Question

I am facing a problem in matching elements in 2 matrices. The first element can be matched using ismember but the second element should be within a range. Please see the example below:

% col1 is integerID, col2 is a number.       -->col1 is Countrycode, col2 is date
bigmat = [ 600 2  
           600 4  
           800 1  
           800 5  
           900 1] ;

% col1 is integerID, col2 is VALUE, col2 is a range    -->value is Exchange rate
datamat = {...
           50   0.1   [2 3 4 5 6]       % 2:6
           50   0.2   [9 10 11]         % 9:11
           600  0.01  [1 2 3 4]         % 1:4
           600  0.2   [8 9 10]          % 8:10
           800  0.12  [1]               % 1:1
           800  0.13  [3 4]             % 3:4
           900  0.15  [1 2]      } ;    % 1:2

I need the answer as:
    ansmat = [ 600 2  0.01
               600 4  0.01
               800 1  0.12
               800 5  nan               % even deleting this row is fine
               930 1  0.15 ] ;

For simplicity:

1. All intIDs from matrix_1 exist in matrix_2.
2. The numbers in range are dates! Within a range, these numbers are consecutive: [1 2…5]
3. For any ID, dates in the next row are not continuous. Eg, you can see [1 2 3 4] and then [8 9 10] in next row.

bigmat is a huge matrix! 300,000-500,000 rows and so a vectorized code would be appreciated. datamat is roughly 5000 rows or less. You can convert the cell to matrix. For each row, I have the minimum and maximum. The 3 column is minimum:maximum. Thanks!

Answer 1

Here is one possible implementation:

%# data matrices
q = [
    600 2  
    600 4  
    800 1  
    800 5  
    900 1
];
M = {
    50   0.1   [2 3 4 5 6]
    50   0.2   [9 10 11]
    600  0.01  [1 2 3 4]
    600  0.2   [8 9 10]
    800  0.12  [1]
    800  0.13  [3 4]
    900  0.15  [1 2]
};

%# build matrix: ID,value,minDate,maxDate
M = [cell2num(M(:,1:2)) cellfun(@min,M(:,3)) cellfun(@max,M(:,3))];

%# preallocate result
R = zeros(size(M,1),3);

%# find matching rows
c = 1;             %# counter
for i=1:size(q,1)
    %# rows indices matching ID
    ind = find( ismember(M(:,1),q(i,:)) );

    %# out of those, keep only those where date number is in range
    ind = ind( M(ind,3) <= q(i,2) & q(i,2) <= M(ind,4) );

    %# check if any
    num = numel(ind);
    if num==0, continue, end

    %# extract matching rows
    R(c:c+num-1,:) = [M(ind,1) repmat(q(i,2),[num 1]) M(ind,2)];
    c = c + num;
end

%# remove excess
R(c:end,:) = [];

The result as expected:

>> R
R =
          600            2         0.01
          600            4         0.01
          800            1         0.12
          900            1         0.15