is there an index maneuver if index is 0?

4 vues (au cours des 30 derniers jours)
Martin
Martin le 28 Sep 2019
Commenté : Martin le 30 Sep 2019
Hello,
I am running a ismember:
[logical idx] = ismember(hund(:,1),dogs(:,2))
('hund' is often around 3,000 rows, and 'dogs' is often around 100,000 rows).
This give me an output that could be something like this
logical = idx =
1 355
1 536
1 746
0 0
1 1042
0 0
Then I need to find the second column in the 'dogs' variable for the index and inset this in the 'hund' variable as the 2. column like this:
hund(:,2) = dogs(idx,2)
The problem is, the above does not account for if the idx is zero. I get this error:
'Index in position 1 is invalid. Array indices must be positive integers or logical values.'
Does anyone know a solution to get arround of this?
  1 commentaire
Walter Roberson
Walter Roberson le 29 Sep 2019
Using logical as the name of a variable can interfere with using logical as the name of a class, and will confuse people.

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the cyclist
the cyclist le 28 Sep 2019
Modifié(e) : the cyclist le 28 Sep 2019
hund(logical,2) = dogs(idx(logical),2)
If hund does not already have a second column, you'll need to do something like
hund = [hund, zeros(numel(hund),1)];
beforehand.
  4 commentaires
the cyclist
the cyclist le 30 Sep 2019
Yeah, although I hope the camel case is a strong hint that this is a variable.
I guess hundIsInDog would do the trick here.
Martin
Martin le 30 Sep 2019
Yeah I agree with you guys, the output names however was just to be as descriptive as possible. However, thanks for the follow up and solution. Take care

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