将OCRed非结构化文本转换为正确的文本

Question

我在MODI中使用Microsoft VB6来映像图片。 （我知道其他OCR工具，如tesseract等，但我发现MODI比其他更准确）

要OCR的图像就像这样

并且，我在OCR之后得到的文字如下所示

Text1
Text2
Text3
Number1
Number2
Number3

这里的问题是不保持来自相对列的相应文本。如何将Number1与Text1映射？

我只能想到这样的解决方案。

MODI提供所有OCRed单词的坐标，如

LeftPos = Img.Layout.Words(0).Rects(0).Left
TopPos = Img.Layout.Words(0).Rects(0).Top

因此，为了对齐同一行中的单词，我们可以匹配每个单词的TopPos，然后按LeftPos对它们进行排序。我们将获得完整的产品线。所以我循环遍历所有单词并将它们的文本以及左侧和顶部存储在mysql表中。然后运行此查询

SELECT group_concat(word ORDER BY `left` SEPARATOR ' ')
FROM test_copy
GROUP BY `top`

我的问题是，每个单词的顶部位置并不完全相同，显然会出现几个像素差异。

我尝试添加DIV 5，用于合并5像素范围内的单词但在某些情况下不起作用。我也尝试在node.js中通过计算每个单词的容差然后按LeftPos进行排序，但我仍然觉得这不是最好的方法。

更新： js代码完成了这项工作，但除了Number1有5个像素差异且Text2在该行中没有对应的情况之外。

有没有更好的主意呢？

Answer 1

我不是100％确定你如何识别“左”栏中的那些单词，但是一旦你识别出那个单词，你就可以通过不只是顶部坐标而是整个单位来查找其中的其他单词。横跨（顶部和底部）的矩形。确定与其他单词的重叠（交集）。请注意下面标有红色的区域。

这是您可以用来检测某些内容是否在同一行中的容差。如果某些东西仅与一个像素重叠，那么它可能来自较低或较高的线。但如果它与高度“Text1”的50％或更多重叠，那么它可能在同一条线上。

---

示例SQL，用于根据atop和bottom coord查找“line”中的所有单词

select 
    word.id, word.Top, word.Left, word.Right, word.Bottom 
from 
    word
where 
    (word.Top >= @leftColWordTop and word.Top <= @leftColWordBottom)
    or (word.Bottom >= @leftColWordTop  and word.Bottom <= @leftColWordBottom)

---

用于计算行数的psuedo VB6代码示例。

'assume words is a collection of WordInfo objects with an Id, Top, 
'   Left, Bottom, Right properties filled in, and a LineAnchorWordId 
'   property that has not been set yet.

'get the words in left-to-right order
wordsLeftToRight = SortLeftToRight(words) 

'also get the words in top-to-bottom order
wordsTopToBottom = SortTopToBottom(words) 

'pass through identifying a line "anchor", that being the left-most 
'   word that starts (and defines) a line
for each anchorWord in wordsLeftToRight

    'check if the word has been mapped to aline yet by checking if 
    '   its anchor property has been set yet.  This assumes 0 is not 
    '   a valid id, use -1 instead if needed
    if anchorWord.LineAnchorWordId = 0 then 

        'not locate every word on this line, as bounded by the 
        '   anchorWord.  every word determined to be on this line 
        '   gets its LineAnchorWordId property set to the Id of the 
        '   anchorWord
        for each lineWord in wordsTopToBottom

            if lineWord.Bottom < anchorWord.Top Then

                'skip it,it is above the line (but keep searching down
                '   because we haven't reached the anchorWord location yet)

            else if lineWord.Top > anchorWord.Bottom Then

                'skip it,it is below the line, and exit the search 
                '   early since all the rest will also be below the line
                exit for

            else if OverlapsWithinTolerance(anchorWord, lineWord) then

                lineWord.LineAnchorWordId = anchorWord.Id

            endif

        next

    end if

next anchorWord

'at this point, every word has been assigned a LineAnchorWordId, 
'   and every word on the same line will have a matching LineAnchorWordId
'   value.  If stored in a DB you can now group them by LineAnchorWordId 
' and sort them by their Left coord to get your output.