Storage of first few values in an array but with an if condition

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

[KSSV]

unique(B)

[Sandip Ghatge]

Thank you.

I am going through the command.

Answer 1

[Rik]

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

[Sandip Ghatge]

Thanks for the answer.

I cannot modify this code to exclude 0.19.

I just want a set when the values of the matrix A first get <= 0.20 and >=0.12, the moment it gets out of these limits(which is at A(4,1) here) the storage into a different matrix say B should stop.

Answer 2

[Guillaume Le Goc]

You could use the find function, where you can specify the first n elements you want :

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));

[Sandip Ghatge]

Thanks for the answer.

But, everytime i wont know, that it would be first 2 values, in such cases?

i was just looking for the set when the values of the matrix A first get <= 0.20 and >=0.12, the moment it gets out of these limits(which is at A(4,1) here) the storage into a different matrix say B should stop.

[Rik]

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?

[Sandip Ghatge]

@Rik, I am sorry, i wanted to accept your answer, but by mistake i accepted this answer. Thanks for bringing it into notice. i am accepting your answer as it takes me to the closest what i was aiming for and also have learnt a new command.