JavaScript-将OCRed非结构化文本转换为正确的文本

我在VB6中使用Microsoft MODI对图像进行OCR. (我知道其他OCR工具,例如tesseract等,但我发现MODI比其他工具更准确)

像是OCRed的图像是这样的

而且,OCR之后收到的文本如下

Text1
Text2
Text3
Number1
Number2
Number3

这里的问题是,来自相反列的相应文本无法保留.如何将Number1与Text1映射？

我只能想到这样的解决方案.

MODI这样提供所有OCRed单词的坐标

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

因此,要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对其进行排序.我们将获得完整的生产线.因此,我遍历了所有单词,并将它们的文本以及左和顶部存储在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都不完全相同,显然会有几个像素差异.

我尝试添加DIV 5,以合并5像素范围内的单词,但在某些情况下不起作用.我还尝试通过计算每个单词的容差然后按LeftPos排序来在node.js中做到这一点,但我仍然觉得这不是最好的方法.

更新：js代码可以完成此工作,但Number1的像素差异为5像素而Text2的该行中没有对应像素的情况除外.

有什么更好的主意吗？

解决方法:

我不确定100％确定如何识别“左”列中的单词,但是一旦识别出该单词,您不仅可以投影“顶”坐标,还可以投影整个矩形(顶部和底部).用其他词确定重叠(相交).注意下面红色标记的区域.

这是您可以用来检测某物是否在同一行中的公差.如果某些东西仅重叠一个像素,那么它可能来自较低或较高的线.但是,如果重叠的高度等于“文字1”的高度的50％或更多,那么它很可能在同一行上.

示例SQL,根据上下坐标查找“行”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)

示例psuedo VB6代码也可以计算行数.

'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.

标签：node-js,ocr,vb6,javascript,modi
来源： https://codeday.me/bug/20191029/1963077.html