# find row with certain values

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Daniel on 20 Dec 2011
Commented: Soyy Tuffjefe on 27 Aug 2019
Hello I am looking for a (simple) way to get the index of a row in which two (or n) values exist
example: looking for 4 and 5 in
[1 5 6; 5 4 3; 9 4 2]
will give me 2, because only row 2 has both 4 and 5 in it
Thanks
Daniel
##### 2 CommentsShowHide 1 older comment
Naz on 22 Dec 2011
I deleted my answer so it will be easier for you to make a decision.

Robert Cumming on 21 Dec 2011
similar to the intersect answer - but I recoded intersect as its quite slow:
x=[1 2 3;4 5 6;3 2 1];
[a b]=find(x==4);
[c d]=find(x==5);
index = c.*NaN;
for kk=1:length(c)
check = a(a==c(kk))';
if ~isempty ( check )
index(kk) = check;
end
end
output = index(~isnan(index));
##### 2 CommentsShowHide 1 older comment
the cyclist on 6 Dec 2017
Edited: the cyclist on 6 Dec 2017
What do you mean by "doesn't show"?
That code calculates the output (at least for me). Nothing is displayed to the screen because the line ends with a semicolon, which suppresses this display.
You can removed that semicolon, or just type
output
to see the result.
If that code is not even calculating the output for you, then please post the full error message you are getting.

### More Answers (5)

Jan on 20 Dec 2011
X = [1 5 6; 5 4 3; 9 4 2]
index = and(any(X == 4, 2), any(X == 5, 2));
[EDITED: Apply ANY along 2nd dim, thanks Cyclist!]
##### 1 CommentShowHide None
the cyclist on 20 Dec 2011
I think the "any" here should be over dimension 2, not dimension 1.

Malcolm Lidierth on 20 Dec 2011
>> x=[1 5 6; 5 4 3; 9 4 2];
>> [a b]=find(x==4);
>> [c d]=find(x==5);
>> intersect(a,c)
ans =
2

the cyclist on 20 Dec 2011
Trust but verify this code:
x = [1 5 6; 5 4 3; 9 4 2]
want(1,1,:) = [4 5];
indexToDesiredRows = all(any(bsxfun(@eq,x,want),2),3)
rowNumbers = find(indexToDesiredRows)

Sean de Wolski on 20 Dec 2011
How about ismember with a for-loop?
doc ismember
Example
A = [1 5 6; 5 4 3; 9 4 2];
want = [4 5];
szA = size(A,1);
idx = false(szA,1);
for ii = 1:szA
idx(ii) = all(ismember(want,A(ii,:)));
end
idx will be a logical vector of rows with 4 and 5. If you want the numeric values:
find(idx)
This will be the most scalable method if say you want 10 different numbers to be present in each row. Calling any/ intersect / all that many times and/or using that many dimensions is not really feasible.
##### 1 CommentShowHide None
Soyy Tuffjefe on 27 Aug 2019
Suppose that A = [1 5 6 13 22; 9 5 4 6 37; 7 1 4 22 37];
and want = [5 6; 1 22; 4,37];
Please, Can you modify your code for find two o more rows; or any idea for me to try this job with your code I have a want matriz of 100x4 and A matrix of 1000x5 order.
Thanks

Ankit on 20 Apr 2013
>> x = [1 2 3 4 1 2 2 1]; find(sum(bsxfun(@eq,x',[1:3]),2)==1) ans =
1
2
3
5
6
7
8
Similar things can be done for an array rather than just a vector (x above).

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