I have a set of 100 datapoints that I have split into 10 bins and displayed as a histogram, however I would like to display it as a relative frequency histogram so I need to scale the height of each bin by 100. How do I do this?
M = readmatrix('Data.csv');
y= M(:,1);
histogram(y,10)
would it be possible to divide the resulting bins in some simple way?
Use the probability normalization,
histogram(y,10,'Normalization','probability')
See the histogram documentation for the other normalizations available.
Related
I have three variables, e.g., latitude, longitude and temperature. For each latitude and longitude, I have corresponding temperature value. I want to plot latitude v/s longitude plot in 5 degree x 5 degree grid , with mean temperature value inserted in that particular grid instead of occurring frequency.
Data= [latGrid,lonGrid] = meshgrid(25:45,125:145);
T = table(latGrid(:),lonGrid(:),randi([0,35],size(latGrid(:))),...
'VariableNames',{'lat','lon','temp'});
At the end, I need it somewhat like the following image:
Sounds to me like you want to scale your grid. The easiest way to do this is to smooth and downsample.
While 2d histograms also bin values into a grid, using a histogram is not the way to find the mean of datapoints in a smooth grid. A histogram counts the occurrence of values in a set of ranges. In a 2d example, a histogram would take the input measurements [1, 3, 3, 5] and count the number of ones, the number of threes, etc. A 2d histogram will count occurrences of pairs of numbers. (You might want to use histogram to help organize a measurements taken at irregular intervals, but that would be a different question)
How to smooth and downsample without the Image Processing Toolbox
Keep your data in the 2d matrix format rather than reshaping it into a table. This makes it easier to find the neighbors of each grid location.
%% Sample Data
[latGrid,lonGrid] = meshgrid(25:45,125:145);
temp = rand(size(latGrid));
There are many tools in Matlab for smoothing matrices. If you want to have the mean of a 5x5 window. You can write a for-loop, use a convolution, or use filter2. My example uses convolution. For more on convolutional filters, I suggest the wikipedia page.
%% Mean filter with conv2
M = ones(5) ./ 25; % 5x5 mean or box blur filter
C_temp = conv2(temp, M, 'valid');
C_temp is a blurry version of the original temperature variable with a slightly smaller size because we can't accurately take the mean of the edges. The border is reduced by a frame of 2 measurements. Now, we just need to take every fifth measurement from C_temp to scale down the grid.
%% Subsample result
C_temp = C_temp(1:5:end, 1:5:end);
% Because we removed a border from C_temp, we also need to remove a border from latGrid and lonGrid
[h, w] = size(latGrid)
latGrid = latGrid(5:5:h-5, 5:5:w-5);
lonGrid = lonGrid(5:5:h-5, 5:5,w-5);
Here's what the steps look like
If you use a slightly more organized, temp variable. It's easier to see that the result is correct.
With Image Processing Toolbox
imresize has a box filter method option that is equivalent to a mean filter. However, you have to do a little calculation to find the scaling factor that is equivalent to using a 5x5 window.
C_temp = imresize(temp, scale, 'box');
I am using the below code to calculate the probabilities of pixel intensities for the image given below. However, the total sum of probabilities sum(sum(probOfPixelIntensities)) is greater than 1.
I'm not sure where the mistake may be. Any help in figuring this out would be greatly appreciated. Thanks in advance.
clear all
clc
close all
I = imread('Images/cameraman.jpg');
I = rgb2gray(I);
imshow(I)
muHist = 134;
sigmaHist = 54;
Iprob = normpdf(double(I), muHist, sigmaHist);
sum(sum(Iprob))
What you are doing is computing the PDF values for every pixel in the image. Iprob is not a normal distribution but you are simply using the image pixels to sample from the distribution of a known mean and standard deviation.
Essentially, you are just performing a data transformation where the image pixel intensities get mapped to values on a normal PDF with a known mean and standard deviation. This is not the same as a PDF and that's why the sum is not 1. On top of this, the image pixel intensities don't even follow a normal distribution itself so there wouldn't be any way that the sum of the distribution is 1.
Not much more to say other than the output of normpdf is not what you are expecting it to be. You should opt to read the documentation of normpdf more carefully: http://www.mathworks.com/help/stats/normpdf.html
If it is your desire to determine the actual PDF of the image, what you need to do is find the histogram of the image, and not do a data transformation. You can do that with imhist. Once you do that, assuming that encountering the intensities is equiprobable, you would divide each histogram entry by the total size of the image and then sum along all bins. You should get the sum to be 1 in this case.
