Dimensions of arrays being concatenated are not consistent.

8 vues (au cours des 30 derniers jours)
gjashta
gjashta le 8 Mai 2019
Commenté : Matt J le 9 Mai 2019
I want to bin the price values in 5 bins and for each bin to group the demand values AND
then calculate the mean and the standard deviation of the demand data for each bin.
Could you please help me to fix the code?
I am trying with the following code but I got this error:
Dimensions of arrays being concatenated are not consistent.
p=Data(:,1);
d=Data(:,2);
partitions = [0 40; 40 55; 55 65; 65 max(x)+1];
for k1 = 1:size(partitions,1)
ypartmean(k1) = mean(di((pi >= partitions(k1,1)) & (pi < partitions(k1,2))));
binct(k1) = numel(di((pi >= partitions(k1,1)) & (pi < partitions(k1,2))));
end

Réponse acceptée

Matt J
Matt J le 8 Mai 2019
Modifié(e) : Matt J le 8 Mai 2019
No need to loop,
E=[0,40,65,55,inf];
[binct,~,G]=histcounts(p,E);
ypartmean=splitapply(@mean,d,G);
  10 commentaires
gjashta
gjashta le 8 Mai 2019
I got an error with scatter plot because EPSs above are vectors with different length.
Matt J
Matt J le 9 Mai 2019
You certainly should reorganize your data into a form that scatter() expects.

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