How to index a matrix in matlab

Assume input matrix I as follows:

I = [
100  56  1
100  54  1
100  65  1
101  5  0
101  10  1
101  15  1
101  20  0
101  30  1
101  20  1
101  50  1
198  30  0
198  20  1
203  10  0
203  5  1
203  60  1
203  20  1
203  15  1
44  70  0
44  65  1
44  45  1
44  50  0
44  35  1
44  35  1
44  50  0
44  70  1
44  75  1
44  65  1
];

I want to create 3 matrix based on the I matrix:

Matrix A:

Based on the unique ID in first column of matrix A, I want to generate numbers if ID is same then corresponding cell get same number and if the ID changed, the corresponding ID reset numbers. Also, if in the same ID number, from the third column in matrix I, there was an interrupt with value 0, then number reset and continue (see the figure below).

Matrix B:

Based on the out put of matrix A, I want to count how many number generated for every ID (and if there was an interrupt with 0). See the figure below.

Matrix C:

Based on the out put of matrix A, I want to sum up the corresponding cells from second column in matrix I. See the figure below.

A = [
100  1
100  1
100  1
101  1
101  1
101  1
101  2
101  2
101  2
101  2
198  1
198  1
203  1
203  1
203  1
203  1
203  1
44  1
44  1
44  1
44  2
44  2
44  2
44  3
44  3
44  3
44  3
];

B = [
100  3
101  3
101  4
198  2
203  5
44  3
44  3
44  4
];

C = [
100  175
101  30
101  120
198  50
203  110
44  180
44  120
44  260
];

[Jan]

What have you tried so far? Which problems occur?

[Amy Xu]

The difficult part is when in the same ID, there is an interrupt with 0. I can't figure out how to reset numbers!

[Jan]

If you post the code, we can suggest an improvement easily. This is more efficient than writing the code from scratch.

I have problems to understand the sentence: " I want to generate numbers if ID is same then corresponding cell get same number and if the ID changed, the corresponding ID reset numbers. Also, if in the same ID number, from the third column in matrix I, there was an interrupt with value 0, then number reset and continue."

[Amy Xu]

Let me give you an example. Consider ID number 44 in the figure. In matrix I, look at the third column. There are three 0. It means that in output matrix A, we should have three set numbers {1,2,3}. If you look at matrix A for ID 44 and second column, you can see that the first three is given by 1, the second three is given by 2 and the third part is given by 3.

Is that clear?

Thanks!

[Amy Xu]

Any solution? @Jan Simon

Answer 1

[Stephen23]

Array A

>> [U,X,Y] = unique(I(:,1));
>> [~,X] = sort(X);
>> [~,X] = sort(X);
>> fun = @(v){cumsum(v==0)+(v(1)==1)};
>> Z = accumarray(X(Y),I(:,3),[],fun);
>> A = [I(:,1),cell2mat(Z)]
A =
 100     1
 100     1
 100     1
 101     1
 101     1
 101     1
 101     2
 101     2
 101     2
 101     2
 198     1
 198     1
 203     1
 203     1
 203     1
 203     1
 203     1
  44     1
  44     1
  44     1
  44     2
  44     2
  44     2
  44     3
  44     3
  44     3
  44     3

Array B

>> V = [true;any(diff(A,1,1),2)]
>> W = accumarray(cumsum(V),ones(size(V)));
>> B = [I(V,1),W]
B =
 100     3
 101     3
 101     4
 198     2
 203     5
  44     3
  44     3
  44     4

Array C

>> T = accumarray(cumsum(V),I(:,2));
>> C = [I(V,1),T]
C =
 100   175
 101    30
 101   120
 198    50
 203   110
  44   180
  44   120
  44   260