finding common elements in a matrix

I believe there should be a simple answer but I could not get it myself.

Assume there is a matrix say A =

    1.3693    1.6266
    0.2054    1.3820
    1.4125    0.4844
    0.0734    0.7480
    1.0499    1.0451

There is one more matrix

B =

1.3693    2.0920
1.9255    1.3820
1.4125    1.8541
0.0734    1.4454
1.0499    1.4112

Now, How do I find common elements in every row of these two matrices?? I must get a final coloumn vector of only the 'common element' of every row of these two matrices.

Any quick reply is highly appreciated

[Matt J]

You cannot organize the result into a vector unless you are sure that there is going to be exactly 1 common element in each row, which is not the case in your example.

[Matt J]

Also, will the "common element" be exactly equal in both A and B or do you need to worry about floating point errors?

[mahesh bvm]

Yes we need to consider the floating point numbers in each row too....

Answer 1

[Jos (10584)]

I have to make a few assumptions as your question and comments are not clear.

Case A - assumptions: (1) exactly one common element per row (2) in the same column in A and B

tf = abs(A-B) < tol
C = A(tf)

Case B - assumptions: (1) exactly one common element per row (2) that element may be at different columns between A and B

tf = abs(A-B) < tol || abs(fliplr(A)-B) < tol
C = A(tf)

Case C - assumptions: (1) zero or one common element per row (2) in the same column in A and B

tf = abs(A-B) < tol
[r,c] = find(tf) 
C = zeros(size(A,1),1) 
C(r) = A(tf)

Case D - Assumptions: (1) zero or one common element per row (2) that element may be at different columns between A and B

...

[Salar]

Hello,

I'm trying to use your code for Case B, but it keeps giving me this error

" Operands to the || and && operators must be convertible to logical scalar
values.

Error in Script_HW5 (line 71)
tf = abs(LA-LB) < 1e-4 || abs(fliplr(LA)-LB) < 1e-4 "

I'd really appreciate it if you could help me with this.

[Stephen23]

@Salar: that is a little bug in Jos' answer. Use or | instead of short-circuit or ||:

tf = abs(A-B) < tol | abs(fliplr(A)-B) < tol
                    ^

[Salar]

Stephen,

Thank you! but this code still doesn't spit out the common ones in order. Can you tell by looking at the code why is that? and is there any way I could get them all in order? because I need to graph them vs Time.

Thank you in advance

[Salar]

And also to be honest, I think this code doesn't do the work see :

a =

    90    50
    20    11
     0    87
    11   110

>> b= [ 41 90; 11 31;87 0; 11 -2]

b =

    41    90
    11    31
    87     1
    11    -2

>> tf = abs(a-b) < 1e-1 | abs(fliplr(a)-b) < 1e-1; >> LambdaS = a(tf)

LambdaS =

    20
     0
    11
    50
    87

The output that I need to get is

LambdaS : 90 11 87 11

[Stephen23]

You need to take into account that MATLAB operates column-wise:

>> a = [90,50;20,11;0,87;11,110]
>> b = [41,90;11,31;87,1;11,-2]
>> at = a.';
>> bt = b.';
>> xt = at==bt | at==bt([2,1],:)
>> at(xt)
ans =
   90
   11
   87
   11

Answer 2

[Patrik Ek]

You may use the function ismember. The syntax is

[member,ind] = ismember(A,B);

The first output shows which elements in A that belongs to B and the second output shows at which linear index in B it appears. Then it would be easy to determine at which row it appears since the element numbering in a matrix in matlab is columnwise. The row should just be

 rows = mod(ind,nrows);
 rows(rows==0 && ind>0) = nrows;

This may however not result in a column vector unless there are exactly 1 common index per row.

BR Patrik

[Matt J]

See also ISMEMBERF on the File Exchange if you need to apply a tolerance for floating point errors.

[Patrik Ek]

That would work I guess, but it is also possible to do a round of on A and B eg,

 A = round(10*A)/10; (one decimal)

and analouge on B. Where 10 is here 1/tol. This will work for any tolerance.

 A = round(1/tol*A)*tol;

However, this is more like a truncation. It truncates the values in A to a value that is a multiple of 1/tol.

[Salar]

Hello,

What are we supped to put for "nrows" ? when I use your code as it is it says undefined function "nrows" and when I plug in my number of rows for it the matrix rows that I get is just not correct. thanks

Answer 3

[David Sanchez]

A = [    1.3693    1.6266;
    0.2054    1.3820;
    1.4125    0.4844;
    0.0734    0.7480;
    1.0499    1.0451];
B = [1.3693    2.0920;
1.9255    1.3820;
1.4125    1.8541;
0.0734    1.4454;
1.0499    1.4112];

C=(A==B);
D=A.*C;
E=sum(D,2);
E =

    1.3693
    1.3820
    1.4125
    0.0734
    1.0499

% single line code
E2 = sum(A.*(A==B),2);
E2 =

    1.3693
    1.3820
    1.4125
    0.0734
    1.0499

[mahesh bvm]

Your answer is very impressive. Thanks a ton. However we found an error in our program in which some of the numbers is showing as '0' in final output... Check out this program

n =5;
x1=exprnd(1,n,1);
x1
z1 = normrnd(0,1,n,1);
z1
x11=x1+z1;
x11
k1=x11
x2=exprnd(1,n,2);
x2
z2 = normrnd(0,1,n,2);
z2
x22=x2+z2;
x22
k2=max(x22')
k2
k=k2'
k
k(:,2) = k
s=k-z2
C =(s==x2);
D = s.*C;
E = sum(D,2);
E

Let me know what exactly is going wrong in the program. Thanks a lot in advance

[Matt J]

If there is no common element in a row, you will get 0 in that row.

[Patrik Ek]

This code snippet does only consider elements that has the exact location. It will not react on for example,

 A = [1 2;
      3 4];
 B = [2 1;
      4 3];

Which will yield a vector,

 E = [0;
      0];

Also this is also requires that each row has exactly one common value.