在扫描的文档中拆分文本行
- 2025-03-11 08:50:00
- admin 原创
- 90
问题描述:
我正在尝试找到一种方法来打破已自适应阈值化的扫描文档中文本行的分割。现在,我将文档的像素值存储为 0 到 255 之间的无符号整数,并取每行像素的平均值,然后根据像素值的平均值是否大于 250 将行分割成范围,然后取符合该范围的每行的中位数。但是,这种方法有时会失败,因为图像上可能会有黑色斑点。
有没有更耐噪音的方法来完成这项任务?
编辑:这里有一些代码。“warped”是原始图像的名称,“cuts”是我想要分割图像的地方。
warped = threshold_adaptive(warped, 250, offset = 10)
warped = warped.astype("uint8") * 255
# get areas where we can split image on whitespace to make OCR more accurate
color_level = np.array([np.sum(line) / len(line) for line in warped])
cuts = []
i = 0
while(i < len(color_level)):
if color_level[i] > 250:
begin = i
while(color_level[i] > 250):
i += 1
cuts.append((i + begin)/2) # middle of the whitespace region
else:
i += 1
编辑 2:添加了示例图像
解决方案 1:
从输入图像中,您需要将文本设为白色,将背景设为黑色
然后,您需要计算钞票的旋转角度。一种简单的方法是找到minAreaRect
所有白点的 ( findNonZero
),然后您得到:
然后你可以旋转你的钞票,使文字水平:
现在您可以计算水平投影(reduce
)。您可以取每行的平均值。th
在直方图上应用一个阈值来解释图像中的一些噪声(这里我使用了0
,即没有噪声)。只有背景的行将有一个值>0
,文本行将0
在直方图中具有值。然后取直方图中每个连续的白色箱序列的平均箱坐标。这将是y
您的线条的坐标:
这是代码。它是用 C++ 编写的,但由于大部分工作都是使用 OpenCV 函数,因此应该很容易转换为 Python。至少,你可以将其用作参考:
#include <opencv2/opencv.hpp>
using namespace cv;
using namespace std;
int main()
{
// Read image
Mat3b img = imread("path_to_image");
// Binarize image. Text is white, background is black
Mat1b bin;
cvtColor(img, bin, COLOR_BGR2GRAY);
bin = bin < 200;
// Find all white pixels
vector<Point> pts;
findNonZero(bin, pts);
// Get rotated rect of white pixels
RotatedRect box = minAreaRect(pts);
if (box.size.width > box.size.height)
{
swap(box.size.width, box.size.height);
box.angle += 90.f;
}
Point2f vertices[4];
box.points(vertices);
for (int i = 0; i < 4; ++i)
{
line(img, vertices[i], vertices[(i + 1) % 4], Scalar(0, 255, 0));
}
// Rotate the image according to the found angle
Mat1b rotated;
Mat M = getRotationMatrix2D(box.center, box.angle, 1.0);
warpAffine(bin, rotated, M, bin.size());
// Compute horizontal projections
Mat1f horProj;
reduce(rotated, horProj, 1, CV_REDUCE_AVG);
// Remove noise in histogram. White bins identify space lines, black bins identify text lines
float th = 0;
Mat1b hist = horProj <= th;
// Get mean coordinate of white white pixels groups
vector<int> ycoords;
int y = 0;
int count = 0;
bool isSpace = false;
for (int i = 0; i < rotated.rows; ++i)
{
if (!isSpace)
{
if (hist(i))
{
isSpace = true;
count = 1;
y = i;
}
}
else
{
if (!hist(i))
{
isSpace = false;
ycoords.push_back(y / count);
}
else
{
y += i;
count++;
}
}
}
// Draw line as final result
Mat3b result;
cvtColor(rotated, result, COLOR_GRAY2BGR);
for (int i = 0; i < ycoords.size(); ++i)
{
line(result, Point(0, ycoords[i]), Point(result.cols, ycoords[i]), Scalar(0, 255, 0));
}
return 0;
}
解决方案 2:
基本步骤如@Miki,
阅读源代码
脱粒
找到最小区域矩形
通过旋转矩阵扭曲
找到并绘制上限和下限
Python 中的代码如下:
#!/usr/bin/python3
# 2018.01.16 01:11:49 CST
# 2018.01.16 01:55:01 CST
import cv2
import numpy as np
## (1) read
img = cv2.imread("img02.jpg")
gray = cv2.cvtColor(img, cv2.COLOR_BGR2GRAY)
## (2) threshold
th, threshed = cv2.threshold(gray, 127, 255, cv2.THRESH_BINARY_INV|cv2.THRESH_OTSU)
## (3) minAreaRect on the nozeros
pts = cv2.findNonZero(threshed)
ret = cv2.minAreaRect(pts)
(cx,cy), (w,h), ang = ret
if w>h:
w,h = h,w
ang += 90
## (4) Find rotated matrix, do rotation
M = cv2.getRotationMatrix2D((cx,cy), ang, 1.0)
rotated = cv2.warpAffine(threshed, M, (img.shape[1], img.shape[0]))
## (5) find and draw the upper and lower boundary of each lines
hist = cv2.reduce(rotated,1, cv2.REDUCE_AVG).reshape(-1)
th = 2
H,W = img.shape[:2]
uppers = [y for y in range(H-1) if hist[y]<=th and hist[y+1]>th]
lowers = [y for y in range(H-1) if hist[y]>th and hist[y+1]<=th]
rotated = cv2.cvtColor(rotated, cv2.COLOR_GRAY2BGR)
for y in uppers:
cv2.line(rotated, (0,y), (W, y), (255,0,0), 1)
for y in lowers:
cv2.line(rotated, (0,y), (W, y), (0,255,0), 1)
cv2.imwrite("result.png", rotated)
最终结果:
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