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Offline babalus  
#1 Posted : 25 February 2021 11:11:56(UTC)
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If the value in the second column of the matrix is greater than the specified constant, then you need to replace the row containing this value by the number of rows equal to quotient of division of this number by constant + 1. In the first rows, the value in the second column must be equal to a constant, and in the last row the value in the second column must be equal to a remainder of the integer division of the number by a constant.

On the left is the original matrix, on the right is what should be obtained.

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Offline Davide Carpi  
#2 Posted : 25 February 2021 13:55:32(UTC)
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That's a rough implementation, although I probably misunderstood some part.

2021-02-25 11_53_19-SMath Studio - [Page1_].png
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on 25/02/2021(UTC)
Offline babalus  
#3 Posted : 25 February 2021 14:06:24(UTC)
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It should be something like this but probably there is a better decision
Offline Davide Carpi  
#4 Posted : 25 February 2021 15:10:01(UTC)
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Seems I missed the physical meaning, this should be a better implementation:

2021-02-25 13_13_59-SMath Studio - [split layers.sm].png

split layers.sm (11kb) downloaded 15 time(s).

Edited by user 25 February 2021 15:14:47(UTC)  | Reason: Not specified

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Offline babalus  
#5 Posted : 25 February 2021 16:24:29(UTC)
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It is magic! Thank you!
Can you do the same whith distance units?
Offline Davide Carpi  
#6 Posted : 25 February 2021 17:42:14(UTC)
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Originally Posted by: babalus Go to Quoted Post
It is magic! Thank you!
Can you do the same whith distance units?


There's not much to change, just add the units to:
  • threshold
  • const
  • rem, here you have to reintroduce the unit (automated by UnitsOf() or directly)


2021-02-25 15_37_39-SMath Studio - [split layers.sm_].png

Edited by user 25 February 2021 19:01:24(UTC)  | Reason: Not specified

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thanks 2 users thanked Davide Carpi for this useful post.
on 25/02/2021(UTC),  on 26/02/2021(UTC)
Offline Richard Smith  
#7 Posted : 25 March 2021 13:43:03(UTC)
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Hint: You can get a submatrix like this:

subMatrix = fullMatrix(1:someRow, : );
subMatrix = fullMatrix(row1:row2, : );
subMatrix = fullMatrix(someRow:end, : ); % End has a special meaning

(pick one), and you can concatenate matrices vertically like this:
tallMatrix = [shortMatrix1; shortMatrix2]; % Use as many as you want
They all have to have the same number of columns of course.

I hope this helps, thanks!

Edited by user 25 March 2021 13:45:05(UTC)  | Reason: Emoticons

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