I have 4 files (with different extensions) in a directory as shown below.
Test1234_`enter code here`Dir (Directory)
FileA_1234_test.dat
FileAAB_1234_test.log
FileABCD_1234_test.dbm
FileABCDE_1234_test.jdt
I want to replace part of directory and file names as shown below
Test6789_Dir (Directory)
FileA_6789_test.dat
FileAAB_6789_test.log
FileABCD_6789_test.dbm
FileABCDE_6789_test.jdt
How can we achieve this in Python3? I have no idea to do this.
I am not sure what you want to do.
maybe something like this
def replace_names(path, str_to_replace='1234', replacement='6789'):
import os
for filename in os.listdir(first_path):
if '_'+str_to_replace+'_' in filename:
new_filename = filename.replace(str_to_replace, replacement)
os.rename(filename, new_filename )
os.rename(path, .replace(str_to_replace, replacement))
#now, run it...
replace_names('Test1234_Dir')
Related
I was trying to create a file without a name in python (only filetype)
I tried this -
open(".gitignore","w+").close()
But it does not work.
edit - it does work real issue is in getting file through glob.glob
classify_folder_name = #path of the folder which contain .gitignore file
rel_paths = glob.glob(classify_folder_name + '/**', recursive=True)
for local_file in rel_paths:
print(local_file)
it does not print .gitignore file.
Any help will be appreciated.
Note -: don't want to use os.listdir()
There are few things that you might check:
files with dot at the beginning are hidden so whatever OS you are using, make sure you have hidden files visibility enabled
It might be saved in different directory
open(".gitignore","w+").close()
It would be better if you do this:
To create a file:
with open('.gitignore', 'w') as fp:
pass
I'll start by mentioning that I've no knowledge in Python but read online that it could help me with my situation.
I'd like to do a few things using (I believe?) a Python script.
I have a bunch of .yml files that I want to transfer the contents into one main .yml file (let's call it Main.yml). However, I'd also like to be able to take the name of each individual .yml and add it before it's content into Main.yml as "##Name". If possible, the script would look like each file in a directory, instead of having to list every .yml file I want it to look for (my directory in question only contains .yml files). Not sure if I need to specify, but just in case: I want to append the contents of all files into Main.yml & keep the indentation (spacing). P.S. I'm on Windows
Example of what I want:
File: Apes.yml
Contents:
Documentation:
"Apes":
year: 2009
img: 'link'
After running the script, my Main.yml would like like:
##Apes.yml
Documentation:
"Apes":
year: 2009
img: 'link'
I'm just starting out in Python too so this was a great opportunity to see if my newly learned skills work!
I think you want to use the os.walk function to go through all of the files and folders in the directory.
This code should work - it assumes your files are stored in a folder called "Folder" which is a subfolder of where your Python script is stored
# This ensures that you have the correct library available
import os
# Open a new file to write to
output_file = open('output.txt','w+')
# This starts the 'walk' through the directory
for folder , sub_folders , files in os.walk("Folder"):
# For each file...
for f in files:
# create the current path using the folder variable plus the file variable
current_path = folder+"\\"+f
# write the filename & path to the current open file
output_file.write(current_path)
# Open the file to read the contents
current_file = open(current_path, 'r')
# read each line one at a time and then write them to your file
for line in current_file:
output_file.write(line)
# close the file
current_file.close()
#close your output file
output_file.close()
using
import sys
folder = sys.argv[1]
for i in folder:
for file in i:
if file == "test.txt":
print (file)
would this access a file in the folder of a subfolder? For Example 1 main folder, with 20 subfolders, and each subfolder has 35 files. I want to pass the folder in commandline and access the first subfolder and the second file in it
Neither. This doesn't look at files or folders.
sys.argv[1] is just a string. i is the characters of that string. for file in i shouldn't work because you cannot iterate a character.
Maybe you want to glob or walk a directory instead?
