Question

这样做的最佳方法是什么？这是我到目前为止所得到的

(defn line-segment [start end]
  (let [x-direction (abs (- (first end) (first start)))
        y-direction (abs (- (last end) (last start)))]
    (cond
      (= 0 x-direction) (something ...)
      (= 0 y-direction) (something ...))))

这是我的最终目标

user=> (line-segment [5 6] [5 8])
([5 6] [5 7] [5 8])

是的，没有对角线，只有x或y移动。

感谢。

Answer 1

我认为这是一个非常优雅的解决方案：

(defn line-segment [start end]
  (let [x1 (first start) x2 (first end)
        y1 (last start)  y2 (last end)
        dx (if (> x1 x2) -1 1)
        dy (if (> y1 y2) -1 1)]
    (for [x (range x1 (+ dx x2) dx)
          y (range y1 (+ dy y2) dy)]
      [x y])))

REPL会议：

user> (line-segment [5 6] [5 8])
([5 6] [5 7] [5 8])
user> (line-segment [5 8] [5 6])
([5 8] [5 7] [5 6])
user> (line-segment [-2 7] [1 7])
([-2 7] [-1 7] [0 7] [1 7])
user> (line-segment [1 7] [-2 7])
([1 7] [0 7] [-1 7] [-2 7])

即使您的示例输出被格式化为向量，此函数也会返回LazySeq。我认为这并不重要。

Answer 2

这是一个简单的解决方案，也允许对角线：

(use 'clojure.contrib.math)

(defn line-segment [start end]
  (let [x1 (first start) x2 (first end)
        y1 (last start)  y2 (last end)
        xdiff (- x2 x1)
        ydiff (- y2 y1)
        maxdiff (max (abs xdiff) (abs ydiff))
        dx (/ xdiff maxdiff)
        dy (/ ydiff maxdiff)]
    (for [i (range (inc maxdiff))]
      [(round (+ x1 (* i dx))) (round (+ y1 (* i dy)))])))

与dbryne的解决方案一样，这会返回一个懒的点序列而不是一个向量：我认为这是最有用的形式，假设您随后想要依次对线段上的每个点执行某些操作。

在Clojure中从两个点创建线段

2 个答案: