Find a local maximum in list without using built-in function FindPeaks











up vote
3
down vote

favorite
1












I am interested finding the peaks in data without using the function FindPeaks. Any suggestion on how to do it effectively?



data1=RandomInteger[{1, 3997670}, 3936];
peaks = FindPeaks[data1]
ListLinePlot[
data1,
Epilog -> {Red, PointSize[0.01], Point[peaks]}
]









share|improve this question




























    up vote
    3
    down vote

    favorite
    1












    I am interested finding the peaks in data without using the function FindPeaks. Any suggestion on how to do it effectively?



    data1=RandomInteger[{1, 3997670}, 3936];
    peaks = FindPeaks[data1]
    ListLinePlot[
    data1,
    Epilog -> {Red, PointSize[0.01], Point[peaks]}
    ]









    share|improve this question


























      up vote
      3
      down vote

      favorite
      1









      up vote
      3
      down vote

      favorite
      1






      1





      I am interested finding the peaks in data without using the function FindPeaks. Any suggestion on how to do it effectively?



      data1=RandomInteger[{1, 3997670}, 3936];
      peaks = FindPeaks[data1]
      ListLinePlot[
      data1,
      Epilog -> {Red, PointSize[0.01], Point[peaks]}
      ]









      share|improve this question















      I am interested finding the peaks in data without using the function FindPeaks. Any suggestion on how to do it effectively?



      data1=RandomInteger[{1, 3997670}, 3936];
      peaks = FindPeaks[data1]
      ListLinePlot[
      data1,
      Epilog -> {Red, PointSize[0.01], Point[peaks]}
      ]






      list-manipulation






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited Nov 24 at 17:14









      kglr

      175k9197402




      175k9197402










      asked Nov 24 at 16:21









      Kiril Danilchenko

      729316




      729316






















          2 Answers
          2






          active

          oldest

          votes

















          up vote
          4
          down vote













          You could try something like:



          pks = {};

          If[data1[[1]] > data1[[2]],
          AppendTo[pks, {1, data1[[1]]}]];

          For[i = 2, i <= Length[data1] - 1, i++,
          If[data1[[i - 1]] < data1[[i]] > data1[[i + 1]],
          AppendTo[pks, {i, data1[[i]]}]]];

          If[data1[[-1]] > data1[[-2]],
          AppendTo[pks, {Length[data1], data1[[-1]]}]];


          It's a simple and rather inelegant way of finding peaks. Basically, if a point is higher than the two points on either side, it's considered to be a peak. In the case of the first and last point, it's a peak if it's higher than the closest point. I'm not sure if that's what you mean by finding all the peaks.



          Is there a reason you can't use FindPeaks? It's almost always the best way. I didn't use it in my answer because you asked for an answer without, but I can get the exact same output with the following code:



          pks2 = FindPeaks[data1, 0, 0, -Infinity];


          If you want to find the local maximum over some range of values, you could do something like:



          Max[data1[[100;;200]]]


          It's position in the list can be found with:



          Position[data1, Max[data1[[100;;200]]]]


          Since you have a list, this will return the absolute local maximum. Calling Max on the entire list would of course return the global max.






          share|improve this answer





















          • If your data has noise, defining a peak as a point that is larger than its neighbors might not be useful. In the presence of noise, you could a) smooth the data, b) have a min threshold between the peak point and the closest neighbor, and/or c) require a peak point be a peak in some window in which points on the left increase while points on the right decrease (with some heuristic for constant sequences). Note that (c) is different from MassDefect's maximum over some range of values method. Other methods exist.
            – Robert Jacobson
            Nov 24 at 19:00










          • @RobertJacobson You're absolutely right. My algorithm is pretty weak. I'm not entirely clear on what kind of peaks the poster is looking for.
            – MassDefect
            Nov 25 at 22:53


















          up vote
          4
          down vote













          The documentation describes a lot of how FindPeaks works, especially the section Scope > Parameters. It can all be done very efficiently with convolutions and vectorized operations.



