將附近的邊界框合並為一個

Question

我是 Python 新手，我正在使用快速入門：使用計算機視覺中的 REST API 和 Python 提取打印文本 (OCR)，以便在 Sales Fliers 中進行文本檢測。為每行文本繪制一個邊界框，它顯示在下一張圖像中。

但我想將附近的文本分組以具有單個分隔框架。 所以對於上圖的情況，它將有 2 個包含最接近文本的邊界框。

下面的代碼提供坐標 Ymin、XMax、Ymin 和 Xmax，並為每行文本繪制一個邊界框。

import requests
# If you are using a Jupyter notebook, uncomment the following line.
%matplotlib inline
import matplotlib.pyplot as plt
from matplotlib.patches import Rectangle
from PIL import Image
from io import BytesIO

# Replace <Subscription Key> with your valid subscription key.
subscription_key = "f244aa59ad4f4c05be907b4e78b7c6da"
assert subscription_key

vision_base_url = "https://westcentralus.api.cognitive.microsoft.com/vision/v2.0/"

ocr_url = vision_base_url + "ocr"

# Set image_url to the URL of an image that you want to analyze.
image_url = "https://cdn-ayb.akinon.net/cms/2019/04/04/e494dce0-1e80-47eb-96c9-448960a71260.jpg"

headers = {'Ocp-Apim-Subscription-Key': subscription_key}
params  = {'language': 'unk', 'detectOrientation': 'true'}
data    = {'url': image_url}
response = requests.post(ocr_url, headers=headers, params=params, json=data)
response.raise_for_status()

analysis = response.json()

# Extract the word bounding boxes and text.
line_infos = [region["lines"] for region in analysis["regions"]]
word_infos = []
for line in line_infos:
    for word_metadata in line:
        for word_info in word_metadata["words"]:
            word_infos.append(word_info)
word_infos

# Display the image and overlay it with the extracted text.
plt.figure(figsize=(100, 20))
image = Image.open(BytesIO(requests.get(image_url).content))
ax = plt.imshow(image)
texts_boxes = []
texts = []
for word in word_infos:
    bbox = [int(num) for num in word["boundingBox"].split(",")]
    text = word["text"]
    origin = (bbox[0], bbox[1])
    patch  = Rectangle(origin, bbox[2], bbox[3], fill=False, linewidth=3, color='r')
    ax.axes.add_patch(patch)
    plt.text(origin[0], origin[1], text, fontsize=2, weight="bold", va="top")
#     print(bbox)
    new_box = [bbox[1], bbox[0], bbox[1]+bbox[3], bbox[0]+bbox[2]]
    texts_boxes.append(new_box)
    texts.append(text)
#     print(text)
plt.axis("off")
texts_boxes = np.array(texts_boxes)
texts_boxes

輸出邊界框

array([[  68,   82,  138,  321],
       [ 202,   81,  252,  327],
       [ 261,   81,  308,  327],
       [ 364,  112,  389,  182],
       [ 362,  192,  389,  305],
       [ 404,   98,  421,  317],
       [  92,  421,  146,  725],
       [  80,  738,  134, 1060],
       [ 209,  399,  227,  456],
       [ 233,  399,  250,  444],
       [ 257,  400,  279,  471],
       [ 281,  399,  298,  440],
       [ 286,  446,  303,  458],
       [ 353,  394,  366,  429]]

但我想通過近距離合並。

Answer 1

您可以在運行代碼之前使用openCV以及Apply dilation和blackhat轉換來處理圖像

Answer 2

謝謝@recnac，您的算法可以幫助我解決它。

我的解決方案是這樣。 生成一個新框，合並距離很近的文本框以獲得一個新框。 在其中有一個封閉的文字。

#Distance definition  between text to be merge
dist_limit = 40

#Copy of the text and object arrays
texts_copied = copy.deepcopy(texts)
texts_boxes_copied = copy.deepcopy(texts_boxes)


#Generate two text boxes a larger one that covers them
def merge_boxes(box1, box2):
    return [min(box1[0], box2[0]), 
         min(box1[1], box2[1]), 
         max(box1[2], box2[2]),
         max(box1[3], box2[3])]



