Question

如何在K（或Q）等基于数组的功能语言中表达此命令功能？

在草率的C ++中：

vector<int> x(10), y(10); // Assume these are initialized with some values.

// BTW, 4 is just a const -- it's part of the algorithm and is arbitrarily chosen.

vector<int> result1(x.size() - 4 + 1); // A place to hold a resulting array.
vector<int> result2(x.size() - 4 + 1); // A place to hold another resulting array.

// Here's the code I want to express functionally.
for (int i = 0; i <= x.size() - 4; i++) {
    int best = x[i + 0] - y[i + 0];
    int bad = best;
    int worst = best;
    for(int j = 0; j < 4; j++) {
        int tmp = x[i + j] - y[i + 0];
        bad = min(bad, tmp);
        if(tmp > best) {
            best = tmp;
            worst = bad;
        }
    }
    result1[i] = best
    result2[i] = worst
}

我最希望在kdb和Q中看到这一点，但欢迎使用其他功能语言。

Answer 1

在Kona（开源K方言）：

首先，设置一些示例值（使用与Clojure解决方案相同）：

a:1+!8;b:8#0        / a is 1..8, b is eight 0s

然后：

{(|/x;&/x)}@+{4#y _ x}[a+b;]'!#a

a 和 b 是上面的x和y变量。（K为变量x，y和z提供了特殊情况。）

要打破这一点：

maxmin:{(|/x;&/x)}  / (max;min) pairs of x
get4:{4#y _ x}      / next 4 from x, starting at y
                    / with <4 remaining, will repeat; doesn't matter for min or max
/ maxmin applied to flipped results of get4(a-b) at each index 0..(length a)-1
maxmin@+get4[a-b;]'!#a

/ result
(4 5 6 7 8 8 8 8
 1 2 3 4 5 6 7 8)

Answer 2

将@silentbicycle的k直接移植到q yield

q)a:1+til 8
q)b:8#0
q){(max x;min x)}flip{4#y _ x}[a+b;]each til count a
4 5 6 7 8 8 8 8
1 2 3 4 5 6 7 8

另一种方法，稍微更加矢量化（imao）：

q){(max;min)@\:flip 4#'(til count x)_\:x+y}[a;b]
4 5 6 7 8 8 8 8
1 2 3 4 5 6 7 8

Answer 3

Clojure（方言Lisp）：

(defn minmax [x y](map #(vector (- (apply max %1) %2) (- (apply min %1) %2)))(partition-all 4 1 x) y)

(minmax [1 2 3 4 5 6 7 8] [0 0 0 0 0 0 0 0])

将给出

[（[4 1] [5 2] [6 3] [7 4] [8 5] [8 6] [8 7] [8 8]）`（结果1，结果2）作为输出..

然后

(map #(first %1) result) is result1
(map #(last %1) result) is result2

如何在基于数组的功能语言（如K（或Q））中表达这种命令功能？

3 个答案: