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i'm trying to extract HOG_features for mathematical symbols classification (i will use SVM classifier). I get a 1xn vector then i have to put all the vectors in a single matrix. The problem is that the size of the feature vector is different for each image so I can't concatenate them.
Is there a way to make all vectors having the same size ?
Thank you in advance.
Here is the code:
rep1 = 'D:\mémoire MASTER\data';
ext = '*.tif' ;
chemin = fullfile(rep1, ext);
list = dir(chemin);
for i=1:length(list)
I = imread(fullfile(rep1, list(i).name), ext(3:end));
if size(I,3)==3 % RGB image
I = rgb2gray(I);
end
I1 = imbinarize(I);
% Extract HOG features data
HOG_feat = extractHOGFeatures(I1,'CellSize', [2 2]);
HOG_feat1 = HOG_feat';
end
You can pad each one with zeros to be as long as the longest one:
e.g. to put two vectors, v1 and v2, into a matrix M:
M = zeros(2,max(length(v1),length(v2)));
M(1,1:length(v1)) = v1;
M(2,1:length(v2)) = v2;
You have the problem that all your vectors are a different size. Instead of trying to coerce them to be the sesame size by zero-padding or interpolating (both I think are bad ideas), change your computation so that the length of the output vector does not depend on the size of the image.
This is your current code:
HOG_feat = extractHOGFeatures(I1,'CellSize', [2 2]);
% ^^^
% the image is split in cells of 2x2 pixels
2x2 cells are way too small for this method anyway. You could instead divide your image into a set number of cells, say 100 cells:
cellSize = ceil(size(I1)/10);
HOG_feat = extractHOGFeatures(I1,'CellSize', cellSize);
(I’m using ceil in the division because I figure it’s necessary to have an integer size. But I’m not sure whether ceil or floor or round is needed here, and I don’t have access to this function to test it. A bit of trial and error should show which method gives consistent output size.)
i am a beginner an have an easy question. I have a signal on y-axis and time signal on x-axis. I need to change boundaries of the time signal. It's between 0 and 18 seconds, but i want to change in between 5 and 10. I used already "xlim", it work for plot but actually i want to create a new time signal.
Any idea? Thank you!
Since you didn't post your code I'll need to make some assumptions. I'll assume you have your data stored in row vectors x and y and that x is uniform and monotonically increasing.
1. Construct a truncated signal using logical indexing.
index = x >= 5 & x <= 10;
x_new = x(index);
y_new = y(index);
plot(x_new, y_new);
The above only takes a subset of the data, if x doesn't contain 5 and 10 then the plot will be truncated. If you're dealing with time series data this is probably the most reasonable approach since it doesn't change the sampling rate.
2. Re-sampling the signal between 5 and 10 using interpolation.
num_samples = 100;
x_new = linspace(5, 10, num_samples);
y_new = interp1(x, y, x_new);
plot(x_new, y_new);
This may not exactly match the original plot since the original samples aren't guaranteed to be included. However it will exactly span the desired domain.
3. If you don't care that x is uniform but want to create a plot that exactly matches the original then you can append the bounds of x to the subset from method 1 and use interp1 to sample y.
x_min = 5; x_max = 10;
index = x > x_min & x < x_max;
x_new = [x_min, x(index), x_max];
y_new = interp1(x, y, x_new);
plot(x_new, y_new);
Example
Example demonstrating the differences between the different methods, plotted with additional offset and markings at samples for clarity.
If you want to delete the elements n from the back of a vector y and store the result in y_cut, you should be able to do that with:
y_cut = y(1:end-n);
It would be important to know in which form you stored the time signal.
If you have one value for each second the solution would be:
y_cut = y(5:10);
But I assume you're storing your y-values as samples with a given sample rate fs
One second would then be equal to fs (for example 44100 for a CD audio file, resulting in 44100 samples per second) and the solution would be:
y_cut = y(5*fs:10*fs);
I hope I could help.
Cheers,
Simon
I need to compute a moving average over a data series, within a for loop. I have to get the moving average over N=9 days. The array I'm computing in is 4 series of 365 values (M), which itself are mean values of another set of data. I want to plot the mean values of my data with the moving average in one plot.
