I am trying to create a vector with the average of every 48 elements in eddyCO2.
Tweedle = eddyCO2(1:47:end);
Tweedle(1) = mean(eddyCO2(1):eddyCO2(48));
for i = 2:length(Tweedle)
Tweedle(i) = mean(eddyCO2((i-1)*48):eddyCO2(i*48)); (ERROR: Index exceeds matrix dimensions)
end
I've tried reshaping and the only thing that seems to work is entering values manually but the size is too large to work without a loop. Why is this error appearing?
Does the following modified version of your script do what you intend? Hope it helps.
Tweedle = eddyCO2(1:48:end);
sz = length(eddyCO2);
for i = 1:length(Tweedle)-1
Tweedle(i) = mean(eddyCO2((i-1)*48+1):eddyCO2(i*48)); % averages elements 1-48, 49-96, etc.
end
Tweedle(i+1) = mean(eddyCO2(i*48+1):sz); % averages remaining items at end of vector
Related
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 would like to divide a vector in many vectors and put all of them in a matrix. I got this error "Subscripted assignment dimension mismatch."
STEP = zeros(50,1);
STEPS = zeros(50,length(locate));
for i = 1:(length(locate)-1)
STEP = filtered(locate(i):locate(i+1));
STEPS(:,i) = STEP;
end
I take the value of "filtered" from (1:50) at the first time for example and I would like to stock it in the first row of a matrix, then for iterations 2, I take value of "filtered from(50:70) for example and I stock it in row 2 in the matrix, and this until the end of the loop..
If someone has an idea, I don't get it! Thank you!
As mentioned in the comments, to make it work you can edit the loopy code at the end with -
STEPS(1:numel(STEP),i) = STEP;
Also, output array STEPS doesn't seem to use the last column. So, the initialization could use one less column, like so -
STEPS = zeros(50,length(locate)-1);
All is good with the loopy code, but in the long run with a high level language like MATLAB, you might want to look for faster codes and one way to achieve that would be vectorized codes. So, let me suggest a vectorized solution using bsxfun's masking capability to process such ragged-arrays. The implementation to cover generic elements in locate would look something like this -
% Get differentiation, which represent the interval lengths for each col
diffs = diff(locate)+1;
% Initialize output array
out = zeros(max(diffs),length(locate)-1);
% Get elements from filtered array for setting into o/p array
vals = filtered(sort([locate(1):locate(end) locate(2:end-1)]));
% Use bsxfun to create a mask that are to be set in o/p array and set thereafter
out(bsxfun(#ge,diffs,(1:max(diffs)).')) = vals;
Sample run for verification -
>> % Inputs
locate = [6,50,70,82];
filtered = randi(9,1,120);
% Get extent of output array for number of rows
N = max(diff(locate))+1;
>> % Original code with corrections
STEP = zeros(N,1);
STEPS = zeros(N,length(locate)-1);
for i = 1:(length(locate)-1)
STEP = filtered(locate(i):locate(i+1));
STEPS(1:numel(STEP),i) = STEP;
end
>> % Proposed code
diffs = diff(locate)+1;
out = zeros(max(diffs),length(locate)-1);
vals = filtered(sort([locate(1):locate(end) locate(2:end-1)]));
out(bsxfun(#ge,diffs,(1:max(diffs)).')) = vals;
>> max_error = max(abs(out(:)-STEPS(:)))
max_error =
0
I have a vector Cycle() that can contain several elements with a variable size.
I want to extract from this vector all the values which are in the odd columns, i.e. Cycle(1), Cycle(3), Cycle(5) ... and save them into a new vector Rcycle.
That's my code:
Rcycle = zeros(1, length(cycle)/2);
Rcycle(1) = cycle(1);
for j=3:length(cycle);
for i=2:length(Rcycle);
Rcycle(i) = cycle(j);
j = j+2;
end
end
Also I want to extract from Cycle() the even columns and save them in a vector Lcycle. My code:
Lcycle = zeros(1, length(cycle)/2);
Lcycle(1) = cycle(2);
for k=4:length(cycle);
for i=2:length(cycle);
Lcycle(i) = cycle(k);
k = k+2;
end
end
By running this for a sample Cycle() with 12 elements I get the right results for Lcycle, but the wrong ones for Rcycle. Also I get the error that my matrix have exceeded its dimension.
