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Conditionally select from array of struct by membership of a list in struct element

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Here's my array of structs:
X(1).Members=[1 2]
X(1).Name="Group 1"
X(2).Members=[2 3]
X(2).Name="Group 2"
Here's the conditional selection form I'm familiar with. I'm hoping there's a better way to express this:
A=[X(ismember(1,[X.Members])).Name]
A = "Group 1"
A=[X(ismember(2,[X.Members])).Name]
A = "Group 1"
A=[X(ismember(3,[X.Members])).Name]
A = "Group 1"
A=[X(ismember(4,[X.Members])).Name]
A = []
Since the 'ismember' operator returns a logical, the expression as formed returns the index of X to 0 or 1.
I would like to apply an expression to X and return each struct in X where a member is in the list of Members in the struct.

  1 Comment

Stephen Cobeldick
Stephen Cobeldick on 22 Jan 2021
The 2nd and 3rd example outputs seem a bit strange:
  • 2nd: the value 2 is a member of both "Group 1" and "Group 2", why does the output only show "Group 1"?
  • 3rd: the value 3 is only a member of "Group 2", why does the output show "Group 1"?

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Accepted Answer

Stephen Cobeldick
Stephen Cobeldick on 22 Jan 2021
Edited: Stephen Cobeldick on 22 Jan 2021
X(1).Members = [1,2];
X(1).Name = 'Group 1';
X(2).Members = [2,3];
X(2).Name = 'Group 2';
F = @(n)arrayfun(@(s)any(s.Members==n),X);
F(1)
ans = 1x2 logical array
1 0
F(2)
ans = 1x2 logical array
1 1
F(3)
ans = 1x2 logical array
0 1
F(4)
ans = 1x2 logical array
0 0
If you really want to use ismember, replace the anonymous function with this:
@(s)ismember(n,s.Members)

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More Answers (1)

Rick Nickle
Rick Nickle on 22 Jan 2021
Thanks Stephen Cobeldick, that function works well, very much appreciated!
Example usage:
A=[X(F(2)).Name]
A =
1×2 string array
"Group 1" "Group 2"
B=[X(F(2))]
B =
1×2 struct array with fields:
Members
Name
% Checking actual content...
struct2table(B)
ans =
2×2 table
Members Name
_______ _________
1 2 "Group 1"
2 3 "Group 2"
C=[X(F(3))]
C =
struct with fields:
Members: [2 3]
Name: "Group 2"
struct2table(C)
ans =
1×2 table
Members Name
_______ _________
2 3 "Group 2"

  1 Comment

Rick Nickle
Rick Nickle on 22 Jan 2021
Adding a variant to show selecting by multiple fields with Stephen's suggestion:
X(1).Members=1
X(1).Owners=2
X(1).Name="Group 1"
X(2).Members=1
X(2).Owners=1
X(2).Name="Group 2"
Q=@(n,o)arrayfun(@(s)any(s.Members==n && s.Owners==o),X)
Y=[X(Q(1,1))]

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