How to find the rectangle with those text

In this image:

I want to find this rectangle with all contents(without those grid's lines)

I think the total of the pixels can help a little like:

img = Import["http://i.stack.imgur.com/lnd5w.png"];

data = Total[ImageData[ImageRotate[ColorNegate[img]]]];

Show[ImageRotate[img], 

 ListLinePlot[data, PlotRange -> All, PlotStyle -> Red]]

And I have such images:

image2

image3

image4

image5

I have no idea to do this thing. Does anyone have an idea?

edited Jan 3 at 10:52

asked Dec 28 '18 at 13:57

yode

10.4k234103

1

$begingroup$
Have you asked this question before? It seems very familiar to me...
$endgroup$
– Carl Lange
Dec 28 '18 at 14:36

$begingroup$
Ah, I was thinking of this and this. Just for context purposes :)
$endgroup$
– Carl Lange
Dec 28 '18 at 14:37

$begingroup$
@CarlLange It's not very similar...
$endgroup$
– yode
Dec 28 '18 at 14:37

$begingroup$
The question isn't but the images are, so I wanted to make sure it wasn't a duplicate somehow. That's all!
$endgroup$
– Carl Lange
Dec 28 '18 at 15:10

add a comment |

In this image:

I want to find this rectangle with all contents(without those grid's lines)

I think the total of the pixels can help a little like:

img = Import["http://i.stack.imgur.com/lnd5w.png"];

data = Total[ImageData[ImageRotate[ColorNegate[img]]]];

Show[ImageRotate[img], 

 ListLinePlot[data, PlotRange -> All, PlotStyle -> Red]]

And I have such images:

image2

image3

image4

image5

I have no idea to do this thing. Does anyone have an idea?

edited Jan 3 at 10:52

asked Dec 28 '18 at 13:57

yode

10.4k234103

1

$begingroup$
Have you asked this question before? It seems very familiar to me...
$endgroup$
– Carl Lange
Dec 28 '18 at 14:36

$begingroup$
Ah, I was thinking of this and this. Just for context purposes :)
$endgroup$
– Carl Lange
Dec 28 '18 at 14:37

$begingroup$
@CarlLange It's not very similar...
$endgroup$
– yode
Dec 28 '18 at 14:37

$begingroup$
The question isn't but the images are, so I wanted to make sure it wasn't a duplicate somehow. That's all!
$endgroup$
– Carl Lange
Dec 28 '18 at 15:10

add a comment |

In this image:

I want to find this rectangle with all contents(without those grid's lines)

I think the total of the pixels can help a little like:

img = Import["http://i.stack.imgur.com/lnd5w.png"];

data = Total[ImageData[ImageRotate[ColorNegate[img]]]];

Show[ImageRotate[img], 

 ListLinePlot[data, PlotRange -> All, PlotStyle -> Red]]

And I have such images:

image2

image3

image4

image5

I have no idea to do this thing. Does anyone have an idea?

edited Jan 3 at 10:52

asked Dec 28 '18 at 13:57

yode

10.4k234103

In this image:

I want to find this rectangle with all contents(without those grid's lines)

I think the total of the pixels can help a little like:

img = Import["http://i.stack.imgur.com/lnd5w.png"];

data = Total[ImageData[ImageRotate[ColorNegate[img]]]];

Show[ImageRotate[img], 

 ListLinePlot[data, PlotRange -> All, PlotStyle -> Red]]

And I have such images:

image2

image3

image4

image5

I have no idea to do this thing. Does anyone have an idea?

image-processing signal-processing

edited Jan 3 at 10:52

asked Dec 28 '18 at 13:57

yode

10.4k234103

edited Jan 3 at 10:52

asked Dec 28 '18 at 13:57

yode

10.4k234103

edited Jan 3 at 10:52

asked Dec 28 '18 at 13:57

yode

10.4k234103

asked Dec 28 '18 at 13:57

yode

10.4k234103

asked Dec 28 '18 at 13:57

yode

10.4k234103

1

$begingroup$
Have you asked this question before? It seems very familiar to me...
$endgroup$
– Carl Lange
Dec 28 '18 at 14:36

