find index of cell array
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Mohamuud hassan
am 11 Apr. 2015
Kommentiert: Stephen23
am 27 Mai 2015
Hello all, suppose i have cell array 6x1 which contains abstracts of articles, so each row of the cell contains 800 character. if i wan to find the position of str='b e h i n d' from that cell array,only b will be generated, so what is wrong my code. the cell arrays data is in the attachment or myabs.mat
load ('myabs.mat');
mydata=X;
[row,column]=size(mydata);
distance=zeros(row,column);
str='behind using a cover';
j=1;
for n=1:row
gec=char(mydata{n});
%for j=1:length(str)
for m=1:800
while j<length(str)
%for j=1:length(str)
if (gec(m)==str(j))
indexx(j)=m;
distance(n,m)=indexx(j);
end
break;
j=j+1;
end
end
end
Akzeptierte Antwort
Stephen23
am 11 Apr. 2015
Bearbeitet: Stephen23
am 11 Apr. 2015
There might be simpler ways to achieve this task, but this seems to work:
str = 'behind';
load myabs
for m = 1:numel(X)
Y = bsxfun(@eq,X{m}(:),str);
for n = 2:size(Y,2)
Y(:,n) = Y(:,n) & cumsum(Y(:,n-1))>0;
end
Z = find(any(diff([false(size(str));cumsum(Y)>0],1),2))
end
and when run it prints this in the command window:
Z =
135
149
235
236
239
266
Z =
72
73
127
136
160
174
Z =
163
170
172
173
177
221
Z =
40
53
65
66
76
112
Z =
216
217
267
279
295
319
Z =
170
175
178
196
197
234
3 Kommentare
Stephen23
am 12 Apr. 2015
Bearbeitet: Stephen23
am 12 Apr. 2015
To store the value of Z just preallocate the output array before the loop and us indexing to allocate the values on each loop iteration:
str = 'behind';
load myabs
Z = nan(size(X,1),numel(str));
for m = 1:numel(X)
Y = bsxfun(@eq,X{m}(:),str);
for n = 2:size(Y,2)
Y(:,n) = Y(:,n) & cumsum(Y(:,n-1))>0;
end
tmp = find(any(diff([false(size(str));cumsum(Y)>0],1),2));
Z(m,1:numel(tmp)) = tmp;
end
and then each row of Z corresponds to one cell of X:
>> Z
Z =
135 149 235 236 239 266
72 73 127 136 160 174
163 170 172 173 177 221
40 53 65 66 76 112
216 217 267 279 295 319
170 175 178 196 197 234
Or you might like to consider using my third solution using regexp, which detects all instances of the pattern string, not just the first one.
Weitere Antworten (1)
Stephen23
am 11 Apr. 2015
Bearbeitet: Stephen23
am 11 Apr. 2015
This task could be performed much faster and more robustly by using inbuilt functions. Why not just use strfind instead of these slow and buggy nested loops:
>> A = {'some line of words.','more text to search','some words with other words too','different information here'};
>> B = strfind(A,'words');
>> B{:}
ans =
14
ans =
[]
ans =
6 23
ans =
[]
which returns a cell array of indices giving the locations in each string where the search text 'words' was found. It correctly found the one instance in the first string, and two instances in the third string.
3 Kommentare
Stephen23
am 11 Apr. 2015
Bearbeitet: Stephen23
am 11 Apr. 2015
It is not clear what the difference is: you want to match every character sequentially, whereas strfind matches the whole string at once. What is the difference in your eyes? Do these characters have to be sequential?
Can you please give an example of one string to be searched and the pattern to be found, and show us how the pattern should be matched to the string.
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