I'm working on a certain program where I need to do different things depending on the extension of the file. Could I just use this?
if m == *.mp3
...
elif m == *.flac
...
Assuming m is a string, you can use endswith:
if m.endswith('.mp3'):
...
elif m.endswith('.flac'):
...
To be case-insensitive, and to eliminate a potentially large else-if chain:
m.lower().endswith(('.png', '.jpg', '.jpeg'))
os.path provides many functions for manipulating paths/filenames. (docs)
os.path.splitext takes a path and splits the file extension from the end of it.
import os
filepaths = ["/folder/soundfile.mp3", "folder1/folder/soundfile.flac"]
for fp in filepaths:
# Split the extension from the path and normalise it to lowercase.
ext = os.path.splitext(fp)[-1].lower()
# Now we can simply use == to check for equality, no need for wildcards.
if ext == ".mp3":
print fp, "is an mp3!"
elif ext == ".flac":
print fp, "is a flac file!"
else:
print fp, "is an unknown file format."
Gives:
/folder/soundfile.mp3 is an mp3!
folder1/folder/soundfile.flac is a flac file!
Use pathlib From Python3.4 onwards.
from pathlib import Path
Path('my_file.mp3').suffix == '.mp3'
If you are working with folders that contain periods, you can perform an extra check using
Path('your_folder.mp3').is_file() and Path('your_folder.mp3').suffix == '.mp3'
to ensure that a folder with a .mp3 suffix is not interpreted to be an mp3 file.
Look at module fnmatch. That will do what you're trying to do.
import fnmatch
import os
for file in os.listdir('.'):
if fnmatch.fnmatch(file, '*.txt'):
print file
or perhaps:
from glob import glob
...
for files in glob('path/*.mp3'):
do something
for files in glob('path/*.flac'):
do something else
one easy way could be:
import os
if os.path.splitext(file)[1] == ".mp3":
# do something
os.path.splitext(file) will return a tuple with two values (the filename without extension + just the extension). The second index ([1]) will therefor give you just the extension. The cool thing is, that this way you can also access the filename pretty easily, if needed!
An old thread, but may help future readers...
I would avoid using .lower() on filenames if for no other reason than to make your code more platform independent. (linux is case sensistive, .lower() on a filename will surely corrupt your logic eventually ...or worse, an important file!)
Why not use re? (Although to be even more robust, you should check the magic file header of each file...
How to check type of files without extensions in python? )
import re
def checkext(fname):
if re.search('\.mp3$',fname,flags=re.IGNORECASE):
return('mp3')
if re.search('\.flac$',fname,flags=re.IGNORECASE):
return('flac')
return('skip')
flist = ['myfile.mp3', 'myfile.MP3','myfile.mP3','myfile.mp4','myfile.flack','myfile.FLAC',
'myfile.Mov','myfile.fLaC']
for f in flist:
print "{} ==> {}".format(f,checkext(f))
Output:
myfile.mp3 ==> mp3
myfile.MP3 ==> mp3
myfile.mP3 ==> mp3
myfile.mp4 ==> skip
myfile.flack ==> skip
myfile.FLAC ==> flac
myfile.Mov ==> skip
myfile.fLaC ==> flac
You should make sure the "file" isn't actually a folder before checking the extension. Some of the answers above don't account for folder names with periods. (folder.mp3 is a valid folder name).
