Question

问题

let x = (0..<10).splitEvery( 3 )
XCTAssertEqual( x, [(0...2),(3...5),(6...8),(9)], "implementation broken" )

我遇到了计算Range等元素数量的问题......

extension Range
{
    func splitEvery( nInEach: Int ) -> [Range]
    {
        let n = self.endIndex - self.startIndex // ERROR - cannot invoke '-' with an argument list of type (T,T)
    }
}

Answer 1

范围内的值为ForwardIndexType，因此您只能advance()个，或计算distance()，但未定义减法-。提前金额必须是相应的输入T.Distance。所以这将是一个可能的实现：

extension Range {
    func splitEvery(nInEach: T.Distance) -> [Range] {
        var result = [Range]() // Start with empty array
        var from  = self.startIndex
        while from != self.endIndex {
            // Advance position, but not beyond the end index:
            let to = advance(from, nInEach, self.endIndex)
            result.append(from ..< to)
            // Continue with next interval:
            from = to
        }
        return result
    }
}

示例：

println( (0 ..< 10).splitEvery(3) )
// Output: [0..<3, 3..<6, 6..<9, 9..<10]

但请注意，0 ..< 10不是整数的列表（或数组）。要将数组拆分为子数组，您可以定义类似的扩展名：

extension Array {
    func splitEvery(nInEach: Int) -> [[T]] {
        var result = [[T]]()
        for from in stride(from: 0, to: self.count, by: nInEach) {
            let to = advance(from, nInEach, self.count)
            result.append(Array(self[from ..< to]))
        }
        return result
    }
}

示例：

println( [1, 1, 2, 3, 5, 8, 13].splitEvery(3) )
// Output: [[1, 1, 2], [3, 5, 8], [13]]

更通用的方法可能是拆分所有可切片对象。但是Sliceable 是协议，协议无法扩展。你可以做的是定义一个函数，它将可切片对象作为第一个参数：

func splitEvery<S : Sliceable>(seq : S, nInEach : S.Index.Distance) -> [S.SubSlice] { 
    var result : [S.SubSlice] = []

    var from  = seq.startIndex
    while from != seq.endIndex {
        let to = advance(from, nInEach, seq.endIndex)
        result.append(seq[from ..< to])
        from = to
    }
    return result
}

（请注意，此功能与（扩展名）方法完全无关以上定义。）

示例：

println( splitEvery("abcdefg", 2) )
// Output: [ab, cd, ef, g]
println( splitEvery([3.1, 4.1, 5.9, 2.6, 5.3], 2) )
// Output: [[3.1, 4.1], [5.9, 2.6], [5.3]]

范围不可切片，但您可以定义一个单独的函数，它需要一个范围论证：

func splitEvery<T>(range : Range<T>, nInEach : T.Distance) -> [Range<T>] { 
    var result : [Range<T>] = []

    var from  = range.startIndex
    while from != range.endIndex {
        let to = advance(from, nInEach, range.endIndex)
        result.append(from ..< to)
        from = to
    }
    return result
}

示例：

println( splitEvery(0 ..< 10, 3) )
// Output: [0..<3, 3..<6, 6..<9, 9..<10]

如何在Swift中实现Haskell的splitEvery？

1 个答案: