Related
I'm trying to upload all my files to a time_name directory (such as "170905_161330") in my ftp server.
# -*- coding:utf-8 -*-
import ftplib
import os
import datetime
CWD = os.getcwd()
NOW_STR = str(datetime.datetime.now())
LOCAL_STR = "HDD1/AutoBackUp/" + NOW_STR[2:4]+NOW_STR[5:7]+NOW_STR[8:10]+"_"+NOW_STR[11:13]+NOW_STR[14:16]+NOW_STR[17:19]+"/"
FTP_ADDRESS = 'FTP_ADDRESS'
FTP_ID = '1'
FTP_PW = '1'
ftp = ftplib.FTP(FTP_ADDRESS, FTP_ID, FTP_PW)
for i in os.listdir(CWD):
FILENAME = str(i)
print(FILENAME)
ftp.mkd(LOCAL_STR)
LOCAL_NAME = LOCAL_STR + FILENAME
print(str(LOCAL_NAME))
with open(FILENAME, 'rb') as fileOBJ:
ftp.storlines('STOR ' + str(LOCAL_NAME), fileOBJ)
ftp.quit()
But the error
ftplib.error_perm: 550 HDD1/AutoBackUp/170905_160635/: Directory not empty
continues to appear, while the first file is uploaded correctly. After that, it doesn't work.
I can check my first file in ftp server, but yeah. second file doesn't exist.
I guess... storlines function only works when upload folder is empty.
How can I solve this problem?
From a very rapid read of your code, I suspect that the problem is in ftp.mkd. You already created the directory at the first iteration of the for loop.
To test this error on your local system, open the terminal:
write a mkdir test command
write a mkdir test again
You'll see an error: File Exist. I think the directory not empty is gnerated from this error in the server.
Modify your code to put ftp.mkd before the for loop:
ftp.mkd(LOCAL_STR)
for i in os.listdir(CWD):
FILENAME = str(i)
print(FILENAME)
LOCAL_NAME = LOCAL_STR + FILENAME
print(str(LOCAL_NAME))
with open(FILENAME, 'rb') as fileOBJ:
ftp.storlines('STOR ' + str(LOCAL_NAME), fileOBJ)
ftp.quit()
and test it again. Please remember to remove the directory from the server before testing it.
I am new to Python scripting. I need to copy few folders from my local machine (windows) to Linux server. As of now, I am copying the folders by opening WinSCP console. I need to automate this process. I have written a below code in Python using Paramiko module library.
import paramiko
import os
transport = paramiko.Transport(('10.10.10.10', 22))
transport.connect(username='weblogic', password='weblogic')
sftp = paramiko.SFTPClient.from_transport(transport)
filepath = '/apps/logs'
localpath = 'C:\\Users\\Public\\test'
sftp.put(localpath,filepath)
Above is not working properly and giving below error. Can you please help me to copy the folder present in the windows path C:\Users\Public\test to Linux server path /apps/logs?
Traceback (most recent call last):
File "C:\Users\Desktop\python\execute_script.py", line 28, in <module>
sftp.put(localpath,filepath)
File "C:\Python27\lib\paramiko\sftp_client.py", line 548, in put
fl = file(localpath, 'rb')
IOError: [Errno 13] Permission denied: 'C:\\Users\\Public\\test'
Please check the below code from the link https://gist.github.com/johnfink8/2190472. I have used put_all method in the snippet.
import paramiko
import socket
import os
from stat import S_ISDIR
class SSHSession(object):
# Usage:
# Detects DSA or RSA from key_file, either as a string filename or a
# file object. Password auth is possible, but I will judge you for
# using it. So:
# ssh=SSHSession('targetserver.com','root',key_file=open('mykey.pem','r'))
# ssh=SSHSession('targetserver.com','root',key_file='/home/me/mykey.pem')
# ssh=SSHSession('targetserver.com','root','mypassword')
# ssh.put('filename','/remote/file/destination/path')
# ssh.put_all('/path/to/local/source/dir','/path/to/remote/destination')
# ssh.get_all('/path/to/remote/source/dir','/path/to/local/destination')
# ssh.command('echo "Command to execute"')
def __init__(self,hostname,username='root',key_file=None,password=None):