          This is basic peak finding:



          findPeaks1[data_] := Module[{rightDiff, leftDiff, peaks},
          rightDiff = ListConvolve[{0, 1, -1}, data];
          leftDiff = ListConvolve[{-1, 1, 0}, data];
          Unitize[Sign[leftDiff] + Sign[rightDiff] - 2]
          ]


          The reason for the 1 at the end of the name will become clear later. The convolutions in this function perform almost the same job as Differences, in the end, we get a list that is 0 in those positions that correspond to values that are larger than both its neighbors.



          findPeaks1[data_, sigma_] := Module[{blurred},
          blurred = GaussianFilter[data, {3, sigma}];
          Unitize[findPeaks1[data] + findPeaks1[blurred]]
          ]


          In this step, we add the second argument, which corresponds to smoothing the data using a Gaussian filter with variance sigma. A zero in this list means that the corresponding element is a peak, using the definition in the previous function, in both the smoothed data and the original data.



          findPeaks1[data_, sigma_, s_] := Module[{sharpness},
          sharpness = ListConvolve[{1, -2, 1}, data];
          Unitize[findPeaks1[data, sigma] + 1 - Unitize[sharpness, s]]
          ]


          In this piece of code, we implemented the third argument, which says that the peaks that don't have a negative second derivative that is larger than s should be discarded. The convolution we use here is the definition for the negative discrete second derivative.



          findPeaks1[data_, sigma_, s_, t_] := Unitize[
          findPeaks1[data, sigma, s] + Unitize[UnitStep[Most@Rest@data - t] - 1]
          ]


          Finally, in this code block, we implement the fourth argument, which is a threshold. It says that the value of a peak must be larger than t.



          For efficiency reasons and because it made the implementation short and simple, these functions all pass lists of integers to each other. They have a 1 suffixed to their names to distinguish them from the interface which we shall now define:



          outpformat[data_, sel_] := Module[{pos, values},
          pos = Position[sel, 0] + 1;
          values = Extract[data, pos];
          Transpose[{Flatten[pos], values}]
          ]

          findPeaks[data_] := outpformat[data, findPeaks1[data]]
          findPeaks[data_, sigma_] := outpformat[data, findPeaks1[data, sigma]]
          findPeaks[data_, sigma_, s_] := outpformat[data, findPeaks1[data, sigma, s]]
          findPeaks[data_, sigma_, s_, t_] := outpformat[data, findPeaks1[data, sigma, s, t]]


          We now have a function findPeaks that does more or less the same thing as FindPeaks. I find that the result differs a bit for the Gaussian filter, they seem to do it a bit differently, but this still captures the idea.



          Of course, there are many other signal processing techniques that we could use to find peaks, but there is no universal answer to how the signal should be processed. If you find that this function is not enough, perhaps because the signal is noisy or something else, then you have to look at the signal to decide what filters and such might be appropriate to make the peaks you are looking for more distinguishable.






          share|improve this answer























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            2 Answers
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            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes








            up vote
            4
            down vote













            You could try something like:



            pks = {};

            If[data1[[1]] > data1[[2]],
            AppendTo[pks, {1, data1[[1]]}]];

            For[i = 2, i <= Length[data1] - 1, i++,
            If[data1[[i - 1]] < data1[[i]] > data1[[i + 1]],
            AppendTo[pks, {i, data1[[i]]}]]];

            If[data1[[-1]] > data1[[-2]],
            AppendTo[pks, {Length[data1], data1[[-1]]}]];


            It's a simple and rather inelegant way of finding peaks. Basically, if a point is higher than the two points on either side, it's considered to be a peak. In the case of the first and last point, it's a peak if it's higher than the closest point. I'm not sure if that's what you mean by finding all the peaks.