Just to verify, let's use the image you provided in your post. We'll read this in from StackOverflow. Once we do that, compute the PDF and then sum over all bins:
%// Load in image
im = rgb2gray(imread('http://i.stack.imgur.com/0XiU5.jpg'));
%// Compute PDF
h = imhist(im) / numel(im);
%// Sum over all bins
fprintf('Total sum over all bins is: %f\n', sum(h));
We get:
Total sum over all bins is: 1.000000
Just to be absolutely sure you understand, this is the PDF of the image. What you did before was perform a data transformation where you transformed all image pixel intensities that conforms to a Gaussian distribution with a known mean and standard deviation. This will not give you a sum of 1 as you expect.
Remember that PDF is only the probability density function $p(x)$. Function which is restricted to range $[0, 1]$ is the integral over all domain of that function $\int_D p(x)dx$.
Refer to the Matlab manual, Y = normpdf(X,mu,sigma) computes the pdf at each of the values in X using the normal distribution with mean mu and standard deviation sigma.
The sum of the pdf is equal to 1.
The sum of the output is not.
I need to calculate the normalized colour histigram (in HSV colourspace) of an image. Using histcn in Matlab with 8 bins for Hue and 4 bins for each of Saturation and Value I get a 8x4x4 histogram. How could I normalize it?
Yes, #ASantosRibeiro is right, you just need to divide by number of elements like
HSV_hist ./ length(InitialData);
or
HSV_hist ./ sum(HSV_hist(:)); % To count the number of element use to create your hist.
I have a vector that I want to print a histogram of of data for. This data ranges from -100 to +100. The amount of data around the outer edges is insignificant and therefore I don't want to see it. I am most interested in showing data from -20 to +20.
1.) How can I limit that window to print on my histogram?
The amount of data I have at 0 outnumbers of the amount of data I have anywhere in the dataset by a minimum of 10:1. When I print the histogram, the layout of element frequency is lost because it is outnumbered by 0.
2.) Is there a way that I can scale the number of 0 values to be three times the number of -1 entries?
I am expecting an exponential drop of this dataset (in general) and therefore three times the frequency of -1 would easily allow me to see the frequency of the other data.
You can use something like
binCenters = -20:5:20;
[N,X] = hist(V,binCenters);
N = N./scalingVector;
bar(X(2:end-1),N(2:end-1));
Note that the code excludes the extremes of N and X from the bar plot, since they contain the number of values smaller than -20 and larger than 20. Also, by building scalingVector appropriately, you can scale N as you please.
You could also just toss out any values outside the [-20,20] range by using
subsetData=data(abs(data)<=20)
1) You can limit the histogram range you see on the plot by just setting the X axes limits:
xlim([-20 20])
Setting bins in hist command is good, but remember thatall the values outside the bins will fall into the most left and right bin. So you will need to set the axes limits anyway.
2) If there is a big difference between values in different bins, one way is to transform values on Y axes to log scale. Unfortunately just setting Y axes to log (set(gca,'YScale','log')) does not work for bar plot. Calculate the histogram with hist or histc (depending on whether you want to specify bins centers or edges) and log2 the values:
[y, xbin] = hist(data);
bar(xbin, log2(y) ,'hist')
Histogram has a few different methods of calling it. I strongly recommend you read the documentation on the function (doc hist)
What you are looking for is to put in a custom range in the histogram bin. It depends a bit on how many bins you want, but something like this will work.
Data=randn(1000,1)*20;
hist(Data,-20:20);
You could, if you want to, change the frequency of the binning as well. You could also change the axis so that you only focus on the range from -20 to 20, using a xaxis([-20 20]) command. You could also ignore the bin at 0, by using an yaxis and limiting the values to exclude the 0 bin. Without knowing what you want exactly, I can only give you suggestions.
I have raw observations of 500 numeric values (ranging from 1 to 25000) in a text file, I wish to make a frequency distribution in MATLAB. I did try the histogram (hist), however I would prefer a frequency distribution curve than blocks and bars.
Any help is appreciated !
If you pass two output parameters to HIST, you will get both the x-axis and y-axis values. Then you can plot the data as you like. For instance,
[counts, bins] = hist(mydata);
plot(bins, counts); %# get a line plot of the histogram
You could try Kernel smoothing density estimate