for loop to scan matrix and output a new matrix

I would like to take an e.g. 8x1 matrix such as the one below:

0

1

0

1

1

1

0

1

and I would like to scan through it and generate a new matrix of the same dimensions, following the rules below.

• Make the first value of the new matrix the same as the first value of the original matrix.
• Then from here on:
• If the value is 0, add a 0 to the new matrix.
• If the value is 1 AND the value above it is 1, assign a 0 to the new matrix.
• However, if the value is 1 but the value above it is 0, then assign a 1 to the new matrix.

So, the resulting matrix should be:

0

1

0

1

0

0

0

1

Any help, with clear notation of what each part of the for loop is doing, would be much appreciated! Many thanks.

[Rik]

Why don't you write the first version?

[Laura Steel]

I think I may have worked something out! But I am not sure it is the best/simplest way of doing it?

matrix = [0; 1; 0; 0; 1; 1; 1; 1; 0; 0; 1; 1; 0];
matrix_new = zeros(13,1);

for i = 1:length(matrix)
    if i == 1
        matrix_new(i,1) = matrix(1,1);
    end
    if matrix(i,1) == 0
        matrix_new(i,1) = 0;
    end
    if matrix(i,1) == 1 & matrix(i-1,1) ==0
        matrix_new(i,1) = 1
    end
    if matrix(i,1) == 1 & matrix(i-1,1) ==1
        matrix_new(i,1) = 0
    end
end

Answer 1

[Walter Roberson]

matrix = [0; 1; 0; 0; 1; 1; 1; 1; 0; 0; 1; 1; 0];

matrix_new = [matrix(1); matrix(2:end) & ~matrix(1:end-1)]

matrix_new = 13×1

0 1 0 0 1 0 0 0 0 0

[Laura Steel]

Hi,

Thank you, this works - and is much cleaner than my method. However, I do not understand how it works - would you be able to explain the "matrix(2:end) & ~matrix(1:end-1)]" part?

Thanks

[Walter Roberson]

matrix(2:end) is false for locations that are 0. With the & operator there, no matter what is in the other part of the expression, 0 in the first expression is going to lead to 0 outcome. This satisfies the "If the value is 0, add a 0 to the new matrix." rule.

matrix(2:end) is true for locations that are 1. But there is a & operator, so the whole thing will not evaluate to true (1) unless the second part evaluates to non-zero as well. The second part is ~matrix(1:end-1) . That looks at the previous entry in each case, and the ~ says that if the previous entry was 1 then the ~ should be 0; & of 1 and 0 is 0, so 0 would be output for that case. That satisfies the "If the value is 1 AND the value above it is 1, assign a 0 to the new matrix" rule.

If the previous entry was 0 then the ~ result is 1. 1&1 is 1, so 1 would be output for that case. This satisfies the rule that if the value is 1 and the value above it is 0, assign a 1 to the new matrix.

[Rik]

Which operator don't you understand?

The & operator performs a logical AND operation on two arrays.

The ~ (tilde) operator performs a logical NOT operation on an array (so flipping 0 and 1, or actually true and false).

For logical operations, 0 is treated as false, while all other numbers are treated as true. There are some edge cases to watch out for (like NaN and inf), but those probably don't matter in this specific case.

[Walter Roberson]

inf and -inf are not a problem, but nan is a problem.

~[-inf inf]

ans = 1×2 logical array
   0   0

~nan

Error using ~
NaN values cannot be converted to logicals.

[Laura Steel]

Brilliant, thank you both very much.