I googled a bit about moving averages and the "conv" command and found something which i tried implementing in my code.:
hold on
for ii=1:4;
M=mean(C{ii},2)
wts = [1/24;repmat(1/12,11,1);1/24];
Ms=conv(M,wts,'valid')
plot(M)
plot(Ms,'r')
end
hold off
So basically, I compute my mean and plot it with a (wrong) moving average. I picked the "wts" value right off the mathworks site, so that is incorrect. (source: http://www.mathworks.nl/help/econ/moving-average-trend-estimation.html) My problem though, is that I do not understand what this "wts" is. Could anyone explain? If it has something to do with the weights of the values: that is invalid in this case. All values are weighted the same.
And if I am doing this entirely wrong, could I get some help with it?
My sincerest thanks.
There are two more alternatives:
1) filter
From the doc:
You can use filter to find a running average without using a for loop.
This example finds the running average of a 16-element vector, using a
window size of 5.
data = [1:0.2:4]'; %'
windowSize = 5;
filter(ones(1,windowSize)/windowSize,1,data)
2) smooth as part of the Curve Fitting Toolbox (which is available in most cases)
From the doc:
yy = smooth(y) smooths the data in the column vector y using a moving
average filter. Results are returned in the column vector yy. The
default span for the moving average is 5.
%// Create noisy data with outliers:
x = 15*rand(150,1);
y = sin(x) + 0.5*(rand(size(x))-0.5);
y(ceil(length(x)*rand(2,1))) = 3;
%// Smooth the data using the loess and rloess methods with a span of 10%:
yy1 = smooth(x,y,0.1,'loess');
yy2 = smooth(x,y,0.1,'rloess');
In 2016 MATLAB added the movmean function that calculates a moving average:
N = 9;
M_moving_average = movmean(M,N)
Using conv is an excellent way to implement a moving average. In the code you are using, wts is how much you are weighing each value (as you guessed). the sum of that vector should always be equal to one. If you wish to weight each value evenly and do a size N moving filter then you would want to do
N = 7;
wts = ones(N,1)/N;
sum(wts) % result = 1
Using the 'valid' argument in conv will result in having fewer values in Ms than you have in M. Use 'same' if you don't mind the effects of zero padding. If you have the signal processing toolbox you can use cconv if you want to try a circular moving average. Something like
N = 7;
wts = ones(N,1)/N;
cconv(x,wts,N);
should work.
You should read the conv and cconv documentation for more information if you haven't already.
I would use this:
% does moving average on signal x, window size is w
function y = movingAverage(x, w)
k = ones(1, w) / w
y = conv(x, k, 'same');
end
ripped straight from here.
To comment on your current implementation. wts is the weighting vector, which from the Mathworks, is a 13 point average, with special attention on the first and last point of weightings half of the rest.
I wonder whether it would be possible to extract only hands from a video with matlab. In the video hands perform some gesture. Because first frames are only background I tried in this way:
readerObj = VideoReader('VideoWithHands.mp4');
nFrames = readerObj.NumberOfFrames;
fr = get(readerObj, 'FrameRate');
writerObj = VideoWriter('Hands.mp4', 'MPEG-4');
set(writerObj, 'FrameRate', fr);
open(writerObj);
bg = read(readerObj, 1); %background
for k = 1 : nFrames
frame = read(readerObj, k);
hands = imabsdiff(frame,bg);
writeVideo(writerObj,hands);
end
close(writerObj);
But I realized that colors of the hands are not "real" and they are transparent. Is there a better way to extract them from video keeping colors and opacity level exploiting the first frames (background)?