Checking the extension of a file:
import os
file_path = "C:/folder/file.mp3"
if os.path.isfile(file_path):
file_extension = os.path.splitext(file_path)[1]
if file_extension.lower() == ".mp3":
print("It's an mp3")
if file_extension.lower() == ".flac":
print("It's a flac")
Output:
It's an mp3
Checking the extension of all files in a folder:
import os
directory = "C:/folder"
for file in os.listdir(directory):
file_path = os.path.join(directory, file)
if os.path.isfile(file_path):
file_extension = os.path.splitext(file_path)[1]
print(file, "ends in", file_extension)
Output:
abc.txt ends in .txt
file.mp3 ends in .mp3
song.flac ends in .flac
Comparing file extension against multiple types:
import os
file_path = "C:/folder/file.mp3"
if os.path.isfile(file_path):
file_extension = os.path.splitext(file_path)[1]
if file_extension.lower() in {'.mp3', '.flac', '.ogg'}:
print("It's a music file")
elif file_extension.lower() in {'.jpg', '.jpeg', '.png'}:
print("It's an image file")
Output:
It's a music file
import os
source = ['test_sound.flac','ts.mp3']
for files in source:
fileName,fileExtension = os.path.splitext(files)
print fileExtension # Print File Extensions
print fileName # It print file name
#!/usr/bin/python
import shutil, os
source = ['test_sound.flac','ts.mp3']
for files in source:
fileName,fileExtension = os.path.splitext(files)
if fileExtension==".flac" :
print 'This file is flac file %s' %files
elif fileExtension==".mp3":
print 'This file is mp3 file %s' %files
else:
print 'Format is not valid'
if (file.split(".")[1] == "mp3"):
print "its mp3"
elif (file.split(".")[1] == "flac"):
print "its flac"
else:
print "not compat"
If your file is uploaded then
import os
file= request.FILES['your_file_name'] #Your input file_name for your_file_name
ext = os.path.splitext(file.name)[-1].lower()
if ext=='.mp3':
#do something
elif ext=='.xls' or '.xlsx' or '.csv':
#do something
else:
#The uploaded file is not the required format
file='test.xlsx'
if file.endswith('.csv'):
print('file is CSV')
elif file.endswith('.xlsx'):
print('file is excel')
else:
print('none of them')
I'm surprised none of the answers proposed the use of the pathlib library.
Of course, its use is situational but when it comes to file handling or stats pathlib is gold.
Here's a snippet:
import pathlib
def get_parts(p: str or pathlib.Path) -> None:
p_ = pathlib.Path(p).expanduser().resolve()
print(p_)
print(f"file name: {p_.name}")
print(f"file extension: {p_.suffix}")
print(f"file extensions: {p_.suffixes}\n")
if __name__ == '__main__':
file_path = 'conf/conf.yml'
arch_file_path = 'export/lib.tar.gz'
get_parts(p=file_path)
get_parts(p=arch_file_path)
and the output:
/Users/hamster/temp/src/pro1/conf/conf.yml
file name: conf.yml
file extension: .yml
file extensions: ['.yml']
/Users/hamster/temp/src/pro1/conf/lib.tar.gz
file name: lib.tar.gz
file extension: .gz
file extensions: ['.tar', '.gz']
Related
I tried to make a program which delete all of the empty files ( whose size is zero ). Then, i run the program by dragging the script file in "command prompt" and run it .
However, no empty files had deleted (but i have some of them).
Please help me to find the error in my code.
import os
a = os.listdir('C:\\Python27')
for folder in a :
sizes = os.stat('C:\\Python27')
b = sizes.st_size
s = folder
if b == 0 :
remove('C:\\Python27\s')
You're assigning the values iterator os.listdir returns to folder and yet you aren't using it at all in os.stat or os.remove, but instead you are passing to them fixed values that you don't need.
You should do something like this:
import os
dir = 'C:\\Python27'
for file_name in os.listdir(dir):
file_path = os.path.join(dir, file_name)
if os.stat(file_path).st_size == 0:
os.remove(file_path)
You can delete something like the following code and you need to add some exception handling. I have used a test folder name to demonstrate.
import os
import sys
dir = 'c:/temp/testfolder'
for root, dirs, files in os.walk(dir):
for file in files:
fname = os.path.join(root, file)
try:
if os.path.getsize(fname) == 0:
print("Removing file %s" %(fname))
os.remove(fname)
except:
print("error: unable to remove 0 byte file")
raise
I want to write a program for this: In a folder I have n number of files; first read one file and perform some operation then store result in a separate file. Then read 2nd file, perform operation again and save result in new 2nd file. Do the same procedure for n number of files. The program reads all files one by one and stores results of each file separately. Please give examples how I can do it.
I think what you miss is how to retrieve all the files in that directory.
To do so, use the glob module.