            Is there a reason you can't use FindPeaks? It's almost always the best way. I didn't use it in my answer because you asked for an answer without, but I can get the exact same output with the following code:



            pks2 = FindPeaks[data1, 0, 0, -Infinity];


            If you want to find the local maximum over some range of values, you could do something like:



            Max[data1[[100;;200]]]


            It's position in the list can be found with:



            Position[data1, Max[data1[[100;;200]]]]


            Since you have a list, this will return the absolute local maximum. Calling Max on the entire list would of course return the global max.






            share|improve this answer





















            • If your data has noise, defining a peak as a point that is larger than its neighbors might not be useful. In the presence of noise, you could a) smooth the data, b) have a min threshold between the peak point and the closest neighbor, and/or c) require a peak point be a peak in some window in which points on the left increase while points on the right decrease (with some heuristic for constant sequences). Note that (c) is different from MassDefect's maximum over some range of values method. Other methods exist.
              – Robert Jacobson
              Nov 24 at 19:00










            • @RobertJacobson You're absolutely right. My algorithm is pretty weak. I'm not entirely clear on what kind of peaks the poster is looking for.
              – MassDefect
              Nov 25 at 22:53















            up vote
            4
            down vote













            You could try something like:



            pks = {};

            If[data1[[1]] > data1[[2]],
            AppendTo[pks, {1, data1[[1]]}]];

            For[i = 2, i <= Length[data1] - 1, i++,
            If[data1[[i - 1]] < data1[[i]] > data1[[i + 1]],
            AppendTo[pks, {i, data1[[i]]}]]];

            If[data1[[-1]] > data1[[-2]],
            AppendTo[pks, {Length[data1], data1[[-1]]}]];


            It's a simple and rather inelegant way of finding peaks. Basically, if a point is higher than the two points on either side, it's considered to be a peak. In the case of the first and last point, it's a peak if it's higher than the closest point. I'm not sure if that's what you mean by finding all the peaks.



            Is there a reason you can't use FindPeaks? It's almost always the best way. I didn't use it in my answer because you asked for an answer without, but I can get the exact same output with the following code:



            pks2 = FindPeaks[data1, 0, 0, -Infinity];


            If you want to find the local maximum over some range of values, you could do something like:



            Max[data1[[100;;200]]]


            It's position in the list can be found with:



            Position[data1, Max[data1[[100;;200]]]]


            Since you have a list, this will return the absolute local maximum. Calling Max on the entire list would of course return the global max.






            share|improve this answer





















            • If your data has noise, defining a peak as a point that is larger than its neighbors might not be useful. In the presence of noise, you could a) smooth the data, b) have a min threshold between the peak point and the closest neighbor, and/or c) require a peak point be a peak in some window in which points on the left increase while points on the right decrease (with some heuristic for constant sequences). Note that (c) is different from MassDefect's maximum over some range of values method. Other methods exist.
              – Robert Jacobson
              Nov 24 at 19:00










            • @RobertJacobson You're absolutely right. My algorithm is pretty weak. I'm not entirely clear on what kind of peaks the poster is looking for.
              – MassDefect
              Nov 25 at 22:53













            up vote
            4
            down vote










            up vote
            4
            down vote









            You could try something like:



            pks = {};

            If[data1[[1]] > data1[[2]],
            AppendTo[pks, {1, data1[[1]]}]];

            For[i = 2, i <= Length[data1] - 1, i++,
            If[data1[[i - 1]] < data1[[i]] > data1[[i + 1]],
            AppendTo[pks, {i, data1[[i]]}]]];

            If[data1[[-1]] > data1[[-2]],
            AppendTo[pks, {Length[data1], data1[[-1]]}]];


            It's a simple and rather inelegant way of finding peaks. Basically, if a point is higher than the two points on either side, it's considered to be a peak. In the case of the first and last point, it's a peak if it's higher than the closest point. I'm not sure if that's what you mean by finding all the peaks.



            Is there a reason you can't use FindPeaks? It's almost always the best way. I didn't use it in my answer because you asked for an answer without, but I can get the exact same output with the following code:



            pks2 = FindPeaks[data1, 0, 0, -Infinity];


            If you want to find the local maximum over some range of values, you could do something like:



            Max[data1[[100;;200]]]


            It's position in the list can be found with:



            Position[data1, Max[data1[[100;;200]]]]


            Since you have a list, this will return the absolute local maximum. Calling Max on the entire list would of course return the global max.






            share|improve this answer












            You could try something like:



            pks = {};

            If[data1[[1]] > data1[[2]],
            AppendTo[pks, {1, data1[[1]]}]];

            For[i = 2, i <= Length[data1] - 1, i++,
            If[data1[[i - 1]] < data1[[i]] > data1[[i + 1]],
            AppendTo[pks, {i, data1[[i]]}]]];

            If[data1[[-1]] > data1[[-2]],
            AppendTo[pks, {Length[data1], data1[[-1]]}]];


            It's a simple and rather inelegant way of finding peaks. Basically, if a point is higher than the two points on either side, it's considered to be a peak. In the case of the first and last point, it's a peak if it's higher than the closest point. I'm not sure if that's what you mean by finding all the peaks.



