all function B = all(A < 0.5,3)
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This is straight forward to me:
A = [0.53 0.67 0.01 0.38 0.07 0.42 0.69]
B = all(A < 0.5,1)
this also makes sense -
A = [0.53 0.67 0.01; 0.38 0.07 0.42 ]
B = all(A < 0.5,2)
But why does this yield the same result?
A = [0.53 0.67 0.01 1.1; 0.38 0.07 0.42 0.01]
B = all(A < 0.5,3)
OR
B = all(A < 0.5,4)
1 Kommentar
Stephen23
am 15 Jun. 2023
Bearbeitet: Stephen23
am 15 Jun. 2023
"But why does this yield the same result?"
Consider the sub-arrays (i.e. vectors) along each of the specified dimensions: what values do they have in them? Remember that all arrays implicitly have infnite trailing singleton dimensions: the fact that each of those vectors for dimensions 3 and 4 happen to have one element in them is irrelevent, ALL applies its algorithm on that one element. It might help to revise this: https://www.mathworks.com/help/matlab/math/multidimensional-arrays.html
Lets consider the top left corner of your data:
A = [0.53 0.67 0.01 0.38 0.07 0.42 0.69];
A(1,1,:) % top left, all elements along 3rd dimension
A(1,1,1,:) % top left page1, all elements along 4th dimension
You can repeat this yourself for every other element in your matrix.
Question: what do you expect the result to be? Why?
Antworten (2)
the cyclist
am 14 Jun. 2023
Bearbeitet: the cyclist
am 14 Jun. 2023
The reason is that all MATLAB arrays have implied singleton dimension beyond the 2nd dimension, if they have not been actually specified.
A = [0.53 0.67 0.01 1.1;
0.38 0.07 0.42 0.01];
size(A,47)
0 Kommentare
Torsten
am 14 Jun. 2023
Verschoben: Torsten
am 14 Jun. 2023
But why does this yield the same result?
Because A has only 2 dimensions, not 3 or 4. Thus "all" operates on all elements separately in both cases.
1 Kommentar
VBBV
am 15 Jun. 2023
Thus "all" operates on all elements separately in both cases., Yes, thats correct. But if one uses 'all' option it would do only once.
A = [0.53 0.67 0.01 1.1; 0.38 0.07 0.42 0.01]
B = all(A < 0.5,'all')
B = all(A < 0.5,3)
B = all(A < 0.5,4)
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