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| Dear Mr. Bill, Thanks for your prompt reply. It works fine. There was additional square brackets in my code after the manipulate. Kind Regards. |
| The attached notebook should help you get started. Check the `Manipulate` documentation to see how to control labels and slider visibility. |
| From the documentation: > If an explicit threshold value is given, Binarize ignores the Method option. » so either use Manipulate[Binarize[yourImage, binOnly], {binOnly, 0, 1}] or Manipulate[ Binarize[yourImage, ... |
| Sorry, I don't understand your question. |
| I am not sure what you mean, but here is my attempt: data = {{1, 1, 0.45098039215686275`}, {1, 2, 0.47058823529411764`}, {1, 3, 0.49019607843137253`}, {1, 4, 0.5019607843137255`}, {1, 5, 0.5333333333333333`}, {1, 6, ... |
| Using the example from `ImageData_x_y_z_Coordinates.nb` {yRange, xRange} = imData // Dimensions Table[{x, y, imData[[y, x]]}, {x, xRange}, {y, yRange}] // Flatten[#, 1] & evaluates to a list of `{x, y, z}` where `z` is the pixel value... |
| First your image is added: img = ![enter image description here][1] Looking at the Dimensions you can see it has 3 color channels: ImageData[img] // Dimensions Therefore it is converted to Grayscale. img2 = ColorConvert[img,... |
| Dear Members, I’m trying to get the coordinates of the ImageData, in term of X and Y separately I have paste my trials and would gratefully need your help. cc = ColorConvert[imageOfPalmTree, "Grayscale"]; ImageDimensions[cc] imageCoordinates =... |
| Welcome to Wolfram Community! Please make sure you know the rules: https://wolfr.am/READ-1ST > It appears but both are sliding together... **Please post complete reproducible code of what you are trying to do.** The rules explain how to... |