Subscripted assignment dimension mismatch.

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Dhipikaa Sri Dhamotharan
编辑： James Tursa 2015-10-1
for i=1:i
var(1,i) = find(m1==indexes(1,i));
end
error is var(1,i) = find(m1==indexes(1,i));

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James Tursa
James Tursa 2015-9-30
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?
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Dhipikaa Sri Dhamotharan
Dhipikaa Sri Dhamotharan 2015-9-30
can u plz explain me with syntax...kindly help me with syntax to change
James Tursa
James Tursa 2015-10-1
编辑：James Tursa 2015-10-1
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}

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