If A is a 3x4 matrix, how do you find B if B=A(1:2)?

For example if A=[1 2 3 4; 5 6 7 8; 9 10 11 12]

B = A(1:2) = ???

And also how come the value of 1:2 in A(1:2, 2:3) is different than A(1:2)?

Answer 1

[Niels]

Hi Nate,

you have to be carefull by calling the contents of a matrix:

>> A=[1 2 3 4; 5 6 7 8; 9 10 11 12]
  A =
       1     2     3     4
       5     6     7     8
       9    10    11    12

1. if you refer to only 1 index such as A(1:2) the Matrix A is seen as vector whose colums are set one below another

>> A(:)
ans =
     1
     5
     9
     2
     6
    10
     3
     7
    11
     4
     8
    12
>> A(1:2)
ans =
     1     5

2. if you refer to both index, you would just get what is to be expected: first index for row, and 2. for column

>> A(1:2,2:3)
ans =
     2     3
     6     7

if you intended to get the first two rows:

>> A(1:2,:)
ans =
     1     2     3     4
     5     6     7     8

to check if B is part of A, i cant think of a simple solution, but you could try to use ismember

but care the first output containing the true/false expression does not say if the exact same matrix is to be found in A but if each value in B is to be found in A:

>> A
A =
     1     2     3     4
     5     6     7     8
     9    10    11    12
>> [yes_or_no,indices]=ismember(B,A)
yes_or_no =
  2×2 logical array % only if this matrix contains only 1s its possible
   1   1            % that B can be inside of A
   1   1            % on the other side only 1s does not necessarily means
indices =           % that B is part of A (look at the example below)
     4     7
     5     8

2. example

>> B(2,1)=12
  B =
       2     3
      12     7
  >> [yes_or_no,indices]=ismember(B,A)
  yes_or_no =
    2×2 logical array
     1   1
     1   1
  indices =
       4     7
      12     8

this is no perfect solution, and if you use ismember you still are not done at this point, its just an idea

Answer 2

[Albert]

use function find, and your problem is solved. A=[1 2 3 4; 5 6 7 8; 9 10 11 12]; B = A(1:2); find(A==B(1)) >>1 which is the location of A==B(1) in matrix A.

A(1:2, 2:3) is different than A(1:2) A(1:2, 2:3) is short for A(1, 2), A(1, 3),A(2, 2), A(2, 3). A(1:2) is short for A(1, 1) and A(2, 1)

These are very basic and introductory knowledge, which you don't want to ask here.

[Niels]

for this Matrix A your solution may work, but imagine A had more than one 1 => the result wouldnt be unequivocal

Answer 3

[Star Strider]

MATLAB uses linear indexing for arrays, because the arrays are stored as consecutive elements in memory in column-dominant representation. So in your matrix, the first element in the vector is 1, the second is 5, the third is 9, the fourth is 2, and so on.

So:

B = A(1:2) = [1  5]

If you choose to explore this convention:

A = [1 2 3 4; 5 6 7 8; 9 10 11 12];
B = A(1:2)
[row,col] = ind2sub(size(A), (1:2))

row =
     1     2

col =
     1     1

so the [row,column] indices for ‘A(1:2)’ are [1,1] (row #1, column #1) and [2,1] (row #2, column #1) respectively.