ismember(a, b) function problem

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Jeffrey DG
Jeffrey DG 2022-5-29
评论： Voss 2022-5-29
Hello,
In brief, I am trying to compare two seperate arrays that decrease/increase till they are equal one another.
a = [1, 1, 0];
b = [0, 1, 0];
logic = ismember(a, b); % or ismembertol(a, b, 0.001);
if all(logic(:))
% Do somthing
end
However, I seem to be getting all true, [1, 1, 1], when I expect a [0, 1, 1] for this particular set of logical comaprison using ismember(). I am not good at coding so will appreciate if anyone could explain why this is happening.
Thank you!

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采纳的回答

Voss
Voss 2022-5-29
编辑：Voss 2022-5-29
Ran in:
ismember(a,b) tells you whether each element of a exists somewhere in b
ismember([1 2 3 4 5],[2 4 6]) % 2 and 4 exist in [2 4 6], but 1, 3, and 5 do not
ans = 1×5 logical array
0 1 0 1 0
With your a and b:
a = [1, 1, 0];
b = [0, 1, 0];
logic = ismember(a,b)
ans = 1×3 logical array
1 1 1
you get logic is all true because all elements of a (i.e., 0 and 1) appear in b.
It seems that you actually want to compare each element of a and b, which you can do using ==
logic = a == b
logic = 1×3 logical array
0 1 1
  2 个评论
Jeffrey DG
Jeffrey DG 2022-5-29
Wow, thanks. I did not know you could also compare arrays just like that. You learn something new everyday!
Voss
Voss 2022-5-29
Ran in:
You're welcome!
FYI, here are some examples of using == with arrays:
x = [1 2 3 4]; % 1-by-4 array
y = [1 2 1 2]; % 1-by-4 array
x == y % compares each element of x to corresponding element of y
ans = 1×4 logical array
1 1 0 0
x = [1 2 3 4]; % 1-by-4 array
y = 2; % scalar
x == y % compares each element of x to y
ans = 1×4 logical array
0 1 0 0
x = [1 2 3 4]; % 1-by-4 array (row vector)
y = [2; 3; 5]; % 3-by-1 array (column vector)
x == y % also works! (compares each element of x to every element of y)
ans = 3×4 logical array
0 1 0 0 0 0 1 0 0 0 0 0
x = [1 2 3 4]; % 1-by-4 array
y = [2 5 8]; % 1-by-3 array
x == y % error
Arrays have incompatible sizes for this operation.

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更多回答（1 个）

Bjorn Gustavsson
Bjorn Gustavsson 2022-5-29
The ismember function checks if elements in a are found in b, not that each element match. In your case both "1" in a have a value found in b and the same is true for the "0". Maybe you want your conditional to be:
if all(a==b)
% Do somthing
end
or perhaps
your_tol = 0.01;
if all(abs(a-b)<=your_tol)
% Do somthing
end
HTH

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