How to read all CSV files from specific folders?

156 visualizaciones (últimos 30 días)
Roozbeh Yousefnejad
Roozbeh Yousefnejad el 1 de Jun. de 2018
Comentada: Behtom el 9 de Nov. de 2023
Hi, I want to open all CSV files and do a calculation on them. I used the code below and it worked
files = subdir('C:\Users\roozm\Desktop\New folder\*.csv');
Subdir function can do it easily. Now I want to be more specific and only open folder with the name of BIN and then read CSV files in only BIN folders.
How can I do that?
  11 comentarios
Roozbeh Yousefnejad
Roozbeh Yousefnejad el 1 de Jun. de 2018
I am just entering it in the command window. Honestly, I do not know how to check if the file is in my current directory.
Paolo
Paolo el 1 de Jun. de 2018
I'm not sure I understand what you mean by 'I do not know how to check if the file is in my current directory.'
Lets say your current working directory is :
C:\Users\roozm\
as from your OP. Could you list the full path of the .csv files you want, this way we can build the dir command?

Iniciar sesión para comentar.

Respuestas (3)

Walter Roberson
Walter Roberson el 1 de Jun. de 2018
files = dir('C:\Users\roozm\Desktop\New folder\**\BIN*\*.csv');
fullpaths = fullfile({files.folder}, {files.name});
Now fullpaths is a cell array of fully qualified .csv files that are directly under a BIN* folder anywhere under "New folder"
  10 comentarios
Roozbeh Yousefnejad
Roozbeh Yousefnejad el 4 de Jun. de 2018
Editada: Roozbeh Yousefnejad el 4 de Jun. de 2018
I am afraid that there is another reason behind it as I am running the code in the debugging mode and step by step (I've attached a pic) So I believe xlsread() doesn't have a problem with the loop. Also, the files are closed.
As you can see in the picture, the curser is on the subset line and it is only the first xlsread command that I get error.
Behtom
Behtom el 9 de Nov. de 2023
Thank You !

Iniciar sesión para comentar.


Roozbeh Yousefnejad
Roozbeh Yousefnejad el 1 de Jun. de 2018
I mean, I don't know how to find my directory path. The file that I want to open is in this path
C:\Users\roozm\Desktop\New folder\F21802010055\BIN\CSV
which is on my desktop.
  4 comentarios
Paolo
Paolo el 1 de Jun. de 2018
Editada: Paolo el 1 de Jun. de 2018
I was referring to this command:
dir('*/BIN*/*.csv')
Use this one if you are running from 'New folder'
Roozbeh Yousefnejad
Roozbeh Yousefnejad el 1 de Jun. de 2018
same error ...

Iniciar sesión para comentar.


Roozbeh Yousefnejad
Roozbeh Yousefnejad el 4 de Jun. de 2018
I replaced filename = filepaths{i}; instead of filename=files(i).name but got this error
Undefined variable "filepaths" or class "filepaths".
Error in new (line 8)
filename = filepaths{i};
  1 comentario
Roozbeh Yousefnejad
Roozbeh Yousefnejad el 4 de Jun. de 2018
is filepaths a function? should I download it and add it to my code?

Iniciar sesión para comentar.

Etiquetas

Productos


Versión

R2018a

Community Treasure Hunt

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

Start Hunting!

Translated by