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.
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...
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.
Related
I need one figure with multiple graphs within two loops.
for i=1:length(state)
[block]
for j=1:length(channel)
[block]
subplot(length(state),length(channel)),j)
plot(a,b)% a and b are arrays of doubles.
end
end
I want one figure with size =length(state)*length(channel); for instance I need all the graphs of state(1)within all channels in the first row etc...
But what I get is multiple figures (the length of state).
If I understand well enough here is a way to do it :
figure()
lx = 2;
ly = 3;
for ii = 1:lx
for jj = 1:ly
subplot(lx,ly,ly*(ii-1)+jj)
plot(ii,jj,'o')
end
end
Why ly*(ii-1)+jj?
The syntax of subplot is the following : subplot(nbRows,nbCols,position) and the position is given by an unique index going over all available subplots (see image) which is ly*(ii-1)+jj.
I wrote this matlab code in order to concatenate the results of the integration of all the columns of a matrix extracted form a multi matrix array.
"datimf" is a matrix composed by 100 matrices, each of 224*640, vertically concatenated.
In the first loop i select every single matrix.
In the second loop i integrate every single column of the selected matrix
obtaining a row of 640 elements.
The third loop must concatenate vertically all the lines previously calculated.
Anyway i got always a problem with the third loop. Where is the error?
singleframe = zeros(224,640);
int_frame_all = zeros(1,640);
conc = zeros(100,640);
for i=0:224:(22400-224)
for j = 1:640
for k = 1:100
singleframe(:,:) = datimf([i+1:(i+223)+1],:);
int_frame_all(:,j) = trapz(singleframe(:,j));
conc(:,k) = vertcat(int_frame_all);
end
end
end
An alternate way to do this without using any explicit loops (edited in response to rayryeng's comment below. It's also worth noting that using cellfun may not be more efficient than explicitly looping.):
nmats = 100;
nrows = 224;
ncols = 640;
datimf = rand(nmats*nrows, ncols);
% convert to an nmats x 1 cell array containing each matrix
cellOfMats = mat2cell(datimf, ones(1, nmats)*nrows, ncols);
% Apply trapz to the contents of each cell
cellOfIntegrals = cellfun(#trapz, cellOfMats, 'UniformOutput', false);
% concatenate the results
conc = cat(1, cellOfIntegrals{:});
Taking inspiration from user2305193's answer, here's an even better "loop-free" solution, based on reshaping the matrix and applying trapz along the appropriate dimension:
datReshaped = reshape(datimf, nrows, nmats, ncols);
solution = squeeze(trapz(datReshaped, 1));
% verify solutions are equivalent:
all(solution(:) == conc(:)) % ans = true
I think I understand what you want. The third loop is unnecessary as both the inner and outer loops are 100 elements long. Also the way you have it you are assigning singleframe lots more times than necessary since it does not depend on the inner loops j or k. You were also trying to add int_frame_all to conc before int_frame_all was finished being populated.
On top of that the j loop isn't required either since trapz can operate on the entire matrix at once anyway.
I think this is closer to what you intended:
datimf = rand(224*100,640);
singleframe = zeros(224,640);
int_frame_all = zeros(1,640);
conc = zeros(100,640);
for i=1:100
idx = (i-1)*224+1;
singleframe(:,:) = datimf(idx:idx+223,:);
% for j = 1:640
% int_frame_all(:,j) = trapz(singleframe(:,j));
% end
% The loop is uncessary as trapz can operate on the entire matrix at once.
int_frame_all = trapz(singleframe,1);
%I think this is what you really want...
conc(i,:) = int_frame_all;
end
It looks like you're processing frames in a video.
The most efficent approach in my experience would be to reshape datimf to be 3-dimensional. This can easily be achieved with the reshape command.
something along the line of vid=reshape(datimf,224,640,[]); should get you far in this regard, where the 3rd dimension is time. vid(:,:,1) then would display the first frame of the video.