#
# Accepts a file-like object (anything with a readlines() function)
# in either dss_key or rsa_key with a private key. Since I don't
# ever intend to leave a server open to a password auth.
#
self.sock = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
self.sock.connect((hostname,22))
self.t = paramiko.Transport(self.sock)
self.t.start_client()
keys = paramiko.util.load_host_keys(os.path.expanduser('~/.ssh/known_hosts'))
key = self.t.get_remote_server_key()
# supposed to check for key in keys, but I don't much care right now to find the right notation
if key_file is not None:
if isinstance(key,str):
key_file=open(key,'r')
key_head=key_file.readline()
key_file.seek(0)
if 'DSA' in key_head:
keytype=paramiko.DSSKey
elif 'RSA' in key_head:
keytype=paramiko.RSAKey
else:
raise Exception("Can't identify key type")
pkey=keytype.from_private_key(key_file)
self.t.auth_publickey(username, pkey)
else:
if password is not None:
self.t.auth_password(username,password,fallback=False)
else: raise Exception('Must supply either key_file or password')
self.sftp=paramiko.SFTPClient.from_transport(self.t)
def command(self,cmd):
# Breaks the command by lines, sends and receives
# each line and its output separately
#
# Returns the server response text as a string
chan = self.t.open_session()
chan.get_pty()
chan.invoke_shell()
chan.settimeout(20.0)
ret=''
try:
ret+=chan.recv(1024)
except:
chan.send('\n')
ret+=chan.recv(1024)
for line in cmd.split('\n'):
chan.send(line.strip() + '\n')
ret+=chan.recv(1024)
return ret
def put(self,localfile,remotefile):
# Copy localfile to remotefile, overwriting or creating as needed.
self.sftp.put(localfile,remotefile)
def put_all(self,localpath,remotepath):
# recursively upload a full directory
os.chdir(os.path.split(localpath)[0])
parent=os.path.split(localpath)[1]
for walker in os.walk(parent):
try:
self.sftp.mkdir(os.path.join(remotepath,walker[0]))
except:
pass
for file in walker[2]:
self.put(os.path.join(walker[0],file),os.path.join(remotepath,walker[0],file))
def get(self,remotefile,localfile):
# Copy remotefile to localfile, overwriting or creating as needed.
self.sftp.get(remotefile,localfile)
def sftp_walk(self,remotepath):
# Kindof a stripped down version of os.walk, implemented for
# sftp. Tried running it flat without the yields, but it really
# chokes on big directories.
path=remotepath
files=[]
folders=[]
for f in self.sftp.listdir_attr(remotepath):
if S_ISDIR(f.st_mode):
folders.append(f.filename)
else:
files.append(f.filename)
print (path,folders,files)
yield path,folders,files
for folder in folders:
new_path=os.path.join(remotepath,folder)
for x in self.sftp_walk(new_path):
yield x
def get_all(self,remotepath,localpath):
# recursively download a full directory
# Harder than it sounded at first, since paramiko won't walk
#
# For the record, something like this would gennerally be faster:
# ssh user#host 'tar -cz /source/folder' | tar -xz
self.sftp.chdir(os.path.split(remotepath)[0])
parent=os.path.split(remotepath)[1]
try:
os.mkdir(localpath)
except:
pass
for walker in self.sftp_walk(parent):
try:
os.mkdir(os.path.join(localpath,walker[0]))
except:
pass
for file in walker[2]:
self.get(os.path.join(walker[0],file),os.path.join(localpath,walker[0],file))
def write_command(self,text,remotefile):
# Writes text to remotefile, and makes remotefile executable.
# This is perhaps a bit niche, but I was thinking I needed it.
# For the record, I was incorrect.
self.sftp.open(remotefile,'w').write(text)
self.sftp.chmod(remotefile,755)
In addition to the answer #user1041177, but here a way to do it when you are on windows to linux host (not really sure which kind of host actually).
