Indexing matrix with multiplication

Hi,

it is the same to apply a mask as an index or multiplying?

Say, I've a 2D matrix T. When I need the values that obey some condition I normally use an index as:

mask = T > 20;
T(mask)

Recently, I came across the problem that I need to keep the dimensions of the original matrix when I apply the index. Hence my question, is the previous code example the same as the following?

mask = T > 20;
T.*mask

Thanks

Answer 1

[Cris LaPierre]

It's not exactly the same. To keep the dimensions, the second option places a zero in each location that does not meet the mask criteria.

T = randi(50,5)

T = 5×5

48 26 42 16 11 13 13 2 3 3 21 38 19 42 19 27 4 17 34 39 35 45 19 21 26

mask = T>20

mask = 5×5 logical array
   1   1   1   0   0
   0   0   0   0   0
   1   1   0   1   0
   1   0   0   1   1
   1   1   0   1   1

T(mask)

ans = 13×1

48 21 27 35 26 38 45 42 42 34

T.*mask

ans = 5×5

48 26 42 0 0 0 0 0 0 0 21 38 0 42 0 27 0 0 34 39 35 45 0 21 26

[user20912]

I see. Just a quick question. If I convert the second option to a vector without zeros, I'll have the same as the other one. Right?

[Walter Roberson]

T = randi(50,5)

T = 5×5

22 9 38 30 43 17 45 42 47 37 16 47 1 14 31 39 20 45 31 20 48 15 33 12 7

mask = T>20

mask = 5×5 logical array
   1   0   1   1   1
   0   1   1   1   1
   0   1   0   0   1
   1   0   1   1   0
   1   0   1   0   0

out1 = T(mask)

out1 = 15×1

22 39 48 45 47 38 42 45 33 30

out2 = T.*mask

out2 = 5×5

22 0 38 30 43 0 45 42 47 37 0 47 0 0 31 39 0 45 31 0 48 0 33 0 0

out3 = nonzeros(out2)

out3 = 15×1

22 39 48 45 47 38 42 45 33 30

isequal(out1, out3)

ans = logical
   1

[user20912]

Thank you.

Answer 2

[Matt J]

No. You can easily see they are not the same by comparing them yourself:

T=randi(30,5)

T = 5×5

13 7 1 23 19 24 3 26 4 24 17 18 11 10 18 7 6 5 17 30 2 8 16 12 6

mask=T>20;

T(mask)

ans = 5×1

24 26 23 24 30

T.*mask

ans = 5×5

0 0 0 23 0 24 0 26 0 24 0 0 0 0 0 0 0 0 0 30 0 0 0 0 0