How to binning 2-d data ?

I have a data of 29459*13 and I want to bin the data by taking 1500 rows means one bin will be 1500*13. Please can anyone help me out with this regard ?

Thank you for your time.

[Walter Roberson]

To confirm, you are looking for a result that is

ceil(29459 / 1500)

ans = 20

20 rows and 1 column, and the value in each of those 20 rows has to somehow "represent" the 1500 by 13 rows it is summarizing?

[ARUNIMA DAS]

Yes, my question is likely to be your comment. Can you help me with this ?

Thank you for your time.

[Christopher McCausland]

A very very low resolotion FFT maybe? However so much data will be lost that I cannot see the point of this @ARUNIMA DAS. What is the purpose of this?

Christopher

[ARUNIMA DAS]

Extracting static features from each bin and do classification.

Thank you.

Answer 1

[Adam Danz]

This approach uses repelem() to generate group IDs by rows.

data is the input matrix.

nRows specifies the number of consecutive rows for each bin (=1500)

binID is the bin ID numbers for each row of data. (1;1;1;....(1500 1s); 2;2;2;...;etc). The last group will have less than 1500 rows since the height of your matrix is not divisible by 1500.

% Create demo data
rng('default')
data = rand(29459,13);

% Specify number of rows per bin
nRows = 1500; 

% Create bin IDs
nGroups = ceil(size(data,1)/nRows);
binID = repelem((1:nGroups)', nRows, 1);
binID(size(data,1)+1:end) = []; 

Data from bin n is isolated by,

data(binID==n, :)

Compute the average of all data within columns of each bin.

binAvg will be a 20x13 matrix for 20 groups and 13 columns.

binAvg = groupsummary(data, binID(:), 'mean')

binAvg = 20×13

0.5000 0.5159 0.4873 0.5007 0.4993 0.4872 0.4819 0.4938 0.4956 0.4993 0.5045 0.4846 0.5154 0.5077 0.5094 0.5004 0.5163 0.4956 0.5026 0.5085 0.4905 0.4869 0.4879 0.4972 0.4950 0.4860 0.5060 0.4965 0.5001 0.5070 0.4909 0.4989 0.5038 0.5077 0.4988 0.5053 0.4802 0.5021 0.4994 0.4934 0.5038 0.5002 0.4930 0.5084 0.5150 0.5034 0.4975 0.5125 0.4915 0.4923 0.5056 0.5001 0.4976 0.5018 0.5007 0.4973 0.4926 0.5023 0.5014 0.5006 0.5037 0.4944 0.5168 0.4980 0.5061 0.4881 0.5065 0.4885 0.5037 0.4915 0.5004 0.5001 0.4989 0.5018 0.5043 0.5044 0.4988 0.4972 0.5043 0.4882 0.5054 0.4983 0.5083 0.5010 0.5066 0.5024 0.5058 0.4988 0.4827 0.5111 0.5172 0.4997 0.4947 0.4979 0.5099 0.5018 0.5054 0.4957 0.5020 0.5097 0.4956 0.5082 0.4995 0.4991 0.4964 0.4993 0.5043 0.5088 0.5049 0.5020 0.4979 0.4948 0.4879 0.4965 0.4955 0.4974 0.4985 0.4942 0.4950 0.5115 0.4819 0.4928 0.5164 0.5041 0.5058 0.4998 0.4979 0.4961 0.4990 0.4977

Confirm results by selecting a bin and comparing its average with the results above

testBin = 8; 
testBinAvg = mean(data(binID==testBin,:));
isequal(testBinAvg, binAvg(testBin,:))

ans = logical
   1

Compute the average of all data combined within each bin

binAvgAll will be 20x1 for 20 groups.

binAvgAll = arrayfun(@(g)mean(data(binID==g,:),'all'), unique(binID))

binAvgAll = 20×1

0.4974 0.4988 0.4997 0.5013 0.5010 0.4988 0.5023 0.5015 0.4988 0.4994

Confirm results by selecting a bin and comparing its average with the results above

testBin = 14;
testBinAvg = mean(data(binID==testBin,:),'all');
isequal(testBinAvg, binAvgAll(testBin))

ans = logical
   1

Answer 2

[Walter Roberson]

blockproc() https://www.mathworks.com/help/images/ref/blockproc.html

You will have to decide how you want to handle the last partial bin that is only 959 rows instead of 1500.

You can return a row vector of the extracted features, and the result would be put together into a 20 x whatever array of features. Which you would transpose or not transpose depending on which classification routine you use.