = 17.93846938645960695
Text("\(myDouble, specifier: "%.2f")")
///Device display: 17.94
方法1:如果您不想为此创建任何新函数,您可以直接按以下方式获取四舍五入后的值。
var roundedValue = (decimalValue * pow(10.0, Double(numberOfPlaces))).rounded())/pow(10.0, Double(numberOfPlaces)
例子:
var numberOfPlaces = 2
var decimalValue = 13312.2423423523523
print("\(((decimalValue * pow(10.0, Double(numberOfPlaces))).rounded())/pow(10.0, Double(numberOfPlaces)))")
方式二:如果你只想打印,可以使用以下代码:
print(String(format: "%.\(numberOfPlaces)f",decimalValue))
var numberOfPlaces = 4
var decimalValue = 13312.2423423523523
print(String(format: "%.\(numberOfPlaces)f",decimalValue))
Swift 5.2的答案
我查看了很多答案,但在截断时总是遇到转换问题。根据我的数学知识,通过截断,如果我有3.1239并且我想要3个小数位,那么我将得到3.123而不是四舍五入(!= 3.1234)。
也许由于过程的性质,我总是成功地使用Double,但总是在Float上遇到问题。
我的方法是创建一个BinaryFloatingPoint的扩展,以便我可以重用它来处理Float、CGFLoat、Double等。
以下扩展获取BinaryFloatingPoint,并可以返回具有给定numberOfDecimals的String或BinaryFloatingPoint值,并处理不同类型的情况:
extension Numeric where Self: BinaryFloatingPoint {
/// Retruns the string value of the BinaryFloatingPoint. The initiaiser
var toString: String {
return String(describing: self)
}
/// Returns the number of decimals. It will be always greater than 0
var numberOfDecimals: Int {
return toString.count - String(Int(self)).count - 1
}
/// Returns a Number with a certain number of decimals
/// - Parameters:
/// - Parameter numberOfDecimals: Number of decimals to return
/// - Returns: BinaryFloatingPoint with number of decimals especified
func with(numberOfDecimals: Int) -> Self {
let stringValue = string(numberOfDecimals: numberOfDecimals)
if self is Double {
return Double(stringValue) as! Self
} else {
return Float(stringValue) as! Self
}
}
/// Returns a string representation with a number of decimals
/// - Parameters:
/// - Parameter numberOfDecimals: Number of decimals to return
/// - Returns: String with number of decimals especified
func string(numberOfDecimals: Int) -> String {
let selfString = toString
let selfStringComponents = selfString.components(separatedBy: ".")
let selfStringIntegerPart = selfStringComponents[0]
let selfStringDecimalPart = selfStringComponents[1]
if numberOfDecimals == 0 {
return selfStringIntegerPart
} else {
if selfStringDecimalPart.count == numberOfDecimals {
return [selfStringIntegerPart,
selfStringDecimalPart].joined(separator: ".")
} else {
if selfStringDecimalPart.count > numberOfDecimals {
return [selfStringIntegerPart,
String(selfStringDecimalPart.prefix(numberOfDecimals))].joined(separator: ".")
} else {
let difference = numberOfDecimals - selfStringDecimalPart.count
let addedCharacters = [Character].init(repeating: "0", count: difference)
return [selfStringIntegerPart,
selfStringDecimalPart+addedCharacters].joined(separator: ".")
}
}
}
}
}
它可能看起来有些老派,但我的所有测试都通过了:
func test_GivenADecimalNumber_ThenAssertNumberOfDecimalsWanted() {
//No decimals
XCTAssertEqual(Float(3).with(numberOfDecimals: 0), 3)
XCTAssertEqual(Float(3.09).with(numberOfDecimals: 0), 3)
XCTAssertEqual(Float(3.999).with(numberOfDecimals: 0), 3)
XCTAssertEqual(Double(3).with(numberOfDecimals: 0), 3)
XCTAssertEqual(Double(3.09).with(numberOfDecimals: 0), 3)
XCTAssertEqual(Double(3.999).with(numberOfDecimals: 0), 3)
//numberOfDecimals == totalNumberOfDecimals
XCTAssertEqual(Float(3.00).with(numberOfDecimals: 2), 3.00)
XCTAssertEqual(Float(3.09).with(numberOfDecimals: 2), 3.09)
XCTAssertEqual(Float(3.01).with(numberOfDecimals: 2), 3.01)
XCTAssertEqual(Float(3.999).with(numberOfDecimals: 3), 3.999)
XCTAssertEqual(Float(3.991).with(numberOfDecimals: 3), 3.991)
XCTAssertEqual(Double(3.00).with(numberOfDecimals: 2), 3.00)
XCTAssertEqual(Double(3.09).with(numberOfDecimals: 2), 3.09)
XCTAssertEqual(Double(3.01).with(numberOfDecimals: 2), 3.01)
XCTAssertEqual(Double(3.999).with(numberOfDecimals: 3), 3.999)
XCTAssertEqual(Double(3.991).with(numberOfDecimals: 3), 3.991)
//numberOfDecimals < totalNumberOfDecimals
XCTAssertEqual(Float(3.00).with(numberOfDecimals: 1), 3.0)
