How to calculate surface of specific color?












3














I have black-and white picture like this
enter image description here
Can anyone tell me is it possible to calculate surface that black color occupies in this picture, so I know what's the percentage that black color has in whole picture.






color image







share|improve this question






















  • What have you tried?
    – C. E.
    Dec 8 '18 at 16:50










  • Maybe ImageMeasurements and if not then ImageData
    – Michael E2
    Dec 8 '18 at 17:09






  • 1




    Nitpick: you don't seek a surface (like a hyperboloid of one sheet), but an area (like 1.2 square millimetres).
    – Andreas Rejbrand
    Dec 8 '18 at 23:41
















3














I have black-and white picture like this
enter image description here
Can anyone tell me is it possible to calculate surface that black color occupies in this picture, so I know what's the percentage that black color has in whole picture.






color image







share|improve this question






















  • What have you tried?
    – C. E.
    Dec 8 '18 at 16:50










  • Maybe ImageMeasurements and if not then ImageData
    – Michael E2
    Dec 8 '18 at 17:09






  • 1




    Nitpick: you don't seek a surface (like a hyperboloid of one sheet), but an area (like 1.2 square millimetres).
    – Andreas Rejbrand
    Dec 8 '18 at 23:41














3












3








3







I have black-and white picture like this
enter image description here
Can anyone tell me is it possible to calculate surface that black color occupies in this picture, so I know what's the percentage that black color has in whole picture.






color image







share|improve this question













I have black-and white picture like this
enter image description here
Can anyone tell me is it possible to calculate surface that black color occupies in this picture, so I know what's the percentage that black color has in whole picture.






color image




color image






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Dec 8 '18 at 16:13









Cro Simpson2.0

784




784












  • What have you tried?
    – C. E.
    Dec 8 '18 at 16:50










  • Maybe ImageMeasurements and if not then ImageData
    – Michael E2
    Dec 8 '18 at 17:09






  • 1




    Nitpick: you don't seek a surface (like a hyperboloid of one sheet), but an area (like 1.2 square millimetres).
    – Andreas Rejbrand
    Dec 8 '18 at 23:41


















  • What have you tried?
    – C. E.
    Dec 8 '18 at 16:50










  • Maybe ImageMeasurements and if not then ImageData
    – Michael E2
    Dec 8 '18 at 17:09






  • 1




    Nitpick: you don't seek a surface (like a hyperboloid of one sheet), but an area (like 1.2 square millimetres).
    – Andreas Rejbrand
    Dec 8 '18 at 23:41
















What have you tried?
– C. E.
Dec 8 '18 at 16:50




What have you tried?
– C. E.
Dec 8 '18 at 16:50












Maybe ImageMeasurements and if not then ImageData
– Michael E2
Dec 8 '18 at 17:09




Maybe ImageMeasurements and if not then ImageData
– Michael E2
Dec 8 '18 at 17:09




1




1




Nitpick: you don't seek a surface (like a hyperboloid of one sheet), but an area (like 1.2 square millimetres).
– Andreas Rejbrand
Dec 8 '18 at 23:41




Nitpick: you don't seek a surface (like a hyperboloid of one sheet), but an area (like 1.2 square millimetres).
– Andreas Rejbrand
Dec 8 '18 at 23:41










3 Answers
3






active

oldest

votes


















6














After reading Michael E2's answer, I realized that one can simply do



1 - Mean[img]



0.106198




There are several other solutions as well. There is a function called ImageLevels that counts the channels:



img = Import["https://i.stack.imgur.com/PdMDk.png"];

levels = ImageLevels[img]



{{0, 521982}, {1, 4393218}}




levels[[1, 2]]/(levels[[1, 2]] + levels[[2, 2]]) // N



0.106198




One could also use



neg = ColorNegate[img];

Total[neg, 2]/(Total[img, 2] + Total[neg, 2])



0.106198




or



{w, h} = ImageDimensions[img];

1 - Total[img, 2]/(w h)



0.106198




One could also explicitly get the matrix of image pixels:



pixels = Flatten@ImageData[img];

1 - Total[pixels]/Length[pixels] // N



0.106198







share|improve this answer































    5














    Just use ImageHistogram with two levels:



    ImageHistogram[myImage,2, FrameTicks->True]


    enter image description here



    or



    Dimensions[SplitBy[ImageData[myImage], First]][[2;;3]]