            Is there a reason you can't use FindPeaks? It's almost always the best way. I didn't use it in my answer because you asked for an answer without, but I can get the exact same output with the following code:



            pks2 = FindPeaks[data1, 0, 0, -Infinity];


            If you want to find the local maximum over some range of values, you could do something like:



            Max[data1[[100;;200]]]


            It's position in the list can be found with:



            Position[data1, Max[data1[[100;;200]]]]


            Since you have a list, this will return the absolute local maximum. Calling Max on the entire list would of course return the global max.







            share|improve this answer












            share|improve this answer



            share|improve this answer










            answered Nov 24 at 16:52









            MassDefect

            37115




            37115












            • If your data has noise, defining a peak as a point that is larger than its neighbors might not be useful. In the presence of noise, you could a) smooth the data, b) have a min threshold between the peak point and the closest neighbor, and/or c) require a peak point be a peak in some window in which points on the left increase while points on the right decrease (with some heuristic for constant sequences). Note that (c) is different from MassDefect's maximum over some range of values method. Other methods exist.
              – Robert Jacobson
              Nov 24 at 19:00










            • @RobertJacobson You're absolutely right. My algorithm is pretty weak. I'm not entirely clear on what kind of peaks the poster is looking for.
              – MassDefect
              Nov 25 at 22:53


















            • If your data has noise, defining a peak as a point that is larger than its neighbors might not be useful. In the presence of noise, you could a) smooth the data, b) have a min threshold between the peak point and the closest neighbor, and/or c) require a peak point be a peak in some window in which points on the left increase while points on the right decrease (with some heuristic for constant sequences). Note that (c) is different from MassDefect's maximum over some range of values method. Other methods exist.
              – Robert Jacobson
              Nov 24 at 19:00










            • @RobertJacobson You're absolutely right. My algorithm is pretty weak. I'm not entirely clear on what kind of peaks the poster is looking for.
              – MassDefect
              Nov 25 at 22:53
















            If your data has noise, defining a peak as a point that is larger than its neighbors might not be useful. In the presence of noise, you could a) smooth the data, b) have a min threshold between the peak point and the closest neighbor, and/or c) require a peak point be a peak in some window in which points on the left increase while points on the right decrease (with some heuristic for constant sequences). Note that (c) is different from MassDefect's maximum over some range of values method. Other methods exist.
            – Robert Jacobson
            Nov 24 at 19:00




            If your data has noise, defining a peak as a point that is larger than its neighbors might not be useful. In the presence of noise, you could a) smooth the data, b) have a min threshold between the peak point and the closest neighbor, and/or c) require a peak point be a peak in some window in which points on the left increase while points on the right decrease (with some heuristic for constant sequences). Note that (c) is different from MassDefect's maximum over some range of values method. Other methods exist.
            – Robert Jacobson
            Nov 24 at 19:00












            @RobertJacobson You're absolutely right. My algorithm is pretty weak. I'm not entirely clear on what kind of peaks the poster is looking for.
            – MassDefect
            Nov 25 at 22:53




            @RobertJacobson You're absolutely right. My algorithm is pretty weak. I'm not entirely clear on what kind of peaks the poster is looking for.
            – MassDefect
            Nov 25 at 22:53










            up vote
            4
            down vote













            The documentation describes a lot of how FindPeaks works, especially the section Scope > Parameters. It can all be done very efficiently with convolutions and vectorized operations.



