How to access a field of a struct by indexing?

1,673 ビュー (過去 30 日間)
Rightia Rollmann
Rightia Rollmann 2017 年 2 月 26 日
コメント済み: Voss 2023 年 12 月 19 日
I have a 1-by-1 struct that possesses 3 fields named B, C, and D. Is there any way to call D by its index (i.e., D is the third field of struct A, so call the third field of struct A without mentioning the field name D) rather than its name (i.e, A.D)?
A.B = 1;
A.C = 2;
A.D = 3;
  1 件のコメント
Stephen23
Stephen23 2017 年 2 月 26 日
編集済み: Stephen23 2017 年 2 月 26 日
@Rightia Rollmann: you might like to consider using a non-scalar structure, which lets you use indexing to access structures:

サインインしてコメントする。

採用された回答

Jan
Jan 2017 年 2 月 26 日
編集済み: Jan 2017 年 2 月 26 日
A_cell = struct2cell(A);
D = A_cell{3}
Keep in mind that the order of the fields of structs is not necessarily constant:
A.B = 1;
A.C = 2;
A.D = 3;
B.B = 1;
B.D = 3;
B.C = 2;
isequal(A, B) % >> TRUE!
  4 件のコメント
Richard Crozier
Richard Crozier 2019 年 8 月 1 日
If struct2cell results in copying the data in the structure, Guillaume's answer below is superior.
James Tursa
James Tursa 2019 年 8 月 1 日
struct2cell creates shared-data-copies of the field variables. So, while there is overhead involved in creating the variable header info, the data itself is not copied.

サインインしてコメントする。

その他の回答 (1 件)

Guillaume
Guillaume 2017 年 2 月 26 日
Yes, there is a way to get the nth field directly:
fns = fieldnames(A);
A.(fns{3})
But be aware that the order of the fields depends solely on the order in which they were created. As Jan pointed out, two structures may be indentical, yet have different field order.
Usually, you would only access fields by their index when you're doing some structure metaprogramming
  5 件のコメント
Vaishnavi
Vaishnavi 2023 年 12 月 19 日
To add on to this question, can someone explain why A.(subsref(fieldnames(A),substruct('{}',{:}))) would not work?
Voss
Voss 2023 年 12 月 19 日
@Vaishnavi: One reason that doesn't work is that you need single quotes around the colon in substruct. But even then, it may not do what you expect. Here's what it does (gets the value of the first field of A):
A = struct('field1',1,'field2',2)
A = struct with fields:
field1: 1 field2: 2
A.(subsref(fieldnames(A),substruct('{}',{':'})))
ans = 1
In order to know whether that "works", one would need to know what you expected it to do.

サインインしてコメントする。

カテゴリ

Help Center および File ExchangeStructures についてさらに検索

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by