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 Android教程網 >> Android技術 >> 關於Android編程 >> android L Palette 實現原理

android L Palette 實現原理

編輯:關於Android編程

有時候,看到一些界面上的色彩,心情可能會很舒暢,有時候,看到一些其他色彩,就覺得很討厭,不爽,看到android L Palette 從圖片中提取篩選出來的顏色,覺得都挺好看的,就去了解了下Palette調色板。

看了代碼,根據我的理解,大概說說主要的步驟:

第一步,將圖片縮小,再整個過程中,可以降低計算量和減少內存的使用,跟不縮小也能達到一樣的效果

 

/**
     * Scale the bitmap down so that it's smallest dimension is
     * {@value #CALCULATE_BITMAP_MIN_DIMENSION}px. If {@code bitmap} is smaller than this, than it
     * is returned.
     */
    private static Bitmap scaleBitmapDown(Bitmap bitmap) {
        final int minDimension = Math.min(bitmap.getWidth(), bitmap.getHeight());

        if (minDimension <= CALCULATE_BITMAP_MIN_DIMENSION) {
            // If the bitmap is small enough already, just return it
            return bitmap;
        }

        final float scaleRatio = CALCULATE_BITMAP_MIN_DIMENSION / (float) minDimension;
        return Bitmap.createScaledBitmap(bitmap,
                Math.round(bitmap.getWidth() * scaleRatio),
                Math.round(bitmap.getHeight() * scaleRatio),
                false);
    }

 

 

第二步,將縮小後的圖片數據,放在一個int 數組裡

 

 /**
     * Factory-method to generate a {@link ColorCutQuantizer} from a {@link Bitmap} object.
     *
     * @param bitmap Bitmap to extract the pixel data from
     * @param maxColors The maximum number of colors that should be in the result palette.
     */
    static ColorCutQuantizer fromBitmap(Bitmap bitmap, int maxColors) {
        final int width = bitmap.getWidth();
        final int height = bitmap.getHeight();


        final int[] pixels = new int[width * height];
        bitmap.getPixels(pixels, 0, width, 0, 0, width, height);


        return new ColorCutQuantizer(new ColorHistogram(pixels), maxColors);
    }

 

 

第三步,將這個int 數組由小到大排序,就相當於,將一張圖片一樣的顏色堆在一起,然後計算共有多少種顏色,每種顏色它是多大,這些是在一個叫ColorHistogram(顏色直方圖)類裡面計算的,用顏色直方圖來說,就是共有多少柱顏色,每柱顏色有多高

 

 

/**
 * Class which provides a histogram for RGB values.
 */
final class ColorHistogram {

    private final int[] mColors;
    private final int[] mColorCounts;
    private final int mNumberColors;

    /**
     * A new {@link ColorHistogram} instance.
     *
     * @param pixels array of image contents
     */
    ColorHistogram(final int[] pixels) {
        // Sort the pixels to enable counting below
        Arrays.sort(pixels);

        // Count number of distinct colors
        mNumberColors = countDistinctColors(pixels);

        // Create arrays
        mColors = new int[mNumberColors];
        mColorCounts = new int[mNumberColors];

        // Finally count the frequency of each color
        countFrequencies(pixels);
    }

    /**
     * @return 獲取共用多少柱不同顏色 number of distinct colors in the image.
     */
    int getNumberOfColors() {
        return mNumberColors;
    }

    /**
     * @return 獲取排好序後的不同顏色的數組  an array containing all of the distinct colors in the image.
     */
    int[] getColors() {
        return mColors;
    }

    /**
     * @return 獲取保存每一柱有多高的數組 an array containing the frequency of a distinct colors within the image.
     */
    int[] getColorCounts() {
        return mColorCounts;
    }

    //計算共用多少柱不同顏色
    private static int countDistinctColors(final int[] pixels) {
        if (pixels.length < 2) {
            // If we have less than 2 pixels we can stop here
            return pixels.length;
        }

        // If we have at least 2 pixels, we have a minimum of 1 color...
        int colorCount = 1;
        int currentColor = pixels[0];

