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

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.

[Stephen Cobeldick]

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"?

Answer 1

[Stephen Cobeldick]

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)

Answer 2

[Rick Nickle]

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"

[Rick Nickle]

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))]