Extract a portion of vector

Hi, I have a problem that I can not solve.

I need to extract a portion of my vector, this vector has 2 columns (first column is time and second is wave height) and 2500 rows, the values of the second column are number and NaN, the data are sampled every 3 hours. The problem is: I need to create a new vector that contains only the registration longer than 20 hours, and for the rest I must to replace with NaN.

Here an example:

 Initial vector= [0  0.2
  3  0.35
  6  NaN
  9  NaN
  12  NaN
  15  NaN
  18  NaN
  21  NaN
  24  0.29
  27  0
  30  0
  33  0
  36  0
  39  0
  42  0
  45  0.18
  48  0.33
  51  NaN
  54  NaN
  57  NaN]

The result should be 
[0  NaN
3  NaN
6  NaN
9  NaN
12  NaN
15  NaN
18  NaN
21  NaN
24  0.29
27  0
30  0
33  0
36  0
39  0
42  0
45  0.18
48  0.33
51  NaN
54  NaN
57  NaN]

Can anyone suggest an approach to solve this problem?

Thanking you in advance

Alessandro Antonini

[Matt Fig]

What do the * mean in your vectors? This will error if I try to use it...

[Alessandro Antonini]

No, * doesn't mean anything. I tried to write in bold the first two rows of the vector and * it appeared. It is wrong, there is not any *.

[Walter Roberson]

Reformatted to make it clearer.

http://www.mathworks.com/matlabcentral/answers/13205-tutorial-how-to-format-your-question-with-markup

Answer 1

[Alessandro Antonini]

I solve the problem with the help of my friend, I needed two while cicle.

to solution was:

C=impuct vector and e1=C(:,2), the code is:

i=1;

while i<length(e1)

      j=0;
      i
      while isnan(e1(i+j))==0 && (i+j)<length(e1)
          j=j+1;

      end
      if j<7
          e1(i:(i+j-1))=NaN;
      end
      i=i+j+1;
  end

Answer 2

[Matt Fig]

Say this is your vector:

A = [3 4;5 6;8 7;9 8;7 5;5 3;3 9;12 4];

And you want all rows from 5 to the end to be nan:

A(5:end,:) = nan;

[Matt Fig]

Alessandro comments:

"no, I think is different. This is my imput: A=[0 0.3; 3 23; 6 NaN; 9 NaN; 12 NaN; 15 NaN; 18 NaN; 21 NaN; 24 0.29; 27 0; 30 0; 33 0; 36 0; 39 0; 42 0; 45 0.18; 48 0.33; 51 NaN;]

The output should be: B=[0 NaN; 3 NaN; 6 NaN; 9 NaN; 12 NaN; 15 NaN; 18 NaN; 21 NaN; 24 0.29; 27 0; 30 0; 33 0; 36 0; 39 0; 42 0; 45 0.18; 48 0.33; 51 NaN;] In which, the position (1,2) and (2,2), becomes NaN, because I need to take into account only the registration more long than 20 hours. The hours are in the first column. As it can be noticed in the vector B, from position (9,2) till (17,2) the values are the same than in vector A, because the registration is longer than 20 hours. I am sorry for my poor exposition, I hope that this can help to understand my problem."

[Matt Fig]

So you simply use the same concept....

B = A;  % Make a copy
B(B(:,1)<=20,2) = nan

Answer 3

[Alessandro Antonini]

I need to take into account only the partial measure more long than 20 hours, that is if I have a vector like that:

A=[0  2;
3  NaN;
6  3;
9  NaN;
12  4;
15  NaN;
18  5;
21  NaN;
24  6;
27  NaN;
30  7;
33  NaN;
36  8;
39  NaN;
42  9;
45  NaN;
48  10;
51  NaN;]

the result will be:

 A=[0  NaN;
3  NaN;
6  NaN;
9  NaN;
12  NaN;
15  NaN;
18  NaN;
21  NaN;
24  NaN;
27  NaN;
30  NaN;
33  NaN;
36  NaN;
39  NaN;
42  NaN;
45  NaN;
48  NaN;
51  NaN;]

I need to consider only if the partial and continues duration is more long than 20 hours, as in the sequent vector from row n°4 to row n° 11, (partial duration of the registration (C(1,11)-C(1,4))=30-9=21 hours), only in this case I need to take into account. All the other values registered I don't want to take into account, they should be replaced with NaN

C=[0  0.3
3  23
6  NaN
9  0.2
12  0.53
15  0.7
18  0.1
21  0.02
24  0.53
27  0.18
30  0.25
33  NaN
36  NaN
39  NaN
42  NaN
45  NaN
48  0.33
51  NaN]

If the vector C is the imput, the result of the operation should be (vector D):

D=[0  NaN;
3  NaN;
6  NaN;
9  0.2;
12  0.53;
15  0.7;
18  0.1;
21  0.02;
24  0.53;
27  0.18;
30  0.25;
33  NaN;
36  NaN;
39  NaN;
42  NaN;
45  NaN;
48  NaN;
51  NaN];

In which, the values in the position D(1,2) and D(2,2) now are NaN, because the partial duration of the registration (continues and different from NaN)is lower than 20 hours, i.e. C(2,1)-C(1,1)=3-0=3.

At the end, the goal would be, take into account only those values registered that have a continuous registration more long than 20 hours.

[Walter Roberson]

http://www.mathworks.com/matlabcentral/answers/13205-tutorial-how-to-format-your-question-with-markup

Answer 4

[Jonathan Epperl]

How about

 vector(vector(:,1)<20, 2) = NaN;

Logical indexing...