用PIL修剪掃描圖像？

Question

什麼是修剪目前正使用被輸入掃描儀，因此有一個大的白色/黑色區域的圖像的方法呢？

Answer 1

熵的解決方案似乎是有問題的，過於密集的計算。爲什麼不檢測邊緣？

我剛纔寫的Python代碼來解決這個同樣的問題我自己。我的背景是髒白的，所以我使用的標準是黑暗和顏色。我簡化了這個標準，只是取每個像素的R，B或B值中的最小值，這樣黑色或飽和紅色都顯得相同。我還使用了每行或每列最多最黑的像素的平均值。然後，我開始在每個邊緣，一直工作，直到我超過了門檻。

這裏是我的代碼：

#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

對於初學者來說，Here is a similar question。 Here is a related question。 And a another related question。

這裏只是一個想法，當然也有其他的方法。我想選擇任意作物邊緣，然後測量entropy *上線的任一側上，然後進行重新選擇作物線（可能使用類似一平分法），直到裁剪出部分的熵下降到低於定義的閾值。正如我想的那樣，您可能需要訴諸粗暴的尋根方法，因爲您不會很好地指示何時裁剪得太少。然後重複其餘3個邊緣。

*我記得發現，在引用網站的熵值法是不完全精確，但我找不到我的筆記

編輯（我敢肯定，這是一個SO後，但是。）： 對於圖像部分的「空白」（熵除外）的其他標準可能是邊緣檢測結果上的對比度或對比度比率。