Why is NaN inserted in wrong position?

I have a matrix

b = [1 3 0;-2 -1 5]

b =

1 3 0

-2 -1 5

When I perform the following operation

b(b(:,3)==5) = NaN;

the NaN is placed a the postion of -2. How come?

[Dyuman Joshi]

"the NaN is placed a the postion of -2. How come?"

Are you sure about that? The output from the code says otherwise -

b = [1 3 0;-2 -1 5];
b(b(1,:)==5) = NaN

b = 2×3

1 3 0 -2 -1 5

No element in the 1st row of b equals to 5, so no assignment will take place.

Answer 1

[Star Strider]

It should not do anything, because 5 is in row 2, not row 1.

That aside, you need to index into ‘b’ correctly to get the desired result —

b = [1 3 0;-2 -1 5]

b = 2×3

1 3 0 -2 -1 5

Check = b(1,:)==5                               % Test Row #1

Check = 1×3 logical array
   0   0   0

Check = b(2,:)==5                               % Test Row #2

Check = 1×3 logical array
   0   0   1

b(2,b(2,:)==5) = NaN                            % Use The Correct Indexing, Specifying The Correct Rows As Well As The Correct Columns

b = 2×3

1 3 0 -2 -1 NaN

.

[Sam]

Thank you for your remark and suggestions. The goal is to replace any entry in and only in the third column that is equal to 5. How could this be performed?

Thanks in advance.

[dpb]

As @Les Beckham showed (or @Star Strider if he had generalized the row index expression).

To amplify, you wrote the LHS index as single value for a 2D array which is linear addressing on the assignment but did the search on 2D expression b(:,3) which returns a 1D (column) vector. The five was in the second row so that logical vector is [0;1] or the result of find() would return the numerical index of '2' which is the correct index into that vector.

Hence, when you wrote b(.)=nan; for the assignment, the "." placeholder was the logical vector [0;1] which is the second element in the array with linear indexing and since MATLAB is column-major storage order, that is position b(1,2) in the 2D array. Ergo, the NaN showed up where the -2 was originally, just like you asked it to! :) Of course, that wasn't what you meant, but MATLAB doesn't know that...

The correct syntax is that you must use the same addressing expression on the LHS as in the RHS to make the two positions commensurate in the portion of the array they refer to; in this case repeating the reference explicitly to column 3.

[Star Strider]

@Sam —

‘The goal is to replace any entry in and only in the third column that is equal to 5.’

Change the conditon statement to specify the third column, similar to the previous example —

b = [1 3 0;-2 -1 5]

b = 2×3

1 3 0 -2 -1 5

b(b(:,3)==5, 3) = NaN

b = 2×3

1 3 0 -2 -1 NaN

This is essentially the same as the original example, however specifying the third column.

To expand on this idea —

b = [1 3 0;-2 -1 5;7 4 6;5 9 5;2 5 8]

b = 5×3

1 3 0 -2 -1 5 7 4 6 5 9 5 2 5 8

b(b(:,3)==5, 3) = NaN

b = 5×3

1 3 0 -2 -1 NaN 7 4 6 5 9 NaN 2 5 8

So it replaces only the ‘5’ values in the third column, leaving all others unchanged.

.

Answer 2

[Les Beckham]

b = [1 3 0;-2 -1 5];
b(b(:,3)==5,3) = NaN % add ,3 to select only the third column for assignment

b = 2×3

1 3 0 -2 -1 NaN