        // Now iterate from the second pixel to the end, counting distinct colors
        for (int i = 1; i < pixels.length; i++) {
            // If we encounter a new color, increase the population
            if (pixels[i] != currentColor) {
                currentColor = pixels[i];
                colorCount++;
            }
        }

        return colorCount;
    }
   
    //計算每一柱有多高
    private void countFrequencies(final int[] pixels) {
        if (pixels.length == 0) {
            return;
        }

        int currentColorIndex = 0;
        int currentColor = pixels[0];

        mColors[currentColorIndex] = currentColor;
        mColorCounts[currentColorIndex] = 1;

        Log.i(pixels.length,+ pixels.length);
       
        if (pixels.length == 1) {
            // If we only have one pixel, we can stop here
            return;
        }
       
        // Now iterate from the second pixel to the end, population distinct colors
        for (int i = 1; i < pixels.length; i++) {
            if (pixels[i] == currentColor) {
                // We've hit the same color as before, increase population
                mColorCounts[currentColorIndex]++;
            } else {
                // We've hit a new color, increase index
                currentColor = pixels[i];

                currentColorIndex++;
                mColors[currentColorIndex] = currentColor;
                mColorCounts[currentColorIndex] = 1;
            }
        }
    }

}

 

第四步,將各種顏色,根據RGB轉HSL算法,得出對應的HSL(H: Hue 色相,S:Saturation 飽和度L Lightness 明度),根據特定的條件,比如是明度L是否接近白色,黑色,還有一個判斷叫isNearRedILine,解釋是@return true if the color lies close to the red side of the I line(接近紅色私密區域附近?).,然後根據這三個條件,過濾掉這些顏色,什麼是HSL和RGB轉HSL算法可以查看下百科,比較有詳細說明

 

 

 /**
     * Private constructor.
     *
     * @param colorHistogram histogram representing an image's pixel data
     * @param maxColors The maximum number of colors that should be in the result palette.
     */
    private ColorCutQuantizer(ColorHistogram colorHistogram, int maxColors) {
        final int rawColorCount = colorHistogram.getNumberOfColors();
        final int[] rawColors = colorHistogram.getColors();//顏色數組
        final int[] rawColorCounts = colorHistogram.getColorCounts();//對應rawColors每一個顏色數組的大小

        // First, lets pack the populations into a SparseIntArray so that they can be easily
        // retrieved without knowing a color's index
        mColorPopulations = new SparseIntArray(rawColorCount);
        for (int i = 0; i < rawColors.length; i++) {
            mColorPopulations.append(rawColors[i], rawColorCounts[i]);
        }

        // Now go through all of the colors and keep those which we do not want to ignore
        mColors = new int[rawColorCount];
        int validColorCount = 0;
        for (int color : rawColors) {
            if (!shouldIgnoreColor(color)) {
                mColors[validColorCount++] = color;
            }
        }
        Log.d(mColors length, +mColors.length);
        if (validColorCount <= maxColors) {
            // The image has fewer colors than the maximum requested, so just return the colors
            mQuantizedColors = new ArrayList();
           
            for (final int color : mColors) {
                mQuantizedColors.add(new Swatch(color, mColorPopulations.get(color)));
            }
        } else {
            // We need use quantization to reduce the number of colors
            mQuantizedColors = quantizePixels(validColorCount - 1, maxColors);
        }
    }

這裡截了張圖看看

style=font-weight:

 

 

第五步,根據是各種亮度,飽和度的取值范圍,比如有活力的暗色,有活力的亮色,柔和的顏色,柔和的暗色,柔和的亮色,找到對應的顏色

 

 private Swatch findColor(float targetLuma, float minLuma, float maxLuma,
                             float targetSaturation, float minSaturation, float maxSaturation) {
        Swatch max = null;
        float maxValue = 0f;

        for (Swatch swatch : mSwatches) {
            final float sat = swatch.getHsl()[1];
            final float luma = swatch.getHsl()[2];

            if (sat >= minSaturation && sat <= maxSaturation &&
                    luma >= minLuma && luma <= maxLuma &&
                    !isAlreadySelected(swatch)) {
                float thisValue = createComparisonValue(sat, targetSaturation, luma, targetLuma,
                        swatch.getPopulation(), mHighestPopulation);
                if (max == null || thisValue > maxValue) {
                    max = swatch;
                    maxValue = thisValue;
                }
            }
        }

        return max;
    }

看下效果圖

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