Has anyone any idea how to solve this in a more smooth way?
Use vector indexing!
Rcyle=cycle(1:2:end); %// Take from cycle starting from 1, each 2, until the end
Lcycle=cycle(2:2:end);%// same, but start at 2.
I am trying to find the max value and its location. Following is the example of the programme,
fname = dir('*.mat');
nfiles = length(fname);
vals = cell(nfiles,1);
phen = cell(nfiles,1);
for i = 1:nfiles
vals{i} = load(fname(i).name);
phen{i} = (vals{i}.phen);
[M, position] = max(phen{i},[],3);
clear vals
end
After the program is executed, all the position is showing 1. There are total 15 files and M is taking the values of the last file.
How to overcome this prpoblem? Any help will be appreciated
I am not sure I understand your question.
However, at every iteration you are computing the max value and position and overwriting them in the next iteration (i.e. not storing them anywhere). So at the end of the loop M and position would correspond to the last entry phen{nfiles}.
Each time you run through your for loop, you are overwriting M with the max from the most recently loaded phen from the dimension of 3. Since your data is only two dimensional, you probably should be using a dimension of 1 or 2 instead of 3. Because you are using 3, max is returning 1 to position. Fix the dimension issue and position should then be the correct value.
What you could do is make M and position the size of nfiles. So instead of
[M, position] = max(phen{i},[],3);
do
%create M and positions arrays here
%ex. M(nfiles) = 0; or a smaller value if your values are negative
%do the same for positions
[M(i), positions(i)] = max(phen{i},[],1); %1 or 2 correction here here!
then after your for loop
...
end
[maxM, maxMposition] = max(M);
position = positions(maxMposition);
I want to create a MATLAB function to import data from files in another directory and fit them to a given model, but because the data need to be filtered (there's "thrash" data in different places in the files, eg. measurements of nothing before the analyzed motion starts).
So the vectors that contain the data used to fit end up having different lengths and so I can't return them in a matrix (eg. x in my function below). How can I solve this?
I have a lot of datafiles so I don't want to use a "manual" method. My function is below. All and suggestions are welcome.
datafit.m
function [p, x, y_c, y_func] = datafit(pattern, xcol, ycol, xfilter, calib, p_calib, func, p_0, nhl)
datafiles = dir(pattern);
path = fileparts(pattern);
p = NaN(length(datafiles));
y_func = [];
for i = 1:length(datafiles)
exist(strcat(path, '/', datafiles(i).name));
filename = datafiles(i).name;
data = importdata(strcat(path, '/', datafiles(i).name), '\t', nhl);
filedata = data.data/1e3;
xdata = filedata(:,xcol);
ydata = filedata(:,ycol);
filter = filedata(:,xcol) > xfilter(i);
x(i,:) = xdata(filter);
y(i,:) = ydata(filter);
y_c(i,:) = calib(y(i,:), p_calib);
error = #(par) sum(power(y_c(i,:) - func(x(i,:), par),2));
p(i,:) = fminsearch(error, p_0);
y_func = [y_func; func(x(i,:), p(i,:))];
end
end
sample data: http://hastebin.com/mokocixeda.md
There are two strategies I can think of:
I would return the data in a vector of cells instead, where the individual cells store vectors of different lengths. You can access data the same way as arrays, but use curly braces: Say c{1}=[1 2 3], c{2}=[1 2 10 8 5] c{3} = [ ].
You can also filter the trash data upon reading a line, if that makes your vectors have the same length.
If memory is not an major issue, try filling up the vectors with distinct values, such as NaN or Inf - anything, that is not found in your measurements based on their physical context. You might need to identify the longest data-set before you allocate memory for your matrices (*). This way, you can use equally sized matrices and easily ignore the "empty data" later on.
(*) Idea ... allocate memory based on the size of the largest file first. Fill it up with e.g. NaN's
matrix = zeros(length(datafiles), longest_file_line_number) .* NaN;
Then run your function. Determine the length of the longest consecutive set of data.
new_max = length(xdata(filter));
if new_max > old_max
old_max = new_max;
end
matrix(i, length(xdata(filter))) = xdata(filter);
Crop your matrix accordingly, before the function returns it ...
matrix = matrix(:, 1:old_max);