I would like to plot multiple frequency histograms on one graph, however, my frequency plot is jagged and not pretty. As shown below with this code:
mmin = min([Data]);
mmax = max([Data]);
figure(1);n = hist(Data, x);
f = n/sum(n);
plot(x,f,'r','LineWidth',3)
To make it smooth, I looked into ksdensity and created the graph below based on this code:
[f,xi] = ksdensity(data);
figure(1)
plot(xi,f);
However, I noticed that my graph is no longer plotting frequency on the y-axis. Is there anyway to correct for this change using ksdensity? I really like how the graph looks as opposed to my frequency histogram and would like to keep using ksdensity, unless there is a better suggestion.
Thank you!
Data Sample:
https://www.dropbox.com/s/4ax2cuvugvqxjh6/splicing_attempt2_normalized_combined.txt?dl=0
The trick is here that I don't think you are calculating the frequency correctly in your histogram. You are neglecting the bin width. Your frequency should be the number of SNPs per position, which requires dividing by the number of (possibly fractional) positions per bin.
Try this:
Data = rand(1, 1e4);
figure(1);
[n, c] = hist(Data, 20);
dc = abs(c(2) - c(1));
f = n./(dc * sum(n));
plot(c,f,'r','LineWidth',3)
[~,f_kde,xi] = kde(Data);
line(xi,f_kde);
I don't have the Statistics Toolbox, so I'm using the File Exchange kde function instead, but both work the same way.
If the first graph is indeed what you are after, then do a little algebra-fu, and instead of dividing the histogram values by the bin width, multiply the kdensity values by the same bin width.
As I mention in my other histogram answer, there are numerous methods for choosing optimal bin widths for a histogram. I chose 20 here for expediency.
Related
I have to plot 10 frequency distributions on one graph. In order to keep things tidy, I would like to avoid making a histogram with bins and would prefer having lines that follow the contour of each histogram plot.
I tried the following
[counts, bins] = hist(data);
plot(bins, counts)
But this gives me a very inexact and jagged line.
I read about ksdensity, which gives me a nice curve, but it changes the scaling of my y-axis and I need to be able to read the frequencies from the y-axis.
Can you recommend anything else?
You're using the default number of bins for your histogram and, I will assume, for your kernel density estimation calculations.
Depending on how many data points you have, that will certainly not be optimal, as you've discovered. The first thing to try is to calculate the optimum bin width to give the smoothest curve while simultaneously preserving the underlying PDF as best as possible. (see also here, here, and here);
If you still don't like how smooth the resulting plot is, you could try using the bins output from hist as a further input to ksdensity. Perhaps something like this:
[kcounts,kbins] = ksdensity(data,bins,'npoints',length(bins));
I don't have your data, so you may have to play with the parameters a bit to get exactly what you want.
Alternatively, you could try fitting a spline through the points that you get from hist and plotting that instead.
Some code:
data = randn(1,1e4);
optN = sshist(data);
figure(1)
[N,Center] = hist(data);
[Nopt,CenterOpt] = hist(data,optN);
[f,xi] = ksdensity(data,CenterOpt);
dN = mode(diff(Center));
dNopt = mode(diff(CenterOpt));
plot(Center,N/dN,'.-',CenterOpt,Nopt/dNopt,'.-',xi,f*length(data),'.-')
legend('Default','Optimum','ksdensity')
The result:
Note that the "optimum" bin width preserves some of the fine structure of the distribution (I had to run this a couple times to get the spikes) while the ksdensity gives a smooth curve. Depending on what you're looking for in your data, that may be either good or bad.
How about interpolating with splines?
nbins = 10; %// number of bins for original histogram
n_interp = 500; %// number of values for interpolation
[counts, bins] = hist(data, nbins);
bins_interp = linspace(bins(1), bins(end), n_interp);
counts_interp = interp1(bins, counts, bins_interp, 'spline');
plot(bins, counts) %// original histogram
figure
plot(bins_interp, counts_interp) %// interpolated histogram
Example: let
data = randn(1,1e4);
Original histogram:
Interpolated:
Following your code, the y axis in the above figures gives the count, not the probability density. To get probability density you need to normalize:
normalization = 1/(bins(2)-bins(1))/sum(counts);
plot(bins, counts*normalization) %// original histogram
plot(bins_interp, counts_interp*normalization) %// interpolated histogram
Check: total area should be approximately 1:
>> trapz(bins_interp, counts_interp*normalization)
ans =
1.0009
I'm using pwelch to plot a power spectral density. I want to use the format
pwelch=(x,window,noverlap,nfft,fs,'onesided')
but with a log scale on the x axis.