I don't know why, but if I keep backslash onto remote path, I get a FileNotFoundException. The only way to work was to replace all '\' by '/'
Maybe someone could tell me the proper way to avoid this situation at all ?
Here a part of the exact same code above to give you breadcrumbs if you encounter the same issue :
def sftp_walk(socket, remotepath):
remotepath = remotepath.replace('\\', '/')
path = remotepath
files = []
folders = []
for f in socket.listdir_attr(remotepath.replace('\\', '/')):
if S_ISDIR(f.st_mode):
folders.append(f.filename)
else:
files.append(f.filename)
print(path, folders, files)
yield path, folders, files
for folder in folders:
new_path = os.path.join(remotepath.replace('\\', '/'), folder)
for x in sftp_walk(socket, new_path):
yield x
def get_all(socket, remotepath, localpath):
remotepath = remotepath.replace('\\', '/')
socket.chdir(os.path.split(remotepath)[0])
parent = os.path.split(remotepath)[1]
try:
os.mkdir(localpath)
except:
pass
for walker in sftp_walk(socket, parent):
try:
os.mkdir(os.path.join(localpath, walker[0]).replace('\\', '/'))
except:
pass
for file in walker[2]:
socket.get(os.path.join(walker[0], file).replace('\\', '/'), os.path.join(localpath, walker[0], file).replace('\\', '/'))
BTW, I am not using those function inside an object, that's why their is 'socket' instead of 'self' because I call those function by passing the SFTP socket to them.
Finally have to say thank you to #user1041177, working like a charm.
I was trying to copy from a windows box to a linux box and got the same error as #Apex above. I was using the put_all method and I had to do some "replace" on parts of the code.
def put_all(self,localpath,remotepath):
remotepath = remotepath.replace('\\', '/')
# recursively upload a full directory
os.chdir(os.path.split(localpath)[0])
parent=os.path.split(localpath)[1]
for walker in os.walk(parent):
try:
self.sftp.mkdir(os.path.join(remotepath,walker[0]).replace('\\', '/'))
except:
pass
for file in walker[2]:
self.put(os.path.join(walker[0],file).replace('\\', '/'),os.path.join(remotepath,walker[0],file).replace('\\', '/'))
I found a few shortcomings with the above methods - first, the putter/getter doesn't function in the way you'd expect - if you want to put /foo/bar into /some/folder, you can't as it won't let you put files from a source folder to a different destination folder - the only thing you can do is put /foo/bar into /some/bar. In addition, you have to specify the source as /foo/bar and the destination as /some to end up with /some/bar - I find this confusing as it's not how most operating/ftp systems handle putting/getting/copying/etc. So, I improved on the answers listed above:
If you're going from Windows to Linux:
def put_dir(source, dest):
source = os.path.expandvars(source).rstrip('\\').rstrip('/')
dest = os.path.expandvars(dest).rstrip('\\').rstrip('/')
for root, dirs, files in os.walk(source):
for dir in dirs:
try:
sftp.mkdir(posixpath.join(dest, ''.join(root.rsplit(source))[1:].replace('\\', '/'), dir))
except:
pass
for file in files:
sftp.put(os.path.join(root, file), posixpath.join(dest, ''.join(root.rsplit(source))[1:].replace('\\', '/'), file))
source = '%USERPROFILE%\\Downloads\\'
dest = '/foo/bar'
put_dir(source, dest)
If you're just doing Windows then swap out posixpath.join with os.path.join and remove .replace('\\', '/'):
def put_dir(source, dest):
source = os.path.expandvars(source).rstrip('\\').rstrip('/')
dest = os.path.expandvars(dest).rstrip('\\').rstrip('/')
for root, dirs, files in os.walk(source):
for dir in dirs:
try:
sftp.mkdir(os.path.join(dest, ''.join(root.rsplit(source))[1:], dir))
except:
pass
for file in files:
sftp.put(os.path.join(root, file), os.path.join(dest, ''.join(root.rsplit(source))[1:], file))
source = '%USERPROFILE%\\Downloads\\'
dest = 'foo\\bar'
put_dir(source, dest)
The reason for the try statement is that sftp.mkdir errors out if the folder already exists.
