Create a histogram of data that is already "bincounts"

Question

Au. 2022-10-11

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1822123-create-a-histogram-of-data-that-is-already-bincounts

评论： Au. 2022-10-14

Dear Matlab community,

I have a set of counts, X_counts, supposed to reprensent the projection of counted particles over one dimension (X).

I would like to use the histogram function because i have in mind to fit then a Poisson distribution to this data set with histfit.

But as X_counts contains mainly "empty" counts, so zeros, my distribution and my histogram peak at 0 as one can see by plotting:

histogram(X_counts)

I tried defining the edges such as:

ctrX = edgesX(1:end-1)+ diff(edgesX)/2;
figure();plot(ctrX, histcounts(X_counts',edgesX));

At the end, i just think i should create another vector that has edgesX values * Xcounts (e.g., there is 400 times the value 31, 0 times the value 27, etc), but i was wondering if there would be something straightforward I could do with the current data and the histogram function, and that i would be missing. Meanwhile i will try fitting this data with fittype and "hold on" on the bar chart i created (attached image).

Thank you for your help!

4 个评论
显示 2更早的评论隐藏 2更早的评论

Star Strider 2022-10-11

在 MATLAB Online 中打开

I am not certain what you want to do, however I seriously doubt that fitting the Poisson distribution to your data is going to get you there —

counts = readmatrix('https://www.mathworks.com/matlabcentral/answers/uploaded_files/1151908/X_counts.txt');

edges = readmatrix('https://www.mathworks.com/matlabcentral/answers/uploaded_files/1151928/edgesX.txt');

poismdl = fitnlm(edges, counts.*edges, @(lambda,x)poisspdf(x,lambda), 10)

poismdl =

Nonlinear regression model: y ~ F(lambda,x) Estimated Coefficients: Estimate SE tStat pValue ________ __________ __________ _______ b1 57.82 6.8683e+06 8.4183e-06 0.99999 Number of observations: 121, Error degrees of freedom: 120 Root Mean Squared Error: 3.76e+03 R-Squared: -0.133, Adjusted R-Squared -0.133 F-statistic vs. zero model: 7.01e-08, p-value = 1

[ypred,yci] = predict(poismdl,edges);

figure

hb = bar(edges, counts.*edges, 'DisplayName','Histogram');

hold on

hyp = plot(edges, ypred, '-r', 'DisplayName','Nonlinear Fit');

hyci = plot(edges, yci, '--r', 'DisplayName', 'Confidence intervals');

hold off

grid

xlabel('Bins')

ylabel('Bins \times Counts')

legend([hb; hyp;hyci(1)],'Location','best')

It is only possible to use histfit on the actual data, not a histogram of it, so I use fitnlm here.

.

TADA 2022-10-11

histfit uses fitdist to fit distribution functions.

fitdist uses the actual data and not the histogram counts (like you have).

I'm afraid you're gonna have to either use curve fitting on the histogram you (you'll probably have to scale the histogram though), which is probably a bit more accurate than "simmulating" the original data like you suggested and using histfit.

Your best bet is if you can get a hold of the original measurements (not the summary), then use histfit like you planned originally.

Au. 2022-10-12

@Matt J, @Star Strider sorry if that wasn't clear. The bar chart I used is the expected outcome. I thought using the histogram and the histfit functions would be a better/simpler way to proceed, because I need to extract the most probable position of the hits. As @TADA said, this will work for the original data set, a collection of coordinates, where i can use the histogram function and probably the histfit then, but not on the counts (this is what I wasn't sure about).

@Star Strider I understand that the b1 coefficent is the 'lambda', so the most probable value of the distribution. Using counts*bins is for sure the most correct way to extract it, but the extracted value then cannot be used as the most probable position.

Thank you for taking the time to answer.

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

Steven Lord 2022-10-11

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1822123-create-a-histogram-of-data-that-is-already-bincounts#answer_1071448

在 MATLAB Online 中打开

Addressing just the question of plotting a histogram given bin counts and bin edges rather than the raw data, you can do this by specifying the BinCounts and BinEdges name-value arguments in your histogram call. I'm going to use histcounts to bin the data but if you have another way to bin the data you could use that instead.

x = randn(1, 1e5);

[counts, edges] = histcounts(x);

figure

histogram(x)

title('Let histogram bin the data')

figure

histogram('BinCounts', counts, 'BinEdges', edges)

title('Use the counts and edges from histcounts')

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

请先登录，再进行评论。

Answer 2

Image Analyst 2022-10-11

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1822123-create-a-histogram-of-data-that-is-already-bincounts#answer_1071798

Several of us (Matt and especially @Star Strider) have expressed doubts. I also do. I question the whole premise. I think this is a good example of a scenario like this

You ask "how do I do X?"

