Why does inequality give different values even though its evaluating the exact same matrix ??

Howcome these two yield different values ??

load td
version1 = sort(td(td(:,5) < 11.29))

version1 = 8×1

11.1700 11.1750 11.1750 11.1901 11.2000 11.2200 11.2848 11.3312

t = (td(:,5));
version2 = sort(t(t < 11.29))

version2 = 8×1

11.1700 11.1750 11.1750 11.2000 11.2200 11.2486 11.2800 11.2848

version1 - version2

ans = 8×1

0 0 0 -0.0099 -0.0200 -0.0286 0.0048 0.0464

Answer 1

[Walter Roberson]

sort(td(td(:,5) < 11.29))

td is a 2D array with multiple columns. You construct a mask from one of the columns, and you use it to index the entire td array -- which by linear indexing is going to effectively have it operate on the first column.

t = (td(:,5));
sort(t(t < 11.29))

but here you extract the column and your linear indexing is against the extracted version that only contains the one column.

[HabenG]

Thanks that explains it. I did that to avoid the binary out put from the inequality. what would be a good way to evalueate the inequality and get the actual values?

[Walter Roberson]

version3 = sort(td(td(:,5) < 11.29, 5))

is a possible way. But your second version with a temporary variable works well and is easier to read.