Create matrix using hashtable MATLAB

Question

I am using hashtable (containers.map ) in MATLAB, and now I want to create a matrix with that information. So when I run my hashtable and then insert my text files for each type I get on my command window something like that:

edit

F = listdlg('PromptString','Different types', 'SelectionMode',...
    'single', 'ListString',E, 'Name','Select a type','ListSize',[230 130]);

        [files,path] = uigetfile ('*.txt','Select your text files',...
            'MultiSelect','on');

where E is just the user's input which in this case is pink, purple and yellow.

%save for each type the user enters the corresponding text files he
            %wants to train
            %A Map object is a data structure that allows you to retrieve values 
            %using a corresponding key. Keys can be real numbers or text strings
            %and provide more flexibility for data access than array indices, 
            %which must be positive integers. Values can be scalar or nonscalar arrays.



handles.map (E(F,:)) = files;
handles.map
a = handles.map.values
b = handles.map.keys
handles.map.size
b = 

    {1x3 cell}    {1x6 cell}    {1x4 cell}


a = 

    'pink  '    'purple'    'yellow'

So I want now to the the total number of b to be the number of rows on my matrix m ; so a total number of rows 14 and each bit from a to be a column; so a total of 3 columns. But I want to create a binary matrix where each column will identify different type. Finally I will have created a matrix like that:

Where the first 3 rows of the matrix says that there are 3text files of type pink, the next 6 rows : 6 text files of type purple and the last 4: 4 text files of type yellow.

I hope that now is more clear. :)

Any help would be appreciated! :) x

Still not entirely clear; could you explain in more detail what the keys and values in your Map are? It would be best if you could create a minimal working example where you populate the Map with values, and then explain what output you would like. — devrobf
– devrobf, Commented Aug 14, 2013 at 10:48
thanks.. I have just edited it, I hope I was more clear now :) — Chrysovalando
– Chrysovalando, Commented Aug 14, 2013 at 11:09

Dan · Accepted Answer · 2013-08-14 11:35:21Z

2

So something like this?

val = cellfun(@length, b)';

m = 0;
for v = 1:size(val)
  m(end:end+val(v)-1,v) = 1;
end

edited Aug 14, 2013 at 11:35

answered Aug 14, 2013 at 11:10

Dan

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4 Comments

Dan Over a year ago

@Chrysovalando Wait... I thought you were saying b is your input and m is your desired output? Is that not the case

Chrysovalando Over a year ago

yes that is the case.. but when I use it in my code I only get one column of 1 but if I run it as it is I get what I want.. So probably I have something crazy going on my code :/

Chrysovalando Over a year ago

please see now my original post :)

Dan Over a year ago

@Chrysovalando Try and transpose my original val calc, see my edit

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