iterating through the same list with two indices

[enderw88@gmail.com]

I need to run through a list comparing each member to every member after that member for a condition. A very naiive approach would be:

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(loop for i from 0 to (1- (length THE-LIST)) do
   (loop for j from (1+ i) to (1- (length THE-LIST)) do
      (test-the-items  (nth i THE-LIST) (nth j THE-LIST))


This is easily understood, but very ugly.

Another lispier, but probably just as ugly, way would be

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(mapcar #'(lambda (item1)
                     (mapcar #'(lambda (item2)
                                         (test-the-items item1 item2))
                                  (nthcdr (position item1 THE-LIST))))  THE-LIST)


Is there a cleaner way to do this?

Edit, after some more thinking, this:

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(loop for i in THE-LIST do
     (loop for j in (nthcdr (1+ (position i THE-LIST)) THE-LIST) do
                 (test-the-items i j))


[gugamilare]

I don't think the solution with MAPCAR is ugly at all, it's only verbose, but lispy at the same time

Another option would be:

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(loop for (elt1 . rest) on THE-LIST do
     (loop for elt2 in rest do
          (test-the-items elt1 elt2)))

[enderw88@gmail.com]

I don't think the solution with MAPCAR is ugly at all, it's only verbose, but lispy at the same time

Another option would be:

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(loop for (elt1 . rest) on THE-LIST do
     (loop for elt2 in rest do
          (test-the-items elt1 elt2)))

That dotted list is perfect, it is easy to understand and efficient. Thanks!

[wvxvw]

Just a side note:

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to (1- (length THE-LIST))

is equivalent to:

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below (length THE-LIST)