Subscripted assignment dimension mismatch.

1 Ansicht (letzte 30 Tage)
Dhipikaa Sri Dhamotharan
Dhipikaa Sri Dhamotharan am 30 Sep. 2015
Bearbeitet: James Tursa am 1 Okt. 2015
for i=1:i
var(1,i) = find(m1==indexes(1,i));
end
error is var(1,i) = find(m1==indexes(1,i));

Akzeptierte Antwort

James Tursa
James Tursa am 30 Sep. 2015
The result of the find is likely a vector, and it appears you are trying to stuff this vector into a scalar spot. Do you need all of the find results, or maybe just one of them? If you need all of them, can you use a cell array for var? What is var being used for downstream in your code?
  3 Kommentare
Dhipikaa Sri Dhamotharan
Dhipikaa Sri Dhamotharan am 30 Sep. 2015
can u plz explain me with syntax...kindly help me with syntax to change
James Tursa
James Tursa am 1 Okt. 2015
Bearbeitet: James Tursa am 1 Okt. 2015
Not knowing what how var is going to be used downstream, I am just going to make a guess as to what might work for you with a cell array. Note that I changed your upper loop index to n ... it didn't make sense for the loop indexing to be i=1:i
n = whatever
var = cell(1,n);
for i=1:n
var{i} = find(m1==indexes(1,i));
end
Downstream in your code, when you need to get at the results of the find you would use (note the use of the curly braces): var{i}

Melden Sie sich an, um zu kommentieren.

Weitere Antworten (0)

Kategorien

Mehr zu Matrix Indexing finden Sie in Help Center und File Exchange

Tags

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by