This question already has answers here:
Get the indices of the n largest elements in a matrix
(4 answers)
Closed 6 years ago.
When using a binary image with several lines I know that this code displays the longest line:
lineStats = regionprops(imsk, {'Area','PixelIdxList'});
[length, index] = max([lineStats.Area]);
longestLine = zeros(size(imsk));
longestLine(lineStats(index).PixelIdxList)=1;
figure
imshow(longestLine)
Is there a way to display the second longest line? I need to display a line that is a little shorter than the longest line in order to connect them.
EDIT: Is there a way to display both lines on the binary image figure?
Thank you.
I would set the longest line to zero and use max again, after I copy the original vector.
lineStats = regionprops(imsk, {'Area','PixelIdxList'});
[length, index] = max([lineStats.Area]);
lineAreas = [lineStats.Area]; %copy all lineStats.Area values into a new vector
lineAreas(index) = NaN; %remove the longest line by setting it to not-a-number
[length2, index2] = max(lineAreas);
EDIT: Response to new question
sort may be a more straight forward approach for multiples, but you can still use max.
lineAreas = [lineStats.Area]; %copy all lineStats.Area values into a new vector
% add a for loop that iteratively stores the desired indices
nLines = 3;
index = zeros(1,nLines);
for iLines = 1:nLines
[length, index(iLines)] = max(lineAreas);
lineAreas(index) = NaN; %remove the longest line by setting it to not-a-number
end
longestLine = zeros(size(imsk));
% I cannot be certain this will work since your example is not reproducible
longestLine([lineStats(index).PixelIdxList]) = 1;
figure
imshow(longestLine)
Instead of using max use sort in descending order and take the second element. Like max, sort also provides the indexes of the returned values, so the two functions are pretty compatible.
eStats = regionprops(imsk, {'Area','PixelIdxList'});
[length, index] = sort([lineStats.Area], 'descend');
longestLine = zeros(size(imsk));
longestLine(lineStats(index(2)).PixelIdxList)=1; % here take the second largest
figure
imshow(longestLine)
As an alternative with focus on performance and ease of use, here's one approach using bwlabel instead of regionprops -
[L, num] = bwlabel(imsk, 8);
count_pixels_per_obj = sum(bsxfun(#eq,L(:),1:num));
[~,sidx] = sort(count_pixels_per_obj,'descend');
N = 3; % Shows N biggest objects/lines
figure,imshow(ismember(L,sidx(1:N))),title([num2str(N) ' biggest blobs'])
On the performance aspect, here's one post that does some benchmarking on snowflakes and coins images from MATLAB's image gallery.
Sample run -
imsk = im2bw(imread('coins.png')); %%// Coins photo from MATLAB Library
N = 2:
N = 3:
I have 3 vectors: Y=rand(1000,1), X=Y-rand(1000,1) and ACTid=randi(6,1000,1).
I'd like to create boxplots by groups of Y and X corresponding to their group value 1:6 (from ACTid).
This is rather ad-hoc and looks nasty
for ii=
dummyY(ii)={Y(ACTid==ii)};
dummyX(ii)={X(ACTid==ii)}
end
Now I have the data in a cell but can't work out how to group it in a boxplot. Any thoughts?
I've found aboxplot function that looks like this but I don't want that, I'd like the builtin boxplot function because i'm converting it to matlab2tikz and this one doesn't do it well.
EDIT
Thanks to Oleg: we now have a grouped boxplot... but the labels are all skew-whiff.
xylabel = repmat({'Bleh','Blah'},1000,1); % need a legend instead, but doesn't appear possible
boxplot([Y(:,end); cfu], {repmat(ACTid,2,1), xylabel(:)} ,'factorgap',10,'color','rk')
set(gca,'xtick',1.5:3.2:50)
set(gca,'xticklabel',{'Direct care','Housekeeping','Mealtimes','Medication','Miscellaneous','Personal care'})
>> ylabel('Raw CFU counts (Y)')
How to add a legend?
