Storage of first few values in an array but with an if condition
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Sandip Ghatge
el 18 de Mayo de 2020
Editada: Sandip Ghatge
el 21 de Mayo de 2020
I am stuck here, because when i write the following code to take in values lesser than or equal to 0.20 and greater than or equal to 0.12, it takes all the values of the first column <= 0.20 and >=0.12. The desired result is taking in first set of values, lesser than or equal to 0.2 and greater than or equal to 0.12 and not all the values.
The code is,
A = [0.25;0.12;0.18;0.21;0.26;0.18;0.19;0.25;0.26;0.12;0.18;0.21];
B = [A(A(:,1) <= 0.20 & A(:,1) >= 0.12,1)];
The output of this is,
B =
0.1200
0.1800
0.1800
0.1900
0.1200
0.1800
What i am desiring is
B =
0.12
0.18
2 comentarios
Respuesta aceptada
Rik
el 18 de Mayo de 2020
Editada: Rik
el 18 de Mayo de 2020
If you want to find the first block of true in L, you can use this code:
A = [0.25;0.12;0.18;0.21;0.26;0.18;0.19;0.25;0.26;0.12;0.18;0.21];
L=A(:,1) <= 0.20 & A(:,1) >= 0.12;
ind1=find(L,1);%first location within L
ind2=find(diff(L)==-1,1);%last location within L
if isempty(ind2),ind2=numel(A);
B=A(ind1:ind2)
Original answer:
Assuming you also wanted the 0.19:
A = [0.25;0.12;0.18;0.21;0.26;0.18;0.19;0.25;0.26;0.12;0.18;0.21];
L=A(:,1) <= 0.20 & A(:,1) >= 0.12;%put in a different variable for readability
B = A(L);
B=unique(B,'stable');%don't sort values
Más respuestas (1)
Guillaume Le Goc
el 18 de Mayo de 2020
A = [0.25;0.12;0.18;0.21;0.26;0.18;0.19;0.25;0.26;0.12;0.18;0.21];
ids = find(A<=0.2 & A>=0.12, 2); % second argument specifies you want only the first 2 elements that match the condition
B = A(ids);
Or directly :
A = [0.25;0.12;0.18;0.21;0.26;0.18;0.19;0.25;0.26;0.12;0.18;0.21];
B = A(find(A<=0.2 & A>=0.12, 2));
3 comentarios
Rik
el 20 de Mayo de 2020
If this answer doesn't solve your question, why did you accept it?
After your comment I have edited my answer. Did you see that?
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