I've also tried
[P,F]=(x,window,noverlap,nfft,fs);
plot(F,P)
but it doesn't give the same resulting plot as above. Therefore,
semilogx(F,P)
isn't a good solution.
OK, so to start, I've never heard of this function or this method. However, I was able to generate the same plot that the function produced using output arguments instead. I ran the example from the help text.
EXAMPLE:
Fs = 1000; t = 0:1/Fs:.296;
x = cos(2*pi*t*200)+randn(size(t)); % A cosine of 200Hz plus noise
pwelch(x,[],[],[],Fs,'twosided'); % Uses default window, overlap & NFFT.
That produces this plot:
I then did: plot(bar,10*log10(foo)); grid on; to produce the linear version (same exact plot, minus labels):
or
semilogx(bar,10*log10(foo)); grid on; for the log scale on the x-axis.
I don't like that the x-scale is sampled linearly but displayed logarithmically (that's a word right?), but it seems to look ok.
Good enough?
I 'm having a problem with creating a joint density function from data. What I have is queue sizes from a stock as two vectors saved as:
X = [askQueueSize bidQueueSize];
I then use the hist3-function to create a 3D histogram. This is what I get:
http://dl.dropbox.com/u/709705/hist-plot.png
What I want is to have the Z-axis normalized so that it goes from [0 1].
How do I do that? Or do someone have a great joint density matlab function on stock?
This is similar (How to draw probability density function in MatLab?) but in 2D.
What I want is 3D with x:ask queue, y:bid queue, z:probability.
Would greatly appreciate if someone could help me with this, because I've hit a wall over here.
I couldn't see a simple way of doing this. You can get the histogram counts back from hist3 using
[N C] = hist3(X);
and the idea would be to normalise them with:
N = N / sum(N(:));
but I can't find a nice way to plot them back to a histogram afterwards (You can use bar3(N), but I think the axes labels will need to be set manually).
The solution I ended up with involves modifying the code of hist3. If you have access to this (edit hist3) then this may work for you, but I'm not really sure what the legal situation is (you need a licence for the statistics toolbox, if you copy hist3 and modify it yourself, this is probably not legal).
Anyway, I found the place where the data is being prepared for a surf plot. There are 3 matrices corresponding to x, y, and z. Just before the contents of the z matrix were calculated (line 256), I inserted:
n = n / sum(n(:));
which normalises the count matrix.
Finally once the histogram is plotted, you can set the axis limits with:
xlim([0, 1]);
if necessary.
With help from a guy at mathworks forum, this is the great solution I ended up with:
(data_x and data_y are values, which you want to calculate at hist3)
x = min_x:step:max_x; % axis x, which you want to see
y = min_y:step:max_y; % axis y, which you want to see
[X,Y] = meshgrid(x,y); *%important for "surf" - makes defined grid*
pdf = hist3([data_x , data_y],{x y}); %standard hist3 (calculated for yours axis)
pdf_normalize = (pdf'./length(data_x)); %normalization means devide it by length of
%data_x (or data_y)
figure()
surf(X,Y,pdf_normalize) % plot distribution
This gave me the joint density plot in 3D. Which can be checked by calculating the integral over the surface with:
integralOverDensityPlot = sum(trapz(pdf_normalize));
When the variable step goes to zero the variable integralOverDensityPlot goes to 1.0
Hope this help someone!
There is a fast way how to do this with hist3 function:
[bins centers] = hist3(X); % X should be matrix with two columns
c_1 = centers{1};
c_2 = centers{2};
pdf = bins / (sum(sum(bins))*(c_1(2)-c_1(1)) * (c_2(2)-c_2(1)));
If you "integrate" this you will get 1.
sum(sum(pdf * (c_1(2)-c_1(1)) * (c_2(2)-c_2(1))))
I have a Matlab figure I want to use in a paper. This figure contains multiple cdfplots.
Now the problem is that I cannot use the markers because the become very dense in the plot.
If i want to make the samples sparse I have to drop some samples from the cdfplot which will result in a different cdfplot line.
How can I add enough markers while maintaining the actual line?
