Develop Reference

How to alter file type and then save to a new directory?

I have been attempting to change all files in a folder of a certain type to another and then save them to another folder I have created.
In my example the files are being changed from '.dna' files to '.fasta' files. I have successfully completed this via this code:
files = Path(directory).glob('*.dna')
for file in files:
record = snapgene_file_to_seqrecord(file)
fasta = record.format("fasta")
print(fasta)
My issue is now with saving these files to a new folder. My attempt has been to use this:
save_path = Path('/Users/user/Documents...')
for file in files:
with open(file,'w') as a:
record = snapgene_file_to_seqrecord(a)
fasta = record.format("fasta").read()
with open(save_path, file).open('w') as f:
f.write(fasta)
No errors come up but it is definitely not working. I can see that there may be an issue with how I am writing this but I can't currently think of a better way to do it.
Thank you in advance!

Hi, You can use os lib to rename the file with the new extension (type)
import os
my_file = 'my_file.txt'
base = os.path.splitext(my_file)[0]
os.rename(my_file, base + '.bin')
And you can use shutil lib to move the file to a new directory.
import shutil
# absolute path
src_path = r"E:\pynative\reports\sales.txt"
dst_path = r"E:\pynative\account\sales.txt"
shutil.move(src_path, dst_path)
Hope that can be of help.

how can i adding custom path for saving the zip_file

I am creating a zip file using the zipfile module. It works like a charm. but that's file, saved in the executed script place.
my script path is a:
[b]c:/User/Administrator/[/b]script.py
and the zipfile saved in:
[b]c:/User/Administrator/[/b]backup.zip
but I want, [b]creating a zipfile, in another path[/b], like this:
[b]d:/backups/[/b]backup.zip
my code like this:
import zipfile
zip_file = zipfile.ZipFile("backup.zip", 'w')
with zip_file:
for file in filePaths:
zip_file.write(file)
my question is a how can I adding custom path for saving the zip_file. because I have not an enough space in C:
tnx a lot.

Give the path you want to ZipFile function.
When you give only the name of the file, it will save the file in the current directory which the program is running.
Do this instead:
import zipfile
# For example you want to save it in drive 'D'
path = "D:\\PathToYourDir\\backup.zip"
zip_file = zipfile.ZipFile(path, 'w')
with zip_file:
for file in filePaths:
zip_file.write(file)

How can I extract all .zip extension in a folder without retaining directory using python?

Here is my code I don't know how can I loop every .zip in a folder, please help me: I want all contents of 5 zip files to extracted in one folder, not including its directory name
import os
import shutil
import zipfile
my_dir = r"C:\\Users\\Guest\\Desktop\\OJT\\scanner\\samples_raw"
my_zip = r"C:\\Users\\Guest\\Desktop\\OJT\\samples\\001-100.zip"
with zipfile.ZipFile(my_zip) as zip_file:
zip_file.setpassword(b"virus")
for member in zip_file.namelist():
filename = os.path.basename(member)
# skip directories
if not filename:
continue
# copy file (taken from zipfile's extract)
source = zip_file.open(member)
target = file(os.path.join(my_dir, filename), "wb")
with source, target:
shutil.copyfileobj(source, target)

repeated question, please refer below link.
How to extract zip file recursively in Pythonn

What you are looking for is glob. Which can be used like this:
#<snip>
import glob
#assuming all your zip files are in the directory below.
for my_zip in glob.glob(r"C:\\Users\\Guest\\Desktop\\OJT\\samples\\*.zip"):
with zipfile.ZipFile(my_zip) as zip_file:
zip_file.setpassword(b"virus")
for member in zip_file.namelist():
#<snip> rest of your code here.

Zip single file

I am trying to zip a single file in python. For whatever reason, I'm having a hard time getting down the syntax. What I am trying to do is keep the original file and create a new zipped file of the original (like what a Mac or Windows would do if you archive a file).
Here is what I have so far:
import zipfile
myfilepath = '/tmp/%s' % self.file_name
myzippath = myfilepath.replace('.xml', '.zip')
zipfile.ZipFile(myzippath, 'w').write(open(myfilepath).read()) # does not zip the file properly

The correct way to zip file is:
zipfile.ZipFile('hello.zip', mode='w').write("hello.csv")
# assume your xxx.py under the same dir with hello.csv
The python official doc says:
ZipFile.write(filename, arcname=None, compress_type=None)
Write the file named filename to the archive, giving it the archive name arcname
You pass open(filename).read() into write(). open(filename).read() is a single string that contains the whole content of file filename, it would throw FileNotFoundError because it is trying to find a file named with the string content.

If the file to be zipped (filename) is in a different directory called pathname, you should use the arcname parameter. Otherwise, it will recreate the full folder hierarchy to the file folder.
from zipfile import ZipFile
import os
with ZipFile(zip_file, 'w') as zipf:
zipf.write(os.path.join(pathname,filename), arcname=filename)

Try calling zipfile.close() afterwards?
from zipfile import ZipFile
zipf = ZipFile("main.zip","w", zipfile.ZIP_DEFLATED)
zipf.write("main.json")
zipf.close()

Since you also want to specify the directory try using os.chdir:
#!/usr/bin/python
from zipfile import ZipFile
import os
os.chdir('/path/of/target/and/destination')
ZipFile('archive.zip', 'w').write('original_file.txt')
Python zipfile : Work with Zip archives
Python Miscellaneous operating system interfaces

How to open a file only using its extension?

I have a Python script which opens a specific text file located in a specific directory (working directory) and perform some actions.
(Assume that if there is a text file in the directory then it will always be no more than one such .txt file)
with open('TextFileName.txt', 'r') as f:
for line in f:
# perform some string manipulation and calculations
# write some results to a different text file
with open('results.txt', 'a') as r:
r.write(someResults)
My question is how I can have the script locate the text (.txt) file in the directory and open it without explicitly providing its name (i.e. without giving the 'TextFileName.txt'). So, no arguments for which text file to open would be required for this script to run.
Is there a way to achieve this in Python?

You could use os.listdir to get the files in the current directory, and filter them by their extension:
import os
txt_files = [f for f in os.listdir('.') if f.endswith('.txt')]
if len(txt_files) != 1:
raise ValueError('should be only one txt file in the current directory')
filename = txt_files[0]

You Can Also Use glob Which is easier than os
import glob
text_file = glob.glob('*.txt')
# wild card to catch all the files ending with txt and return as list of files
if len(text_file) != 1:
raise ValueError('should be only one txt file in the current directory')
filename = text_file[0]
glob searches the current directory set by os.curdir
You can change to the working directory by setting
os.chdir(r'cur_working_directory')

Since Python version 3.4, it is possible to use the great pathlib library. It offers a glob method which makes it easy to filter according to extensions:
from pathlib import Path
path = Path(".") # current directory
extension = ".txt"
file_with_extension = next(path.glob(f"*{extension}")) # returns the file with extension or None
if file_with_extension:
with open(file_with_extension):
...