Here is an example which will duplicate all the files with extension *.txt to files with extension *.out
import glob
list_of_files = glob.glob('./*.txt') # create the list of file
for file_name in list_of_files:
FI = open(file_name, 'r')
FO = open(file_name.replace('txt', 'out'), 'w')
for line in FI:
FO.write(line)
FI.close()
FO.close()
import sys
# argv is your commandline arguments, argv[0] is your program name, so skip it
for n in sys.argv[1:]:
print(n) #print out the filename we are currently processing
input = open(n, "r")
output = open(n + ".out", "w")
# do some processing
input.close()
output.close()
Then call it like:
./foo.py bar.txt baz.txt
You may find the fileinput module useful. It is designed for exactly this problem.
I've just learned of the os.walk() command recently, and it may help you here.
It allows you to walk down a directory tree structure.
import os
OUTPUT_DIR = 'C:\\RESULTS'
for path, dirs, files in os.walk('.'):
for file in files:
read_f = open(os.join(path,file),'r')
write_f = open(os.path.join(OUTPUT_DIR,file))
# Do stuff
Combined answer incorporating directory or specific list of filenames arguments:
import sys
import os.path
import glob
def processFile(filename):
fileHandle = open(filename, "r")
for line in fileHandle:
# do some processing
pass
fileHandle.close()
def outputResults(filename):
output_filemask = "out"
fileHandle = open("%s.%s" % (filename, output_filemask), "w")
# do some processing
fileHandle.write('processed\n')
fileHandle.close()
def processFiles(args):
input_filemask = "log"
directory = args[1]
if os.path.isdir(directory):
print "processing a directory"
list_of_files = glob.glob('%s/*.%s' % (directory, input_filemask))
else:
print "processing a list of files"
list_of_files = sys.argv[1:]
for file_name in list_of_files:
print file_name
processFile(file_name)
outputResults(file_name)
if __name__ == '__main__':
if (len(sys.argv) > 1):
processFiles(sys.argv)
else:
print 'usage message'
from pylab import *
import csv
import os
import glob
import re
x=[]
y=[]
f=open("one.txt",'w')
for infile in glob.glob(('*.csv')):
# print "" +infile
csv23=csv2rec(""+infile,'rb',delimiter=',')
for line in csv23:
x.append(line[1])
# print len(x)
for i in range(3000,8000):
y.append(x[i])
print ""+infile,"\t",mean(y)
print >>f,""+infile,"\t\t",mean(y)
del y[:len(y)]
del x[:len(x)]
I know I saw this double with open() somewhere but couldn't remember where. So I built a small example in case someone needs.
""" A module to clean code(js, py, json or whatever) files saved as .txt files to
be used in HTML code blocks. """
from os import listdir
from os.path import abspath, dirname, splitext
from re import sub, MULTILINE
def cleanForHTML():
""" This function will search a directory text files to be edited. """
## define some regex for our search and replace. We are looking for <, > and &
## To replaced with &ls;, > and &. We might want to replace proper whitespace
## chars to as well? (r'\t', ' ') and (f'\n', '<br>')
search_ = ((r'(<)', '<'), (r'(>)', '>'), (r'(&)', '&'))
## Read and loop our file location. Our location is the same one that our python file is in.
for loc in listdir(abspath(dirname(__file__))):
## Here we split our filename into it's parts ('fileName', '.txt')
name = splitext(loc)
if name[1] == '.txt':
## we found our .txt file so we can start file operations.
with open(loc, 'r') as file_1, open(f'{name[0]}(fixed){name[1]}', 'w') as file_2:
## read our first file
retFile = file_1.read()
## find and replace some text.
for find_ in search_:
retFile = sub(find_[0], find_[1], retFile, 0, MULTILINE)
## finally we can write to our newly created text file.
file_2.write(retFile)
This thing also works for reading multiple files, my file name is fedaralist_1.txt and federalist_2.txt and like this, I have 84 files till fedaralist_84.txt
And I'm reading the files as f.
