Finding contents of cell in another one

조회 수: 24(최근 30일)
babak
babak 2012년 9월 19일
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
  댓글 수: 10
Matt Tearle
Matt Tearle 2012년 9월 24일
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)

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답변(4개)

Jan
Jan 2012년 9월 19일
편집: Jan 2012년 9월 19일
file1 = randi(1000, 1, 1109); % Test data
rxn = randi(1000, 1170, 7);
match = ismember(rxn(:, 5), file1);
result = rxn(match, :);
  댓글 수: 6
Jan
Jan 2012년 9월 19일
편집: Jan 2012년 9월 19일
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.

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Shane
Shane 2012년 9월 19일
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
  댓글 수: 6
Jan
Jan 2012년 9월 19일
Shane's code assumes also, that your data are double matrices. If you are talking about cell strings, please specify this. A small set of example data would be a good idea also.

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Matt Tearle
Matt Tearle 2012년 9월 19일
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),:);
  댓글 수: 7
Matt Tearle
Matt Tearle 2012년 9월 20일
From your description of the data, it sounds a lot like you have a cell array (as Jan says, "2.11.7.8" could not be stored as a numeric type, so it is probably a string, and you mentioned that the contents of rxn are "some words"). If rxn is a 1170-by-7 cell array, and file1 is a 1109-by-1 cell array, then the code I gave should still work.

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Javier
Javier 2012년 9월 21일
편집: Javier 2012년 9월 21일
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

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