#Computer a Matrix similarity of distances of the text and object
def calc_sim(text, obj):
    # text: ymin, xmin, ymax, xmax
    # obj: ymin, xmin, ymax, xmax
    text_ymin, text_xmin, text_ymax, text_xmax = text
    obj_ymin, obj_xmin, obj_ymax, obj_xmax = obj

    x_dist = min(abs(text_xmin-obj_xmin), abs(text_xmin-obj_xmax), abs(text_xmax-obj_xmin), abs(text_xmax-obj_xmax))
    y_dist = min(abs(text_ymin-obj_ymin), abs(text_ymin-obj_ymax), abs(text_ymax-obj_ymin), abs(text_ymax-obj_ymax))

    dist = x_dist + y_dist
    return dist

#Principal algorithm for merge text 
def merge_algo(texts, texts_boxes):
    for i, (text_1, text_box_1) in enumerate(zip(texts, texts_boxes)):
        for j, (text_2, text_box_2) in enumerate(zip(texts, texts_boxes)):
            if j <= i:
                continue
            # Create a new box if a distances is less than disctance limit defined 
            if calc_sim(text_box_1, text_box_2) < dist_limit:
            # Create a new box  
                new_box = merge_boxes(text_box_1, text_box_2)            
             # Create a new text string 
                new_text = text_1 + ' ' + text_2

                texts[i] = new_text
                #delete previous text 
                del texts[j]
                texts_boxes[i] = new_box
                #delete previous text boxes
                del texts_boxes[j]
                #return a new boxes and new text string that are close
                return True, texts, texts_boxes

    return False, texts, texts_boxes


need_to_merge = True

#Merge full text 
while need_to_merge:
    need_to_merge, texts_copied, texts_boxes_copied = merge_algo(texts_copied, texts_boxes_copied)

texts_copied

Answer 3

您可以檢查兩個方框（ x_min ， x_max ， y_min ， y_max ）的y_max ，如果差異小於close_dist ，則應將它們合並到一個新的方框中。 然后在兩個for循環中連續執行此操作：

from itertools import product

close_dist = 20

# common version
def should_merge(box1, box2):
    for i in range(2):
        for j in range(2):
            for k in range(2):
                if abs(box1[j * 2 + i] - box2[k * 2 + i]) <= close_dist:
                    return True, [min(box1[0], box2[0]), min(box1[1], box2[1]), max(box1[2], box2[2]),
                                  max(box1[3], box2[3])]
    return False, None


# use product, more concise
def should_merge2(box1, box2):
    a = (box1[0], box1[2]), (box1[1], box1[3])
    b = (box2[0], box2[2]), (box2[1], box2[3])

    if any(abs(a_v - b_v) <= close_dist for i in range(2) for a_v, b_v in product(a[i], b[i])):
        return True, [min(*a[0], *b[0]), min(*a[1], *b[1]), max(*a[0], *b[0]), max(*a[1], *b[1])]

    return False, None

def merge_box(boxes):
    for i, box1 in enumerate(boxes):
        for j, box2 in enumerate(boxes[i + 1:]):
            is_merge, new_box = should_merge(box1, box2)
            if is_merge:
                boxes[i] = None
                boxes[j] = new_box
                break

    boxes = [b for b in boxes if b]
    print(boxes)

測試代碼：

boxes = [[68, 82, 138, 321],
         [202, 81, 252, 327],
         [261, 81, 308, 327],
         [364, 112, 389, 182],
         [362, 192, 389, 305],
         [404, 98, 421, 317],
         [92, 421, 146, 725],
         [80, 738, 134, 1060],
         [209, 399, 227, 456],
         [233, 399, 250, 444],
         [257, 400, 279, 471],
         [281, 399, 298, 440],
         [286, 446, 303, 458],
         [353, 394, 366, 429]]

print(merge_box(boxes))

輸出：

[[286, 394, 366, 458], [261, 81, 421, 327], [404, 98, 421, 317], [80, 738, 134, 1060], [353, 394, 366, 429]]

無法做視覺測試，請為我測試。

希望對您有所幫助，如有其他問題，請發表評論。 :)

Answer 4

嗨，我想你的問題會用easyocr解決

導入easyocr

reader = easyocr.Reader(['en'])

result = reader.readtext('image_name.jpg',paragraph=True)

打印（結果）