How to make a statement true for column 5-6?
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I have a matrix
M= [1 50 60 70 50 40
2 NaN 10 20 10 10
3 NaN 20 NaN NaN NaN
1 NaN 60 30 40 50
2 10 20 10 20 NaN
1 30 20 40 NaN 50
2 NaN 50 50 1 NaN];
I want to check that, for each row, whether it is ture that the element in column 2 is NaN and the elements in column 5-6 are both non-NaN. I tried to code like this:
P=isnan(M(:,2))&isfinite(M(:,5:6))
However, the result is in 2 columns, column "NaN in column 2 and non-NaN in column 5" and column "NaN in column 2 and non-NaN in column 6".
Could anyone please tell me how to have the answer in 1 column?
I think this code works,
P=isnan(M(:,2))&isfinite(M(:,5))&isfinite(M(:,6))
but is there a better way to express "column 5-6" instead of making a isfinite statement for each column seperately?
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Accepted Answer
Voss
on 7 Feb 2023
Edited: Voss
on 10 Feb 2023
'is there a better way to express "column 5-6"'
Yes.
You can use all(). Specifically, all(~isnan(__),2) where the 2 tells all() to operate over the columns (2nd dimension). ("All are non-NaN.")
Or you can use any(). Specifically, ~any(isnan(__),2). ("Not any are NaN.")
M = [1 50 60 70 50 40
2 NaN 10 20 10 10
3 NaN 20 NaN NaN NaN
1 NaN 60 30 40 50
2 10 20 10 20 NaN
1 30 20 40 NaN 50
2 NaN 50 50 1 NaN];
These three are equivalent:
P = isnan(M(:,2)) & ~isnan(M(:,5)) & ~isnan(M(:,6))
P = isnan(M(:,2)) & all(~isnan(M(:,[5 6])),2)
P = isnan(M(:,2)) & ~any(isnan(M(:,[5 6])),2)
Note that I'm using ~isnan() where you used isfinite(). The difference is how Inf and -Inf are treated.
~isnan(Inf)
isfinite(Inf)
And since the rules you describe talk about NaN vs non-NaN and say nothing about finte vs infinite, I think it makes sense to use isnan and ~isnan.
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