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How to read all CSV files from specific folders?

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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 Commenti
Roozbeh Yousefnejad
Roozbeh Yousefnejad il 1 Giu 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 il 1 Giu 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?

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Risposte (3)

Walter Roberson
Walter Roberson il 1 Giu 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 Commenti
Roozbeh Yousefnejad
Roozbeh Yousefnejad il 4 Giu 2018
Modificato: Roozbeh Yousefnejad il 4 Giu 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.

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Roozbeh Yousefnejad
Roozbeh Yousefnejad il 1 Giu 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 Commenti
Paolo
Paolo il 1 Giu 2018
Modificato: Paolo il 1 Giu 2018
I was referring to this command:
dir('*/BIN*/*.csv')
Use this one if you are running from 'New folder'

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Roozbeh Yousefnejad
Roozbeh Yousefnejad il 4 Giu 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};

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