EDIT: Well, I have found a good setting for vision.ForegroundDetector object, now hands are white logical regions but when I try to visualize them with:
videoSource = vision.VideoFileReader('VideoWithHands.mp4', 'VideoOutputDataType', 'uint8');
detector = vision.ForegroundDetector('NumTrainingFrames', 46, 'InitialVariance', 4000, 'MinimumBackgroundRatio', 0.2);
videoplayer = vision.VideoPlayer();
hands = uint8(zeros(720,1280,3));
while ~isDone(videoSource)
frame = step(videoSource);
fgMask = step(detector, frame);
[m,n] = find(fgMask);
a = [m n];
if isempty(a)==true
hands(:,:,:) = uint8(zeros(720,1280,3));
else
hands(m,n,1) = frame(m,n,1);
hands(m,n,2) = frame(m,n,2);
hands(m,n,3) = frame(m,n,3);
end
step(videoplayer, hands)
end
release(videoplayer)
release(videoSource)
or put them into a videofile with:
eaderObj = VideoReader('Video 9.mp4');
nFrames = readerObj.NumberOfFrames;
fr = get(readerObj, 'FrameRate');
writerObj = VideoWriter('hands.mp4', 'MPEG-4');
set(writerObj, 'FrameRate', fr);
detector = vision.ForegroundDetector('NumTrainingFrames', 46, 'InitialVariance', 4000, 'MinimumBackgroundRatio', 0.2);
open(writerObj);
bg = read(readerObj, 1);
frame = uint8(zeros(size(bg)));
for k = 1 : nFrames
frame = read(readerObj, k);
fgMask = step(detector, frame);
[m,n] = find(fgMask);
hands = uint8(zeros(720,1280));
if isempty([m n]) == true
hands(:,:) = uint8(zeros(720,1280));
else
hands(m,n) = frame(m,n);
end
writeVideo(writerObj,mani);
end
close(writerObj);
...my PC crashes. Some suggestion?
So you're trying to cancel out the background, making it black, right?
The easiest way to do this should be to filter it, you can do that by comparing your difference data to a threshold value and then using the result as indices to set a custom background.
filtered = imabsdiff(frame,bg);
bgindex = find( filtered < 10 );
frame(bgindex) = custombackground(bgindex);
where custombackground is whatever image file you want to put into the background. If you want it to be just black or white, use 0 or 255 instead of custombackground(bgindex). Note that the numbers depend on your video data's format and could be inaccurate (except 0, this one should always be right). If too much gets filtered out, lower the 10 above, if too much remains unfiltered, increase the 10.
At the end, you write your altered frame back into the video, so it just replaces the hands variable in your code.
Also, depending on your format, you might have to do the comparison across RGB values. This is slightly more complicated as it involves checking 3 values at the same time and doing some magic with the indices. This is the RGB version (works with anything containing 3 color bands):
filtered = imabsdiff(frame,bg); % differences at each pixel in each color band
totalfiltered = sum(filtered,3); % sums up the differences
% in each color band (RGB)
bgindex = find( totalfiltered < 10 ); % extracts indices of pixels
% with color close to bg
allind = sub2ind( [numel(totalfiltered),3] , repmat(bgindex,1,3) , ...
repmat(1:3,numel(bgindex),1) ); % index magic
frame(allind) = custombackground(allind); % copy custom background into frame
EDIT :
Here's a detailed explanation of the index magic.
Let's assume a 50x50 image. Say the pixel at row 2, column 5 is found to be background, then bgindex will contain the number 202 (linear index corresponding to [2,5] = (5-1)*50+2 ). What we need is a set of 3 indices corresponding to the matrix coordinates [2,5,1], [2,5,2] and [2,5,3]. That way, we can change all 3 color bands corresponding to that pixel. To make calculations easier, this approach actually assumes linear indexing for the image and thus converts it to a 2500x1 image. Then it expands the 3 color bands, creating a 2500x3 matrix. We now construct the indices [202,1], [202,2] and [202,3] instead.
To do that, we first construct a matrix of indices by repeating our values. repmat does this for us, it creates the matrices [202 202 202] and [1 2 3]. If there were more pixels in bgindex, the first matrix would contain more rows, each repeating the linear pixel coordinates 3 times. The second matrix would contain additional [1 2 3] rows. The first argument to sub2ind is the size of the matrix, in this case, 2500x3, so we calculate the number of pixels with numel applied to the sum vector (which collapses the image's 3 bands into 1 value and thus has 1 value per pixel) and add a static 3 in the second dimension.
sub2ind now takes each element from the first matrix as a row index, each corresponding element from the second matrix as a column index and converts them to linear indices into a matrix of the size we determined earlier. In our example, this results in the indices [202 2702 5202]. sub2ind preserves the shape of the inputs, so if we had 10 background pixels, this result would have the size 10x3. But since linear indexing doesn't care about the shape of the index matrix, it just takes all of those values.
