# How to calculate index of minimum value in cell array

26 views (last 30 days)

Show older comments

##### 0 Comments

### Accepted Answer

Stephen
on 9 May 2017

Edited: Stephen
on 9 May 2017

>> m1{1,1} = [0.34463,0.3276,0.3615,0.3446,0.3559,0.3389,0.3389,0.3446];

>> m1{1,2} = [0.36723,0.3333,0.3615,0.3615,0.3446,0.3559,0.3163,0.3333];

>> m1{2,1} = [0.3390];

>> m1{2,2} = 5:9;

An efficient method is to use min once:

>> [val,idx] = min([m1{:}]);

>> len = [0;cumsum(cellfun('length',m1(:)))];

>> idc = find(len>=idx,1,'first')-1; % cell index

>> idv = idx-len(idc); % vector index

The minimum is thus:

>> val

val = 0.31630

and the minimum can be obtained using those indices:

>> m1{idc}(idv)

ans = 0.31630

You can get the cell array row and column indices by using ind2sub:

>> [row,col] = ind2sub(size(m1),idc)

row = 1

col = 2

##### 8 Comments

Stephen
on 9 May 2017

"if the value of idc,idv is 2,7 means i.e 2nd column 7th cell...if idc,idv value is 3,1 means 3rd column ..."

No they do not.

Both idc and idv are linear indices, which means that if idc is seven it will be the seventh cell in the call array (assuming numel>=7). The value of idc does not tell you the column of the cell containing the minimum! (But I did show you how to get the row and column indices, by using ind2sub. Personally I would not bother).

And the index idv is the index of position in the numeric vector within that cell, so this is certainly not the "7th cell" or "1st cell", because idv does not refer to cells at all but elements of the numeric vector inside that cell.

My code does not use row or column indices at all, so be careful that you do not incorrectly interpret those linear indices. You can learn more about linear indices by reading the documentation (note that row and column indices are called subscript indices):

### More Answers (2)

KSSV
on 9 May 2017

m1{1,1}=[0.34463 0.3276 0.3615 0.3446 0.3559 0.3389 0.3389 0.3446] ;

m1{1,2}=[0.36723 0.3333 0.3615 0.3615 0.3446 0.3559 0.3163 0.3333] ;

m1{1,3}=[0.3390] ;

iwant = cell(size(m1)) ;

for i = 1:length(m1) ;

[idx,val] = min(m1{i}) ;

iwant{i} = [idx,val] ;

end

##### 3 Comments

Jan
on 9 May 2017

[MinValue, MaxValue, MinIndex, MaxIndex, MinArg, MaxArg] = MinMaxElem(m1{:});

In older Matlab version this was faster than calling min and max separately, but modern Matlab versions are more efficient.

##### 0 Comments

### See Also

### Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!