I had the same problem with grouping data in a box plot. A further constraint of mine was that different groups have different amounts of data points. Based on a tutorial I found, this seems to be a nice solution I wanted to share with you:
x = [1,2,3,4,5,1,2,3,4,6];
group = [1,1,2,2,2,3,3,3,4,4];
positions = [1 1.25 2 2.25];
boxplot(x,group, 'positions', positions);
set(gca,'xtick',[mean(positions(1:2)) mean(positions(3:4)) ])
set(gca,'xticklabel',{'Direct care','Housekeeping'})
color = ['c', 'y', 'c', 'y'];
h = findobj(gca,'Tag','Box');
for j=1:length(h)
patch(get(h(j),'XData'),get(h(j),'YData'),color(j),'FaceAlpha',.5);
end
c = get(gca, 'Children');
hleg1 = legend(c(1:2), 'Feature1', 'Feature2' );
Here is a link to the tutorial.
A two-line approach (although if you want to retain two-line xlables and center those in the first line, it's gonna be hackish):
Y = rand(1000,1);
X = Y-rand(1000,1);
ACTid = randi(6,1000,1);
xylabel = repmat('xy',1000,1);
boxplot([X; Y], {repmat(ACTid,2,1), xylabel(:)} ,'factorgap',10)
The result:
EDIT
To center labels...
% Retrieve handles to text labels
h = allchild(findall(gca,'type','hggroup'));
% Delete x, y labels
throw = findobj(h,'string','x','-or','string','y');
h = setdiff(h,throw);
delete(throw);
% Center labels
mylbl = {'this','is','a','pain','in...','guess!'};
hlbl = findall(h,'type','text');
pos = cell2mat(get(hlbl,'pos'));
% New centered position for first intra-group label
newPos = num2cell([mean(reshape(pos(:,1),2,[]))' pos(1:2:end,2:end)],2);
set(hlbl(1:2:end),{'pos'},newPos,{'string'},mylbl')
% delete second intra-group label
delete(hlbl(2:2:end))
Exporting as .png will cause problems...
Okay so this sounds easy but no matter how many times I have tried I still cannot get it to plot correctly. I need only 3 lines on the same graph however still have an issue with it.
iO = 2.0e-6;
k = 1.38e-23;
q = 1.602e-19;
for temp_f = [75 100 125]
T = ((5/9)*temp_f-32)+273.15;
vd = -1.0:0.01:0.6;
Id = iO*(exp((q*vd)/(k*T))-1);
plot(vd,Id,'r',vd,Id,'y',vd,Id,'g');
legend('amps at 75 F', 'amps at 100 F','amps at 125 F');
end;
ylabel('Amps');
xlabel('Volts');
title('Current through diode');
Now I know the plot function that is currently in their isn't working and that some kind of variable needs setup like (vd,Id1,'r',vd,Id2,'y',vd,Id3,'g'); however I really can't grasp the concept of changing it and am seeking help.
You can use the "hold on" function to make it so each plot command plots on the same window as the last.
It would be better to skip the for loop and just do this all in one step though.
iO = 2.0e-6;
k = 1.38e-23;
q = 1.602e-19;
temp_f = [75 100 125];
T = ((5/9)*temp_f-32)+273.15;
vd = -1.0:0.01:0.6;
% Convert this 1xlength(vd) vector to a 3xlength(vd) vector by copying it down two rows.
vd = repmat(vd,3,1);
% Convert this 1x3 array to a 3x1 array.
T=T';
% and then copy it accross to length(vd) so each row is all the same value from the original T
T=repmat(T,1,length(vd));
%Now we can calculate Id all at once.
Id = iO*(exp((q*vd)./(k*T))-1);
%Then plot each row of the Id matrix as a seperate line. Id(1,:) means 1st row, all columns.
plot(vd,Id(1,:),'r',vd,Id(2,:),'y',vd,Id(3,:),'g');
ylabel('Amps');
xlabel('Volts');
title('Current through diode');
And that should get what you want.