To confirm this is correct, let's revert the values in the example. The original image data would have the size 50x50x3. For an NxMxP matrix, a linear index to the subscript [n m p] can be calculated as ind = (p-1)*M*N + (m-1)*N + n. Using our values, we get the following:
[2 5 1] => 202
[2 5 2] => 2702
[2 5 3] => 5202
ind2sub confirms this.
Yes, there is a better way. The computer vision system toolbox includes a vision.ForegroundDetector object that does what you need. It implements the Gaussian Mixture Model algorithm for background subtraction.
I have a vector of data, which contains integers in the range -20 20.
Bellow is a plot with the values:
This is a sample of 96 elements from the vector data. The majority of the elements are situated in the interval -2, 2, as can be seen from the above plot.
I want to eliminate the noise from the data. I want to eliminate the low amplitude peaks, and keep the high amplitude peak, namely, peaks like the one at index 74.
Basically, I just want to increase the contrast between the high amplitude peaks and low amplitude peaks, and if it would be possible to eliminate the low amplitude peaks.
Could you please suggest me a way of doing this?
I have tried mapstd function, but the problem is that it also normalizes that high amplitude peak.
I was thinking at using the wavelet transform toolbox, but I don't know exact how to reconstruct the data from the wavelet decomposition coefficients.
Can you recommend me a way of doing this?
One approach to detect outliers is to use the three standard deviation rule. An example:
%# some random data resembling yours
x = randn(100,1);
x(75) = -14;
subplot(211), plot(x)
%# tone down the noisy points
mu = mean(x); sd = std(x); Z = 3;
idx = ( abs(x-mu) > Z*sd ); %# outliers
x(idx) = Z*sd .* sign(x(idx)); %# cap values at 3*STD(X)
subplot(212), plot(x)
EDIT:
It seems I misunderstood the goal here. If you want to do the opposite, maybe something like this instead:
%# some random data resembling yours
x = randn(100,1);
x(75) = -14; x(25) = 20;
subplot(211), plot(x)
%# zero out everything but the high peaks
mu = mean(x); sd = std(x); Z = 3;
x( abs(x-mu) < Z*sd ) = 0;
subplot(212), plot(x)
If it's for demonstrative purposes only, and you're not actually going to be using these scaled values for anything, I sometimes like to increase contrast in the following way:
% your data is in variable 'a'
plot(a.*abs(a)/max(abs(a)))
edit: since we're posting images, here's mine (before/after):
You might try a split window filter. If x is your current sample, the filter would look something like:
k = [L L L L L L 0 0 0 x 0 0 0 R R R R R R]
For each sample x, you average a band of surrounding samples on the left (L) and a band of surrounding samples on the right. If your samples are positive and negative (as yours are) you should take the abs. value first. You then divide the sample x by the average value of these surrounding samples.
y[n] = x[n] / mean(abs(x([L R])))
Each time you do this the peaks are accentuated and the noise is flattened. You can do more than one pass to increase the effect. It is somewhat sensitive to the selection of the widths of these bands, but can work. For example:
Two passes:
What you actually need is some kind of compression to scale your data, that is: values between -2 and 2 are scale by a certain factor and everything else is scaled by another factor. A crude way to accomplish such a thing, is by putting all small values to zero, i.e.
x = randn(1,100)/2; x(50) = 20; x(25) = -15; % just generating some data
threshold = 2;
smallValues = (abs(x) <= threshold);
y = x;
y(smallValues) = 0;
figure;
plot(x,'DisplayName','x'); hold on;
plot(y,'r','DisplayName','y');
legend show;
Please do not that this is a very nonlinear operation (e.g. when you have wanted peaks valued at 2.1 and 1.9, they will produce very different behavior: one will be removed, the other will be kept). So for displaying, this might be all you need, for further processing it might depend on what you are trying to do.
To eliminate the low amplitude peaks, you're going to equate all the low amplitude signal to noise and ignore.
If you have any apriori knowledge, just use it.
if your signal is a, then
a(abs(a)<X) = 0
where X is the max expected size of your noise.
If you want to get fancy, and find this "on the fly" then, use kmeans of 3. It's in the statistics toolbox, here:
http://www.mathworks.com/help/toolbox/stats/kmeans.html
Alternatively, you can use Otsu's method on the absolute values of the data, and use the sign back.
Note, these and every other technique I've seen on this thread is assuming you are doing post processing. If you are doing this processing in real time, things will have to change.