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Histogram function plots different colours from those requested

159 visualizaciones (últimos 30 días)
I"m trying to color-code each of 4 histograms using a predefined color scheme. Here's a minimal working example:
colours = [0 32 96;
192 0 0;
71 207 255;
255 143 143;
] / 255;
for i=1:4
x = randn(1,100);
subplot(1,4, i)
values = histogram(x, 'FaceColor', colours (i, :));
end
However, in the image I'm getting, the colors are actually (slightly) different, for instance for the first histogram I get (102,121,160) instead of (0,32,96):

Respuesta aceptada

Steven Lord
Steven Lord el 28 de Mzo. de 2022
The histogram object has a FaceAlpha property, so that if you overlay two histogram objects you can see the one underneath. If you change that FaceAlpha property the color appears slightly different and may be closer to what you want.
x = randn(1, 100);
figure
h = histogram(x, 'FaceColor', 'r');
title("Histograph with FaceAlpha " + h.FaceAlpha)
figure
h = histogram(x, 'FaceColor', 'r', 'FaceAlpha', 1);
title("Histograph with FaceAlpha " + h.FaceAlpha)

Más respuestas (1)

Scott MacKenzie
Scott MacKenzie el 28 de Mzo. de 2022
Editada: Scott MacKenzie el 28 de Mzo. de 2022
The histogram function uses a face alpha of 0.6 by default. That's why the colors appear a bit lighter than the values from your eye dropper app. If you set the face alpha to 1.0 (see below), then the displayed colors will agree with the eye dropper values.
colours = [0 32 96;
192 0 0;
71 207 255;
255 143 143;
] / 255;
for i=1:4
x = randn(1,100);
subplot(2,4, i)
values = histogram(x,'FaceColor', colours(i, :));
values.FaceAlpha = 1.0; % NOTE: default is 0.6
values.FaceColor * 255
end
ans = 1×3
0 32 96
ans = 1×3
192 0 0
ans = 1×3
71 207 255
ans = 1×3
255 143 143
  3 comentarios
Scott MacKenzie
Scott MacKenzie el 28 de Mzo. de 2022
Editada: Scott MacKenzie el 28 de Mzo. de 2022
@z8080, hmm, right your are. I dug a bit deeper and discovered why there is a discrepancy between the set colors and the eye dropper colors. See my modified answer for the details.
Seems as I was modifying my answer, @Steven Lord posted an answer. Same story.
z8080
z8080 el 29 de Mzo. de 2022
Indeed, but thanks again Scott!

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