I have a list of elements and I would like to create a new list by applying a function to successive overlapping pairs from the original list.

Specifically, I am trying compute the Pythagorean sum of each pair. So, if the original list is {$a,b,c,d,...$} I want {$\sqrt{a^2+b^2},\sqrt{b^2+c^2},\sqrt{c^2+d^2},...$}

Since this is similar to the built-in Differences function, (which turns the original list into {$b-a,c-b,d-c,...$}), I expected to find a built-in function along the lines of BuiltIn[f,{a,b,c,d,...},options] where f is defined by the user. But alas, my search has been fruitless...

Thanks in advance!


Right after posting this, I came up with

Sqrt[#1^2 + #2^2] & @@@ Subsequences[#, {2}] &@{a, b, c, d}

which does the trick. I was going to update that here, but you all already came through with your own answers, so I'll try those as well. Thanks!

  • 2
    $\begingroup$ Norm /@ Partition[yourList, 2, 1]? $\endgroup$
    – MarcoB
    Apr 22 '20 at 18:11
  • 3
    $\begingroup$ Perhaps something like Sqrt[#1^2 + #2^2] & @@@ Partition[list, 2, 1]? $\endgroup$
    – J. M.'s torpor
    Apr 22 '20 at 18:11
  • 2
    $\begingroup$ Relevant answers here, in particular PartitionMap. $\endgroup$ Apr 22 '20 at 18:12

You can use BlockMap + Norm

BlockMap[Norm, {a, b, c, d, e}, 2, 1]
{Sqrt[Abs[a]^2 + Abs[b]^2], Sqrt[Abs[b]^2 + Abs[c]^2], 
  Sqrt[Abs[c]^2 + Abs[d]^2], Sqrt[Abs[d]^2 + Abs[e]^2]}
TeXForm @ Simplify[%, Thread[{a, b, c, d, e} >= 0]]


  • $\begingroup$ Beautiful. Thanks! $\endgroup$ Apr 22 '20 at 18:28
  • 1
    $\begingroup$ @AaronEiben, my pleasure. Thank you for the accept. $\endgroup$
    – kglr
    Apr 22 '20 at 18:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.