Paramiko does not support recursive operations.
You can use pysftp. It's a wrapper around Paramiko that has more Python-ish look and feel and supports recursive operations. See
pysftp.Connection.put_r()
pysftp.Connection.get_r()
Or you can just base your code on pysftp source code. Or see my answer to Python pysftp get_r from Linux works fine on Linux but not on Windows.
Suppose my python code is executed a directory called main and the application needs to access main/2091/data.txt.
how should I use open(location)? what should the parameter location be?
I found that below simple code will work.. does it have any disadvantages?
file = "\2091\sample.txt"
path = os.getcwd()+file
fp = open(path, 'r+');
With this type of thing you need to be careful what your actual working directory is. For example, you may not run the script from the directory the file is in. In this case, you can't just use a relative path by itself.
If you are sure the file you want is in a subdirectory beneath where the script is actually located, you can use __file__ to help you out here. __file__ is the full path to where the script you are running is located.
So you can fiddle with something like this:
import os
script_dir = os.path.dirname(__file__) #<-- absolute dir the script is in
rel_path = "2091/data.txt"
abs_file_path = os.path.join(script_dir, rel_path)
This code works fine:
import os
def read_file(file_name):
file_handle = open(file_name)
print file_handle.read()
file_handle.close()
file_dir = os.path.dirname(os.path.realpath('__file__'))
print file_dir
#For accessing the file in the same folder
file_name = "same.txt"
read_file(file_name)
#For accessing the file in a folder contained in the current folder
file_name = os.path.join(file_dir, 'Folder1.1/same.txt')
read_file(file_name)
#For accessing the file in the parent folder of the current folder
file_name = os.path.join(file_dir, '../same.txt')
read_file(file_name)
#For accessing the file inside a sibling folder.
file_name = os.path.join(file_dir, '../Folder2/same.txt')
file_name = os.path.abspath(os.path.realpath(file_name))
print file_name
read_file(file_name)
I created an account just so I could clarify a discrepancy I think I found in Russ's original response.
For reference, his original answer was:
import os
script_dir = os.path.dirname(__file__)
rel_path = "2091/data.txt"
abs_file_path = os.path.join(script_dir, rel_path)
This is a great answer because it is trying to dynamically creates an absolute system path to the desired file.
Cory Mawhorter noticed that __file__ is a relative path (it is as well on my system) and suggested using os.path.abspath(__file__). os.path.abspath, however, returns the absolute path of your current script (i.e. /path/to/dir/foobar.py)
To use this method (and how I eventually got it working) you have to remove the script name from the end of the path:
import os
script_path = os.path.abspath(__file__) # i.e. /path/to/dir/foobar.py
script_dir = os.path.split(script_path)[0] #i.e. /path/to/dir/
rel_path = "2091/data.txt"
abs_file_path = os.path.join(script_dir, rel_path)
The resulting abs_file_path (in this example) becomes: /path/to/dir/2091/data.txt
It depends on what operating system you're using. If you want a solution that is compatible with both Windows and *nix something like:
from os import path
file_path = path.relpath("2091/data.txt")
with open(file_path) as f:
<do stuff>
should work fine.
The path module is able to format a path for whatever operating system it's running on. Also, python handles relative paths just fine, so long as you have correct permissions.
Edit:
As mentioned by kindall in the comments, python can convert between unix-style and windows-style paths anyway, so even simpler code will work:
with open("2091/data/txt") as f:
<do stuff>
That being said, the path module still has some useful functions.
I spend a lot time to discover why my code could not find my file running Python 3 on the Windows system. So I added . before / and everything worked fine:
import os
script_dir = os.path.dirname(__file__)
file_path = os.path.join(script_dir, './output03.txt')
print(file_path)
fptr = open(file_path, 'w')
Try this:
from pathlib import Path
data_folder = Path("/relative/path")
file_to_open = data_folder / "file.pdf"
f = open(file_to_open)
print(f.read())
Python 3.4 introduced a new standard library for dealing with files and paths called pathlib. It works for me!
