Assign array in field struct
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class(Sis)
ans =
'struct'
length(Sis)
ans =
82
c=1
2
3
..
82
i want to create new field in struct
i want this:
Sis.b(1)=1;
Sis.b(2)=2;
..
Sis.b(82)=82;
now i want to semplify calculate
c=1:82
c=c'
>> [Sis.b]=c
Insufficient number of outputs from right hand side of equal sign to satisfy assignment.
3 个评论
Walter Roberson
2023-7-9
i want this:
Sis.b(1)=1;
Sis.b(2)=2;
Your Sis is a 1 x 82 struct. Sis.b(1) by structure expansion would be as if you had written
Sis(1).b, Sis(2).b, Sis(3).b, Sis(4).b, Sis(5).b ... Sis(82).b(1) = 1;
Sis(1).b, Sis(2).b, Sis(3).b, Sis(4).b, Sis(5).b ... Sis(82).b(2) = 2;
which is not what you want. You want the equivalent of
Sis(1).b = 1;
Sis(2).b = 2;
...
Sis(82).b = 82;
Vilém Frynta
2023-7-9
oh, i misunderstood. thanks
shamal
2023-7-9
采纳的回答
Example data
Sis(1).a = 1;
Sis(2).a = 2;
Sis(3).a = 3;
c = [30 40 50];
Assign elements of c to new field of elements of Sis
temp = num2cell(c);
[Sis.b] = temp{:};
Sis.b
Or in one line
clear
Sis(1).a = 1;
Sis(2).a = 2;
Sis(3).a = 3;
c = [30 40 50];
[Sis.b] = table2struct(table(c')).Var1;
Sis.b
3 个评论
Walter Roberson
2023-7-9
When you use table() and pass in an expression (instead of a variable name) then table() automatically uses 'Var' followed by the column number as the name of the variable. So for example table([10;20]; [30;40]) would produce a 2 x 2 table with variables 'Var1' and 'Var2'
shamal
2023-7-9
更多回答（1 个）
Sis = struct();
Sis.b = 1:82'
Sis.b(1)
Sis.b(10)
Hope this helps.
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