            This is basic peak finding:



            findPeaks1[data_] := Module[{rightDiff, leftDiff, peaks},
            rightDiff = ListConvolve[{0, 1, -1}, data];
            leftDiff = ListConvolve[{-1, 1, 0}, data];
            Unitize[Sign[leftDiff] + Sign[rightDiff] - 2]
            ]


            The reason for the 1 at the end of the name will become clear later. The convolutions in this function perform almost the same job as Differences, in the end, we get a list that is 0 in those positions that correspond to values that are larger than both its neighbors.



            findPeaks1[data_, sigma_] := Module[{blurred},
            blurred = GaussianFilter[data, {3, sigma}];
            Unitize[findPeaks1[data] + findPeaks1[blurred]]
            ]


            In this step, we add the second argument, which corresponds to smoothing the data using a Gaussian filter with variance sigma. A zero in this list means that the corresponding element is a peak, using the definition in the previous function, in both the smoothed data and the original data.



            findPeaks1[data_, sigma_, s_] := Module[{sharpness},
            sharpness = ListConvolve[{1, -2, 1}, data];
            Unitize[findPeaks1[data, sigma] + 1 - Unitize[sharpness, s]]
            ]


            In this piece of code, we implemented the third argument, which says that the peaks that don't have a negative second derivative that is larger than s should be discarded. The convolution we use here is the definition for the negative discrete second derivative.



            findPeaks1[data_, sigma_, s_, t_] := Unitize[
            findPeaks1[data, sigma, s] + Unitize[UnitStep[Most@Rest@data - t] - 1]
            ]


            Finally, in this code block, we implement the fourth argument, which is a threshold. It says that the value of a peak must be larger than t.



            For efficiency reasons and because it made the implementation short and simple, these functions all pass lists of integers to each other. They have a 1 suffixed to their names to distinguish them from the interface which we shall now define:



            outpformat[data_, sel_] := Module[{pos, values},
            pos = Position[sel, 0] + 1;
            values = Extract[data, pos];
            Transpose[{Flatten[pos], values}]
            ]

            findPeaks[data_] := outpformat[data, findPeaks1[data]]
            findPeaks[data_, sigma_] := outpformat[data, findPeaks1[data, sigma]]
            findPeaks[data_, sigma_, s_] := outpformat[data, findPeaks1[data, sigma, s]]
            findPeaks[data_, sigma_, s_, t_] := outpformat[data, findPeaks1[data, sigma, s, t]]


            We now have a function findPeaks that does more or less the same thing as FindPeaks. I find that the result differs a bit for the Gaussian filter, they seem to do it a bit differently, but this still captures the idea.



            Of course, there are many other signal processing techniques that we could use to find peaks, but there is no universal answer to how the signal should be processed. If you find that this function is not enough, perhaps because the signal is noisy or something else, then you have to look at the signal to decide what filters and such might be appropriate to make the peaks you are looking for more distinguishable.






            share|improve this answer



























              up vote
              4
              down vote













              The documentation describes a lot of how FindPeaks works, especially the section Scope > Parameters. It can all be done very efficiently with convolutions and vectorized operations.



              This is basic peak finding:



              findPeaks1[data_] := Module[{rightDiff, leftDiff, peaks},
              rightDiff = ListConvolve[{0, 1, -1}, data];
              leftDiff = ListConvolve[{-1, 1, 0}, data];
              Unitize[Sign[leftDiff] + Sign[rightDiff] - 2]
              ]


              The reason for the 1 at the end of the name will become clear later. The convolutions in this function perform almost the same job as Differences, in the end, we get a list that is 0 in those positions that correspond to values that are larger than both its neighbors.



              findPeaks1[data_, sigma_] := Module[{blurred},
              blurred = GaussianFilter[data, {3, sigma}];
              Unitize[findPeaks1[data] + findPeaks1[blurred]]
              ]


              In this step, we add the second argument, which corresponds to smoothing the data using a Gaussian filter with variance sigma. A zero in this list means that the corresponding element is a peak, using the definition in the previous function, in both the smoothed data and the original data.



              findPeaks1[data_, sigma_, s_] := Module[{sharpness},
              sharpness = ListConvolve[{1, -2, 1}, data];
              Unitize[findPeaks1[data, sigma] + 1 - Unitize[sharpness, s]]
              ]


              In this piece of code, we implemented the third argument, which says that the peaks that don't have a negative second derivative that is larger than s should be discarded. The convolution we use here is the definition for the negative discrete second derivative.



              findPeaks1[data_, sigma_, s_, t_] := Unitize[
              findPeaks1[data, sigma, s] + Unitize[UnitStep[Most@Rest@data - t] - 1]
              ]


              Finally, in this code block, we implement the fourth argument, which is a threshold. It says that the value of a peak must be larger than t.



