is there an index maneuver if index is 0?
7 Ansichten (letzte 30 Tage)
Ältere Kommentare anzeigen
Martin
am 28 Sep. 2019
Kommentiert: Martin
am 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 Kommentar
Walter Roberson
am 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.
Akzeptierte Antwort
the cyclist
am 28 Sep. 2019
Bearbeitet: the cyclist
am 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 Kommentare
the cyclist
am 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.
Weitere Antworten (0)
Siehe auch
Kategorien
Mehr zu Logical finden Sie in Help Center und File Exchange
Produkte
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!