Indexing for a double in cell array matlab - matlab

Question

I have a cell array called Event of 179X59 entries and each element is a <1X14> double
so for example Event{1,1} is a 14 bit binary number in it.. like 0 1 0 0 0 0 0 0 0 0 1 0 0 0 which is spread out through columns 1 to 14 so each columns has a bit.
Now my task is to index through each element of the cell array and get into the double and assign a particular alphabet say a for the first bit if I see 1.
So if my alphabets are A through N for a binary 1 1 1 1 1 1 1 1 1 1 1 1 1 1, need to get
A B C D E F G H I J K L M N instead of that binary number.
So for the example give in the second line 0 1 0 0 0 0 0 0 0 0 1 0 0 0 I need to get
0 B 0 0 0 0 0 0 0 0 K 0 0 0
and in the end should return B,K removing zeros.
I have tried assigning each element to a matrix and tried using strrep but its not helping me out.

Answer 1

Without removing zeros:
alphabet = [11:24]; %// numbers. Replace values as needed
resultzeros = bsxfun(#times, alphabet, vertcat(Event{:}));
resultzeros = reshape(mat2cell(resultzeros, ones(1, size(resultzeros,1))), size(Event));
If you then need to remove zeros:
result = cellfun(#(x) nonzeros(x).', resultzeros, 'uni', 0);
Or directly:
result = cellfun(#(x) alphabet(x~=0), Event, 'uni', 0);
Example:
>> Event = {[0 0 1 0], [1 1 1 1]; [1 1 0 0], [0 0 1 1]};
>> alphabet = [4 5 6 7];
gives
result{1,1} =
6
result{2,1} =
4 5
result{1,2} =
4 5 6 7
result{2,2} =
6 7

Answer 2

Code
cellfun(#(x) cellstr(char(nonzeros(x.*[65:65+numel(x)-1])))', Event,'uni',0)
Results
For a sample Event:
Event = {
[0 1 0 1 0 0 1] [0 1 0 1 0];
[0 1 1 1 0] [0 0 0 1 1 1 0 0 0 0 1 1 0 1]}
Output -
>> Event{:}
ans =
0 1 0 1 0 0 1
ans =
0 1 1 1 0
ans =
0 1 0 1 0
ans =
0 0 0 1 1 1
>> out{:}
ans =
'B' 'D' 'G'
ans =
'B' 'C' 'D'
ans =
'B' 'D'
ans =
'D' 'E' 'F'
Of course, the number of columns and rows in Event at its "upper level" are maintained with out -
>> Event
Event =
[1x7 double] [1x5 double]
[1x5 double] [1x14 double]
>> out
out =
{1x3 cell} {1x2 cell}
{1x3 cell} {1x6 cell}

Answer 3

I suppose that time is not critical, so you can at least loop. Rough sketch:
alphabet = {A, B, C, D, E...}
j= 0;
for i=1:nBit
if( Event{1,1}(i) == 1)
EventAlphabet{1,1}{j} = alpabet{i};
j = j + 1;
end
end
Your question is a little bit unclear, so my answer is a rough sketch that may help you nontheless.