XCTAssertEqual(Float(3.09).with(numberOfDecimals: 1), 3.0)
XCTAssertEqual(Float(3.01).with(numberOfDecimals: 1), 3.0)
XCTAssertEqual(Float(3.999).with(numberOfDecimals: 2), 3.99)
XCTAssertEqual(Float(3.991).with(numberOfDecimals: 2), 3.99)
XCTAssertEqual(Double(3.00).with(numberOfDecimals: 1), 3.0)
XCTAssertEqual(Double(3.09).with(numberOfDecimals: 1), 3.0)
XCTAssertEqual(Double(3.01).with(numberOfDecimals: 1), 3.0)
XCTAssertEqual(Double(3.999).with(numberOfDecimals: 2), 3.99)
XCTAssertEqual(Double(3.991).with(numberOfDecimals: 2), 3.99)
//numberOfDecimals > totalNumberOfDecimals
XCTAssertEqual(Float(3.00).with(numberOfDecimals: 3), 3.000)
XCTAssertEqual(Float(3.09).with(numberOfDecimals: 3), 3.090)
XCTAssertEqual(Float(3.01).with(numberOfDecimals: 3), 3.010)
XCTAssertEqual(Float(3.999).with(numberOfDecimals: 4), 3.9990)
XCTAssertEqual(Float(3.991).with(numberOfDecimals: 4), 3.9910)
XCTAssertEqual(Double(3.00).with(numberOfDecimals: 3), 3.000)
XCTAssertEqual(Double(3.09).with(numberOfDecimals: 3), 3.090)
XCTAssertEqual(Double(3.01).with(numberOfDecimals: 3), 3.010)
XCTAssertEqual(Double(3.999).with(numberOfDecimals: 4), 3.9990)
XCTAssertEqual(Double(3.991).with(numberOfDecimals: 4), 3.9910)
}
func test_GivenADecimal_ThenAssertStringValueWithDecimalsWanted() {
//No decimals
XCTAssertEqual(Float(3).string(numberOfDecimals: 0), "3")
XCTAssertEqual(Float(3.09).string(numberOfDecimals: 0), "3")
XCTAssertEqual(Float(3.999).string(numberOfDecimals: 0), "3")
XCTAssertEqual(Double(3).string(numberOfDecimals: 0), "3")
XCTAssertEqual(Double(3.09).string(numberOfDecimals: 0), "3")
XCTAssertEqual(Double(3.999).string(numberOfDecimals: 0), "3")
//numberOfDecimals == totalNumberOfDecimals
XCTAssertEqual(Float(3.00).string(numberOfDecimals: 2), "3.00")
XCTAssertEqual(Float(3.09).string(numberOfDecimals: 2), "3.09")
XCTAssertEqual(Float(3.01).string(numberOfDecimals: 2), "3.01")
XCTAssertEqual(Float(3.999).string(numberOfDecimals: 3), "3.999")
XCTAssertEqual(Float(3.991).string(numberOfDecimals: 3), "3.991")
XCTAssertEqual(Double(3.00).string(numberOfDecimals: 2), "3.00")
XCTAssertEqual(Double(3.09).string(numberOfDecimals: 2), "3.09")
XCTAssertEqual(Double(3.01).string(numberOfDecimals: 2), "3.01")
XCTAssertEqual(Double(3.999).string(numberOfDecimals: 3), "3.999")
XCTAssertEqual(Double(3.991).string(numberOfDecimals: 3), "3.991")
//numberOfDecimals < totalNumberOfDecimals
XCTAssertEqual(Float(3.00).string(numberOfDecimals: 1), "3.0")
XCTAssertEqual(Float(3.09).string(numberOfDecimals: 1), "3.0")
XCTAssertEqual(Float(3.01).string(numberOfDecimals: 1), "3.0")
XCTAssertEqual(Float(3.999).string(numberOfDecimals: 2), "3.99")
XCTAssertEqual(Float(3.991).string(numberOfDecimals: 2), "3.99")
XCTAssertEqual(Double(3.00).string(numberOfDecimals: 1), "3.0")
XCTAssertEqual(Double(3.09).string(numberOfDecimals: 1), "3.0")
XCTAssertEqual(Double(3.01).string(numberOfDecimals: 1), "3.0")
XCTAssertEqual(Double(3.999).string(numberOfDecimals: 2), "3.99")
XCTAssertEqual(Double(3.991).string(numberOfDecimals: 2), "3.99")
//numberOfDecimals > totalNumberOfDecimals
XCTAssertEqual(Float(3.00).string(numberOfDecimals: 3), "3.000")
XCTAssertEqual(Float(3.09).string(numberOfDecimals: 3), "3.090")
XCTAssertEqual(Float(3.01).string(numberOfDecimals: 3), "3.010")
XCTAssertEqual(Float(3.999).string(numberOfDecimals: 4), "3.9990")
XCTAssertEqual(Float(3.991).string(numberOfDecimals: 4), "3.9910")
XCTAssertEqual(Double(3.00).string(numberOfDecimals: 3), "3.000")
XCTAssertEqual(Double(3.09).string(numberOfDecimals: 3), "3.090")
XCTAssertEqual(Double(3.01).string(numberOfDecimals: 3), "3.010")
XCTAssertEqual(Double(3.999).string(numberOfDecimals: 4), "3.9990")
XCTAssertEqual(Double(3.991).string(numberOfDecimals: 4), "3.9910")
}
Numeric
协议扩展并限制为 BinaryFloatingPoint
是毫无意义的。 - Leo DabusDouble(stringValue) as! Self
??? 顺便说一下,如果它不是 Double
,你怎么保证它是一个 Float
。试试 Float80(123.456789).with(numberOfDecimals: 2)
// BOOM - Leo DabusSelf?
。 - Leo Dabus