How to find the exact location of a word in a string?

I have a string that 'chemical engineering is a challenge for electrical engineer'. I used to use 'strfind' function to find the exact location of the word‘engineer'. However, there is a problem that word engineering is also included in my results. How can i just get the location of word 'engineer' instead of 'engineering'.

 list='chemical engineering is a challenge for electrical engineer';
 temp=findstr(list,'engineer')

The result is

temp =

      10    52

Answer 1

[Star Strider]

This regexp call will pick up only ‘engineer’:

Str = 'chemical engineering is a challenge for electrical engineer';
idxs = regexp(Str, 'engineer\>')

idxs =
    52

[Yunfei Zhang]

Thank your for your answer. But it seems that regexp only works on known words. If i want to the frequency of dictionary's words in 20 documents, how can i achieve that? In my data, I have 20 documents(strings) and a direction containing 1099 words. The right function is vital for me to calculate the word frequency. Code below is my original plan and it has the problem mentioned above.

docum=zeros(20,1099);
  for i=1:size(pre,2)
      for j=1:size(word,1)
          temp=findstr(pre{i}{1},word{j});    
          if(~isempty(temp))     
          docum(i,j)=size(temp,2);  
          end
      end
  end

[Guillaume]

What's stopping you from building the regular expression for each word?

docum = zeros(numel(pre), numel(word));
for pidx = 1:numel(pre)
   for widx = 1:numel(word)
       docum(pidx, widx) = numel(regexp(pre{pidx}, sprintf('%s\>', word{widx})));
   end
end

[Star Strider]

Wow!

That is a completely different question!

Your ‘data.mat’ file has the ‘pre’ and ‘word’ variables. I have absolutely no idea what to do with them. You need to describe them in some detail, and how they relate to your problem.

Specifically, describe what in your problem (and variables) correspond to ‘Str’ in my code, and what correspond to ‘engineer’.

[Yunfei Zhang]

Sorry for confusion. Before asking this question, i simplified the question. 'Pre' is a cell matrix containing 20 documents and each document is a long string. 'word' is a cell matrix and containing 1099 words from these 20 document after removing stopwords. What I wanted to do is to construct a 20*1099 matrix to show each word's frequency in different documents and it leaded to the problem mentioned above that 'engineer' may have higher frequency than the 'engineering' for the word dictionary. However, I think the function you suggested is the correct way to find the location of each word. After finding the correct location of words like 'enginer', I can calculate the frequency of this word and indicate it at the corresponding location using code below. Guillaume provided me with a method of building the regular expression for each word and it works. However, it is based on the sacrifice of time to achieve higher accuracy and it takes much longer time when processing a large number of articles (when 'pre' contains a large number of long strings.)

if(~isempty(temp))     
        docum(i,j)=size(temp,2);  
end

[Guillaume]

You can prebuild the regular expressions before the loops if you wish.

word = strcat(word, '\>')

[Yunfei Zhang]

Thank you! It helps a lot for controlling the processing time as i also want to do the feature selection and clustering for my data.

Answer 2

[Guillaume]

Another option, since the words you're trying to match are always delimited by spaces or the end of the sentence (other punctuation marks are already embedded in the words), is to add a space to the end of each word and to the end of each sentences. That way 'engineer ' does not match 'engineering ' anymore:

tic
docum = zeros(numel(pre), numel(word));
word2 = strcat(word, {' '}); %strcat removes trailing ' ' if it's not in a cell array
pre2 = strcat(vertcat(pre{:}), {' '}); %why is your pre a cell array of 1x1 cell arrays?
for widx = 1:numel(word)
   docum(:, widx) = cellfun(@numel, strfind(pre2, word2{widx}));
end
toc

I'm not convinced it's going to be faster than regexp:

tic
docum = zeros(numel(pre), numel(word));
word2 = strcat(word, '\>'); 
pre2 =vertcat(pre{:}); %why is your pre a cell array of 1x1 cell arrays?
for widx = 1:numel(word)
   docum(:, widx) = cellfun(@numel, regexp(pre2, word2{widx}));
end
toc

In my testing they take both more or less the same time.

[Yunfei Zhang]

Both options are very fast. I used to build the regular expressions inside the loop so it took longer time. Sorry for confusion.

[Star Strider]

@Guillaume — Thank you. I had to be away for a few minutes.

[Guillaume]

@Yunfei, what is probably having the most effect on the processing speed is that I apply the regexp or strfind to all the sentences at once. There is only one loop, looping over the individual words.