Question

I’m reading The Scheme Programming Language book. I’m trying to do exercise 2.8.7:

Use map to define a procedure, transpose, that takes a list of pairs and returns a pair of lists as follows.

(transpose ‘((a . 1) (b . 2) (c . 3))) ;;=> ((a b c) 1 2 3)

[Hint: ((a b c) 1 2 3) is the same as ((a b c) . (1 2 3)).]

I found out that (map list '(a 1) '(b 2) '(c 3)) gives me '((a b c) (1 2 3)). I guess I could solve the exercise by writing a lot of boiler plate for turning '((a . 1) (b . 2) (c . 3)) into (map list '(a 1) '(b 2) '(c 3)) and '((a b c) (1 2 3)) into ((a b c) 1 2 3). However, I’m sure that’s not the point of the exercise.

Can anyone help me out here? Is there an obvious way to do it using map that I’m missing?

Answer 1

Yes. The obvious solution involves calling map twice and then consing together the two results.