用 PIL 修剪扫描的图像？

Question

如何修剪使用扫描仪输入并因此具有大的白色/黑色区域的图像？

Answer 1

熵解似乎有问题并且计算量过大。为什么不进行边缘检测？

我刚刚编写了这段 python 代码来为自己解决同样的问题。我的背景是肮脏的白色，所以我使用的标准是黑暗和颜色。我通过只为每个像素取最小的 R、B 或 B 值来简化这个标准，这样黑色或饱和红色都一样突出。我还使用了每一行或每一列的最暗像素的平均值。然后我从每个边缘开始，一直努力，直到我越过一个门槛。

这是我的代码：

#these values set how sensitive the bounding box detection is
threshold = 200     #the average of the darkest values must be _below_ this to count (0 is darkest, 255 is lightest)
obviousness = 50    #how many of the darkest pixels to include (1 would mean a single dark pixel triggers it)

from PIL import Image

def find_line(vals):
    #implement edge detection once, use many times 
    for i,tmp in enumerate(vals):
        tmp.sort()
        average = float(sum(tmp[:obviousness]))/len(tmp[:obviousness])
        if average <= threshold:
            return i
    return i    #i is left over from failed threshold finding, it is the bounds

def getbox(img):
    #get the bounding box of the interesting part of a PIL image object
    #this is done by getting the darekest of the R, G or B value of each pixel
    #and finding were the edge gest dark/colored enough
    #returns a tuple of (left,upper,right,lower)

    width, height = img.size    #for making a 2d array
    retval = [0,0,width,height] #values will be disposed of, but this is a black image's box 

    pixels = list(img.getdata())
    vals = []                   #store the value of the darkest color
    for pixel in pixels:
        vals.append(min(pixel)) #the darkest of the R,G or B values

    #make 2d array
    vals = np.array([vals[i * width:(i + 1) * width] for i in xrange(height)])

    #start with upper bounds
    forupper = vals.copy()
    retval[1] = find_line(forupper)

    #next, do lower bounds
    forlower = vals.copy()
    forlower = np.flipud(forlower)
    retval[3] = height - find_line(forlower)

    #left edge, same as before but roatate the data so left edge is top edge
    forleft = vals.copy()
    forleft = np.swapaxes(forleft,0,1)
    retval[0] = find_line(forleft)

    #and right edge is bottom edge of rotated array
    forright = vals.copy()
    forright = np.swapaxes(forright,0,1)
    forright = np.flipud(forright)
    retval[2] = width - find_line(forright)

    if retval[0] >= retval[2] or retval[1] >= retval[3]:
        print "error, bounding box is not legit"
        return None
    return tuple(retval)

if __name__ == '__main__':
    image = Image.open('cat.jpg')
    box = getbox(image)
    print "result is: ",box
    result = image.crop(box)
    result.show()

Answer 2

对于初学者，Here is a similar question。 Here is a related question。 And a another related question.

这只是一个想法，当然还有其他方法。我会选择任意裁切边缘，然后测量线两侧的entropy*，然后继续重新选择裁切线（可能使用类似二等分的方法），直到裁切部分的熵下降低于定义的阈值。正如我认为的那样，您可能需要采用粗暴的寻根方法，因为您无法很好地指示何时收割得太少。然后对剩余的 3 条边重复此操作。

*我记得发现引用网站中的熵方法并不完全准确，但我找不到我的笔记（不过我确信它是在 SO 帖子中。）

编辑： 图像部分“空白”的其他标准（熵除外）可能是对比度或边缘检测结果的对比度。