Retaining relative sizes of local maxima using MaxDetect[List]

Question

I would like to be able to find the position of the two greatest local maxima in a list.

For the example: list={1,2,3,4,2,1,2,6,5,4,3,2,}

I would like to return {4,8}

Initially I was using the Ordering[list,-2] to successfully do this, however, the above scenario shows that this fails quite easily (here 5>4).

I then tried using Ordering[MaxDetect[list],-2] which almost does the job; however, I have the problem that my list is long and has many (smaller) local maxima, which MaxDetect makes equal in value to the 2 larger local maxima, so Ordering fails to select the appropriate peaks.

Is there a way of retaining the relative sizes of the peaks after using MaxDetect, so that this method will work?

Thanks!

**Using the option MaxDetect[list,value] does not work, as different runs change the value that these maxima have.

Answer 1

You can also do it explicitly, as follows,

findTwoMaxima[list_List] := Block[{l, p1, p2},
  l = list;
 (* find position of maximum:*)
  p1 = Flatten[Position[l, Max[l]]][[1]]; 
 (* Replace this entry by a very negative number:*)
  l[[p1]] = -Infinity;
 (* now find maximum again:*)
  p2 = Flatten[Position[l, Max[l]]][[1]];
  {p1, p2}
  ]

I always prefer writing your own code for doing simple stuff, so you don't have to rely on the fancy algorithm, which is a black box to you, that was used in FindPeaks or whatever other function you use.

Answer 2

Following ubpdqn comment, this is the solution that worked for my problem:

Peaks = Transpose@ FindPeaks[Take[Delete[Abs[Fourier[CoM]], 1], RunTime/2], 0, 0];

Find the position in Peaks of the two highest values (two greatest local peaks)

MaxTwoPosition = Ordering[Peaks[[2]], -2];

Find the corresponding position in my list of data:

locationofbothpeaks = {Peaks[[1, MaxTwoPosition[[1]]]], Peaks[[1, MaxTwoPosition[[2]]]]}