How to filter a cell array filled with strings?

6 visualizaciones (últimos 30 días)
Maximilian Stopfer
Maximilian Stopfer el 4 de Dic. de 2019
Editada: Stephen23 el 12 de Abr. de 2024
Dear all,
I have a dataset with 576 cells [1x576] and all cells contain strings [1x1 struct]. These strings contain two colums (field, value), e. g.:
A01 = struct('a',1,'b',2,'c',0)
A02 = struct('a',0,'b',2,'c',1)
A03 = struct('a',2,'b',2,'c',0)
A04 = struct('a',0,'b',1,'c',1)
A05 = struct('a',1,'b',0,'c',2)
A = {A01,A02,A03,A04,A05}
How can I filter all cells with a specific field name and extract them (as example: how can I filter all structs with a=1 and save the b values in new array)?
Sincerely yours, Max
  1 comentario
Stephen23
Stephen23 el 12 de Abr. de 2024
Editada: Stephen23 el 12 de Abr. de 2024
Storing lots of scalar structures in a cell array is inefficient data design, which leads to inefficient code.
Assuming that all of the scalar structures have the samew fields (as your example shows), then much better data design would be to use one non-scalar structure array:
A = struct('a',{1,0,2,0,1},'b',{2,2,2,1,0},'c',{0,1,0,1,2})
A = 1x5 struct array with fields:
a b c
This has more efficient memory and access, as well as offering some convenient syntaxes for accessing the data:

Iniciar sesión para comentar.

Respuesta aceptada

Adam Danz
Adam Danz el 4 de Dic. de 2019
Editada: Adam Danz el 6 de Dic. de 2019
This will return a logical array the same size as 'A' where TRUE values indicate elements of A that meet the following requirements:
  • A{i} is a structure
  • A{i} contains a field named 'a'
  • The value of field 'a' is equal to 1
keyField = 'a';
keyValue = 1;
idx = cellfun(@(s)isstruct(s) && any(strcmp(fieldnames(s),keyField)) && isequal(s.(keyField),keyValue), A)
To filter those structures from the cell array A,
A(idx)

Más respuestas (0)

Categorías

Más información sobre Structures en Help Center y File Exchange.

Productos


Versión

R2019b

Community Treasure Hunt

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

Start Hunting!

Translated by