# Binning data into a new matrix

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Allan Miller le 6 Jan 2020
Commenté : Max Murphy le 13 Jan 2020
I have a three column matrix which consists of three columns (x,y,z) as shown below:
I would like to bin the data in column Y but instead of puting the count in a new column i.e bins, I would like to put the the corresponding value in column Z. For example, I would like to bin the value in column Y (-2.5 in blue cell), but instead of putting the count in bins, I would like to put the value in colum Z (12 in red cell) in that bin.
I have written the code below but its putting the count only:
yy = my_matrix(:,2) % taking the second column
% binning
edges = -2.5:0.3:2.5;
N = histcounts(yy,edges);
new_matrix(:,i)= N;
How can I improve it?
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Allan Miller le 6 Jan 2020
Thanks Guillaume,
So I would like to put an array of values which have a value of 12 in the ith bin and zero elsewhere. i.e, for that example, I would like to put a value of 12 in the bin corresponding to -2.2 : -2.5 and put zeros in other bins from -2.1 : 2.5
Guillaume le 6 Jan 2020
Modifié(e) : Guillaume le 6 Jan 2020
Sorry, I don't understand. Here is a simple exampl, given:
% Y Z
M = [-2.5 12
-2.4 15
-2.3 -1
-2.2 5
-2.1 10]
and
edges = [-2.5 -2.2 -1.9];
What exact output do you want?
result = ????
Please write it out as a cell array or matrix.

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### Réponse acceptée

Max Murphy le 6 Jan 2020
Modifié(e) : Max Murphy le 6 Jan 2020
Your code is returning counts only because you are only requesting the first output of histcounts
yy = my_matrix(:,2); % taking the second column
zz = my_matrix(:,3); % according to the diagram
% binning
edges = -2.5:0.3:2.5;
[N,~,bin] = histcounts(yy,edges);
% match the values of zz to the bins that elements
% of yy went into.
%% EDIT %%
zz_in_bins = cell(size(N));
u = unique(bin); % To avoid dealing with empty bins
binIndex = reshape(u,1,numel(u));
for idx = binIndex
zz_in_bins{idx} = zz(bin==idx);
end
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Guillaume le 13 Jan 2020
I think we understood that. The problems come when you have several X that falls into the same bin. Which of the matching Y values goes into the bin? You never answered my question so we don't know.
If there is only ever one X and Y per bin, then you're not actually doing any binning and there are much simplers ways to achieve what you want.
Max Murphy le 13 Jan 2020
Indeed, if there is one X and Y per bin, I would not follow the code that I've posted above, and instead use discretize as suggested by Guillaume and Steven elsewhere in this thread.

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### Plus de réponses (1)

Steven Lord le 6 Jan 2020
Use the discretize function. See the "Assign Bin Values" example on that documentation page as I believe it does what you're trying to do.
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