operands to the and && operators must be convertible to logical scalar
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Hana
on 5 Aug 2014
Commented: Michael Haderlein
on 7 Aug 2014
why do I get this error? I used single & also,but got the same error.
for ind_x = 1:1000
for ind_y =1:800
[ind_x,ind_y] =find (inc >= 37.5 && inc < 38.5) && (lc == 3);
me = mean(radar(ind_x,ind_y));
end
end
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Accepted Answer
Michael Haderlein
on 6 Aug 2014
Edited: Michael Haderlein
on 6 Aug 2014
You have to use & instead of && (as Sara suggested) AND your last statement must be inside the find brackets.
[ind_x,ind_y] =find (inc >= 37.5 & inc < 38.5 & lc == 3);
2 Comments
Michael Haderlein
on 7 Aug 2014
Yesterday, I obviously didn't check your code very well. Let's go through it:
In each loop iteration, you will get the same result from the find function as its parameters do not change.
Then, you overwrite your loop iterators by the result of the find function (not recommended).
Third, are inc and lc matrices or vectors? In case of vectors, one of the outputs of find will always be 1.
Finally, even if you change the line with the find, me will be overwritten every iteration. I suppose that's also not your goal.
However, to get this code running and doing what it's supposed to do, we need more information. What is inc and lc and me?
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