Code:
import os
script_path = os.path.abspath(__file__)
path_list = script_path.split(os.sep)
script_directory = path_list[0:len(path_list)-1]
rel_path = "main/2091/data.txt"
path = "/".join(script_directory) + "/" + rel_path
Explanation:
Import library:
import os
Use __file__ to attain the current script's path:
script_path = os.path.abspath(__file__)
Separates the script path into multiple items:
path_list = script_path.split(os.sep)
Remove the last item in the list (the actual script file):
script_directory = path_list[0:len(path_list)-1]
Add the relative file's path:
rel_path = "main/2091/data.txt
Join the list items, and addition the relative path's file:
path = "/".join(script_directory) + "/" + rel_path
Now you are set to do whatever you want with the file, such as, for example:
file = open(path)
import os
def file_path(relative_path):
dir = os.path.dirname(os.path.abspath(__file__))
split_path = relative_path.split("/")
new_path = os.path.join(dir, *split_path)
return new_path
with open(file_path("2091/data.txt"), "w") as f:
f.write("Powerful you have become.")
If the file is in your parent folder, eg. follower.txt, you can simply use open('../follower.txt', 'r').read()
Get the path of the parent folder, then os.join your relative files to the end.
# get parent folder with `os.path`
import os.path
BASE_DIR = os.path.dirname(os.path.abspath(__file__))
# now use BASE_DIR to get a file relative to the current script
os.path.join(BASE_DIR, "config.yaml")
The same thing with pathlib:
# get parent folder with `pathlib`'s Path
from pathlib import Path
BASE_DIR = Path(__file__).absolute().parent
# now use BASE_DIR to get a file relative to the current script
BASE_DIR / "config.yaml"
Python just passes the filename you give it to the operating system, which opens it. If your operating system supports relative paths like main/2091/data.txt (hint: it does), then that will work fine.
You may find that the easiest way to answer a question like this is to try it and see what happens.
Not sure if this work everywhere.
I'm using ipython in ubuntu.
If you want to read file in current folder's sub-directory:
/current-folder/sub-directory/data.csv
your script is in current-folder
simply try this:
import pandas as pd
path = './sub-directory/data.csv'
pd.read_csv(path)
When I was a beginner I found these descriptions a bit intimidating. As at first I would try
For Windows
f= open('C:\Users\chidu\Desktop\Skipper New\Special_Note.txt','w+')
print(f)
and this would raise an syntax error. I used get confused alot. Then after some surfing across google. found why the error occurred. Writing this for beginners
It's because for path to be read in Unicode you simple add a \ when starting file path
f= open('C:\\Users\chidu\Desktop\Skipper New\Special_Note.txt','w+')
print(f)
And now it works just add \ before starting the directory.
In Python 3.4 (PEP 428) the pathlib was introduced, allowing you to work with files in an object oriented fashion:
from pathlib import Path
working_directory = Path(os.getcwd())
path = working_directory / "2091" / "sample.txt"
with path.open('r+') as fp:
# do magic
The with keyword will also ensure that your resources get closed properly, even if you get something goes wrong (like an unhandled Exception, sigint or similar)
I'm trying to rename some files in a directory using Python.
Say I have a file called CHEESE_CHEESE_TYPE.*** and want to remove CHEESE_ so my resulting filename would be CHEESE_TYPE
I'm trying to use the os.path.split but it's not working properly. I have also considered using string manipulations, but have not been successful with that either.
Use os.rename(src, dst) to rename or move a file or a directory.
$ ls
cheese_cheese_type.bar cheese_cheese_type.foo
$ python
>>> import os
>>> for filename in os.listdir("."):
... if filename.startswith("cheese_"):
... os.rename(filename, filename[7:])
...
>>>
$ ls
cheese_type.bar cheese_type.foo
Here's a script based on your newest comment.
