# Linear interpolation of elements in a list to generate new list

For a list of any given length, I want to generate another list from it adding elements which are mean of the consecutive elements, in between the elements, of the original list. For example:

list = {a, b, c};
newlist = {a, (a+b)/2, b, (b+c)/2, c}


I have written this function which does the job but I need much more elegant solution:

intFunc[list_] := Module[{list1},
list1 = ConstantArray[0, 2*Length[list] - 1];
Table[list1[[2*i - 1]] = list[[i]], {i, 1, Length[list]}];
Table[list1[[2*i]] = (list[[i]] + list[[i + 1]])/2, {i, 1, Length[list] - 1}];
list1
];

• This is not exactly what you want, but do you know ArrayResample? Try: ArrayResample[{a,b,c},5] Sep 21, 2015 at 12:17
• I added parentheses where I think you forgot to do so. Please correct me if I were wrong. Sep 21, 2015 at 12:19
• @MarkusRoellig I'll look into it. Sep 21, 2015 at 12:33
• @yohbs thanks for the correction Sep 21, 2015 at 12:34

intFunc[list_]:=Interpolation[list, InterpolationOrder->1]@Range[1, Length@list, 1/2]

• Looks good. Thanks! Sep 21, 2015 at 12:35

My take:

list = {a, b, c};
Riffle[list, MovingAverage[list, 2]]
(* {a, (a + b)/2, b, (b + c)/2, c} *)


Equivalently:

MovingAverage[Riffle[list, list], 2]


Finally, can be generalized to multiple points in between original data points. Equivalent to previous solutions (if using n=2):

intList[list_, n_] := MovingAverage[Flatten[Transpose[ConstantArray[list, n]], 1], n]


Demo (offsets in data include for clarity):

list = {1, 4, 9};
ListPlot[Evaluate @ (#/2 + intList[list, #] & /@ Range), DataRange -> {1, 3}] Re the comment by Markus Roellig: using ArrayResample like so gives the same results.

intList2[list_, n_] := ArrayResample[list, (Length[list] - 1)*n + 1]

• Works well. Thanks. Sep 21, 2015 at 13:20
• @mrkbtr added an extension for multiple datapoints in between the original points, in case you might need that. Sep 21, 2015 at 13:23

For integer lists, this s/b nicely faster than others (interpolation and the two moving-average based solutions:

Riffle[#, Divide[Most@# + Rest@#, 2]] & On real/symbolic, it is comparable to the moving-average based solutions. The interpolation based solution performs uniformly poorly on large lists regardless of type.

• Perhaps you meant Riffle[#, Divide...] & rather than Riffle[list, Divide...]& ? Sep 22, 2015 at 9:55
• @LLlAMnYP oops, yes, fixing. Thanks for catching.
– ciao
Sep 22, 2015 at 10:09
• On my machine with a list of 4*^6 reals your solution still performs a factor of 2 better than the second best. Sep 22, 2015 at 10:12