我是python中的新手,我正在使用Quickstart:使用REST API和计算机视觉中的Python提取打印文本(OCR),用于销售飞行器中的文本检测。因此,该算法具有坐标Ymin,XMax,Ymin和Xmax以及为每行文本绘制一个边界框,它显示在下一个图像中。
但我想将附近的文本分组以具有单个分隔框架。所以对于上面图像的情况,它将有2个包含最接近文本的边界框。
下面的代码提供坐标Ymin,XMax,Ymin和Xmax,并为每行文本绘制一个边界框。
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
输出边界框
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]]
但我想近距离合并。
在运行代码之前,您可以使用openCV和Apply dilation和blackhat变换来处理图像
谢谢@recnac你的算法帮我解决了。
我的解决方案是这样。生成一个新框,将文本框与近距离合并以获得新框。其中有一个密切的文本。
#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
你可以检查两个盒子(x_min
,x_max
,y_min
,y_max
)的boudary,如果差异小于close_dist
,他们应该合并到一个新的盒子。然后在两个for
循环中继续这样做:
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)
测试代码:
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))
输出:
[[286, 394, 366, 458], [261, 81, 421, 327], [404, 98, 421, 317], [80, 738, 134, 1060], [353, 394, 366, 429]]
不能做视力测试,请为我测试。
希望对您有所帮助,并在您有其他问题时发表评论。 :)