# Finding contents of cell in another one

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Hi I have a matrix that is called file1=[1109x1] and another matrix that is caled rxn=[1170x7] i need to know if the contents of the 'file1' exist in column#5 of the rxn and if the answer is yes, the whole content of the related row in 'rxn' would be copied to a new matrix, what should i do? Please help me on this issue Thank you alot!

[EDITED, Jan, moved from comment sections]

Here is what my data exactly is: File1 contains 1column x 1109rows and each cell contains numbers like this: 2.11.7.8

Rxn contains 7column x 1170rows. In every cells of column#5 exists numbers like what it was in file1 (2.11.7.8) I am going to compars every cells of file1 with cells of rxn column#5 and if the numbers were exactly the same, the whole content of the related row in rxn would be copied to a new matrix Please note that i need the exact properties of each cell

##### 10 Comments

Matt Tearle
on 24 Sep 2012

rxn is a 2-dimensional cell array, so you should be indexing into the 5th column, which you are not. However, you should not need to do this in a loop. The code I provided earlier should work. If not, please explain what isn't working and why. For starters, what do you get if you do this:

idx = ismember(rxn(:,5),file1);

nnz(idx)

### Answers (4)

Jan
on 19 Sep 2012

Edited: Jan
on 19 Sep 2012

file1 = randi(1000, 1, 1109); % Test data

rxn = randi(1000, 1170, 7);

match = ismember(rxn(:, 5), file1);

result = rxn(match, :);

##### 6 Comments

Jan
on 19 Sep 2012

Edited: Jan
on 19 Sep 2012

In your question I find "rxn=[1170x7]", which usually means that the data is a matrix of type double. If you use a different input, it would be a really good idea to explain this explicitely. Otherwise guessing what you are looking for wastes your and my time.

Of course my method replies numbers. I used numbers as test data as explained in the comment. It should not be to hard to omit the two lines containing randi, such that your data are processed.

Shane
on 19 Sep 2012

You could use ismember and indexing values to determine which rxn columns have corresponding values within file1 but this can also be solved using a set of loops as shown:

count= 0;

for i = 1 : length(file1)

for j = 1 : length(rxn)

if file1(i) == rxn(j,5)

count = count + 1;

finalMatrix(count,:) = rxn(j,:);

end

end

end

The finalMatrix will be a matrix containing all of the rows of rxn that have their column 5 values identified within file1

##### 6 Comments

Jan
on 19 Sep 2012

Matt Tearle
on 19 Sep 2012

If I understand your intent correctly, you want every row of rxn for which the value in the 5th column is one of values in the vector file1. If so:

result = rxn(ismember(rxn(:,5),file1),:);

##### 7 Comments

Matt Tearle
on 19 Sep 2012

As Jan explained, the lines

file1 = randi(1000, 1, 1109);

rxn = randi(1000, 1170, 7);

create some test data (because, funnily enough, we don't have access to your actual data). Given your description of your data (an 1109-element vector and a 1170-by-7 matrix),

result = rxn(ismember(rxn(:,5),file1),:);

does what you asked. If it is not working as you expected, you need to explain what you expected, and exactly what your data looks like and what you are doing to it.

BTW, that line of code should work on virtually any data type. However, if you're working with cell arrays, it will depend on the contents. Again, you need to describe your data precisely.

Matt Tearle
on 20 Sep 2012

Javier
on 21 Sep 2012

Edited: Javier
on 21 Sep 2012

Hello Babak

Ok big issue here but not impossible. The following procedure works in Matlab R2012a.

Step 1

Import 3 files: One will be Rxn (7 columns), File1 and other called "Var1" will be column 5 of Rxn. When you import this to Matlab you will have string information not data. Because the information is in Excel (for example) use the function num2str in the import process.

Step 2

Now, for each string in File1, we begging the search in Var1. This will work for the search of the first string in File1.

for i=1:1170

R(i,1)=sum(find(A(i,:)==File1(1,:)),2);

end

How R works. For 2.11.7.8 we have 8 characters. Then, if R= 36 all the string match and we find what we are looking for. If you remove the sum, an error will appear. At the end I provide a code that you can check. If the number of characters change, establish a condition in Step 3 according to string data search.

Step 3

R is column vector. Find in R the value 36. Remember, 2.11.7.8 we have 8 characters, and the sum of 1 to 8 is 36.

RR=find(R==36) %Now you have the rows

Step 4

New matrix and copy process

for i=1:size(RR,1)

Newm(i,:)=Rxn(RR(i,1),:)

end

Code

%Step 1 Search a value (A(1,:)). This is the first value in A. It could be the first element of File1

a=randn(100,4);

A=[num2str(abs(round(a(:,1)))),repmat('.11.',100,1),num2str(abs(round(a(:,2))))];

B=[num2str(abs(round(a(:,3)))),repmat('.11.',100,1),num2str(abs(round(a(:,4))))];

%My Rxn matrix (2 columns A,B)

Rxn=[A,B];

Var1=A; %search in this column

File1=A(1,:) %search this data

%Step 2

for i=1:size(Rxn,1)

R(i,1)=sum(find(Var1(i,:)==File1(1,:)),2);

end

%Step 3

RR=find(R==21) %Data are like 1.11.1, 6 characters and sum(1 to 6)=21

%Step 4

for i=1:size(RR,1)

Newm(i,:)=Rxn(RR(i,1),:)

end

Feel free to make a comment and verify that the code supplied works before comment.

Regards

Javier

##### 0 Comments

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