I use imtophat to apply a filter to an m x n array. I then find the local max using imextendedmax(). I get mostly 0's everywhere except for 1's in the general areas where I am expecting a local max. The weird thing is, though, that I don't get just one local max. Instead in these places I get MANY elements with 1's such as
00011100000
00111111000
00000110000
yet the values there are close but NOT equal so I would expect that there would be one that is higher than all of the rest. So I'm wondering:
if this is a bug and how I might fix it
how you would choose choose the element of these 1's with the highest value.
a) This is a feature. You are calling imextendedmax with two input arguments. The second input is a measure for how different pixels can be from the maximum and still be counted for the extended maximum.
b) You can choose the elements with the highest value using max on the pixels within the group.
%# for testing, create a mask with two groups and an image of corresponding size
msk = repmat([00011100000;...
00111111000;...
00000110000],1,2);
img = rand(size(msk));
imSize = size(img);
%# to find groups of connected ones, apply connected component labeling
cc = bwconncomp(msk);
%# loop through all groups and find the location of the maximum intensity pixel
%# You could do this without loop, but it would be much less readable
maxCoordList = zeros(cc.NumObjects,2);
for i = 1:cc.numObjects
%# read intensities corresponding to group
int = img(cc.PixelIdxList{i});
%# find which pixel is brightest
[maxInt,maxIdx] = max(int);
%# maxIdx indexes into PixelIdxList, which indexes into the image.
%# convert to [x,y]
maxCoordList = ind2sub(imSize,cc.PixelIdxList{i}(maxIdx));
end
%# confirm by plotting
figure
imshow(img,[])
hold on
plot(maxCoordList(:,2),maxCoordList(:,1),'.r')
Related
The objective is to see if two images, which have one object captured in each image, matches.
The object or image I have stored. This will be used as a baseline:
item1 (This is being matched in the code)
The object/image that needs to matched with-this is stored:
input (Need to see if this matches with what is stored
My method:
Covert images to gray-scale.
Extract SURF interest points.
Obtain features.
Match features.
Get 50 strongest features.
Match the number of strongest features with each image.
Take the ratio of- number of features matched/ number of strongest
features (which is 50).
If I have two images of the same object (two images taken separately on a camera), ideally the ratio should be near 1 or near 100%.
However this is not the case, the best ratio I am getting is near 0.5 or even worse, 0.3.
I am aware the SURF detectors and features can be used in neural networks, or using a statistics based approach. I believe I have approached the statistics based approach to some extent by using 50 of the strongest features.
Is there something I am missing? What do I add onto this or how do I improve it? Please provide me a point to start from.
%Clearing the workspace and all variables
clc;
clear;
%ITEM 1
item1 = imread('Loreal.jpg');%Retrieve order 1 and digitize it.
item1Grey = rgb2gray(item1);%convert to grayscale, 2 dimensional matrix
item1KP = detectSURFFeatures(item1Grey,'MetricThreshold',600);%get SURF dectectors or interest points
strong1 = item1KP.selectStrongest(50);
[item1Features, item1Points] = extractFeatures(item1Grey, strong1,'SURFSize',128); % using SURFSize of 128
%INPUT : Aquire Image
input= imread('MakeUp1.jpg');%Retrieve input and digitize it.
inputGrey = rgb2gray(input);%convert to grayscale, 2 dimensional matrix
inputKP = detectSURFFeatures(inputGrey,'MetricThreshold',600);%get SURF dectectors or interest
strongInput = inputKP.selectStrongest(50);
[inputFeatures, inputPoints] = extractFeatures(inputGrey, strongInput,'SURFSize',128); % using SURFSize of 128
pairs = matchFeatures(item1Features, inputFeatures, 'MaxRatio',1); %matching SURF Features
totalFeatures = length(item1Features); %baseline number of features
numPairs = length(pairs); %the number of pairs
percentage = numPairs/50;
if percentage >= 0.49
disp('We have this');
else
disp('We do not have this');
disp(percentage);
end
The baseline image
The input image
I would try not doing selectStrongest and not setting MaxRatio. Just call matchFeatures with the default options and compare the number of resulting matches.
The default behavior of matchFeatures is to use the ratio test to exclude ambiguous matches. So the number of matches it returns may be a good indicator of the presence or absence of the object in the scene.
If you want to try something more sophisticated, take a look at this example.
I am attempting to create a model whereby there is a line - represented as a 1D matrix populated with 1's - and points on the line are generated at random. Every time a point is chosen (A), it creates a 'zone of exclusion' (based on an exponential function) such that choosing another point nearby has a much lower probability of occurring.
Two main questions:
(1) What is the best way to generate an exponential such that I can multiply the numbers surrounding the chosen point to create the zone of exclusion? I know of exppdf however i'm not sure if this allows me to create an exponential which terminates at 1, as I need the zone of exclusion to end and the probability to return to 1 eventually.