One method is to get XData/YData properties from your curves follow solution (1) from #ephsmith and set it back. Here is an example for one curve.
y = evrnd(0,3,100,1); %# random data
%# original data
subplot(1,2,1)
h = cdfplot(y);
set(h,'Marker','*','MarkerSize',8,'MarkerEdgeColor','r','LineStyle','none')
%# reduced data
subplot(1,2,2)
h = cdfplot(y);
set(h,'Marker','*','MarkerSize',8,'MarkerEdgeColor','r','LineStyle','none')
xdata = get(h,'XData');
ydata = get(h,'YData');
set(h,'XData',xdata(1:5:end));
set(h,'YData',ydata(1:5:end));
Another method is to calculate empirical CDF separately using ECDF function, then reduce the results before plotting with PLOT.
y = evrnd(0,3,100,1); %# random data
[f, x] = ecdf(y);
%# original data
subplot(1,2,1)
plot(x,f,'*')
%# reduced data
subplot(1,2,2)
plot(x(1:5:end),f(1:5:end),'r*')
Result
I know this is potentially unnecessary given MATLAB's built-in functions (in the Statistics Toolbox anyway) but it may be of use to other viewers who do not have access to the toolbox.
The empirical CMF (CDF) is essentially the cumulative sum of the empirical PMF. The latter is attainable in MATLAB via the hist function. In order to get a nice approximation to the empirical PMF, the number of bins must be selected appropriately. In the following example, I assume that 64 bins is good enough for your data.
%# compute a histogram with 64 bins for the data points stored in y
[f,x]=hist(y,64);
%# convert the frequency points in f to proportions
f = f./sum(f);
%# compute the cumulative sum of the empirical PMF
cmf = cumsum(f);
Now you can choose how many points you'd like to plot by using the reduced data example given by yuk.
n=20 ; % number of total data markers in the curve graph
M_n = round(linspace(1,numel(y),n)) ; % indices of markers
% plot the whole line, and markers for selected data points
plot(x,y,'b-',y(M_n),y(M_n),'rs')
verry simple.....
try reducing the marker size.
x = rand(10000,1);
y = x + rand(10000,1);
plot(x,y,'b.','markersize',1);
For publishing purposes I tend to use the plot tools on the figure window. This allow you to tweak all of the plot parameters and immediately see the result.
If the problem is that you have too many data points, you can:
1). Plot using every nth sample of the data. Experiment to find an n that results in the look you want.
2). I typically fit curves to my data and add a few sparsely placed markers to plots of the fits to differentiate the curves.
Honestly, for publishing purposes I have always found that choosing different 'LineStyle' or 'LineWidth' properties for the lines gives much cleaner results than using different markers. This would also be a lot easier than trying to downsample your data, and for plots made with CDFPLOT I find that markers simply occlude the stairstep nature of the lines.
I have an image and my aim is to binarize the image. I have filtered the image with a low pass Gaussian filter and have computed the intensity histogram of the image.
I now want to perform smoothing of the histogram so that I can obtain the threshold for binarization. I used a low pass filter but it did not work. This is the filter I used.
h = fspecial('gaussian', [8 8],2);
Can anyone help me with this? What is the process with respect to smoothing of a histogram?
imhist(Ig);
Thanks a lot for all your help.
I've been working on a very similar problem recently, trying to compute a threshold in order to exclude noisy background pixels from MRI data prior to performing other computations on the images. What I did was fit a spline to the histogram to smooth it while maintaining an accurate fit of the shape. I used the splinefit package from the file exchange to perform the fitting. I computed a histogram for a stack of images treated together, but it should work similarly for an individual image. I also happened to use a logarithmic transformation of my histogram data, but that may or may not be a useful step for your application.
[my_histogram, xvals] = hist(reshape(image_volume), 1, []), number_of_bins);
my_log_hist = log(my_histogram);
my_log_hist(~isfinite(my_log_hist)) = 0; % Get rid of NaN values that arise from empty bins (log of zero = NaN)
figure(1), plot(xvals, my_log_hist, 'b');
hold on
breaks = linspace(0, max_pixel_intensity, numberofbreaks);
xx = linspace(0, max_pixel_intensity, max_pixel_intensity+1);
pp = splinefit(xvals, my_log_hist, breaks, 'r');
plot(xx, ppval(pp, xx), 'r');
Note that the spline is differentiable and you can use ppdiff to get the derivative, which is useful for finding maxima and minima to help pick an appropriate threshold. The numberofbreaks is set to a relatively low number so that the spline will smooth the histogram. I used linspace in the example to pick the breaks, but if you know that some portion of the histogram exhibits much greater curvature than elsewhere, you'd want to have more breaks in that region and less elsewhere in order to accurately capture the shape of the histogram.
To smooth the histogram you need to use a 1-D filter. This is easily done using the filter function. Here is an example:
I = imread('pout.tif');
h = imhist(I);
smooth_h = filter(normpdf(-4:4, 0,1),1,h);
Of course you can use any smoothing function you choose. The mean would simply be ones(1,8).
Since your goal here is just to find the threshold to binarize an image you could just use the graythresh function which uses Otsu's method.