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

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HabenG
HabenG 2021-12-11
评论： Walter Roberson 2021-12-11
Ran in:
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

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Walter Roberson
Walter Roberson 2021-12-11
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.
  2 个评论
HabenG
HabenG 2021-12-11
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
Walter Roberson 2021-12-11
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.

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