Changing file extension in Python

Question

Suppose from index.py with CGI, I have post file foo.fasta to display file. I want to change foo.fasta's file extension to be foo.aln in display file. How can I do it?

Dorset

Does this answer your question? How to replace (or strip) an extension from a filename in Python? – Lionellionello

Answer 1

os.path.splitext(), os.rename()

for example:

# renamee is the file getting renamed, pre is the part of file name before extension and ext is current extension
pre, ext = os.path.splitext(renamee)
os.rename(renamee, pre + new_extension)

Incipit

Can you be more specific, i saw documentation before too but didn't work. – Dorset

Use the first function to get the base. Combine it with the new extension and pass the old filename and the new filename to the second function. – Incipit

actually, its better to use this method instead for python3: pathlib.path(pathtofile).with_suffix(".mynewext"). The way, suggested with pathlib.path(pathtofile).stem works but will delete the path before the basename. – Antic

str(pathlib.path(pathtofile).with_suffix(".mynewext")) – Antic

There's no real need for assigning a variable to the original extension. I find this is more straightforward: os.rename(filename, os.path.splitext(filename)[0] + '.aln') – Salad

Answer 2

An elegant way using pathlib.Path:

from pathlib import Path
p = Path('mysequence.fasta')
p.rename(p.with_suffix('.aln'))

Donavon

Although the OP didn't ask to perform a rename, it was in the tags, and if you're going to perform a rename and if it's possible the input might have a path and not just a filename, this technique is the right one. – Perineurium

Regarding .with_suffix(), the properties .suffix and .suffixes should have setters. – Impudence

Oddly enough, it seems like .suffix is read-only, at least as of 3.9.5. – Ibnsaud

@Ibnsaud Paths are immutable docs.python.org/3/library/pathlib.html#general-properties – Donavon

Answer 3

import os
thisFile = "mysequence.fasta"
base = os.path.splitext(thisFile)[0]
os.rename(thisFile, base + ".aln")

Where thisFile = the absolute path of the file you are changing

Martinmas

I like this answer more because it provides an example and not just cites the methods needed to accomplish the task. Thanks @Martinmas – Uneven

base, _ = os.path.splitext(thisFile) is more idiomatic. – Criollo

Answer 4

Starting from Python 3.4 there's pathlib built-in library. So the code could be something like:

from pathlib import Path

filename = "mysequence.fasta"
new_filename = Path(filename).stem + ".aln"

https://docs.python.org/3.4/library/pathlib.html#pathlib.PurePath.stem

I love pathlib :)

Dermatoglyphics

This is even better with python 3.6 string interpolation syntax ( python.org/dev/peps/pep-0498 ) new_filename = f"{Path(filename).stem}.aln" 😎😎😎 – Backchat

Be careful - stem also strips the path if one is present. If you wanted to rename the file and if a path was supplied (which admittedly it wasn't in the question), this technique would fail. – Perineurium

Also, the result is a string, no longer a pathlib Path. p.parent / (p.stem + '.aln') will give you a new Path. – Elaterite

Warning: Path(filename).stem drops the directory (prefix) part of the filename. – Poultice

Answer 5

Use this:

os.path.splitext("name.fasta")[0]+".aln"

And here is how the above works:

The splitext method separates the name from the extension creating a tuple:

os.path.splitext("name.fasta")

the created tuple now contains the strings "name" and "fasta". Then you need to access only the string "name" which is the first element of the tuple:

os.path.splitext("name.fasta")[0]

And then you want to add a new extension to that name:

os.path.splitext("name.fasta")[0]+".aln"

Darbie

Answer 6

As AnaPana mentioned pathlib is more new and easier in python 3.4 and there is new with_suffix method that can handle this problem easily:

from pathlib import Path
new_filename = Path(mysequence.fasta).with_suffix('.aln')

Sacred

Answer 7

Using pathlib and preserving full path:

from pathlib import Path
p = Path('/User/my/path')
new_p = Path(p.parent.as_posix() + '/' + p.stem + '.aln')

Frederiksberg

to make it just a little bit simpler new_p = Path(p.with_suffix('').as_posix() + '.aln') – Nashoma

Answer 8

Sadly, I experienced a case of multiple dots on file name that splittext does not worked well... my work around:

file = r'C:\Docs\file.2020.1.1.xls'
ext = '.'+ os.path.realpath(file).split('.')[-1:][0]
filefinal = file.replace(ext,'')
filefinal = file + '.zip'
os.rename(file ,filefinal)

Disfeature

Answer 9

>> file = r'C:\Docs\file.2020.1.1.xls'
>> ext = '.'+ os.path.realpath(file).split('.')[-1:][0]
>> filefinal = file.replace(ext,'.zip')
>> os.rename(file ,filefinal) 

Bad logic for repeating extension, sample: 'C:\Docs\.xls_aaa.xls.xls'

Faludi