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I am doing my PhD in Clinical Neuropsychology; I had someone helping me analyse images (.jpg pictures) in Mathematica, but it turns out this person made some really bad decisions, and well, now I have to do it myself. But, I have no idea.

Basically, what I want is to analyze the JPEG image, get a RGB histogram, and get the values for the mean and standard deviation of this histogram.

Using ImageHistogram [[*image*], Appearance, -> "Separated"] works fine, but I can't figure out the next step, turning the histogram plot into numbers.

How can I do that?

Thank you for helping!

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    $\begingroup$ You could use ImageData[] to get the values being binned for the histograms... $\endgroup$ Oct 12, 2012 at 13:27
  • $\begingroup$ Thank you for responding!!I am trying it right now! $\endgroup$
    – onemonkey
    Oct 12, 2012 at 13:31
  • $\begingroup$ Hmmm... it says A very large out put was generated, here is a sample of it. "{0., 0., 0.}, {0., 0., 0.}, {0., 0., 0.}," $\endgroup$
    – onemonkey
    Oct 12, 2012 at 13:33
  • $\begingroup$ And if I input: StandardDeviation[Out[37]] I get pretty much the same result. $\endgroup$
    – onemonkey
    Oct 12, 2012 at 13:34
  • $\begingroup$ You might need the Interleaving option... $\endgroup$ Oct 12, 2012 at 13:40

2 Answers 2

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There is one suggestion.

img = ExampleData[{"TestImage", "Lena"}];
imgChannelsRGB = Transpose@Flatten[ImageData[img], 1];

Creating Histogram:

GraphicsRow[Histogram /@ imgChannelsRGB, ImageSize -> 500]

enter image description here

Geting statistics:

TableForm[Through[{Mean, StandardDeviation}[#]] & /@ imgChannelsRGB, 
 TableHeadings -> {{"Red", "Blue", "Green"}, {"Mean", "StDev"}}]

enter image description here

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  • $\begingroup$ Alternatively: imgChannelsRGB = Flatten /@ ImageData[img, Interleaving -> False]. The Lenna example image is easily accessed as ExampleData[{"TestImage", "Lena"}]. $\endgroup$ Oct 12, 2012 at 14:00
  • $\begingroup$ Nice tip. Changed the example. $\endgroup$
    – Murta
    Oct 12, 2012 at 14:03
  • $\begingroup$ Am trying it right now! $\endgroup$
    – onemonkey
    Oct 12, 2012 at 14:09
  • $\begingroup$ O wow, this is very promissing! I'm going to play with this (and then I might be back). Thnx! $\endgroup$
    – onemonkey
    Oct 12, 2012 at 14:43
  • $\begingroup$ You could use ImageData[img] ~Flatten~ {{3}, {1, 2}} in place of Transpose@Flatten[ImageData[img], 1], eliminating Transpose. $\endgroup$
    – Mr.Wizard
    Oct 12, 2012 at 15:19
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I thought it might be interesting to combine the graphs and statistics in small multiple style, but I got stuck... How do you present two numbers as a label without showing the curly brackets?

small multiple

Code:

{r, g, b} = ColorSeparate[img];
Row[
 Labeled[
    ImageHistogram[#, Appearance -> "Transparent",  Joined -> False, 
     Frame -> None],
    {
     Mean[Flatten[ImageData[#]]],
     StandardDeviation[Flatten[ImageData[#]]]
     },
    ImageSize -> 200
    ] & /@ {r, g, b}
 ]

Edit with bill (?!)'s suggestion:

GraphicsRow[
 Labeled[
    ImageHistogram[#, Appearance -> "Transparent", Joined -> False, 
     Frame -> None],
    Style[
     Column[
      {
       Mean[Flatten[ImageData[#]]],
       StandardDeviation[Flatten[ImageData[#]]]
       }
      ], 
     14, Bold], 
    ImageSize -> 500
    ] & /@ {r, g, b}]

better small multiple

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    $\begingroup$ Maybe add a Column ? Row[Labeled[ ImageHistogram[#, Appearance -> "Transparent", Joined -> False, Frame -> None], Column[{Mean[Flatten[ImageData[#]]], StandardDeviation[Flatten[ImageData[#]]]}], ImageSize -> 200] & /@ {r, g, b}] $\endgroup$ Oct 12, 2012 at 19:26
  • $\begingroup$ @b.gatessucks Good idea! As usual, your comments are better than my answers! :) $\endgroup$
    – cormullion
    Oct 12, 2012 at 19:30
  • $\begingroup$ Wow! I have been getting some help from a friend, and we will definitely come back to this! $\endgroup$
    – onemonkey
    Oct 15, 2012 at 9:46
  • $\begingroup$ OK, so we are on a roll, but now I have a new, buyt related qustion. Should i post it here or open a new topic? $\endgroup$
    – onemonkey
    Oct 19, 2012 at 8:54

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