Is it possible to use vertcat with dot notation?
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EDITED Is it possible to use vertcat with dot notation?
This is an example once I extract names from Excel files contained in a folder:
%input
filename = dir(fullfile(directory,'*file*'));
% output
>> filename
filename =
3×1 struct array with fields:
name
folder
date
bytes
isdir
datenum
>> filename.name
ans =
'file_1.xlsx'
ans =
'file_2.xlsx'
ans =
'file_3.xlsx'
% try to concatenate vertically the three files' names
>> vertcat(filename{:}.name)
Brace indexing is not supported for variables of this type.
7 Kommentare
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Stephen23
am 13 Dez. 2022
Verschoben: Stephen23
am 13 Dez. 2022
"The error is being caused by the fact that filename is not a struct."
FILENAME is a struct (as returned by DIR). Curly brace indexing is not defined for structures (see the error message).
"Is it possible to use vertcat with dot notation?"
Of course, you can use any comma-separated list as the inputs to VERTCAT:
S = struct('X',{'hello','world'}) % content have same number of columns...
vertcat(S.X) % so VERTCAT is possible but fragile. Creates a character matrix.
But I would not recommend it because it is very fragile code: using VERTCAT on the structure from DIR() assumes that all of the filenames have exactly the same length, otherwise your code will throw an error:
S = struct('X',{'a','bcd'}) % content have different number of columns...
vertcat(S.X) % which is why VERTCAT is a bad approach.
As Jonas correctly wrote, you should probably use a cell array: {filename.name}
Or, simply refer directly to the content of the structure array.
Weitere Antworten (2)
Peter Perkins
am 13 Dez. 2022
Depending on what you are doing, you may find it convenient to turn that struct into a table (which wil automatically vertcat the file names for you):
s = dir
t = struct2table(s)
2 Kommentare
Jonas
am 14 Dez. 2022
if you just want to have all file names available, you could use { }
e.g. in a cell array
{filename.name}
if you do not need further information from the dir out beside the name, you could also abbreviate it to one line
{dir(fullfile(directory,'*file*')).name}
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