for file in filename:
with open(f'federalist_{file}.txt','r') as f:
f.read()
import os
def rename_files():
# (1) get file names from a folder
file_list = os.listdir(r"C:\Users\USEER\Desktop\Udacity\Udacity - Programming Foundation with Python\Project\prank\prank")
# print(file_list)
saved_path = os.getcwd()
print("Current Working Directory is " + saved_path)
os.chdir(r"C:\Users\USEER\Desktop\Udacity\Udacity - Programming Foundation with Python\Project\prank\prank")
# (2) for each file, rename file name
for file_name in file_list:
print("Old Name - " + file_name)
print("New Name - " + file_name.translate("0123456789"))
os.rename(file_name, file_name.translate("0123456789"))
os.chdir(saved_path)
rename_files()
The code above doesn't rename the file by removing the integers. Can anyone help? (Python 3x)
import re
new_name = re.sub('[0-9]', '', file_name)
In Python 3 the String.translate is gone. Therefore you need to use the str.translate. It needs 'str.maketrans' which normally creates a translation table with the first two arguments supplied(not needed in this example), the third argument supplies the characters to be stripped.
This line should have the desired effect ...
os.rename(file_name, file_name.translate(str.maketrans('','','0123456789'))
Previous suggestions used .strip() however in this case as the numbers are mixed in with the filenames (not before or after) I believe it would not work, another used Regular Expressions which is perfectly valid, however within the context of this particular Udacity course translate was the suggested solution.
Here are the docs for maketrans :
[https://docs.python.org/3/library/stdtypes.html#str.maketrans][1]
The problem is in your translate function that doesn't do anything. There are better options available, but if your want to use translate then the proper syntax is:
#!/usr/bin/env python2
import string
new_name = string.translate(file_name, None, "0123456789")
Here is one way of renaming files.
Use os.renames to rename the files
Use file_name.strip("0123456789") to remove numbers
Code is given below:
import os
def file_rename():
name_list=os.listdir(r"C:\python\prank")
print(name_list)
saved_path=os.getcwd()
print("Current working directory is"+saved_path)
os.chdir(r"C:\python\prank")
for file_name in name_list:
print("old name"+file_name)
print("new name"+file_name.strip("0123456789"))
os.renames(file_name,file_name.strip("0123456789"))
os.chdir(saved_path)
file_rename()
To read more about os.renames check here.
To read more about the strip function, check here.
Another way to do this without import re. Instead of utilizing the .translate, use the .strip.
os.rename(file_name, file_name.strip('0123456789'))
Another observation is that your code wont read the new file name after changing it. At the top of your code you are reading file names and saving those name in file_list
# (1) get file names from a folder
file_list = os.listdir(r"C:\Users\USEER\Desktop\Udacity\Udacity - Programming Foundation with Python\Project\prank\prank")
In the for loop, where you are changing the name of each file, YOU ARE NOT reading the new file's name. You need to do something like this.
# (2) for each file, rename file name
for file_name in file_list:
print("Old Name - " + file_name)
os.rename(file_name, file_name.strip("0123456789"))
# (3) read file's name again... 'file_list' has old names
new_file_list = os.listdir(r"C:\Users\USEER\Desktop\Udacity\Udacity - Programming Foundation with Python\Project\prank\prank")
for file_name in new_file_list:
print("New file's name: " + new_file_name)
os.chdir(saved_path)
import os
def rename_files():
#1 Get file names from the folder
file = os.listdir(r"C:\Web\Python\prank")
print(file)
saved_path = os.getcwd()
print("Current Working Directory is"+saved_path)
os.chdir(r"C:\Web\Python\prank")
#2 For each file name rename file names
for file_name in file:
print("Old Name - " + file_name)
os.rename(file_name,file_name.strip("0123456789"))
print("New Name - " + file_name)
os.chdir(saved_path)
rename_files()
enter code here
import os
dir="/home/lucidvis/myPythonHome/prank/"
def rename_files():
# get file names from a folder
filenames = os.listdir(dir)
# print(filenames)
for file in filenames:
#print(file)
try:
#with os.open(filename for filename in filenames,"r+"):
#read_file = filename.read()
# new_name = file.translate(str.maketrans('','', "0123456789"))
new_filname = (file.translate(str.maketrans('','', "0123456789"))).replace(" ","")
#print(dir+new_file_name)
os.rename(dir+file,dir+new_file_name)
except SyntaxError as e:
print(e)
continue
# for each file, rename filename
rename_files()
import os
dir="/home/lucidvis/myPythonHome/prank/"
def rename_files():
# get file names from a folder
filenames = os.listdir(dir)
# iterate through the list of filenames
for file in filenames:
#print(file)
try:
#assign a variable to new names for easy manipulation
new_filname = (file.translate(str.maketrans('','', "0123456789"))).replace(" ","")
#concatenating the directory name to old and new file names
os.rename(dir+file,dir+new_file_name)
#just to manage errors
except SyntaxError as e:
print(e)
continue
#file renaming function call
rename_files()
will guys i was trying to solve this problem because i see udacity online courses and it require to rename file without numbers thanks to simon for his replay i have to figured it out
this is my code to rename files without numbers, hope it jhelp anyone who stuck
import os
import re
def rename():
#get the list of the photo name
plist = os.listdir(r"D:\pay\prank")
print(plist)
#removing the numbers from the photo names
os.chdir(r"D:\pay\prank")
for pname in plist :
os.rename(pname, re.sub('[0-9]', '' , pname))
print(pname)
rename()
import os
import re
from string import digits
#Get file names
file_list = os.listdir(r"C:\Users\703305981\Downloads\prank\prank")
print(file_list)
#chenage directory.
