Convert space separated string table to cell?

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Bjoern
Bjoern 2014-6-17
评论： Cedric 2014-6-17
Hello :)
I can't seem to figure out how to convert a string table without using a (textscan etc.) loop,
from :
Table = ['A2 6C 33 04 00 81 00 80';'3F 11 65 01 0A '];
to :
Result = {'A2' '6C' '33' '04' '00' '81' '00' '80';'3F' '11' '65' '01' '0A' ' ' ' ' ' '};
Please note that "Table" is fixed-width Nx23 which could simplify things.
FYI, the end-goal is to convert from hex to decimal to cell-table:
Dec = [162 108 51 4 0 129 0 128; 63 17 101 1 10 0 0 0];
  1 个评论
José-Luis
José-Luis 2014-6-17
编辑：José-Luis 2014-6-17
Do you need to distinguish between a zero caused by a space and the string 00?

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采纳的回答

Cedric
Cedric 2014-6-17
编辑：Cedric 2014-6-17
>> Dec = reshape( hex2dec( regexp( reshape( Table', 1, [] ), '..\s?', ...
'match' )), 8, [] )'
Dec =
162 108 51 4 0 129 0 128
63 17 101 1 10 0 0 0

更多回答（3 个）

Azzi Abdelmalek
Azzi Abdelmalek 2014-6-17
Table = ['A2 6C 33 04 00 81 00 80';'3F 11 65 01 0A ']
Table(:,3:3:end)=[];
[n,m]=size(Table);
[bb,aa]=meshgrid(1:2:m,1:n);
out=arrayfun(@(x,y) hex2dec(Table(x,y:y+1)),aa,bb)
  3 个评论
显示 1更早的评论
Bjoern
Bjoern 2014-6-17
Thank you soo much Azzi, very clever :)
However, even though I didn't explicitly mention it, I need it to be scalable. When I use a larger "Table" (60000x23 char) arrayfun takes a very long time to finish.
Instead, by passing a 60000x8 cell hex2dec is very fast. The problem is that I can't figure out how to construct the cell.
As reference, please try the below to see how much faster it is:
Cells = {'A2' '6C' '33' '04' '00' '81' '00' '80';'3F' '11' '65' '01' '0A' ' ' ' ' ' '};
Cells = repmat(Cells, 30000, 1); % Create large cell
out = uint8(hex2dec(Cells)); % Convert hex to dec
out = reshape(out, 60000, 8); % Reshape to desired shape
Azzi Abdelmalek
Azzi Abdelmalek 2014-6-17
Table = ['A2 6C 33 04 00 81 00 80';'3F 11 65 01 0A ']
Table(:,3:3:end)=[];
[n,m]=size(Table);
[bb,aa]=meshgrid(1:2:m,1:n);
out=arrayfun(@(x,y) Table(x,y:y+1),aa,bb,'un',0)

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Andrei Bobrov
Andrei Bobrov 2014-6-17
编辑：Andrei Bobrov 2014-6-17
nn = size(Table,1);
a = cellfun(@(x)regexp(x,'\w*','match'), num2cell(Table,2),'un',0);
n = cellfun(@numel,a);
mm = max(n);
m = mm - n;
b = arrayfun(@(x)[hex2dec(a{x});zeros(m(x),1)]', (1:nn)','un',0);
out = cat(1,b{:});
Azzi Abdelmalek
Azzi Abdelmalek 2014-6-17
编辑：Azzi Abdelmalek 2014-6-17
Table = ['A2 6C 33 04 00 81 00 80';'3F 11 65 01 0A ']
[n,m]=size(Table)
Table(:,3:3:end)='/';
ss=regexp(num2cell(Table,2),'/+','split')
out=cellfun(@hex2dec,reshape([ss{:}],[],n)')
  2 个评论
Bjoern
Bjoern 2014-6-17
Awesome, I really appreciate it!
Comparing this solution to Cedric Wannaz below, his is actually about 3 times faster so I will have to pick that one as "the answer". Thanks again!
Cedric
Cedric 2014-6-17
Careful, "faster" is not always "better"! My solution assumes that you know a priori the max width in terms of number of columns.

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