              For efficiency reasons and because it made the implementation short and simple, these functions all pass lists of integers to each other. They have a 1 suffixed to their names to distinguish them from the interface which we shall now define:



              outpformat[data_, sel_] := Module[{pos, values},
              pos = Position[sel, 0] + 1;
              values = Extract[data, pos];
              Transpose[{Flatten[pos], values}]
              ]

              findPeaks[data_] := outpformat[data, findPeaks1[data]]
              findPeaks[data_, sigma_] := outpformat[data, findPeaks1[data, sigma]]
              findPeaks[data_, sigma_, s_] := outpformat[data, findPeaks1[data, sigma, s]]
              findPeaks[data_, sigma_, s_, t_] := outpformat[data, findPeaks1[data, sigma, s, t]]


              We now have a function findPeaks that does more or less the same thing as FindPeaks. I find that the result differs a bit for the Gaussian filter, they seem to do it a bit differently, but this still captures the idea.



              Of course, there are many other signal processing techniques that we could use to find peaks, but there is no universal answer to how the signal should be processed. If you find that this function is not enough, perhaps because the signal is noisy or something else, then you have to look at the signal to decide what filters and such might be appropriate to make the peaks you are looking for more distinguishable.






              share|improve this answer

























                up vote
                4
                down vote










                up vote
                4
                down vote









                The documentation describes a lot of how FindPeaks works, especially the section Scope > Parameters. It can all be done very efficiently with convolutions and vectorized operations.



                This is basic peak finding:



                findPeaks1[data_] := Module[{rightDiff, leftDiff, peaks},
                rightDiff = ListConvolve[{0, 1, -1}, data];
                leftDiff = ListConvolve[{-1, 1, 0}, data];
                Unitize[Sign[leftDiff] + Sign[rightDiff] - 2]
                ]


                The reason for the 1 at the end of the name will become clear later. The convolutions in this function perform almost the same job as Differences, in the end, we get a list that is 0 in those positions that correspond to values that are larger than both its neighbors.



                findPeaks1[data_, sigma_] := Module[{blurred},
                blurred = GaussianFilter[data, {3, sigma}];
                Unitize[findPeaks1[data] + findPeaks1[blurred]]
                ]


                In this step, we add the second argument, which corresponds to smoothing the data using a Gaussian filter with variance sigma. A zero in this list means that the corresponding element is a peak, using the definition in the previous function, in both the smoothed data and the original data.



                findPeaks1[data_, sigma_, s_] := Module[{sharpness},
                sharpness = ListConvolve[{1, -2, 1}, data];
                Unitize[findPeaks1[data, sigma] + 1 - Unitize[sharpness, s]]
                ]


                In this piece of code, we implemented the third argument, which says that the peaks that don't have a negative second derivative that is larger than s should be discarded. The convolution we use here is the definition for the negative discrete second derivative.



                findPeaks1[data_, sigma_, s_, t_] := Unitize[
                findPeaks1[data, sigma, s] + Unitize[UnitStep[Most@Rest@data - t] - 1]
                ]


                Finally, in this code block, we implement the fourth argument, which is a threshold. It says that the value of a peak must be larger than t.



