deleting matrix rows by criteria?

I have a matrix as follows(this is sample data from a much longer matrix)

sample_position =

       58.044       1.0831
       110.19       244.98
       109.95       255.78
      -150.57       181.66
      -445.85      -160.02
      -570.87       -404.5
      -727.97       0.7062
      -529.34       306.81
       581.03       122.86
       761.24      -350.04
       62.561      0.98683
      -150.74       185.85
       -454.6      -160.39
      -571.29      -402.92
      -730.77      0.38753
       -529.9        305.2
       583.51       120.23
       762.82      -350.56

I need to delete rows that do not fall under some criteria, say: 55<column1<65 & 0<column2<4

So if the row doesn't comply to both criteria, it gets deleted.

now the tricky part(for me anyway) is that the criteria is different for every row.

let me try explaining it a bit more. for the first row, the example criterion applies. for the second row, a new set of criterion exists (say 130<column1<140 & 20<column2<35)

I need to write something that will do the following:

>apply criterion 1 to row 1

>if row 1 is within criterion 1, write row 1 to output variable

>if row 1 is not within criterion 1, delete row 1 and apply criterion 1 to the next row(repeat until a row is within criterion 1 and write that row to output variable)

>apply criterion 2 to next row

PS:I have 8 different criterions. once I reach the 8th, the code needs to go back to the first critetion and repeat the process until it goes through all the rows in the matrix

any ideas?(i think there should be a loop somewhere?)

[Walter Roberson]

Is the criteria of a different form for each row, or is it just with different cutoff values for each row? If so then are the cutoff values to be calculated (e.g., partly based on row number) or are the cutoff values opaque per-row constants (determined through outside information) ?

[Peter]

the cutoff values are determined by outside experimental data, such that it is a simple set of ranges

Answer 1

[Azzi Abdelmalek]

Edit

M=[58.044       1.0831
  110.19       244.98
  109.95       255.78
  -150.57       181.66
  -445.85      -160.02
  -570.87       -404.5
  -727.97       0.7062
  -529.34       306.81
  581.03       122.86
  761.24      -350.04
  62.561      0.98683
  -150.74       185.85
  -454.6      -160.39
  -571.29      -402.92
  -730.77      0.38753
  -529.9        305.2
  583.51       120.23
  762.82      -350.56]
n=3  % number of criterion
out=[];
kk=1
for k=1:size(M,1)
  switch kk
      case 1
          idx=55<M(k,1)& M(k,1)<65;
      case 2
          idx=0<M(k,2) & M(k,2)<4;
      case 0
          idx=50<M(k,1)& M(k,1)<60;
  end
  if idx==1
      out(end+1,:)=M(k,:);
      kk=mod(kk+1,n);
  end
end
disp(out)

In your case, write the variable kk will vary from 1 to 7 then 0

[Peter]

Okay, that works for the first row of the matrix. how would I go about applying that to the whole matrix, considering that the values for the range change?

[Azzi Abdelmalek]

This works for all rows that matches your criterion. Have you tested the code?

[Jan]

@Azzi: The problem might be, that Peter dis not specify, what this exactly means:

 now the tricky part(for me anyway) is that the criteria is different for every row.

@Peter: Please elaborate this by editing the original question.

[Peter]

edited, have a look

[Azzi Abdelmalek]

n=3  % number of criterion
for k=1:size(M,1)
  kk=mod(k,n);
  switch kk
      case 1
          idx=55<M(:,1)& M(:,1)<65;
      case 2
          idx=0<M(:,2) & M(:,2)<4;
      case 0
          idx=50<M(:,1)& M(:,1)<60;
  end
  M(idx,:)=[];
end

[Peter]

if its not asking too much, can you walk me through this code? I dont quite get it

[Peter]

also, I find that if i use your initial suggestion:

out=M(55<M(:,1)& M(:,1)<65 & 0<M(:,2) & M(:,2)<4,:)

I could do it 8 times and get 8 matrices that have my correct rows. is there a way to interlace them this way:

matrix =

Row 1(from matrix 1)
Row 1(from matrix 2)
Row 1(from matrix 3)
Row 1(from matrix 4)
Row 1(from matrix 5)
Row 1(from matrix 6)
Row 1(from matrix 7)
Row 1(from matrix 8)
Row 2(from matrix 1)
Row 2(from matrix 2)
Row 2(from matrix 3)
Row 2(from matrix 4)
and so on

this is just another way of looking at it, tho if you can explain to me how your loop works i will be very grateful

[Azzi Abdelmalek]

Look at edited answer