$begingroup$
Ah, I was thinking of this and this. Just for context purposes :)
$endgroup$
– Carl Lange
Dec 28 '18 at 14:37

$begingroup$
@CarlLange It's not very similar...
$endgroup$
– yode
Dec 28 '18 at 14:37

$begingroup$
The question isn't but the images are, so I wanted to make sure it wasn't a duplicate somehow. That's all!
$endgroup$
– Carl Lange
Dec 28 '18 at 15:10

add a comment |

1

$begingroup$
Have you asked this question before? It seems very familiar to me...
$endgroup$
– Carl Lange
Dec 28 '18 at 14:36

$begingroup$
Ah, I was thinking of this and this. Just for context purposes :)
$endgroup$
– Carl Lange
Dec 28 '18 at 14:37

$begingroup$
@CarlLange It's not very similar...
$endgroup$
– yode
Dec 28 '18 at 14:37

$begingroup$
The question isn't but the images are, so I wanted to make sure it wasn't a duplicate somehow. That's all!
$endgroup$
– Carl Lange
Dec 28 '18 at 15:10

Have you asked this question before? It seems very familiar to me...

– Carl Lange
Dec 28 '18 at 14:36

Ah, I was thinking of this and this. Just for context purposes :)

– Carl Lange
Dec 28 '18 at 14:37

@CarlLange It's not very similar...

– yode
Dec 28 '18 at 14:37

The question isn't but the images are, so I wanted to make sure it wasn't a duplicate somehow. That's all!

– Carl Lange
Dec 28 '18 at 15:10

add a comment |

2 Answers
2

active

oldest

votes

Binarization and ComponentMeasurements seems the straightforward solution:

img = Import["https://i.stack.imgur.com/lnd5w.png"];

Selct all components that are less than half as long as the image:

comp = ComponentMeasurements[MorphologicalBinarize[ColorNegate@img], 

   "BoundingBox", #CaliperLength < Min[ImageDimensions[img]]*0.5 &];

This selects all digits and boxes, but not the grid lines, because they are longer than height/2:

HighlightImage[img, Rectangle @@@ comp[[All, 2]]]

enter image description here

Then combine the individual bounding boxes to one big bounding box:

HighlightImage[img, 

 Rectangle @@ 

  Transpose[MinMax /@ Transpose[Flatten[comp[[All, 2]], 1]]]]

enter image description here

You might have to adjust binarization and the criteria in ComponentMeasurements, but for me, it worked on all the images you posted on the first try.

answered Dec 28 '18 at 15:12

Niki Estner

31.1k376133

$begingroup$
Sorry for catching a cold to delay response. Your method is very simple to understand. But your method cannot deal with the image5. This is why I want to do it with signal-processing method...
$endgroup$
– yode
Jan 3 at 10:54

$begingroup$
@yode: Sure it can, you just need to adjust the criteria function in ComponentMeasurements.
$endgroup$
– Niki Estner
Jan 3 at 11:13

add a comment |

This works quite well:

findRegion[img_] := TextRecognize[img, "BoundingBox", RecognitionPrior -> "SparseText"]



HighlightImage[#, findRegion@#] & /@ imgs

As you can see, the result for the first image is not perfect unfortunately, but hopefully this gives you some ideas.

answered Dec 28 '18 at 14:13

Lukas Lang

7,31511032

$begingroup$
Yes, as your try, the Tesseract not very stable..
$endgroup$
– yode
Dec 28 '18 at 14:40

add a comment |

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

Binarization and ComponentMeasurements seems the straightforward solution:

img = Import["https://i.stack.imgur.com/lnd5w.png"];

Selct all components that are less than half as long as the image:

comp = ComponentMeasurements[MorphologicalBinarize[ColorNegate@img], 

   "BoundingBox", #CaliperLength < Min[ImageDimensions[img]]*0.5 &];