#!/usr/bin/env python
from os import rename, listdir
badprefix = "cheese_"
fnames = listdir('.')
for fname in fnames:
if fname.startswith(badprefix*2):
rename(fname, fname.replace(badprefix, '', 1))
The following code should work. It takes every filename in the current directory, if the filename contains the pattern CHEESE_CHEESE_ then it is renamed. If not nothing is done to the filename.
import os
for fileName in os.listdir("."):
os.rename(fileName, fileName.replace("CHEESE_CHEESE_", "CHEESE_"))
Assuming you are already in the directory, and that the "first 8 characters" from your comment hold true always. (Although "CHEESE_" is 7 characters... ? If so, change the 8 below to 7)
from glob import glob
from os import rename
for fname in glob('*.prj'):
rename(fname, fname[8:])
I have the same issue, where I want to replace the white space in any pdf file to a dash -.
But the files were in multiple sub-directories. So, I had to use os.walk().
In your case for multiple sub-directories, it could be something like this:
import os
for dpath, dnames, fnames in os.walk('/path/to/directory'):
for f in fnames:
os.chdir(dpath)
if f.startswith('cheese_'):
os.rename(f, f.replace('cheese_', ''))
Try this:
import os
import shutil
for file in os.listdir(dirpath):
newfile = os.path.join(dirpath, file.split("_",1)[1])
shutil.move(os.path.join(dirpath,file),newfile)
I'm assuming you don't want to remove the file extension, but you can just do the same split with periods.
This sort of stuff is perfectly fitted for IPython, which has shell integration.
In [1] files = !ls
In [2] for f in files:
newname = process_filename(f)
mv $f $newname
Note: to store this in a script, use the .ipy extension, and prefix all shell commands with !.
See also: http://ipython.org/ipython-doc/stable/interactive/shell.html
Here is a more general solution:
This code can be used to remove any particular character or set of characters recursively from all filenames within a directory and replace them with any other character, set of characters or no character.
import os
paths = (os.path.join(root, filename)
for root, _, filenames in os.walk('C:\FolderName')
for filename in filenames)
for path in paths:
# the '#' in the example below will be replaced by the '-' in the filenames in the directory
newname = path.replace('#', '-')
if newname != path:
os.rename(path, newname)
It seems that your problem is more in determining the new file name rather than the rename itself (for which you could use the os.rename method).
It is not clear from your question what the pattern is that you want to be renaming. There is nothing wrong with string manipulation. A regular expression may be what you need here.
import os
import string
def rename_files():
#List all files in the directory
file_list = os.listdir("/Users/tedfuller/Desktop/prank/")
print(file_list)
#Change current working directory and print out it's location
working_location = os.chdir("/Users/tedfuller/Desktop/prank/")
working_location = os.getcwd()
print(working_location)
#Rename all the files in that directory
for file_name in file_list:
os.rename(file_name, file_name.translate(str.maketrans("","",string.digits)))
rename_files()
This command will remove the initial "CHEESE_" string from all the files in the current directory, using renamer:
$ renamer --find "/^CHEESE_/" *
I was originally looking for some GUI which would allow renaming using regular expressions and which had a preview of the result before applying changes.
On Linux I have successfully used krename, on Windows Total Commander does renaming with regexes, but I found no decent free equivalent for OSX, so I ended up writing a python script which works recursively and by default only prints the new file names without making any changes. Add the '-w' switch to actually modify the file names.
#!/usr/bin/python
# -*- coding: utf-8 -*-
import os
import fnmatch
import sys
import shutil
import re
def usage():
print """
Usage:
%s <work_dir> <search_regex> <replace_regex> [-w|--write]
By default no changes are made, add '-w' or '--write' as last arg to actually rename files
after you have previewed the result.
