How can I make smaller matrices (size unknown) from a large matrix?

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John Hunt 2017-10-3

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评论： Cedric 2017-10-4

I have a matrix that has 51 columns and 46999 rows. The 8th column has values 1 to 36. I want to create a matrix for each value in that column (36 total). I want to stack rows that have the same value to make these matrices.

Thus far I know can use a for loop to set up a matrix 36 times then I was trying to use a while loop to "stack" the rows that share the same value in the 8th column, but I can't get that while loop right. Any suggestions?

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John Hunt 2017-10-4

That worked awesome! Thank you, could you explain it a bit so I can understand a bit better whats going on please?

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Answer 1

James Tursa 2017-10-3

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在 MATLAB Online 中打开

M = your matrix;
result = cellfun(@(x)M(M(:,8)==x,:),num2cell(1:36),'uni',false);

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Cedric 2017-10-4

编辑：Cedric 2017-10-4

在 MATLAB Online 中打开

I had written this elsewhere.

-----------------------------------------

Don't forget to [Accept] his answer if it helped.

I'll explain the ARRAYFUN version, because I think that it is what James had in mind actually.

result = arrayfun( ref_to_function, array, options )

applies the function given by its reference (1st arg) to each element of the array. The default behavior assumes that the function outputs a scalar, and ARRAYFUN outputs an array of all these scalars.

We often use it passing some values of indices as the array in fact, to apply a function on selected parts of a data array. It is not the best way to do it, but if you wanted to compute the mean per column of some data array

 >> data = randi( 10, 4, 5 )
 data =
     3     8     6     6     9
     7     1     7     9     5
     1     4     2     2    10
     6     5     8     1     9

you could do it as follows:

 >> result = arrayfun( @(colId) mean(data(:,colId)), 1:size(data, 2) )
 result =
    4.2500    4.5000    5.7500    4.5000    8.2500

here the array on which ARRAYFUN operates (iterates) is

1:size(data, 2)

which is a vector of all column IDs, and the function is

@(colId) mean(data(:,colId))

which is an anonymous function of one parameter (named colId), which internally calls MEAN on the relevant column of the data array.

This is equivalent to writing the following loop, but it is more concise:

 nCols = size( data, 2 ) ;
 result = zeros( 1, nCols ) ;
 for colId = 1 : nCols
     result(colId) = mean(data(:,colId)) ;
 end

Now you want to group rows of your data array M and not compute the mean. You probably understand that

@(valCol8) M(M(:,8)==valCol8,:)

selects all relevant rows when given an integer in 1 to 36. To test, you can name this function:

>> selector = @(valCol8) M(M(:,8)==valCol8,:)

and check passing a value:

>> selector(36)

This should have selected all rows with column 8 equal to 36. So you understand most of the call to ARRAYFUN (and why we pass the array 1:36). Selected rows are not "a scalar output" though and we must tell ARRAYFUN that it is allowed to output a cell array of all non-scalar outputs, which is what is done with the options

'UniformOutput', false

where the function understands shorter versions of the parameter name (e.g. 'uni').

Jan 2017-10-4

+1 for both of you.

@Cedric: Now you know the reason, why I "boost" sometimes. It is for the cases, where the standard voting system is to rough. :-)

Cedric 2017-10-4

:-)

It has already been proposed, but I really think that allowing voting for comment would help. This would lead people to add much more value to others' answers by taking time to write well developed comments.

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