                For efficiency reasons and because it made the implementation short and simple, these functions all pass lists of integers to each other. They have a 1 suffixed to their names to distinguish them from the interface which we shall now define:



                outpformat[data_, sel_] := Module[{pos, values},
                pos = Position[sel, 0] + 1;
                values = Extract[data, pos];
                Transpose[{Flatten[pos], values}]
                ]

                findPeaks[data_] := outpformat[data, findPeaks1[data]]
                findPeaks[data_, sigma_] := outpformat[data, findPeaks1[data, sigma]]
                findPeaks[data_, sigma_, s_] := outpformat[data, findPeaks1[data, sigma, s]]
                findPeaks[data_, sigma_, s_, t_] := outpformat[data, findPeaks1[data, sigma, s, t]]


                We now have a function findPeaks that does more or less the same thing as FindPeaks. I find that the result differs a bit for the Gaussian filter, they seem to do it a bit differently, but this still captures the idea.



                Of course, there are many other signal processing techniques that we could use to find peaks, but there is no universal answer to how the signal should be processed. If you find that this function is not enough, perhaps because the signal is noisy or something else, then you have to look at the signal to decide what filters and such might be appropriate to make the peaks you are looking for more distinguishable.






                share|improve this answer














                The documentation describes a lot of how FindPeaks works, especially the section Scope > Parameters. It can all be done very efficiently with convolutions and vectorized operations.



                This is basic peak finding:



                findPeaks1[data_] := Module[{rightDiff, leftDiff, peaks},
                rightDiff = ListConvolve[{0, 1, -1}, data];
                leftDiff = ListConvolve[{-1, 1, 0}, data];
                Unitize[Sign[leftDiff] + Sign[rightDiff] - 2]
                ]


                The reason for the 1 at the end of the name will become clear later. The convolutions in this function perform almost the same job as Differences, in the end, we get a list that is 0 in those positions that correspond to values that are larger than both its neighbors.



                findPeaks1[data_, sigma_] := Module[{blurred},
                blurred = GaussianFilter[data, {3, sigma}];
                Unitize[findPeaks1[data] + findPeaks1[blurred]]
                ]


                In this step, we add the second argument, which corresponds to smoothing the data using a Gaussian filter with variance sigma. A zero in this list means that the corresponding element is a peak, using the definition in the previous function, in both the smoothed data and the original data.



                findPeaks1[data_, sigma_, s_] := Module[{sharpness},
                sharpness = ListConvolve[{1, -2, 1}, data];
                Unitize[findPeaks1[data, sigma] + 1 - Unitize[sharpness, s]]
                ]


                In this piece of code, we implemented the third argument, which says that the peaks that don't have a negative second derivative that is larger than s should be discarded. The convolution we use here is the definition for the negative discrete second derivative.



                findPeaks1[data_, sigma_, s_, t_] := Unitize[
                findPeaks1[data, sigma, s] + Unitize[UnitStep[Most@Rest@data - t] - 1]
                ]


                Finally, in this code block, we implement the fourth argument, which is a threshold. It says that the value of a peak must be larger than t.



                For efficiency reasons and because it made the implementation short and simple, these functions all pass lists of integers to each other. They have a 1 suffixed to their names to distinguish them from the interface which we shall now define:



                outpformat[data_, sel_] := Module[{pos, values},
                pos = Position[sel, 0] + 1;
                values = Extract[data, pos];
                Transpose[{Flatten[pos], values}]
                ]

                findPeaks[data_] := outpformat[data, findPeaks1[data]]
                findPeaks[data_, sigma_] := outpformat[data, findPeaks1[data, sigma]]
                findPeaks[data_, sigma_, s_] := outpformat[data, findPeaks1[data, sigma, s]]
                findPeaks[data_, sigma_, s_, t_] := outpformat[data, findPeaks1[data, sigma, s, t]]


                We now have a function findPeaks that does more or less the same thing as FindPeaks. I find that the result differs a bit for the Gaussian filter, they seem to do it a bit differently, but this still captures the idea.



                Of course, there are many other signal processing techniques that we could use to find peaks, but there is no universal answer to how the signal should be processed. If you find that this function is not enough, perhaps because the signal is noisy or something else, then you have to look at the signal to decide what filters and such might be appropriate to make the peaks you are looking for more distinguishable.







                share|improve this answer














                share|improve this answer



                share|improve this answer








                edited Nov 24 at 20:01

























                answered Nov 24 at 19:36









                C. E.

                49.1k395197




                49.1k395197






























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