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Azzera filtri

all function B = all(A < 0.5,3)

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Marie Nancy B
Marie Nancy B il 14 Giu 2023
Modificato: Stephen23 il 15 Giu 2023
This is straight forward to me:
A = [0.53 0.67 0.01 0.38 0.07 0.42 0.69]
A = 1×7
0.5300 0.6700 0.0100 0.3800 0.0700 0.4200 0.6900
B = all(A < 0.5,1)
B = 1×7 logical array
0 0 1 1 1 1 0
this also makes sense -
A = [0.53 0.67 0.01; 0.38 0.07 0.42 ]
A = 2×3
0.5300 0.6700 0.0100 0.3800 0.0700 0.4200
B = all(A < 0.5,2)
B = 2×1 logical array
0 1
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]
A = 2×4
0.5300 0.6700 0.0100 1.1000 0.3800 0.0700 0.4200 0.0100
B = all(A < 0.5,3)
B = 2×4 logical array
0 0 1 0 1 1 1 1
OR
B = all(A < 0.5,4)
B = 2×4 logical array
0 0 1 0 1 1 1 1
  1 Commento
Stephen23
Stephen23 il 15 Giu 2023
Modificato: Stephen23 il 15 Giu 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
ans = 0.5300
A(1,1,1,:) % top left page1, all elements along 4th dimension
ans = 0.5300
You can repeat this yourself for every other element in your matrix.
Question: what do you expect the result to be? Why?

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Risposte (2)

the cyclist
the cyclist il 14 Giu 2023
Modificato: the cyclist il 14 Giu 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)
ans = 1

Torsten
Torsten il 14 Giu 2023
Spostato: Torsten il 14 Giu 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 Commento
VBBV
VBBV il 15 Giu 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]
A = 2×4
0.5300 0.6700 0.0100 1.1000 0.3800 0.0700 0.4200 0.0100
B = all(A < 0.5,'all')
B = logical
0
B = all(A < 0.5,3)
B = 2×4 logical array
0 0 1 0 1 1 1 1
B = all(A < 0.5,4)
B = 2×4 logical array
0 0 1 0 1 1 1 1

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