在Scheme编程中将列表的每个元素与另一个列表的每个元素相乘 [英] Multiplying each element of a list with each element of another list in Scheme programming

问题描述

我正在尝试在Scheme中执行以下操作:

i am trying to do the following in Scheme:

List<int> list = new List<int>();
List<int> list1 = new List<int>();
List<int> list2 = new List<int>();
list.Add(1);
list.Add(2);
list.Add(3);
list.Add(4);
list1.Add(2);
list1.Add(4);
list1.Add(6);
list1.Add(8);

for (int i = 0; i < list.Count; i++)
{
    for (int p = 0; p < list1.Count; p++)
    {
         list2.Add(list[i] * list1[p]);
    }
}

如上面的代码所示，

我试图将第一个列表的每个元素与第二个列表中的每个元素相乘.所以1 * 2、1 * 4、1 * 6、1 * 8，然后转到下一个元素2 * 2,2 * 4 ..等等.

as seen in the code above, I am trying to multiply each element of the first list with every element in the second list. So 1*2, 1*4, 1*6, 1*8, then going to the next element, 2*2,2*4.. etc.

我无法将其实施到Scheme中.我尝试使用map函数，但这似乎无法按照我想要的方式工作.有什么想法吗?

I am having trouble implementing this into Scheme. I tried using the map function but this doesn't seem to work the way I want it to. Any ideas?

推荐答案

我们首先定义两个输入列表，由于list是Scheme中的内置过程，因此不宜对其进行覆盖，因此我将其重命名. ):

We start by defining the two input lists, I renamed them since list is a built-in procedure in Scheme and is not a good idea to overwrite it):

(define l '(1 2 3 4))
(define l1 '(2 4 6 8))

我假设您希望结果列表是平坦的"-例如，它不包含元素列表，仅包含元素(如果您可以在l2中包含列表列表，只需删除将以下内容展平的调用).为此，我们需要定义flatten过程:

I'm assuming that you want your result list to be "flat" - e.g., it doesn't contain lists of elements, only elements (if you're ok with having a list of lists in l2, simply delete the call to flatten below). For that, we need to define the flatten procedure:

(define (atom? x)
  (and (not (pair? x)) (not (null? x))))

(define (flatten lst)
  (cond ((null? lst) empty)
        ((atom? lst) (list lst))
        (else (append (flatten (car lst))
                      (flatten (cdr lst))))))

最后，眼前的问题.一旦您了解了如何嵌套两个map过程，就很简单-看一下 SICP .

Finally, the problem at hand. It's simple once you understand how to nest two map procedures - take a look at the nested mappings section in the book SICP.

(define l2
  (flatten
   (map (lambda (i)
          (map (lambda (j)
                 (* i j))
               l1))
        l)))

此时，l2包含预期的答案:

At this point, l2 contains the expected answer:

(2 4 6 8 4 8 12 16 6 12 18 24 8 16 24 32)

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