I'm trying to check the full range of values on the x-axis of a histogram, especially while I expected the full range to be [0, 255], when I used the following command for the histogram [h, bins] = hist(H), I got the following:
h =
221 20 6 4 1 1 2 0 0 1
bins =
Columns 1 through 7
8.2500 24.7500 41.2500 57.7500 74.2500 90.7500 107.2500
Columns 8 through 10
123.7500 140.2500 156.7500
This implies that the maximum range I got here is up to 165.
If we look at the histogram below, we can see that 165 seems to be the maximum number of frequency value. How do I know the maximum value (range) of the x-axis?
I think you need either one of these 3 options:
max(H)
Or
numel(unique(H))
or
numel(H)
I would start at the top till you find the one you need.
Related
i have a question regarding histc:
I choose the max and min of a sorted signal as my range.
ma = ssigPE(end);
mi = ssigPE(1);
range = mi:ma;
[bincountsO,indO2] = histc(ssigPE, range);
so the range i get back is:
range = [-1.097184703736132 -0.097184703736132 0.902815296263868]
my problem is that just 2 bins get develop, so bincountsO has 2 bins
and indO2 has values as 0, 1 and 2
What am I doing wrong? I guess I m using the range wrong. I read the text here:
http://de.mathworks.com/help/matlab/ref/histc.html#inputarg_binranges
but I don't get it.
The bin ranges tell you where do bins start and stop. So a value of [0 1 2 7]for example, will give 3 bins: [0 1] , [1 2] , [2 7]
In matlab if you do mi:ma it will create an array from the value mi to ma with a step of 1. With your values, that gives just 3 values, hence 2 bins. There are 2 ways of creating a given step size length vectors.
Step size if 100 as an example
range=mi:(ma-mi)/100:ma;
alternatively, and way clearer
range=linspace(mi,ma,100)
I am trying to do a density plot for a data containing two columns with different ranges. The RMSD column is [0-2] and Angle is [0-200] ranges.
My data in the file is like this:
0.0225370 37.088
0.1049553 35.309
0.0710002 33.993
0.0866880 34.708
0.0912664 33.011
0.0932054 33.191
0.1083590 37.276
0.1104145 34.882
0.1027977 34.341
0.0896688 35.991
0.1047578 36.457
0.1215936 38.914
0.1105484 35.051
0.0974138 35.533
0.1390955 33.601
0.1333878 32.133
0.0933365 35.714
0.1200465 33.038
0.1155794 33.694
0.1125247 34.522
0.1181806 37.890
0.1291700 38.871
I want both x and y axis to be binned 1/10th of the range
The 0 of both the axis to be starting in the same
Print the number of elements in each grid of the matrix like this and make a density plot based on these number of elements
0 0.1 0.2 (RMSD)
0 0 1 3
20 2 0 4
40 1 0 5
60 0 0 2
(Angle)
I can find ways to do 1-D binning but then I am stumped about how to make a density plot from those values and havent even dared to attempt2-D binning + plotting.
Thanks for the help
I think you want hist3. Assuming you want to specifty bin edges (not bin centers), use
result = hist3(data, 'Edges', {[0 .1 .2], [0 20 40 60]}).';
where data denotes your data.
From the linked documentation:
hist3(X,'Edges',edges), where edges is a two-element cell array of numeric vectors with monotonically non-decreasing values, uses a 2-D grid of bins with edges at edges{1} in the first dimension and at edges{2} in the second. The (i,j)th bin includes the value X(k,:) if
edges{1}(i) <= X(k,1) < edges{1}(i+1)
edges{2}(j) <= X(k,2) < edges{2}(j+1)
With your example data this gives
result =
0 0 0
8 14 0
0 0 0
0 0 0
For those who don't have Statistics and Machine Learning Toolbox to run bivariate histogram (hist3), it may be more practical using an alternative to solve 2-D hist problem. The following function generates the same output
function N = hist3_alt(x,y,edgesX,edgesY)
N = zeros(length(edgesY)-1,length(edgesX)-1);
[~,~,binX] = histcounts(x,edgesX);
for ii=1:numel(edgesX)-1
N(:,ii) = (histcounts(y(binX==ii),edgesY))';
end
It's simple and efficient. Then you could run the function like this:
N = hist3_alt(x,y,[0:0.1:2],[0:20:200])
Here is my dataset:
pressure(time, case, x, y)
>> size(pressure)
ans =
100 1 289 570
How to get a spatial nanmean pressure for x from 30 to 60 and y from 40 to 70 in each time step?
For example: a nanmean value for that particular region for each timestep from time 1 to time 100.
I tried this, "spatial_mean_pressure = nanmean(pressure(:,:,30:60,40:70))" It averaged the pressure in the timeserie. This is not the result I want.
>> size(spatial_mean_pressure)
ans =
1 1 31 31
I like to get the results like this:
>> size(spatial_mean_pressure)
ans =
100 1 1 1
You are trying to get a mean for an entire block of matrix. Therefore, you should apply nanmean twice and not once. Also, apply it along a particular dimension to get the desired result. I think this is what you want.
x=randi(10,[100 1 10 25]);
First take the mean along the third dimension.
mean_x_3=nanmean(x,3);
You would get an answer of size = [100 1 1 25]. Then take the mean along 4th dimension.
mean_x_4=nanmean(mean_x_3,4);
This should give you the desired answer. You can write this in one line as,
mean_x = nanmean(nanmean(x,3),4);
I have got the following results after using [h, bins] = hist(H) in matlab:
h =
221 20 6 4 1 1 2 0 0 1
bins =
Columns 1 through 7
8.2500 24.7500 41.2500 57.7500 74.2500 90.7500 107.2500
Columns 8 through 10
123.7500 140.2500 156.7500
How do I know the full range of values? Especially that I expected to have up to 255, that is `[0,255], and if we analyze the range of the bins below we will have the following for the ten bins respectively:
0-16.5
16.5-33
33-49.5
49.5-66
66-82.5
82.5-99
99-115.5
115.5-132
132-148.5
148.5-165
So, did I get this range only due to having only 10 bins?
Thanks.
yes, the 10 bins are the default of hist. If you know you might have values between [0,255] you can force whatever bin positions you want, for example:
[h, bins] = hist(H,0:255)
will create 256 bins each for each integer value [0,255]
I got the following results after applying:[h,bins]=hist(data), such that, the data will contain the LBP (Local Binary Pattern) values.
h =
221 20 6 4 1 1 2 0 0 1
bins =
Columns 1 through 7
8.2500 24.7500 41.2500 57.7500 74.2500 90.7500 107.2500
Columns 8 through 10
123.7500 140.2500 156.7500
I want to ask the following:
Does the first bin represent the values 0-8.25 and the second bin the values 8.26-24.75, and so forth?
For the h value 221, does it mean that we have computed 221 an LBP value ranging from 0-8.25?
1) No. The bin location is in the center value of the bin, that is, for the first bin the values are 0-16.5, the second bin is 16.5-33, etc. Use histc if it is more natural to specify bin edges instead of centers.
2) h(1)=221 means that from your entire data set (that has 256 elements according to your question), 221 elements had values ranging between 0-16.5 .