Matrix Operations?

Question

As I am a SCHEME beginner, I am having a tough time figuring out how to work with matrices.

I am unsure on how to do the following:

A) Given a matrix, return the dimensions of said matrix

EX:// '(2 3 4) (1 0 6)

should return (2.3) for 2 rows, three columns

B) Reverse the order of rows in a given matrix

EX:// given '((1 2) (3 5) (9 0))

the reverse should be '((9 0)(3 5)(1 2))

C) Same as part B but reverse order of columns instead

D) Reverse order of columns and rows

Thanks in advance! This would really mean a lot if someone could give any help at all!

If your matrices are represented as lists of lists, you can do all these very simply with standard list operations. — molbdnilo, Apr 09 '13 at 08:31

score 3 · Answer 1 · answered Apr 09 '13 at 02:06

Standard R6RS Scheme doesn't provide matrices, though some implementations may provide them. The usual trick when implementing them yourself is to use a vector of vectors, rather than a list of lists as you showed above, because you generally don't want to access the elements of the matrix in order, and vectors provide constant-time access to each of their elements whereas lists provide linear-time access to each of their elements.

I have a small library of matrix operations at my blog; you can find uses of that library with the search function on the blog.

score 1 · Answer 2 · answered Apr 09 '13 at 14:05

This is very simple to implement in terms of lists operations - given that your matrix representation is nothing more than a list of sublists! This is obviously a homework, so you better try to solve it by your own means. But I can give you some hints and test cases:

(define (dimensions m)
  (cons <???> <???>)) ; length of m and length of m's first row

(dimensions '((2 3 4) (1 0 6)))
; => (2 . 3)

(define (reverse-rows m)
  <???>) ; just reverse the list

(reverse-rows '((1 2) (3 5) (9 0)))
; => '((9 0)(3 5)(1 2))

(define (reverse-columns m)
  <???>) ; reverse each of the sublists. Hint: use `map`

(reverse-columns '((1 2) (3 5) (9 0)))
; => '((2 1) (5 3) (0 9))

(define (reverse-columns-rows m)
  <???>) ; call previous procedures, the ouput of one is the input of the other

(reverse-columns-rows '((1 2) (3 5) (9 0)))
; => '((0 9) (5 3) (2 1))

Ben Kovitz · Answer 3 · 2013-04-12T11:03:10.697

(define dimensions
  (λ (mat)
    (cons (length mat) (length (car mat)))))

This works because the matrix is represented as a list of rows, where each row is a list. So, (length mat) is the number of rows. Since each row is represented as a list of all the items in its columns, you can find out how many columns there are just by finding out how many elements are in one of the rows. For simplicity, you can just look at the first row: (length (car mat)).

(define reverse-rows reverse)

This works because the matrix is just a list of rows. So, you can just bind reverse-rows to the standard procedure for reversing any list.

(define reverse-columns
  (λ (mat)
    (map reverse mat)))

This works because each row is just a list of items. The map applies reverse to each item in mat (that is, each row) and returns a new list, containing the new rows.

(define reverse-rows-columns
  (λ (mat)
    (reverse-rows (reverse-columns mat))))

This works because up above, we've already defined a way to reverse rows and a way to reverse columns. If you want to do both, you can just combine them.

Matrix Operations?

3 Answers3