operands to the and && operators must be convertible to logical scalar

조회 수: 2 (최근 30일)
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

채택된 답변

Michael Haderlein
Michael Haderlein 2014년 8월 6일
편집: Michael Haderlein 2014년 8월 6일
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);
  댓글 수: 2
Hana
Hana 2014년 8월 6일
Thanks Michael,it is solved now. Now the me=mean(ind_x,ind_y) gives error, it should return a single mean value of radar from the indices found from the condition.
Michael Haderlein
Michael Haderlein 2014년 8월 7일
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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Sara
Sara 2014년 8월 5일
In find you need to use & not &&

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