how to find the column number of first zero element of each row

12 vues (au cours des 30 derniers jours)
Jason
Jason le 19 Fév 2016
Modifié(e) : Sunetra CV le 14 Nov 2019
Hi,all. could you please tell me how to find the column number of first zero element of each row. eg. for the matrix in the picture, I want the output will be outputmatrix=[3;4;4;5;5;6;6;6;6;6;1;1;1;1;1]. Thank you!

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MHN
MHN le 19 Fév 2016
Modifié(e) : MHN le 19 Fév 2016
This is your answer:
a =[ -1 3 0 0 0 0
2 3 4 5 0 0
1 2 1 0 0 0]
b = abs(a); % it is a trick which forces min to do not consider your negative numbers as minimum
[~,ind] = min(b')
  2 commentaires
MHN
MHN le 19 Fév 2016
Modifié(e) : MHN le 19 Fév 2016
So, for your matrix it is just one line code:
[~,ind] = (min((abs(SumofSet1))'))
Jason
Jason le 19 Fév 2016
thank you very much!

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Plus de réponses (2)

Adam
Adam le 19 Fév 2016
Modifié(e) : Adam le 19 Fév 2016
sum( abs( cumprod( yourMatrix, 2 ) ) > 0, 2 ) + 1;
should work I think. I tested in on a small matrix I made and it seems to give the correct answer. There may be simpler methods though!
The logic behind it is as follows:
  • cumprod is the cumulative product, run along each row (dimension 2) which will always be 0 for values after the first 0, hence the result of this will be a matrix in which only the elements before the first 0 on a row will be non-zero.
  • abs will ensure that all non-zero elements are positive in order for the following check to work correctly.
  • taking the sum of all the elements greater than 0 will give you the index of the last non-zero element.
  • Adding 1 at the end will give the element of the first 0 which should always be the column immediately after the last non-zero.
If need be you can use a small tolerance e.g. 1e-6 to hard set elements below that absolute tolerance to 0 before doing this as values that are not precisely 0 will likely cause problems.
  2 commentaires
Jason
Jason le 19 Fév 2016
Thank you,but the result is not the same as i wanted. I didn't understand why you use 'sum' function here. because the element in matrix is very small, use 'cumsum' may cause the inaccuracy of the results.
Walter Roberson
Walter Roberson le 19 Fév 2016
sum(cumprod(~YourMatrix, 2), 2) + 1
Note: this will be incorrect / misleading for any row for which there is no 0 (in which case no desired result has been defined by the poster)

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Sunetra CV
Sunetra CV le 14 Nov 2019
Modifié(e) : Sunetra CV le 14 Nov 2019
will this work if all the elements of my matrix are decimals?

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