This selects all digits and boxes, but not the grid lines, because they are longer than height/2:

HighlightImage[img, Rectangle @@@ comp[[All, 2]]]

enter image description here

Then combine the individual bounding boxes to one big bounding box:

HighlightImage[img, 

 Rectangle @@ 

  Transpose[MinMax /@ Transpose[Flatten[comp[[All, 2]], 1]]]]

enter image description here

You might have to adjust binarization and the criteria in ComponentMeasurements, but for me, it worked on all the images you posted on the first try.

answered Dec 28 '18 at 15:12

Niki Estner

31.1k376133

$begingroup$
Sorry for catching a cold to delay response. Your method is very simple to understand. But your method cannot deal with the image5. This is why I want to do it with signal-processing method...
$endgroup$
– yode
Jan 3 at 10:54

$begingroup$
@yode: Sure it can, you just need to adjust the criteria function in ComponentMeasurements.
$endgroup$
– Niki Estner
Jan 3 at 11:13

add a comment |

Binarization and ComponentMeasurements seems the straightforward solution:

img = Import["https://i.stack.imgur.com/lnd5w.png"];

Selct all components that are less than half as long as the image:

comp = ComponentMeasurements[MorphologicalBinarize[ColorNegate@img], 

   "BoundingBox", #CaliperLength < Min[ImageDimensions[img]]*0.5 &];

This selects all digits and boxes, but not the grid lines, because they are longer than height/2:

HighlightImage[img, Rectangle @@@ comp[[All, 2]]]

enter image description here

Then combine the individual bounding boxes to one big bounding box:

HighlightImage[img, 

 Rectangle @@ 

  Transpose[MinMax /@ Transpose[Flatten[comp[[All, 2]], 1]]]]

enter image description here

You might have to adjust binarization and the criteria in ComponentMeasurements, but for me, it worked on all the images you posted on the first try.

answered Dec 28 '18 at 15:12

Niki Estner

31.1k376133

$begingroup$
Sorry for catching a cold to delay response. Your method is very simple to understand. But your method cannot deal with the image5. This is why I want to do it with signal-processing method...
$endgroup$
– yode
Jan 3 at 10:54

$begingroup$
@yode: Sure it can, you just need to adjust the criteria function in ComponentMeasurements.
$endgroup$
– Niki Estner
Jan 3 at 11:13

add a comment |

Binarization and ComponentMeasurements seems the straightforward solution:

img = Import["https://i.stack.imgur.com/lnd5w.png"];

Selct all components that are less than half as long as the image:

comp = ComponentMeasurements[MorphologicalBinarize[ColorNegate@img], 

   "BoundingBox", #CaliperLength < Min[ImageDimensions[img]]*0.5 &];

This selects all digits and boxes, but not the grid lines, because they are longer than height/2:

HighlightImage[img, Rectangle @@@ comp[[All, 2]]]

enter image description here

Then combine the individual bounding boxes to one big bounding box:

HighlightImage[img, 

 Rectangle @@ 

  Transpose[MinMax /@ Transpose[Flatten[comp[[All, 2]], 1]]]]

enter image description here

You might have to adjust binarization and the criteria in ComponentMeasurements, but for me, it worked on all the images you posted on the first try.

answered Dec 28 '18 at 15:12

Niki Estner

31.1k376133

Binarization and ComponentMeasurements seems the straightforward solution:

img = Import["https://i.stack.imgur.com/lnd5w.png"];

Selct all components that are less than half as long as the image:

comp = ComponentMeasurements[MorphologicalBinarize[ColorNegate@img], 

   "BoundingBox", #CaliperLength < Min[ImageDimensions[img]]*0.5 &];

This selects all digits and boxes, but not the grid lines, because they are longer than height/2:

HighlightImage[img, Rectangle @@@ comp[[All, 2]]]

enter image description here

Then combine the individual bounding boxes to one big bounding box:

HighlightImage[img, 

 Rectangle @@ 

  Transpose[MinMax /@ Transpose[Flatten[comp[[All, 2]], 1]]]]

enter image description here

You might have to adjust binarization and the criteria in ComponentMeasurements, but for me, it worked on all the images you posted on the first try.

answered Dec 28 '18 at 15:12

Niki Estner

31.1k376133

answered Dec 28 '18 at 15:12

Niki Estner

31.1k376133

answered Dec 28 '18 at 15:12

Niki Estner

31.1k376133

answered Dec 28 '18 at 15:12

Niki Estner

31.1k376133

$begingroup$
Sorry for catching a cold to delay response. Your method is very simple to understand. But your method cannot deal with the image5. This is why I want to do it with signal-processing method...
$endgroup$
– yode
Jan 3 at 10:54

$begingroup$
@yode: Sure it can, you just need to adjust the criteria function in ComponentMeasurements.
$endgroup$
– Niki Estner
Jan 3 at 11:13

add a comment |

$begingroup$
Sorry for catching a cold to delay response. Your method is very simple to understand. But your method cannot deal with the image5. This is why I want to do it with signal-processing method...
$endgroup$
– yode
Jan 3 at 10:54

$begingroup$
@yode: Sure it can, you just need to adjust the criteria function in ComponentMeasurements.
$endgroup$
– Niki Estner
Jan 3 at 11:13

Sorry for catching a cold to delay response. Your method is very simple to understand. But your method cannot deal with the image5. This is why I want to do it with signal-processing method...

– yode
Jan 3 at 10:54

@yode: Sure it can, you just need to adjust the criteria function in ComponentMeasurements.

– Niki Estner
Jan 3 at 11:13

add a comment |

This works quite well:

findRegion[img_] := TextRecognize[img, "BoundingBox", RecognitionPrior -> "SparseText"]



HighlightImage[#, findRegion@#] & /@ imgs

As you can see, the result for the first image is not perfect unfortunately, but hopefully this gives you some ideas.

answered Dec 28 '18 at 14:13

Lukas Lang

7,31511032

$begingroup$
Yes, as your try, the Tesseract not very stable..
$endgroup$
– yode
Dec 28 '18 at 14:40

add a comment |

This works quite well:

findRegion[img_] := TextRecognize[img, "BoundingBox", RecognitionPrior -> "SparseText"]



HighlightImage[#, findRegion@#] & /@ imgs

As you can see, the result for the first image is not perfect unfortunately, but hopefully this gives you some ideas.

answered Dec 28 '18 at 14:13

Lukas Lang

7,31511032

$begingroup$
Yes, as your try, the Tesseract not very stable..
$endgroup$
– yode
Dec 28 '18 at 14:40

add a comment |

This works quite well:

findRegion[img_] := TextRecognize[img, "BoundingBox", RecognitionPrior -> "SparseText"]



HighlightImage[#, findRegion@#] & /@ imgs

As you can see, the result for the first image is not perfect unfortunately, but hopefully this gives you some ideas.

answered Dec 28 '18 at 14:13

Lukas Lang

7,31511032

This works quite well:

findRegion[img_] := TextRecognize[img, "BoundingBox", RecognitionPrior -> "SparseText"]



HighlightImage[#, findRegion@#] & /@ imgs

As you can see, the result for the first image is not perfect unfortunately, but hopefully this gives you some ideas.

answered Dec 28 '18 at 14:13

Lukas Lang

7,31511032

answered Dec 28 '18 at 14:13

Lukas Lang

7,31511032

answered Dec 28 '18 at 14:13

Lukas Lang

7,31511032

answered Dec 28 '18 at 14:13

Lukas Lang

7,31511032

$begingroup$
Yes, as your try, the Tesseract not very stable..
$endgroup$
– yode
Dec 28 '18 at 14:40

add a comment |

$begingroup$
Yes, as your try, the Tesseract not very stable..
$endgroup$
– yode
Dec 28 '18 at 14:40

Yes, as your try, the Tesseract not very stable..

– yode
Dec 28 '18 at 14:40

add a comment |

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