I have black-and white picture like this:
Can anyone tell me if it's possible to calculate the area that black color occupies in this picture, so I know what's the percentage that black color has in the whole picture?
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$\begingroup$ What have you tried? $\endgroup$– C. E.Dec 8 '18 at 16:50
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$\begingroup$ Maybe ImageMeasurements[] and if not then ImageData[] $\endgroup$– Michael E2Dec 8 '18 at 17:09
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1$\begingroup$ Nitpick: you don't seek a surface (like a hyperboloid of one sheet), but an area (like 1.2 square millimetres). $\endgroup$– Andreas RejbrandDec 8 '18 at 23:41
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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
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Just use ImageHistogram with two levels:
ImageHistogram[myImage,2, FrameTicks->True]
or
Dimensions[SplitBy[ImageData[myImage], First]][[2;;3]]
(*
{886, 1342}
*)
answered Dec 8 '18 at 17:03
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$\begingroup$ So the percentage of black color is (886/1342)*100% $\endgroup$ Dec 8 '18 at 17:19
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For a binary image:
img = Import["https://i.stack.imgur.com/PdMDk.png"];
1 - ImageMeasurements[img, "MeanIntensity"]
(* 0.106198 *)
answered Dec 8 '18 at 18:30
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1$\begingroup$ Thanks to this I realized that
1 - Mean[img]works... $\endgroup$– C. E.Dec 9 '18 at 13:09 -