While solution given by thclpr works it scans only immediate files in the directory and not files in the sub directories if any. Although this is not the requirement but just in case someone wishes to scan sub directories too below is the code that uses os.walk
import os
from glob import glob
PATH = "/home/someuser/projects/someproject"
EXT = "*.csv"
all_csv_files = [file
for path, subdir, files in os.walk(PATH)
for file in glob(os.path.join(path, EXT))]
print(all_csv_files)
Copied from this blog.
Use the python glob module to easily list out the files we need.
import glob
path_csv=glob.glob("../data/subfolrder/*.csv")
By using the combination of filters and lambda, you can easily filter out csv files in given folder.
import os
files = os.listdir("/path-to-dir")
files = list(filter(lambda f: f.endswith('.csv'), files))
# lambda returns True if filename name ends with .csv or else False
# and filter function uses the returned boolean value to filter .csv files from list files.
use Python OS module to find csv file in a directory.
the simple example is here :
import os
# This is the path where you want to search
path = r'd:'
# this is the extension you want to detect
extension = '.csv'
for root, dirs_list, files_list in os.walk(path):
for file_name in files_list:
if os.path.splitext(file_name)[-1] == extension:
file_name_path = os.path.join(root, file_name)
print file_name
print file_name_path # This is the full path of the filter file
from os import listdir
def find_csv_filenames( path_to_dir, suffix=".csv" ):
filenames = listdir(path_to_dir)
return [ filename for filename in filenames if filename.endswith( suffix ) ]
The function find_csv_filenames()
returns a list of filenames as strings, that reside in the directory path_to_dir
with the given suffix (by default, ".csv").
Addendum
How to print the filenames:
filenames = find_csv_filenames("my/directory")
for name in filenames:
print name
You could just use glob
with recursive = true
, the pattern **
will match any files and zero or more directories, subdirectories and symbolic links to directories.
import glob, os
os.chdir("C:\\Users\\username\\Desktop\\MAIN_DIRECTORY")
for file in glob.glob("*/.csv", recursive = true):
print(file)
Many (linked) answers change working directory with os.chdir()
. But you don't have to.
Recursively print all CSV files in /home/project/
directory:
pathname = "/home/project/**/*.csv"
for file in glob.iglob(pathname, recursive=True):
print(file)
Requires python 3.5+. From docs [1]:
pathname
can be either absolute (like /usr/src/Python-1.5/Makefile
) or relative (like ../../Tools/*/*.gif
)pathname
can contain shell-style wildcards.recursive
is true, the pattern **
will match any files and zero or more directories, subdirectories and symbolic links to directoriesimport os
path = 'C:/Users/Shashank/Desktop/'
os.chdir(path)
for p,n,f in os.walk(os.getcwd()):
for a in f:
a = str(a)
if a.endswith('.csv'):
print(a)
print(p)
This will help to identify path also of these csv files
Please use this tested working code. This function will return a list of all the CSV files with absolute CSV file paths in your specified path.
import os
from glob import glob
def get_csv_files(dir_path, ext):
os.chdir(dir_path)
return list(map(lambda x: os.path.join(dir_path, x), glob(f'*.{ext}')))
print(get_csv_files("E:\\input\\dir\\path", "csv"))
I had to get csv
files that were in subdirectories, therefore, using the response from tchlpr I modified it to work best for my use case:
import os
import glob
os.chdir( '/path/to/main/dir' )
result = glob.glob( '*/**.csv' )
print( result )
import os
import glob
path = 'c:\\'
extension = 'csv'
os.chdir(path)
result = glob.glob('*.{}'.format(extension))
print(result)
This solution uses the python function filter. This function creates a list of elements for which a function returns true. In this case, the anonymous function used is partial matching '.csv' on every element of the directory files list obtained with os.listdir('the path i want to look in')
import os
filepath= 'filepath_to_my_CSVs' # for example: './my_data/'
list(filter(lambda x: '.csv' in x, os.listdir('filepath_to_my_CSVs')))
Source: Stackoverflow.com