Effacer les filtres
Effacer les filtres

cell contents in a loop_fangjun's answer

2 vues (au cours des 30 derniers jours)
YOGESH
YOGESH le 15 Août 2011
Hello all, I have a vars = cell(1,9) like following, the code works fine when there is a single row.
vars = {'OKR_evapo','MDF_albedo','HDU_wind','GDf_temp','MPGF_humidity', 'MDF_pressure','YKK_temperature','XRYO_rain','MPIO_radiation'};
files={'myd_11_MDF_albedo_a54576' 'mod_19_GDf_temp_vhk464567' 'mcd_13_MPGF_humidity.sdjfg590856' 'myd_11_MDF_pressure_46358' 'cyd_11_YKK_temperature_a54576' 'dod_13_XRYO_rain_vhk464567' 'ecd_11_MPIO_radiation.sdjfg590856'};
Index=false(size(vars));
for k=1:length(vars)
for j=1:length(files)
if strfind(files{j},vars{k})
Index(k)=true;
break;
end
end
end
How can I read all the rows in a loop if I have data/filenames in cell(10,9) instead of the single row. There are multiple rows in a cell of size 10*9. I should get a matrix of size 10*9 containing 1 and 0s.
  2 commentaires
Oleg Komarov
Oleg Komarov le 15 Août 2011
Please format the code: http://www.mathworks.com/matlabcentral/answers/13205-tutorial-how-to-format-your-question-with-markup#answer_18099
Fangjun Jiang
Fangjun Jiang le 15 Août 2011
Why do you ask the same question in two different places?

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Réponse acceptée

Fangjun Jiang
Fangjun Jiang le 15 Août 2011
You need to change the last part of the code as below.
Index=false(size(files));
for j=1:size(files,1)
for k=1:size(files,2)
for n=1:length(vars)
if strfind(files{j,k},vars{n})
Index(j,k)=true;
break;
end
end
end
end
Or looping this way to guess what is expected.
vars = {'ind' 'ger' 'us' 'swis' 'kor' 'ch'};
files = {'india' ,'germany','usa' ,'korea','','';...
'germany','usa' ,'china','' ,'','';...
'usa' ,'swiss' ,'china','' ,'',''};
N_row=size(files,1);
N_col=size(vars,2);
Index=false(N_row,N_col);
for j=1:N_row
for k=1:N_col
for n=1:size(files,2)
if strfind(files{j,n},vars{k})
Index(j,k)=true;
break;
end
end
end
end
Index =
1 1 1 0 1 0
0 1 1 0 0 1
0 0 1 1 0 1

Plus de réponses (3)

Oleg Komarov
Oleg Komarov le 15 Août 2011
vars = {'ind' 'ger' 'us' 'swis' 'kor' 'ch'};
files = {'india' ,'germany','usa' ,'korea','','';...
'germany','usa' ,'china','' ,'','';...
'usa' ,'swiss' ,'china','' ,'',''};
szF = size(files);
Index = false(szF);
for n = 1:numel(vars)
[r,c] = find(~cellfun('isempty',strfind(files,vars{n})));
c(:) = n;
Index(sub2ind(szF,r,c)) = true;
end
  1 commentaire
YOGESH
YOGESH le 15 Août 2011
Thanks Oleg.
This also works fine.

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Jan
Jan le 15 Août 2011
Replace "for k=1:length(vars)" by "for k=1:numel(vars)".

YOGESH
YOGESH le 15 Août 2011
Thanks. but it seems that it is not doing well. e.g. I created a cell (of size (3,6) like
vars = {'ind' 'ger' 'us' 'swis' 'kor' 'ch'};
files = cell([{'india'},{'germany'},{'usa'},{'korea'},{''},{''};...
{'germany'},{'usa'},{'china'},{''},{''},{''};...
{'usa'},{'swiss'},{'china'},{''},{''},{''}]);
If I run the code, I should get
Index = [ 1 1 1 0 1 0;
0 1 1 0 0 1;
0 0 1 1 0 1 ]
all I am getting is like
[ 1 1 1 1 0 0; 1 1 1 0 0 0; 1 1 1 0 0 0]
@Oleg, thanks for the formating tutorial. cheers
  7 commentaires
Fangjun Jiang
Fangjun Jiang le 15 Août 2011
To be honest, the OP's description and examples are not clear or even mis-leading. Anyway, I added another version which created the results expected. It is a tedious for-loop with strfind(). That's all.
Oleg Komarov
Oleg Komarov le 15 Août 2011
@Fangjun: well, you're rigth about the confusion. In fact my solution works for the specific example just because files is padded with empty strings.
I guess it won't be difficult to preallocate Index with m from files and n from vars.

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