沿3维对3D数组排序

Question

I have a 144x91x92 matrix stored in o3{1} . I'd like to look across the third dimension (representing the days in one season) and sort it. Then, I want to pull out the top 10 and bottom 10 percentile of values including the index of those values. This would find the top 10 and bottom 10 percentile of days for each grid cell (this would be different for each grid cell).

I'm trying the following:

[Y,I] = sort(o3{1},3); % sort o3 along 3rd dimension
o3_sorted = o3{1}(I);
ind_top10 = o3_sorted(90);
ind_bot10 = o3_sorted(10);    

But I know I'm not pulling out the top 10 and bottom 10th percentile correct. Plus, this way does not tell me the indices (different for each of the 144x91 grid cells) for the top and bottom 10 percentile of days. I am hoping to end up with 144x91x10 matrices for the top 10 percentile of days, the bottom 10 percentile of days, and the indices for each.

Answer 1

Try it like this:

[~, I] = sort(o3{1},3); %// sort o3 along 3rd dimension
ind_top10 = I(:,:,end-8:end);
ind_bot10 = I(:,:,1:9);

I3 = cat(3, ind_top10, ind_bot10); %// you might want to skip this but and just work with the top and bottom separately from here on

[I1, I2, ~] = ndgrid(1:size(o3{1},1), 1:size(o3{1},2), 1:size(I3,3));
ind = sub2ind(size(o3{1}),I1,I2,I3)

And now

o3{2}(ind)
o3{3}(ind)
%// etc...

Answer 2

Here's a slightly different idea from what Dan had suggested :

%% // Init
clear variables; clc;

%% // Generate some data:
o3 = cell(3,1);
for indO = 1:3
  o3{indO} = randi(intmax('uint16'),144,91,92,'uint16');
end

%% // Find 10 & 90 percentiles:
percentiles = cat(3,prctile(o3{1},10,3),prctile(o3{1},90,3));

%% // Find indices of relevant values
select_bot10 = bsxfun(@ge,o3{1},percentiles(:,:,1)); %// replace @ge with @gt if needed
select_top10 = bsxfun(@le,o3{1},percentiles(:,:,2)); %// replace @le with @lt if needed
%// Another optional way to index the values if required:
[rb,cb,vb] = ind2sub(size(select_bot10),find(select_bot10));
[rt,ct,vt] = ind2sub(size(select_top10),find(select_top10));

%% // Get values from o3{1..3} etc.
bot10 = o3{1}(select_bot10);
top10 = o3{1}(select_top10);
%// etc.

This solution might not be suitable for your specific needs as it is, but adaptations should be straightforward. Also note that since exact percentiles are taken, the number of elements would probably differ between bot10 and top10 .

Credit for the 3D find goes to Kenneth Eaton / gnovice .