os.chdir(r"C:\Users\703305981\Downloads\prank\prank")
print (os.getcwd())
#Change the Name.
for file_name in file_list:
os.rename(file_name, re.sub(r'[0-9]+', '', file_name))
I have this code that will let the user choose which file he wants to update by passing an argument in the command line, and then it do some more things but I have not included that here:
import sys
import os
from sys import argv
path = "/home/Desktop/python/test"
files = os.walk( path )
filename = argv[1]
if filename in files:
inputFile = open(filename, 'r')
else:
print "no match found"
sys.exit()
inputFile.close()
When I run the script it keeps giving me "no match found" but im pretty sure the file is there. I cant see what Im doing wrong
os.walk() returns a generator, one that produces tuples with (root, directories, files) values for each iteration.
You can't use that generator to test for a single file, not with a simple in membership test.
You'll also need to re-instate the whole path; you can't just open an unclassified filename without the directory it lives in. Just use a for loop here, and break once you found it. The else suite on a for loop only executes when you did not use break (e.g. the file was not found):
path = "/home/Desktop/python/test"
filename = argv[1]
for root, directories, files in os.walk(path):
if filename in files:
full_path = os.path.join(root, filename)
break
else:
print "no match found"
sys.exit()
with open(full_path) as input_file:
# do something with the file
I added a with statement to handle the lifetime of the file object; once the with block is exited the file is automatically closed for you.
Alternatively, you may use following code snippet.
import os.path
filename = argv[1]
path = "/home/Desktop/python/test/"
if os.path.isfile(path + filename):
inputFile = open(path + filename, "r")
else:
print "File Not Found"
In my code, I want to execute import filename for all the files that are present in a directory. I have a file abc.py under workspace.
I am currently doing the following :
for filename in os.listdir(homedir/workspace)
exec "import " + filename
filename = eval(filename + '.function(variable)')
The problem is that instead of doing import abc, it is doing import abc.py, and then showing the error no module named py
How can I resolve this?
Thanks in advance!
You can use os.path.splitext
os.path.splitext(filename)[0]
Of the returned two element array, the first element is just the filename, the second element is the extension. To be extra safe, you could double check that you did indeed grab a .py file:
if os.path.splitext(filename)[1] == '.py':
# do stuff
To run .function(variable) for all Python files in a directory:
import os
import sys
from importlib import import_module
dirpath = 'homedir/workspace'
sys.path.append(dirpath)
for filename in os.listdir(path)
module_name, ext = os.path.splitext(filename)
if ext == '.py':
try:
result = import_module(module_name).function(variable)
except Exception as e:
sys.stderr.write('error: %s: %s\n' % (filename, e))
continue
else:
print('%s: %s' % (filename, result))
See also Building a minimal plugin architecture in Python.
You can use glob:
import glob
import os
os.chdir("/mydir")
for file in glob.glob("*.py"):
# do something with file
Read this to make sure You really want to use exec