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

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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));

Respuesta aceptada

Michael Haderlein
Michael Haderlein el 6 de Ag. de 2014
Editada: Michael Haderlein el 6 de Ag. de 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 comentarios
Hana el 6 de Ag. de 2014
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 el 7 de Ag. de 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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Más respuestas (1)

Sara el 5 de Ag. de 2014
In find you need to use & not &&


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