cell contents in a loop_fangjun's answer

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 Comments

Please format the code: http://www.mathworks.com/matlabcentral/answers/13205-tutorial-how-to-format-your-question-with-markup#answer_18099
Why do you ask the same question in two different places?

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 Accepted Answer

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

More Answers (3)

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
Jan
Jan on 15 Aug 2011
Replace "for k=1:length(vars)" by "for k=1:numel(vars)".
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 Comments

Check again. The result from the code is correct!
No, he want to run strfind(vars,files).
@Oleg, in your solution, you also used strfind(files,vars)?!
Yes because there's no way to find a longer string in a shorter. To emulate the effect of strfind(vars,files) I adjust the column position to n.
No. The issue is not that! The issue is who the logical index should be corresponding to! From all his description and example, I thought the index should be corresponding to the long strings in the variable "files". For any one of the long strings in the variable "files", if any of the short strings is found, the corresponding logical index should be true!
But apparently NOT! What he wants is a logical index with the same number of rows as "files" but same columns as "vars".
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.
@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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