Azzera filtri
Azzera filtri

histogram plot is emplty

5 visualizzazioni (ultimi 30 giorni)
Selina Loh
Selina Loh il 1 Lug 2017
Commentato: Star Strider il 1 Lug 2017
I have errors in displaying the histogram for a matrix (3648x5472). The histogram displayed is blank. The matrix value is in uint8.

Risposta accettata

Image Analyst
Image Analyst il 1 Lug 2017
You can use imhist, or you can also use histogram(). They will both work. What is probably happening is that your V is having every one of its pixels replaced by true or false (1 or 0). So there are only two bins. Now one of them, the zero bin, is so high, and it's at the very left of the chart that it looks like the Y axis. If you use imhist() you will have 256 bins and the 0 and 1 bin will be so small and at the edge of the chart box that you probably don't notice them. If you use histogram, you should see two fat bars - one at 0 and one at 1.

Più risposte (1)

Star Strider
Star Strider il 1 Lug 2017
You probably need to use the imhist function, since your uint8 array may be an image.
This works for me:
M = randi(intmax('uint8'), 3648, 5472, 'uint8'); % Create Matrix
H = imhist(M);
figure(1)
bar(H)
  2 Commenti
Selina Loh
Selina Loh il 1 Lug 2017
V(I3>254) = 255; V(I3<=254) = 0; V = uint8(reshape(V,[3648,5472]));
imhist(V)
I am trying to display a historgram for a black and white image. I cannot directly use the imhist function?
Star Strider
Star Strider il 1 Lug 2017
Knowing it was a binary image would have helped. If you want to know the number of true or ‘1’ values, you do not need a histogram. Simply count the ‘1’ values, and if you want the fraction that are ‘1’, divide by the total number of elements in the binary image:
BinaryImage = randi([0 1], 3648, 5472, 'uint8'); % Create Binary Image
NrEq_1 = nnz(BinaryImage); % Number Of ‘1’ Values
Frac_1 = NrEq_1/numel(BinaryImage); % Fraction Of Total That Are ‘1’ Values

Accedi per commentare.

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by