Remove element from array but want to keep in indices

A=[1 2;1 5;2 3;2 4;2 5;3 4;4 5;4 7;4 9;5 6;6 11;6 12;6 13; 7 8;7 9;9 10;9 14;10 11;12 13;13 14] I omit the second row: A(2,:)=[]; so size A changes to 19*2 and [2,3] changes to my second element. If I want to keep [2,3] as a third element, what should I do?

[KALYAN ACHARJYA]

Can you elaborate your question exactly? Use Maths to explain your problem.

Answer 1

[Andrei Bobrov]

A(2,:)=nan;

[NA]

Thank you. the problem is that after omitting that row I used it as a input of my function. so give me this error: Error using grValidation (line 21) 1st and 2nd columns of the array E must be positive! Is there any other way instead of nan?

[Walter Roberson]

No, that is a validation routine from the Graph Theory Toolbox from the file exchange, and that Toolbox does not support omitted elements.

Answer 2

[Walter Roberson]

To keep that row as being the third row... don't remove the second row.

The alternative is to use containers.map to create mappings between subscripts and values. Then when you delete keys from the map, the remaining items would keep the same keys.

.... But what you probably need to hear is:

Run your loop backwards so that you delete rows starting from the end. The rows after will "fall down" into the earlier slots, but that is okay because next iteration you will be examining a slot that has not had an opportunity to fall down because it was below all slots you looked at before that.

Answer 3

[madhan ravi]

A(2,:)=0

Did it work?

[NA]

No. give me this error Error using grValidation (line 21) 1st and 2nd columns of the array E must be positive!

[madhan ravi]

Can you upload your code!

[Walter Roberson]

That toolbox does not support omitted nodes.