find row with certain values

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

[Daniel]

thank you all for the answers

since all of your answers do what I wanted I'll choose the "winner" by:

1. short code

2. running time

3. running time on large matrices and/or many numbers to find

[Naz]

I deleted my answer so it will be easier for you to make a decision.

Answer 1

[Robert Cumming]

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));

[kd p]

output doesn't show for this!

[the cyclist]

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.

Answer 2

[Jan]

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!]

[the cyclist]

I think the "any" here should be over dimension 2, not dimension 1.

Answer 3

[Malcolm Lidierth]

>> 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

Answer 4

[the cyclist]

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)

Answer 5

[Sean de Wolski]

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.

[Soyy Tuffjefe]

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

Answer 6

[Ankit]

>> 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).