查找2D名單鄰居，聰明的方法

Question

讓我們假設我有一個二維表（或數組，如果你願意的話），它看起來是這樣的：

'((I I I O) 
    (I X I O) 
    (I I I O)) 

現在讓我們假設我d喜歡找到X的所有鄰居。在這種情況下，我的函數將返回一個8個I：s的列表。我將如何以智能的方式實現這個功能？我已經實現了這樣一個功能：

(defun get-neighbours (board x y) 
    (let ((output '())) 
    (push (get-if-possible board (1+ x) y) output) 
    (push (get-if-possible board (1- x) y) output) 
    (push (get-if-possible board x (1+ y)) output) 
    (push (get-if-possible board x (1- y)) output) 
    (push (get-if-possible board (1+ x) (1+ y)) output) 
    (push (get-if-possible board (1- x) (1- y)) output) 
    (push (get-if-possible board (1+ x) (1- y)) output) 
    (push (get-if-possible board (1- x) (1+ y)) output) 
    output)) 

而且就是這樣......醜陋。

Answer 1

首先，你仍然在勢在必行土地。

(defun get-neighbours (board x y) 
    (let ((output '())) 
    (push (get-if-possible board (1+ x) y) output) 
    (push (get-if-possible board (1- x) y) output) 
    (push (get-if-possible board x (1+ y)) output) 
    (push (get-if-possible board x (1- y)) output) 
    (push (get-if-possible board (1+ x) (1+ y)) output) 
    (push (get-if-possible board (1- x) (1- y)) output) 
    (push (get-if-possible board (1+ x) (1- y)) output) 
    (push (get-if-possible board (1- x) (1+ y)) output) 
    output)) 

你這樣做：變量聲明，將其綁定到NIL，變量的變異，返回變量。

很簡單，就是：

(defun get-neighbours (board x y) 
    (list (get-if-possible board (1+ x) y) 
     (get-if-possible board (1- x) y) 
     (get-if-possible board x  (1+ y)) 
     (get-if-possible board x  (1- y)) 
     (get-if-possible board (1+ x) (1+ y)) 
     (get-if-possible board (1- x) (1- y)) 
     (get-if-possible board (1+ x) (1- y)) 
     (get-if-possible board (1- x) (1+ y)))) 

可以 '縮短' 與當地的宏代碼：

(defun get-neighbours (board x y) 
    (macrolet ((all-possible (&rest x-y-combinations) 
       `(list 
       ,@(loop for (a b) on x-y-combinations by #'cddr 
         collect `(get-if-posssible board ,a ,b))))) 
    (all-possible (1+ x) y 
        (1- x) y 
        x  (1+ y) 
        x  (1- y) 
        (1+ x) (1+ y) 
        (1- x) (1- y) 
        (1+ x) (1- y) 
        (1- x) (1+ y)))) 

現在人們可以抽象了一點：

(defmacro invoke-fn-on (fn &rest x-y-combinations) 
    `(funcall (function ,fn) 
      ,@(loop for (a b) on x-y-combinations by #'cddr 
        collect `(get-if-posssible board ,a ,b)))) 

(defun get-neighbours (board x y) 
    (invoke-fn-on list 
       (1+ x) y 
       (1- x) y 
       x  (1+ y) 
       x  (1- y) 
       (1+ x) (1+ y) 
       (1- x) (1- y) 
       (1+ x) (1- y) 
       (1- x) (1+ y))) 

關於LOOP：

> (loop for (a b) on '(1 2 3 4 5 6) by #'cddr collect (list a b)) 
((1 2) (3 4) (5 6)) 

ON將模式（a b）移動到列表上（1 2 3 4 5 6）。它會給（1 2），（2 3），（3 4），（4 5）和（5 6）。每次迭代時，通過使用CDR獲取剩餘列表，將列表向前移動一個。現在說，我們移動CDDR的兩個項目，而不是CDR的一個項目。所以我們得到三次迭代和對（1 2），（3 4）和（5 6）。

另一種方法是通過引入不同的表結構的座標對稍微簡化LOOP：

(defmacro invoke-fn-on (fn x-y-combinations) 
    `(funcall (function ,fn) 
      ,@(loop for (a b) in x-y-combinations 
        collect `(get-if-posssible board ,a ,b)))) 

(defun get-neighbours (board x y) 
    (invoke-fn-on list 
       '(((1+ x) y ) 
        ((1- x) y ) 
        (x  (1+ y)) 
        (x  (1- y)) 
        ((1+ x) (1+ y)) 
        ((1- x) (1- y)) 
        ((1+ x) (1- y)) 
        ((1- x) (1+ y))))) 

Answer 2

準備了這樣的事情：

(defconstant +neighbour-offsets+ 
     '((-1 . -1) (0 . -1) (1 . -1) 
     (-1 . 0)   (1 . 0) 
     (-1 . 1) (0 . 1) (1 . 1))) 

，然後你的函數可以簡單如

(defun get-neighbours (board x y) 
    (mapcar (lambda (offs) 
      (let ((dx (car offs)) 
        (dy (cdr offs))) 
       (get-if-possible board (+ x dx) (+ y dy)))) 
      +neighbour-offsets+))