I want to compose three single exposures into an RGB image.

I have three 2D arrays that represent each exposure and I can combine them into a 3D array, specifying the RGB values for each pixel. My problem is that the third exposure is not bright enough, and I want to make it brighter.

If I want to make an Image brighter, I just use Lighter[image] and that sorts it. But in this case, I want to increase the brightness of only one of the three exposures, so I think I need to work with the array.

In the documentation, they state:

ImageAdjust[Image[data]] is equivalent to Image[Rescale[data]]

How do I implement it?

What I have done:

Code for reference:

 {imageR = array5000/Max[array5000],
  imageG = array1000/Max[array1000],
  imageB = array100/Max[array100],
 rgb = Table[{imageR[[i, j]], imageG[[i, j]], imageB[[i, j]]}, {i, 
    pixels}, {j, pixels}];
 Export["buddhaBrot.jpg", Image[rgb]]

The result of


is too dark, whereas I would like it to be like

  • $\begingroup$ Wait ... if you know that you want Lighter@Ligther@Image[array5000/Max[array5000]]. ... why don't you just do that? $\endgroup$ – Dr. belisarius Mar 22 '16 at 13:38
  • $\begingroup$ Also, what are array5000 etc.? $\endgroup$ – dr.blochwave Mar 22 '16 at 13:39
  • $\begingroup$ Also, you can use the arguments of ImageAdjust[image, {contrast, brightness, gamma}] $\endgroup$ – dr.blochwave Mar 22 '16 at 13:39
  • $\begingroup$ BTW ... Have you seen this mathematica.stackexchange.com/q/9342/193 ? $\endgroup$ – Dr. belisarius Mar 22 '16 at 13:39
  • $\begingroup$ Perhaps you mean ImageExposureCombine $\endgroup$ – yode Mar 22 '16 at 13:41

First, convert your data arrays to images with Image[].

If your imageR and imageG have good brightness, and imageB is too dark:

meanIntensity = Mean[ImageMeasurements[#, "MeanIntensity"]& /@ {imageR, imageG}]
badIntensity = ImageMeasurements[imageB, "MeanIntensity"]
imageBadj = ImageMultiply[imageB, meanIntensity/badIntensity]

combinedImage = ColorCombine[{imageR, imageG, imageBadj}]

should give you a color image with all channels appropriately bright.

  • 1
    $\begingroup$ It's not what I asked, but turns out you knew what I wanted better than me :) $\endgroup$ – Andrea Mar 23 '16 at 15:09

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