""" % (os.path.basename(sys.argv[0]))
def rename_files(directory, search_pattern, replace_pattern, write_changes=False):
pattern_old = re.compile(search_pattern)
for path, dirs, files in os.walk(os.path.abspath(directory)):
for filename in fnmatch.filter(files, "*.*"):
if pattern_old.findall(filename):
new_name = pattern_old.sub(replace_pattern, filename)
filepath_old = os.path.join(path, filename)
filepath_new = os.path.join(path, new_name)
if not filepath_new:
print 'Replacement regex {} returns empty value! Skipping'.format(replace_pattern)
continue
print new_name
if write_changes:
shutil.move(filepath_old, filepath_new)
else:
print 'Name [{}] does not match search regex [{}]'.format(filename, search_pattern)
if __name__ == '__main__':
if len(sys.argv) < 4:
usage()
sys.exit(-1)
work_dir = sys.argv[1]
search_regex = sys.argv[2]
replace_regex = sys.argv[3]
write_changes = (len(sys.argv) > 4) and sys.argv[4].lower() in ['--write', '-w']
rename_files(work_dir, search_regex, replace_regex, write_changes)
Example use case
I want to flip parts of a file name in the following manner, i.e. move the bit m7-08 to the beginning of the file name:
# Before:
Summary-building-mobile-apps-ionic-framework-angularjs-m7-08.mp4
# After:
m7-08_Summary-building-mobile-apps-ionic-framework-angularjs.mp4
This will perform a dry run, and print the new file names without actually renaming any files:
rename_files_regex.py . "([^\.]+?)-(m\\d+-\\d+)" "\\2_\\1"
This will do the actual renaming (you can use either -w or --write):
rename_files_regex.py . "([^\.]+?)-(m\\d+-\\d+)" "\\2_\\1" --write
You can use os.system function for simplicity and to invoke bash to accomplish the task:
import os
os.system('mv old_filename new_filename')
This works for me.
import os
for afile in os.listdir('.'):
filename, file_extension = os.path.splitext(afile)
if not file_extension == '.xyz':
os.rename(afile, filename + '.abc')
What about this :
import re
p = re.compile(r'_')
p.split(filename, 1) #where filename is CHEESE_CHEESE_TYPE.***
I have a few files I want to delete, they have the same name at the start but have different version numbers. Does anyone know how to delete files using the start of their name?
Eg.
version_1.1
version_1.2
Is there a way of delting any file that starts with the name version?
Thanks
import os, glob
for filename in glob.glob("mypath/version*"):
os.remove(filename)
Substitute the correct path (or . (= current directory)) for mypath. And make sure you don't get the path wrong :)
This will raise an Exception if a file is currently in use.
If you really want to use Python, you can just use a combination of os.listdir(), which returns a listing of all the files in a certain directory, and os.remove().
I.e.:
my_dir = # enter the dir name
for fname in os.listdir(my_dir):
if fname.startswith("version"):
os.remove(os.path.join(my_dir, fname))
However, as other answers pointed out, you really don't have to use Python for this, the shell probably natively supports such an operation.
In which language?
In bash (Linux / Unix) you could use:
rm version*
or in batch (Windows / DOS) you could use:
del version*
If you want to write something to do this in Python it would be fairly easy - just look at the documentation for regular expressions.
edit:
just for reference, this is how to do it in Perl:
opendir (folder, "./") || die ("Cannot open directory!");
#files = readdir (folder);
closedir (folder);
unlink foreach (grep /^version/, #files);
import os
os.chdir("/home/path")
for file in os.listdir("."):
if os.path.isfile(file) and file.startswith("version"):
try:
os.remove(file)
except Exception,e:
print e
The following function will remove all files and folders in a directory which start with a common string:
import os
import shutil
def cleanse_folder(directory, prefix):
for item in os.listdir(directory):
path = os.path.join(directory, item)
if item.startswith(prefix):
if os.path.isfile(path):
os.remove(path)
elif os.path.isdir(os.path.join(directory, item)):
shutil.rmtree(path)
else:
print("A simlink or something called {} was not deleted.".format(item))
import os
import re
directory = "./uploaded"
pattern = "1638813371180"
files_in_directory = os.listdir(directory)
filtered_files = [file for file in files_in_directory if ( re.search(pattern,file))]
for file in filtered_files:
path_to_file = os.path.join(directory, file)
os.remove(path_to_file)