ismember has different length depending on argument order

I have 3 arrays with two columns each. Array 2 contains indices of matches between column1 of array 1 and 3.

Now I only want to keep rows in arrays 1 and 3, which are represented in array 2.

I'm using this code right now:

mask1 = ismember(array2(:,1), array1);
mask2 = ismember(array2(:,2), array3);
mask1 = logical(mask1.*mask2)
array2 = array2(mask1,:);

mask3 = ismember(array1, (array2(:,1)));
array1 = array1(mask3,:);
mask4 = ismember(array3, (array2(:,2)));
array3 = array3(mask4,:);

The problem is, that

ismember(array1, (array2(:,1)))

has a different length than

ismember((array2(:,1)), array1)

even I just switched the arguments position.

Is there anyone who can explain to me why this can happen and how to fix it?

Answer 1

[Felix Bertoli]

Thanks! I figured it out. It was, because some numbers were duplicate, so the results were confusing for me.

[Image Analyst]

Again, you can simplify it and just use logical indexing rather than the confusing ismember(). It will be simpler and more intuitive.

Answer 2

[Yongjian Feng]

Type

help ismember 

in the matlab command line window to get quick help.

Basically the answer is the same length as the first input argument. When you have

ismember([a, b], otherArrary)

The answer is an array of length of 2 for this example. Each element in the returned array shows whether the corresponding elelment in the first import array is a member of the second input array.

Answer 3

[Image Analyst]

You don't even need the admittedly confusing ismember() function. Assuming the matrices are regular numerical arrays (not cell arrays of strings), you can use simple logical indexing:

% Create sample data matrices m1 and m3:
m1 = [(1:9)', (11:19)']
m3 = [[0,2,0,4,0,6,0,8,0]', (11:19)']
% Find where column 1 of m1 matches column 1 of m3 and store in m2:
m2 = m1(:, 1) == m3(:, 1)
% Keep only rows in m1 and m2 that are in m3
rowsToKeep = m2;
m1 = m1(rowsToKeep, :)
m3 = m3(rowsToKeep, :)

You'll see:

m1 =

1 11

2 12

3 13

4 14

5 15

6 16

7 17

8 18

9 19

m3 =

0 11

2 12

0 13

4 14

0 15

6 16

0 17

8 18

0 19

m2 =

9×1 logical array

0

1

0

1

0

1

0

1

0

m1 =

2 12

4 14

6 16

8 18

m3 =

2 12

4 14

6 16

8 18

Answer 4

[Matt J]

Wouldn't it just be,

mask = ismember(array2(:,1), array1) & ismember(array2(:,2), array3);

fn=@(z) z(mask,:);

array1=fn(array1);
array2=fn(array2);
array3=fn(array3);