findpeak for every column of a matrix

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Rafi.P
Rafi.P il 27 Lug 2020
Commentato: Rafi.P il 27 Lug 2020
i have one issue, i have matrix 760x5 how to findpeaks for every column without having to write findpeak in each column
this is the example coding that I used :
[pks1_p,loc1_p] = findpeaks(y_p(:,1),'MinPeakHeight',0.09,'MinPeakDistance',100);
[pks2_p,loc2_p] = findpeaks(y_p(:,2),'MinPeakHeight',0.09,'MinPeakDistance',100);
[pks3_p,loc3_p] = findpeaks(y_p(:,3),'MinPeakHeight',0.09,'MinPeakDistance',100);
[pks4_p,loc4_p] = findpeaks(y_p(:,4),'MinPeakHeight',0.09,'MinPeakDistance',100);
[pks5_p,loc5_p] = findpeaks(y_p(:,5),'MinPeakHeight',0.09,'MinPeakDistance',100);
the problem is I have to write findpeak one by one to see the value of each column. how to get findpeak to follow the number of columns that exist without having to be written one by one
but I need values ​​from pks and loc of each column in the workspace :

Risposta accettata

Mohammad Sami
Mohammad Sami il 27 Lug 2020
You can use size(y_p,2) to find how many columns there are in your data.
Thereafter you can either use a for loop or arrayfun to iterate over the columns.
[pks_p,loc_p] = arrayfun(@(col)findpeaks(y_p(:,col),'MinPeakHeight',0.09,'MinPeakDistance',100),1:size(y_p,2),'UniformOutput',false);
% pks_p, loc_p would be cell arrays. index 1 will correspond to column 1 and so on
  3 Commenti
Mohammad Sami
Mohammad Sami il 27 Lug 2020
you mean you want them to appear in workspace ?
i would suggest use indexing to access the data from each col.
% example
for i = 1:length(pks_p)
pksi_p = pks_p{i};
loci_p = loc_p{i};
% do something
end
Rafi.P
Rafi.P il 27 Lug 2020
Thank you Mohammad Sami

Accedi per commentare.

Più risposte (1)

Steven Lord
Steven Lord il 27 Lug 2020
Consider using islocalmax with the dim input argument.

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