(2) How can I modify matrix values plus/minus a specific index (including that index)? I got as far as:
x(1:100) = 1; % Creates a 1D-matrix populated with 1's
p = randi([1 100],1,1);
x(p) =
But am not sure how to go about using the randomly generated number to alter values in the matrix.
Any help would be much appreciated,
Anna
Don't worry about exppdf, pick the width you want (how far away from the selected point does the probability return to 1?) and define some simple function that makes a small vector with zero in the middle and 1 at the edges. So here I'm just modifying a section of length 11 centred on p and doing nothing to the rest of x:
x(1:100)=1;
p = randi([1 100],1,1);
% following just scaled
somedist = (abs(-5:5).^2)/25;
% note - this will fail if p is at edges of data, but see below
x(p-5:p+5)=x(p-5:p+5).*somedist;
Then, instead of using randi to pick points you can use datasample which allows for giving weights. In this case your "data" is just the numbers 1:100. However, to make edges easier I'd suggest initialising with a "weight" vector which has zero padding - these sections of x will not be sampled from but stop you from having to make edge checks.
x = zeros([1 110]);
x(6:105)=1;
somedist = (abs(-5:5).^2)/25;
nsamples = 10;
for n = 1:nsamples
p = datasample(1:110,1,'Weights',x);
% if required store chosen p somewhere
x(p-5:p+5)=x(p-5:p+5).*somedist;
end
For an exponential exclusion zone you could do something like:
somedist = exp(abs(-5:5))/exp(5)-exp(0)/exp(5);
It doesn't quite return to 1 but fairly close. Here's the central region of x (ignoring the padding) after two separate runs:
An issue I've come across multiple times is wanting to take two similar data sets and create histograms from them where the bins are identical, so as to easily calculate things like histogram overlap.
You can define the number of bins (obviously) using
[counts, bins] = hist(data,number_of_bins)
But there's not an obvious way (as far as I can see) to make the bin size equal for several different data sets. If remember when I initially looked finding various people who seem to have the same issue, but no good solutions.
The right, easy way
As pointed out by horchler, this can easily be achieved using either histc (which lets you define your bins vector), or vectorizing your histogram input into hist.
The wrong, stupid way
I'm leaving below as a reminder to others that even stupid questions can yield worthwhile answers
I've been using the following approach for a while, so figured it might be useful for others (or, someone can very quickly point out the correct way to do this!).
The general approach relies on the fact that MATLAB's hist function defines an equally spaced number of bins between the largest and smallest value in your sample. So, if you append a start (smallest) and end (largest) value to your various samples which is the min and max for all samples of interest, this forces the histogram range to be equal for all your data sets. You can then truncate the first and last values to recreate your original data.
For example, create the following data set
A = randn(1,2000)+7
B = randn(1,2000)+9
[counts_A, bins_A] = hist(A, 500);
[counts_B, bins_B] = hist(B, 500);
Here for my specific data sets I get the following results
bins_A(1) % 3.8127 (this is also min(A) )
bins_A(500) % 10.3081 (this is also max(A) )
bins_B(1) % 5.6310 (this is also min(B) )
bins_B(500) % 13.0254 (this is also max(B) )
To create equal bins you can simply first define a min and max value which is slightly smaller than both ranges;
topval = max([max(A) max(B)])+0.05;
bottomval = min([min(A) min(B)])-0.05;
The addition/subtraction of 0.05 is based on knowledge of the range of values - you don't want your extra bin to be too far or too close to the actual range. That being said, for this example by using the joint min/max values this code will work irrespective of the A and B values generated.
Now we re-create histogram counts and bins using (note the extra 2 bins are for our new largest and smallest value)
[counts_Ae, bins_Ae] = hist([bottomval, A, topval], 502);
[counts_Be, bins_Be] = hist([bottomval, B, topval], 502);
Finally, you truncate the first and last bin and value entries to recreate your original sample exactly
bins_A = bins_Ae(2:501)
bins_B = bins_Ae(2:501)
counts_A = counts_Ae(2:501)
counts_B = counts_Be(2:501)
Now
bins_A(1) % 3.7655
bins_A(500) % 13.0735
bins_B(1) % 3.7655
bins_B(500) % 13.0735
From this you can easily plot both histograms again
bar([bins_A;bins_B]', [counts_A;counts_B]')
And also plot the histogram overlap with ease
bar(bins_A,(counts_A+counts_B)-(abs(counts_A-counts_B)))
I've been struggling with a problem for a while:) in Matlab.
I have an image (A.tif) in which I would like to find maxima (with defined threshold) but more specific coordinates of these maxima. My goal is to create short profiles on the image crossing these maxima (let say +- 20 pixels on both sides of the maximum)
I tried this:
[r c]=find(A==max(max(A)));
I suppose that r and c are coordinates of maximum (only one/first or every maximum?)
How can I implement these coordinates into ,for example improfile function?
I think it should be done using nested loops?
