Averaging sequential data if values are the same?

2 vues (au cours des 30 derniers jours)
yz
yz le 10 Août 2017
Commenté : yz le 11 Août 2017
I want to be able to average the values of an array of n rows and 5 columns, based on the repeated values of a specific column.
I only want to average the values that are next to each other and not all that repeat. Such that:
%Original data:
key_column = [-6 -6 -6 -6 -5 -5 -3 -3 -3 -1 -1 -1 0 0 1 1 2 2 2 3 3 5 5 5 5 6 6 6 6 4 4 4 3 3 3 3 3 3 3 2 2 2 1 1 1 1 0 0 -1 -1 -2 -3 -5 -5 -5 -5 -5 -5 -5 -6 -6 -6];
%Ends up as:
key_column_avg = [-6 -5 -3 -1 -0 1 2 3 5 6 4 3 2 1 0 -1 -2 -3 -5 -6]
I'm running into trouble because the values of the repeated numbers appear at multiple points in the column and I only want the to be averaged if they are "next to each other" on the column. Also, I'm having issues because the number of repeated values isn't the same for each number. Sometimes the value repeats for 10 times, sometimes for 3, sometimes for 7, sometimes for 4.
Is it possible to do this?
Thank you.
  4 commentaires
yz
yz le 10 Août 2017
No, I want to average all columns based on the repeated sequential values of this specific column.
yz
yz le 10 Août 2017
Attached is a sample of my data set. Column 4 is the column of interest for the averaging, but I wish to apply the averaging to all columns (based on the n number of values on column 4)

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Réponse acceptée

Andrei Bobrov
Andrei Bobrov le 10 Août 2017
key_column_avg = key_column(diff([0,key_column(:)'],1,2) ~= 0);
  4 commentaires
Andrei Bobrov
Andrei Bobrov le 10 Août 2017
Modifié(e) : Andrei Bobrov le 10 Août 2017
d = data_example;
g = cumsum(diff([0;d(:,4)]) ~= 0);
out = splitapply(@mean,d,g);
or
d = data_example;
g = cumsum(diff([0;d(:,4)]) ~= 0);
[gg,c] = ndgrid(g,1:size(d,2));
out1 = accumarray([gg(:),c(:)],d(:),[],@mean);
yz
yz le 11 Août 2017
Thank you so much!

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Plus de réponses (2)

KL
KL le 10 Août 2017
key_column_avg = key_column(diff([0 key_column])~=0);

José-Luis
José-Luis le 10 Août 2017
key_column = [-6 -6 -6 -6 -5 -5 -3 -3 -3 -1 -1 -1 0 0 1 1 2 2 2 3 3 5 5 5 5 6 6 6 6 4 4 4 3 3 3 3 3 3 3 2 2 2 1 1 1 1 0 0 -1 -1 -2 -3 -5 -5 -5 -5 -5 -5 -5 -6 -6 -6];
result = [key_column(diff(key_column) > 0), key_column(end) ]

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