    (*



    {886, 1342}



    *)






    share|improve this answer























    • So the percentage of black color is (886/1342)*100%
      – Cro Simpson2.0
      Dec 8 '18 at 17:19



















    4














    For a binary image:



    img = Import["https://i.stack.imgur.com/PdMDk.png"];
    
1 - ImageMeasurements[img, "MeanIntensity"]
    
(*  0.106198  *)





    share|improve this answer

















    • 1




      Thanks to this I realized that 1 - Mean[img] works...
      – C. E.
      Dec 9 '18 at 13:09










    • @C.E. Cool. I didn't know that.
      – Michael E2
      Dec 9 '18 at 13:15











    Your Answer





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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    6














    After reading Michael E2's answer, I realized that one can simply do



    1 - Mean[img]



    0.106198




    There are several other solutions as well. There is a function called ImageLevels that counts the channels:



    img = Import["https://i.stack.imgur.com/PdMDk.png"];
    
levels = ImageLevels[img]



    {{0, 521982}, {1, 4393218}}




    levels[[1, 2]]/(levels[[1, 2]] + levels[[2, 2]]) // N



    0.106198




    One could also use



    neg = ColorNegate[img];
    
Total[neg, 2]/(Total[img, 2] + Total[neg, 2])



    0.106198




    or



    {w, h} = ImageDimensions[img];
    
1 - Total[img, 2]/(w h)



    0.106198




    One could also explicitly get the matrix of image pixels:



    pixels = Flatten@ImageData[img];
    
1 - Total[pixels]/Length[pixels] // N



    0.106198







    share|improve this answer




























      6














      After reading Michael E2's answer, I realized that one can simply do



      1 - Mean[img]



      0.106198




      There are several other solutions as well. There is a function called ImageLevels that counts the channels:



      img = Import["https://i.stack.imgur.com/PdMDk.png"];
      
levels = ImageLevels[img]



      {{0, 521982}, {1, 4393218}}




      levels[[1, 2]]/(levels[[1, 2]] + levels[[2, 2]]) // N



      0.106198




      One could also use



      neg = ColorNegate[img];
      
Total[neg, 2]/(Total[img, 2] + Total[neg, 2])



      0.106198




      or



      {w, h} = ImageDimensions[img];
      
1 - Total[img, 2]/(w h)



      0.106198




      One could also explicitly get the matrix of image pixels:



      pixels = Flatten@ImageData[img];
      
1 - Total[pixels]/Length[pixels] // N



      0.106198







      share|improve this answer


























        6












        6








        6






        After reading Michael E2's answer, I realized that one can simply do



        1 - Mean[img]



        0.106198




        There are several other solutions as well. There is a function called ImageLevels that counts the channels:



        img = Import["https://i.stack.imgur.com/PdMDk.png"];
        
levels = ImageLevels[img]



        {{0, 521982}, {1, 4393218}}




        levels[[1, 2]]/(levels[[1, 2]] + levels[[2, 2]]) // N



        0.106198




        One could also use



        neg = ColorNegate[img];
        
Total[neg, 2]/(Total[img, 2] + Total[neg, 2])



        0.106198




        or



        {w, h} = ImageDimensions[img];
        
1 - Total[img, 2]/(w h)



        0.106198




        One could also explicitly get the matrix of image pixels:



        pixels = Flatten@ImageData[img];
        
1 - Total[pixels]/Length[pixels] // N



        0.106198







        share|improve this answer














        After reading Michael E2's answer, I realized that one can simply do



        1 - Mean[img]



        0.106198




        There are several other solutions as well. There is a function called ImageLevels that counts the channels:



        img = Import["https://i.stack.imgur.com/PdMDk.png"];
        
levels = ImageLevels[img]



        {{0, 521982}, {1, 4393218}}




        levels[[1, 2]]/(levels[[1, 2]] + levels[[2, 2]]) // N



        0.106198




        One could also use



        neg = ColorNegate[img];
        
Total[neg, 2]/(Total[img, 2] + Total[neg, 2])



        0.106198




        or



        {w, h} = ImageDimensions[img];
        
1 - Total[img, 2]/(w h)



        0.106198




        One could also explicitly get the matrix of image pixels:



        pixels = Flatten@ImageData[img];
        
1 - Total[pixels]/Length[pixels] // N



        0.106198








        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited Dec 9 '18 at 13:07

























        answered Dec 8 '18 at 17:21









        C. E.

        49.9k397202




        49.9k397202























            5














            Just use ImageHistogram with two levels:



            ImageHistogram[myImage,2, FrameTicks->True]


            enter image description here



            or



            Dimensions[SplitBy[ImageData[myImage], First]][[2;;3]]


            (*



            {886, 1342}



            *)






            share|improve this answer























            • So the percentage of black color is (886/1342)*100%
              – Cro Simpson2.0
              Dec 8 '18 at 17:19
















            5














            Just use ImageHistogram with two levels:



            ImageHistogram[myImage,2, FrameTicks->True]


            enter image description here



            or



            Dimensions[SplitBy[ImageData[myImage], First]][[2;;3]]


            (*



            {886, 1342}



            *)






            share|improve this answer























            • So the percentage of black color is (886/1342)*100%
              – Cro Simpson2.0
              Dec 8 '18 at 17:19














            5












            5








            5






            Just use ImageHistogram with two levels:



            ImageHistogram[myImage,2, FrameTicks->True]


            enter image description here



            or



            Dimensions[SplitBy[ImageData[myImage], First]][[2;;3]]


            (*



            {886, 1342}



            *)






            share|improve this answer














            Just use ImageHistogram with two levels:



            ImageHistogram[myImage,2, FrameTicks->True]


            enter image description here



            or



            Dimensions[SplitBy[ImageData[myImage], First]][[2;;3]]


            (*



            {886, 1342}



            *)







            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited Dec 8 '18 at 17:14

























            answered Dec 8 '18 at 17:03









            David G. Stork

            23.3k22051




            23.3k22051












            • So the percentage of black color is (886/1342)*100%
              – Cro Simpson2.0
              Dec 8 '18 at 17:19


















            • So the percentage of black color is (886/1342)*100%
              – Cro Simpson2.0
              Dec 8 '18 at 17:19
















            So the percentage of black color is (886/1342)*100%
            – Cro Simpson2.0
            Dec 8 '18 at 17:19




            So the percentage of black color is (886/1342)*100%
            – Cro Simpson2.0
            Dec 8 '18 at 17:19











            4














            For a binary image:



            img = Import["https://i.stack.imgur.com/PdMDk.png"];
            
1 - ImageMeasurements[img, "MeanIntensity"]
            
(*  0.106198  *)





            share|improve this answer

















            • 1




              Thanks to this I realized that 1 - Mean[img] works...
              – C. E.
              Dec 9 '18 at 13:09










            • @C.E. Cool. I didn't know that.
              – Michael E2
              Dec 9 '18 at 13:15
















            4














            For a binary image:



            img = Import["https://i.stack.imgur.com/PdMDk.png"];
            
1 - ImageMeasurements[img, "MeanIntensity"]
            
(*  0.106198  *)





            share|improve this answer

















            • 1




              Thanks to this I realized that 1 - Mean[img] works...
              – C. E.
              Dec 9 '18 at 13:09










            • @C.E. Cool. I didn't know that.
              – Michael E2
              Dec 9 '18 at 13:15














            4












            4








            4






            For a binary image:



            img = Import["https://i.stack.imgur.com/PdMDk.png"];
            
1 - ImageMeasurements[img, "MeanIntensity"]
            
(*  0.106198  *)





            share|improve this answer












            For a binary image:



            img = Import["https://i.stack.imgur.com/PdMDk.png"];
            
1 - ImageMeasurements[img, "MeanIntensity"]
            
(*  0.106198  *)






            share|improve this answer












            share|improve this answer



            share|improve this answer










            answered Dec 8 '18 at 18:30









            Michael E2

            145k11195464




            145k11195464








            • 1




              Thanks to this I realized that 1 - Mean[img] works...
              – C. E.
              Dec 9 '18 at 13:09










            • @C.E. Cool. I didn't know that.
              – Michael E2
              Dec 9 '18 at 13:15














            • 1




              Thanks to this I realized that 1 - Mean[img] works...
              – C. E.
              Dec 9 '18 at 13:09










            • @C.E. Cool. I didn't know that.
              – Michael E2
              Dec 9 '18 at 13:15








            1




            1




            Thanks to this I realized that 1 - Mean[img] works...
            – C. E.
            Dec 9 '18 at 13:09




            Thanks to this I realized that 1 - Mean[img] works...
            – C. E.
            Dec 9 '18 at 13:09












            @C.E. Cool. I didn't know that.
            – Michael E2
            Dec 9 '18 at 13:15




            @C.E. Cool. I didn't know that.
            – Michael E2
            Dec 9 '18 at 13:15


















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