retrieve binary data back
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Elysi Cochin
el 17 de Mzo. de 2019
Comentada: Rik
el 18 de Mzo. de 2019
I have a binary data as
00 10 11 01 10 11 11 01
I need to replace this data as if
00 = 0
01 = 1
10 or 11 = *
if i get a new data as below
0 * * 1 * * * 1
can you suggest any way to get back the binary data
00 10 11 01 10 11 11 01
1 comentario
Rik
el 17 de Mzo. de 2019
Because your mapping is not unique, this is not possible. There is no way for the reverser to know if you * was 10 or 11, so it will not be able to return a perfect reconstruction.
Respuesta aceptada
Rik
el 18 de Mzo. de 2019
In case you wanted to accept my answer, I'll repost my code in an actual answer. However, I would suggest you use Madhan's solution. That one ignores the last bit if the length is odd (which might not be what you want), while my code throws an error.
data='0010110110111101';
replace_with={'00','0';'01','1';'10','*';'11','*'};
key_list=replace_with(:,1);
val_list=replace_with(:,2);
%check for even numbers:
if rem(numel(data),2)~=0
error('bits are not paired')
end
%convert to 1 pair per cell
datacell=mat2cell(data,1,2*ones(numel(data)/2,1));
% %loop through the replacer elements
% for n=1:numel(key_list)
% datacell(ismember(datacell,key_list{n}))=val_list(n);
% end
inds=cellfun(@(x) find_indices(x,key_list),datacell);
L= inds~=0;
datacell(L)=val_list(inds(L));
%convert back to a char array
result=cell2mat(datacell);
clc,disp(result)
%local function:
function inds=find_indices(element,key_list)
[~,inds]=ismember(element,key_list);
end
3 comentarios
Rik
el 18 de Mzo. de 2019
No, that reversal is not possible, unless there is a known pattern to your data.
Más respuestas (2)
madhan ravi
el 18 de Mzo. de 2019
Editada: madhan ravi
el 18 de Mzo. de 2019
c=strjoin(regexp(c,'\d{2}+','match')); % for safety measure
% ^----- your original binary data
Actual_datas={'00','01','10|11',' '};
Replace_with={'0','1','*',''};
D=regexprep(c,Actual_datas,Replace_with)
0 comentarios
Image Analyst
el 17 de Mzo. de 2019
Use strrep(). Try this:
str = '00 10 11 01 10 11 11 01'
% Make the length a multiple of three.
while rem(length(str), 3) ~= 0
str = [str, ' '];
end
% Make replacements
% 00 = 0
% 01 = 1
% 10 or 11 = *
str = strrep(str, '00 ', '0 ')
str = strrep(str, '01 ', '1 ')
str = strrep(str, '10 ', '* ')
str = strrep(str, '11 ', '* ')
% If desired, trim off any trailing or leading spaces.
str = strtrim(str)
4 comentarios
Rik
el 17 de Mzo. de 2019
Editada: Rik
el 17 de Mzo. de 2019
Judging by the track record of Elysi I wouldn't call them a beginner anymore, but just in case other/future readers need an example (which assumes the spaces are not in the actual data):
data='0010110110111101';
replace_with={'00','0';'01','1';'10','*';'11','*'};
key_list=replace_with(:,1);
val_list=replace_with(:,2);
%check for even numbers:
if rem(numel(data),2)~=0
error('bits are not paired')
end
%convert to 1 pair per cell
datacell=mat2cell(data,1,2*ones(numel(data)/2,1));
% %loop through the replacer elements
% for n=1:numel(key_list)
% datacell(ismember(datacell,key_list{n}))=val_list(n);
% end
inds=cellfun(@(x) find_indices(x,key_list),datacell);
L= inds~=0;
datacell(L)=val_list(inds(L));
%convert back to a char array
result=cell2mat(datacell);
clc,disp(result)
%local function:
function inds=find_indices(element,key_list)
[~,inds]=ismember(element,key_list);
end
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