# Efficient way of reading cell arrays

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djr on 4 Aug 2014
Edited: per isakson on 4 Dec 2016
Hi
What would be the most efficient way to read content from cells; is it using for loops or cellfun functions. For example, My data have 3 layers of cells (cell in cell in cell). So, I used 3 'for loops' to get to data that are in the third cell. The code looks like this:
for j=1:size(evaldata,1)
for k=1:size(evaldata{j,1},1)
for l=1:size(evaldata{j,1}{k,2},1)
A{l,:}={evaldata{j,1}{k,1},(sscanf(evaldata{j,1}{k,2}{l,1}, '%f').')};
%
if ismember(A{l,1}{1,2}(2),v)
store{j}{l,k}{1,1}=A{l}{1,1};
for w=2:14
store{j}{l,k}(1,w)={A{l}{1,2}(w-1)};
end
else
store{j}{l,k}='not selected';
end
end
end
fz=cellfun(@isempty,evaldata{1,1});
fzz=find(fz(:,2));
for kk=1:size(fzz,1)
store{j}{1,fzz(kk)}=evaldata{1,1}{fzz(kk),1};
end
end
It probably does not mean much without explaining what it does... However, this looks like a brute force. Is there a bettwer way to deal with multiple array cells that are not the same size or declaration (strings, numbers, ...)?
Last few days I asked for help several times. This time I decided to do it by muself and I ended up having 100 loops :( :( :(. Maybe because I only used Fortran when it comes to programing.
Cheers, Djordje
Image Analyst on 5 Aug 2014
Well you're right that it doesn't mean much without any explanation. No comments at all , and it looks so cryptic that I didn't even attempt to figure out what it does. It looks like you're trying to transfer some data from A and evaldata into store, but that's about all I got from it.

per isakson on 5 Aug 2014
Edited: per isakson on 4 Dec 2016
I continue where we left in your last question. I assume
• There are many data-files in one folder
• The names of the data-files match '\d[4}_\d{4}', i.e four digits, underscore followed by four digits. It's hard-coded.
The function, add_met_data_to_lib, can be used repeatedly to read data-files in many folders. Syntax
Create new lib
Example:
lib =
Map with properties:
Count: 2925
KeyType: char
ValueType: any
>> lib('19491020T0600')
ans =
1.0e+03 *
Columns 1 through 9
0.0010 0.0110 0.0804 0.0526 1.0232 -0.0001 0.0026 0.0058 0.4113
0.0020 0.0010 0.0726 0.0449 1.0288 -0.0003 0.0050 0.0057 0.9271
0.0030 0 0.0393 0.0497 1.0288 -0.0004 0.0100 0.0105 0.1410
Columns 10 through 13
0.0020 0.0008 0.0790 0.0537
0.0004 0.0001 0.0747 0.0450
0.0018 0.0005 0.0360 0.0524
>>
where
function lib = add_met_data_to_lib( glob, lib )
narginchk( 1, 2 )
if nargin == 1
lib = containers.Map('KeyType','char','ValueType','any');
end
%
file_list = transpose( dir( glob ) );
%
is_met_file = not( isempty( regexp( [file_list.name] ...
, '\d{4}_\d{4}', 'start' ) ) );
file_list( not( is_met_file ) ) = [];
%
for file = file_list
lib = met2lib( file.name, lib );
end
end
per isakson on 7 Aug 2014
Yes, this becoming too specific for this forum. I sent a mail and blurred your mail-address

Ahmet Cecen on 5 Aug 2014
This really doesn't make much sense without context to me. This is an incredibly inefficient data storage configuration. From what I understand, you can access everything you need in 3 for loops. (for the first eval.mat)
for i=1:14
currentI=eval{i}
for j=1:size(currentI,1)
currentJ=currentI{j,2};
for k=1:size(currentJ,1}
currentdata=currentJ{k,1}
end
end
end
With this, at the evaluation step you have the following information: '0101194900' 1 0 77.20 43.17 1038.47 -0.268 7.442 7.1651644.929 0.000 0.000 75.96 41.45