We show you how to do X.

You say "but that doesn't solve my problem."

We say, "what is your real problem."

You say "Well I want to do Y."

We say "Well if you want to do Y, you should not do X, you should do Z."

So you have multidimensional data. Let's say 2 dimensions, like an image of particles, and you have dimensions X and Y. So you take a projection over X, for example you get the average vertical profile going down the image by summing all pixels in each column for each row. You have a column vector of "rows" by 1 values where each value is the sum horizontally across all columns. So maybe row 1 had a sum of 1000 and column 2 had a sum of 500, and column 3 had a sum of 1400, etc. Now you're saying that you consider this a histogram, which is not an accurate description, and then you say you want to fit this to a Poisson distribution, which I think is the second problem (wrong thing to do). Why do you think this profile should follow a Poisson distribution? I see no justification for that assumption. Secondly, what are your values? You know that for lambda's more than 5 or 10, the Poisson is pretty much the same shape as a Gaussian. If your values are in the hundreds, your distribution would pretty much be Gaussian. If it were skewed, the distribution would probably be log-normal, and there is good theoretical basis for that in particle sizing theory. However I still think that you don't want to get a probability distribution from your profile/projection of your data along one dimension.

So let's say that you were able to turn your projection into a probability distribution (either by using the data itself or by taking a histogram of the projected/summed values), then what? What will you do with that information?

3 个评论
显示 1更早的评论隐藏 1更早的评论

Au. 2022-10-12

@Image Analyst: So what i want is the most probable position of the hits. A couple of comments to what you wrote: the pixel intensity distribution (in real life) or counts (in simulations) has an underlying distibution (that you can see on the bar plot). I called the sum of all counts in the simulated detector array, in each column for each row, X_counts (attached file). The histogram arrises from simulated data, which originally is a collection of coordinates, that I analysed to extract counts - to be able to compare to an intensity in real life. Now the distribution might not be a Poisson distribution, but the Gaussian distribution for sure is not, since the final distribution depends on the initial conditions in simulations. This is also supported by simulations that i did with a higher number of events. As far as i understand, the Poisson distribution applying for counting experiments (a number of events occuring in a fixed space, in my case). But yes, here my lambda is not a rate of events, which is then going agaisnt using the Poisson distrbution.

Image Analyst 2022-10-12

在 MATLAB Online 中打开

Not sure if you have an image or not, but the image itself is a "map" of the probability of hits. So wouldn't the "most probably position" simply be the weighted centroid of the "hits" (pixel gray levels)?

mask = true(size(grayImage));
props = regionprops(mask, grayImage, 'WeightedCentroid');
xCentroid = props.WeightedCentroid(1);
yCentroid = props.WeightedCentroid(2);

Yes, Poisson is used for counts, and if your counts are low, like less than 10 or 20, then I think the histogram of the whole image (taking each pixel value as a count), or whole data set, should give a Poisson distribution. If the counts are in the hundreds, then it will still be Poisson but look pretty much Gaussian, which might be more convenient mathematically for dealing with it.

However if you project (sum) along one dimension I don't know how that is still Poisson. For example, the sum of samples drawn from a Gaussian distribution is no longer a Gaussian distribution - it's chi square distribution. But maybe it is. I'm not a Ph.D. statistician. Maybe it's a compound Poisson distribution

https://en.wikipedia.org/wiki/Compound_Poisson_distribution

Au. 2022-10-14

Yes the weighted centroid is appropriate! I will use this :)

Thank you for the interesting input about the distributions - in my case, the final distribution depends on the initial parameters of the simulation (circular, conic, gaussian, etc. initial distributions).

请先登录，再进行评论。

Create a histogram of data that is already "bincounts"

4 个评论
显示 2更早的评论隐藏 2更早的评论

回答（2 个）

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

3 个评论
显示 1更早的评论隐藏 1更早的评论

另请参阅

类别

标签

产品

版本

Community Treasure Hunt

Create a histogram of data that is already "bincounts"

4 个评论 显示 2更早的评论隐藏 2更早的评论

回答（2 个）

0 个评论 显示 -2更早的评论隐藏 -2更早的评论

3 个评论 显示 1更早的评论隐藏 1更早的评论

另请参阅

类别

标签

产品

版本

Community Treasure Hunt

4 个评论
显示 2更早的评论隐藏 2更早的评论

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

3 个评论
显示 1更早的评论隐藏 1更早的评论