Matrix Manipulation in Racket

Question

I need to write a function that produces the nth column of a matrix in Racket without using recursion. For example (list (list 1 2 3) (list 2 3 4) (list 6 7 9)) if I wanted the 2nd column I would receive (list 2 3 7).

I tried (append (map (lambda (n) (list-ref (list-ref M) n) n)) M)), but it keeps showing me an error.

Does this answer your question? https://stackoverflow.com/questions/10523361/how-do-i-get-an-item-from-a-list-at-a-given-index-in-racket-language — ThisIsAQuestion, Nov 23 '20 at 20:56
Use `map` with a function (i.e. use lambda) that calls `list-ref`. — soegaard, Nov 23 '20 at 22:39

ad absurdum · Accepted Answer · 2020-11-24T03:41:21.677

The attempted solution uses list-ref twice, which is not needed, and one of those calls does not have enough arguments: (list-ref M) has the function taking only one argument, but it requires two arguments -- a list and an integer.

The list-ref function will return an indexed element from an input list, so (list-ref '(1 2 3) 1) will return 2. Consider what map will do here: the matrix is represented as a list of lists (a list of rows), i.e., as ((1 2 3) (2 3 4) (6 7 9)). The map function will act on the members of the input list, which are the lists (1 2 3), (2 3 4), and (6 7 9). By mapping the list-ref function over that input, you can take whichever element you want from the sublists. So, (map (lambda (row) (list-ref row 1)) '((1 2 3) (2 3 4) (6 7 9))) would evaluate to (2 3 7), as desired.

Here is a function that lets you take any column from a matrix:

(define (nth-column M n)
  (map (lambda (row) (list-ref row n)) M))

Sample interactions:

scratch.rkt> (define M '((1 2 3)
                         (2 3 4)
                         (6 7 9)))
scratch.rkt> (nth-column M 1)
'(2 3 7)
scratch.rkt> (nth-column M 0)
'(1 2 6)
scratch.rkt> (nth-column M 2)
'(3 4 9)

Matrix Manipulation in Racket

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