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I have a list of the form:

list = {1,1,1,2,2,2,3,3,3,4,4,4,7,7,7,19,19,19};

Where we can see that there are an ordered set of integer values that repeat themselves some fixed number of times N. Is there a nice one-liner to transform this list into something like:

transformedList = {1,1+1/3,1+2/3,2,2+1/3,2+2/3,3,3+1/3,3+2/3,4,4+1/3,4+2/3,7,7+1/3,7+2/3,19,19+1/3,19+2/3}

Where we essentially choose each new element we come by in the list as an anchorpoint, and replace the repeats of that element as equispaced points along a line the anchor integer value and the next higher integer?

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  • $\begingroup$ You say "equispaced points along a line between each successive pair of anchor points", but you transform $4,4,4,7$ to $4,4+\frac13,4+\frac23,7$ instead of $4,5,6,7$. $\endgroup$
    – user484
    Commented Aug 8, 2013 at 2:00
  • $\begingroup$ Rahul is right; something doesn't add up here. If your sample output is really what you want we should rephrase this question to match. Is it? $\endgroup$
    – Mr.Wizard
    Commented Aug 8, 2013 at 2:06
  • $\begingroup$ @RahulNarain Ahhh... I'm so sorry. $\endgroup$
    – Niniar
    Commented Aug 8, 2013 at 2:06
  • $\begingroup$ @Mr.Wizard I have updated the example, terrible mistake. $\endgroup$
    – Niniar
    Commented Aug 8, 2013 at 2:07
  • $\begingroup$ @Niniar Since I already answered your original question would you mind posting a new one with the revised information, and restoring this one to the original? $\endgroup$
    – Mr.Wizard
    Commented Aug 8, 2013 at 2:10

2 Answers 2

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Here's the first thing that comes to mind:

# + Most@Range[0, 1, 1/Length@#] & /@ Split[list] // Flatten

{1, 4/3, 5/3, 2, 7/3, 8/3, 3, 10/3, 11/3, 4, 13/3, 14/3, 7, 22/3, 23/3, 19, 58/3, 59/3}
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Basically the same approach as MrWizard but a bit shorter and using (except Range) different functions

Join @@ (# + Range[0, 1 - 1/#2, 1/#2] & @@@ Tally[list])

(* {1,4/3,5/3,2,7/3,8/3,3,10/3,11/3,4,13/3,14/3,7,22/3,23/3,19,58/3,59/3} *)
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  • $\begingroup$ Wait-- I really like the style but I didn't make the assumption that the list is monotonically increasing. $\endgroup$
    – Mr.Wizard
    Commented Aug 8, 2013 at 3:11
  • $\begingroup$ You know, I'm being too much of a nitpicker. It's a good answer, just not the same behavior as mine. +1 (again) $\endgroup$
    – Mr.Wizard
    Commented Aug 8, 2013 at 3:23

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