Thanks for every suggestion
Your code is working but it finds only global maximum coordinates.I would like to find multiple maxima (with defined threshold) and properly address its coordinates to create multiple profiles crossing every maximum found. I have little problem with improfile function :
improfile(IMAGE,[starting point],[ending point]) .
Lets say that I get [rows, columns] matrix with coordinates of each maximum and I'm trying to create one direction profile which starts in the same row where maximum is (about 20 pixels before max) and of course ends in the same row (also about 20 pixels from max) .
is this correct expression :improfile(IMAGE,[rows columns-20],[rows columns+20]); It plots something but it seems to only joins maxima rather than making intensity profiles
You're not giving enough information so I had to guess a few things. You should apply the max() to the vectorized image and store the index:
[~,idx] = max(I(:))
Then transform this into x and y coordinates:
[ix,iy] = ind2sub(size(I),idx)
This is your x and y of the maximum of the image. It really depends what profile section you want. Something like this is working:
I = imread('peppers.png');
Ir = I(:,:,1);
[~,idx]=max(Ir(:))
[ix,iy]=ind2sub(size(Ir),idx)
improfile(Ir,[0 ix],[iy iy])
EDIT:
If you want to instead find the k largest values and not just the maximum you can do an easy sort:
[~,idx] = sort(I(:),'descend');
idxk = idx(1:k);
[ix,iy] = ind2sub(size(I),idxk)
Please delete your "reply" and instead edit your original post where you define your problem better
just wondering if anyone has any ideas about an issue I'm having.
I have a fair amount of data that needs to be displayed on one graph. Two theoretical lines that are bold and solid are displayed on top, then 10 experimental data sets that converge to these lines are graphed, each using a different identifier (eg the + or o or a square etc). These graphs are on a log scale that goes up to 1e6. The first few decades of the graph (< 1e3) look fine, but as all the datasets converge (> 1e3) it's really difficult to see what data is what.
There's over 1000 data points points per decade which I can prune linearly to an extent, but if I do this too much the lower end of the graph will suffer in resolution.
What I'd like to do is prune logarithmically, strongest at the high end, working back to 0. My question is: how can I get a logarithmically scaled index vector rather than a linear one?
My initial assumption was that as my data is lenear I could just use a linear index to prune, which lead to something like this (but for all decades):
//%grab indicies per decade
ind12 = find(y >= 1e1 & y <= 1e2);
indlow = find(y < 1e2);
indhigh = find(y > 1e4);
ind23 = find(y >+ 1e2 & y <= 1e3);
ind34 = find(y >+ 1e3 & y <= 1e4);
//%We want ind12 indexes in this decade, find spacing
tot23 = round(length(ind23)/length(ind12));
tot34 = round(length(ind34)/length(ind12));
//%grab ones to keep
ind23keep = ind23(1):tot23:ind23(end);
ind34keep = ind34(1):tot34:ind34(end);
indnew = [indlow' ind23keep ind34keep indhigh'];
loglog(x(indnew), y(indnew));
But this causes the prune to behave in a jumpy fashion obviously. Each decade has the number of points that I'd like, but as it's a linear distribution, the points tend to be clumped at the high end of the decade on the log scale.
Any ideas on how I can do this?
I think the easiest way to do this would be to use the LOGSPACE function to generate a set of indices into your data. For example, to create a set of 100 points logarithmically spaced from 1 to N (the number of points in your data), you can try the following:
indnew = round(logspace(0,log10(N),100)); %# Create the log-spaced index
indnew = unique(indnew); %# Remove duplicate indices
loglog(x(indnew),y(indnew)); %# Plot the indexed data
Creating a logarithmically-spaced index like this will result in fewer values being chosen from the end of the vector relative to the start, thus pruning values more severely towards the end of the vector and improving the appearance of the log plot. It would therefore be most effective with vectors that are sorted in ascending order.
The way I understand the problem is that your x-values are linearly spaced, so that if you plot them logarithmically, there are way more data points in 'higher' decades, so that markers lie extremely close to one another. For example, if x goes from 1 to 1000, there are 10 points in the first decade 90 in the second, and 900 in the third. You want to have, say, 3 points per decade instead.
I see two ways to solve the problem. The easier one is to use differently colored lines instead of different markers. Thus, you don't sacrifice any data points, and you can still distinguish everything.
The second solution is to create an unevenly spaced index. Here's how you can do that.
%# create some data
x = 1:1000;
y = 2.^x;
%# plot the graph and see the dots 'coalesce' very quickly
figure,loglog(x,y,'.')
%# for the example, I use a step size of 0.7, which is `log(1)`
xx = 0.7:0.7:log(x(end)); %# this is where I want the data to be plotted
%# find the indices where we want to plot by finding the closest `log(x)'-values
%# run unique to avoid multiples of the same index
indnew = unique(interp1(log(x),1:length(x),xx,'nearest'));
%# plot with fewer points
figure,loglog(x(indnew),y(indnew),'.')