Here's a short example using the os.walk method.
import os
import sys
input_path = sys.argv[1]
filters = ["test.txt"]
print(f"Searching input path '{input_path}' for matches in {filters}...")
for root, dirs, files in os.walk(input_path):
for file in files:
if file in filters:
print("Found a match!")
match_path = os.path.join(root, file)
print(f"The path is: {match_path}")
If the above file was named file_finder.py, and you wanted to search the directory my_folder, you would call python file_finder.py my_folder from the command line. Note that if my_folder is not in the same directory as file_finder.py, then you have to provide the full path.
No, this won't work, because folder will be a string, so you'll be iterating through the characters of the string. You could use the os module (e.g., the os.listdir() method). I don't know what exactly are you passing to the script, but probably it would be easiest by passing an absolute path. Look at some other methods in the module used for path manipulation.
i have written a python script to rename all the files present in a folder by removing all the numbers from the file name but this doesn't work .
Note :Same code works fine for python2.7
import os
def rename_files():
#(1) get file names from a folder
file_list = os.listdir(r"D:\prank")
print(file_list)
saved_path = os.getcwd()
print("Current working Directory is " + saved_path)
os.chdir(r"D:\prank")
#(2) for each file ,rename filename
for file_name in file_list:
os.rename(file_name, file_name.translate(None,"0123456789"))
rename_files()
Can anyone tell me how to make it work.Is the translate function which is not working properly
The problem is with os.rename() portion of your code.
os.rename() requires you to give it a full path to the file/folder you want to change it to, while you only gave it the file_name and not the full path.
You have to add the full path to the folders/files directory.
so it should look like this:
def rename_files():
# add the folder path
folder_path = "D:\prank\\"
file_list = os.listdir(r"D:\prank")
print(file_list)
saved_path = os.getcwd()
print("Current working Directory is " + saved_path)
os.chdir(r"D:\prank")
# Concat the folder_path with file_name to create the full path.
for file_name in file_list:
full_path = folder_path + file_name
print (full_path) # See the full path here.
os.rename(full_path, full_path.translate(None, "0123456789"))
look up the documentation for os, heres what ive found on rename:
os.rename(src, dst, *, src_dir_fd=None, dst_dir_fd=None)
Rename the file or directory src to dst. If dst is a directory, OSError will be raised. On Unix, if dst exists and is a file, it will be replaced silently if the user has permission. The operation may fail on some Unix flavors if src and dst are on different filesystems. If successful, the renaming will be an atomic operation (this is a POSIX requirement). On Windows, if dst already exists, OSError will be raised even if it is a file.
This function can support specifying src_dir_fd and/or dst_dir_fd to supply paths relative to directory descriptors.
If you want cross-platform overwriting of the destination, use replace().
New in version 3.3: The src_dir_fd and dst_dir_fd arguments.
heres a link to the documentation, hope this helps, thanks
https://docs.python.org/3/library/os.html
Others have pointed out other issues with your code, but as to your use of translate, in Python 3.x, you need to pass a dictionary mapping ordinals to new values (or None). This code would work:
import string
...
file_name.translate(dict(ord(c), None for c in string.digits))
but this seems easier to understand:
import re
...
re.sub(r'\d', '', file_name)
I am working on a search-and-destroy type program which I need it to do is search all directories with a certain file-name and append them to a list. after that delete all those files...not objects in list or the list...
import os
file_list=[]
for root, dirs, files in os.walk(path-to-dir'):
for f_name in files:
if f_name.startswith("file-name"):
file_list.append(f_name)
I could write up to appending part of the code but I don't know next...
Some help please
To remove a file from your computer, use os.remove(). It takes full path to the file as it's parameter, so instead of calling os.remove("infectedFile.dll") you would call os.remove("C:/program files/avira/infectedFile.dll")
So your file_list should contain full paths to the files, and then just call:
for file in file_list:
os.remove(file)
Modify your file_list.append(f_name). The f_name is only a bare name. You need to add the path to the file name in the time of processing, because you do not know where the file was found in the directory hierarchy:
file_list.append(os.path.join(root, f_name))
The root variable contains the path during walking.
To make check whether your code works, just print the content of the list:
print('\n'.join(file_list))
Or you can do it in the loop to get ready for the later part:
for fname in file_list:
print(fname)
Then you just add the os.remove(fname) to remove the file name:
for fname in file_list:
print('removing', fname)
os.remove(fname)