Open a file with a specific extension, regardless of the name-CodePudding

I have the following code:

folder_names = []
spreadsheet_contents = []
all_data = pd.DataFrame()
current_directory = Path.cwd()


for folder in current_directory.iterdir():
    
    folder_names.append(folder.name)
    
    file_n = '*.csv'
         
    spreadsheet_path = folder / file_n
    spreadsheet_contents.append(pd.read_excel(spreadsheet_path, skiprows = 1, header = None, usecols = [5]))

The problem is that the .csv files in each folder are named differently. The '*.csv' method does not work. Does anyone have an idea how to open the .csv file for each subfolder even though they are all named differently?

CodePudding user response：

for the sake of simplicity im not writing the complete code,

import glob

replace file_n = '*.csv' with file_n = glob.glob('*.csv')

and loop over the list of file names.

CodePudding user response：

You can use glob:

import glob, os
os.chdir("/mydir")
for file in glob.glob("*.txt"):
    print(file)

or simply os.listdir:

import os
for file in os.listdir("/mydir"):
    if file.endswith(".txt"):
        print(os.path.join("/mydir", file))

or if you want to traverse directory, use os.walk:

import os
for root, dirs, files in os.walk("/mydir"):
    for file in files:
        if file.endswith(".txt"):
             print(os.path.join(root, file))

source : stackoverflow

Also check this out : Working with CSV file for Data Science

We can also use regex !