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Raster to ASCII Conversion [duplicate]

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?

An elegant way using pathlib.Path:
from pathlib import Path
p = Path('mysequence.fasta')
p.rename(p.with_suffix('.aln'))

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)

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

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 :)

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"

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')

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')

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)

>> 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'

Trouble getting file name with extension using pathlib

I have this code which gives me the filename no problem. It gives it to me without the extension. Any way to get with extension?
from pathlib import Path
file = 'somepath'
path_object = Path(file)
filename = path_object.stem

You can use .name attribute of the Path object.
>>> from pathlib import Path
>>>
>>> p = Path("SomePath/filename.txt")
>>> p.name
'filename.txt'

'/' or '\' for paths?

I am having this problem when sharing files with my friends...
I have windows, some use mac and others Linux. When I share python files that contain commands of creating directories, like for instance:
Path_Results = os.path.dirname(Path_Definition)+'\Results'
the directory is not created because in Windows \ is used, whereas in mac and linux / are used.
Any ideas how can I create a more general script ?
Thanks in advance

Use pathlib.Path. Then you'll stop concerning yourself with / or \
On Windows:
>>> from pathlib import Path
>>> updir = Path("..")
>>> resdir = updir / "Result"
>>> resdir
WindowsPath('../Result')
>>> str(resdir)
'..\\Result'
On Linux, Mac, BSD, and other *nix:
>>> from pathlib import Path
>>> updir = Path("..")
>>> resdir = updir / "Result"
>>> resdir
PosixPath('../Result')
>>> str(resdir)
'../Result'
Nearly all stdlib modules and functions accept Path as is, no need to str() it. Example:
from pathlib import Path
resdir = Path("../Result")
filepath = resdir / "somefilename.txt"
assert isinstance(filepath, Path)
with open(filepath, "rt") as fin:
for ln in fin:
# do things
Provided there's a file ..\Result\somefilename.txt (in Windows) or ../Result/somefilename.txt (in Linux/Mac/BSD), the code will work exactly the same.
Edit: For your particular code:
from pathlib import Path
...
# Assuming `Path_Definition` is not a Path object
Path_Results = Path(Path_Definition) / 'Results'

os.sep exists for this very purpose. You can also use os.path.join.

Its not good idea to use '/' or '' directly. path separator mentioned by #mrks is what we need to use.

Python filename from path with \

I have a list of paths that looks like this C:/Users/myuser/Documents/files\my_file_1.csv and I want to get the file name from that by doing this:
path=['C:/Users/myuser/Documents/files\my_file_1.csv','C:/Users/myuser/Documents/files\my_file_2.csv',...]
filename, file_extension = os.path.splitext(path[0])
and I always get 'C:/Users/myuser/Documents/files\my_file_1' I know it must be for the ' \ ' slash but I haven't been able to replace it. Can anyone give me an idea ?

You can use os.path.basename to get just the filename without the full directory, then os.path.splitext to remove the file extension.
>>> import os
>>> [os.path.splitext(os.path.basename(i))[0] for i in path]
['my_file_1', 'my_file_2']
Or if you want the filename and extension, but no directories
>>> [os.path.basename(i) for i in path]
['my_file_1.csv', 'my_file_2.csv']

As you are using windows and if you are using python 3.4+
>>> from pathlib import PureWindowsPath
>>> path=['C:/Users/myuser/Documents/files\my_file_1.csv','C:/Users/myuser/Documents/files\my_file_2.csv']
>>> print([PureWindowsPath(i).name for i in path])
['my_file_1.csv', 'my_file_2.csv']

Extract a part of the filepath (a directory) in Python

I need to extract the name of the parent directory of a certain path. This is what it looks like:
C:\stuff\directory_i_need\subdir\file.jpg
I would like to extract directory_i_need.

import os
## first file in current dir (with full path)
file = os.path.join(os.getcwd(), os.listdir(os.getcwd())[0])
file
os.path.dirname(file) ## directory of file
os.path.dirname(os.path.dirname(file)) ## directory of directory of file
...
And you can continue doing this as many times as necessary...
Edit: from os.path, you can use either os.path.split or os.path.basename:
dir = os.path.dirname(os.path.dirname(file)) ## dir of dir of file
## once you're at the directory level you want, with the desired directory as the final path node:
dirname1 = os.path.basename(dir)
dirname2 = os.path.split(dir)[1] ## if you look at the documentation, this is exactly what os.path.basename does.

For Python 3.4+, try the pathlib module:
>>> from pathlib import Path
>>> p = Path('C:\\Program Files\\Internet Explorer\\iexplore.exe')
>>> str(p.parent)
'C:\\Program Files\\Internet Explorer'
>>> p.name
'iexplore.exe'
>>> p.suffix
'.exe'
>>> p.parts
('C:\\', 'Program Files', 'Internet Explorer', 'iexplore.exe')
>>> p.relative_to('C:\\Program Files')
WindowsPath('Internet Explorer/iexplore.exe')
>>> p.exists()
True

All you need is parent part if you use pathlib.
from pathlib import Path
p = Path(r'C:\Program Files\Internet Explorer\iexplore.exe')
print(p.parent)
Will output:
C:\Program Files\Internet Explorer
Case you need all parts (already covered in other answers) use parts:
p = Path(r'C:\Program Files\Internet Explorer\iexplore.exe')
print(p.parts)
Then you will get a list:
('C:\\', 'Program Files', 'Internet Explorer', 'iexplore.exe')
Saves tone of time.

First, see if you have splitunc() as an available function within os.path. The first item returned should be what you want... but I am on Linux and I do not have this function when I import os and try to use it.
Otherwise, one semi-ugly way that gets the job done is to use:
>>> pathname = "\\C:\\mystuff\\project\\file.py"
>>> pathname
'\\C:\\mystuff\\project\\file.py'
>>> print pathname
\C:\mystuff\project\file.py
>>> "\\".join(pathname.split('\\')[:-2])
'\\C:\\mystuff'
>>> "\\".join(pathname.split('\\')[:-1])
'\\C:\\mystuff\\project'
which shows retrieving the directory just above the file, and the directory just above that.

import os
directory = os.path.abspath('\\') # root directory
print(directory) # e.g. 'C:\'
directory = os.path.abspath('.') # current directory
print(directory) # e.g. 'C:\Users\User\Desktop'
parent_directory, directory_name = os.path.split(directory)
print(directory_name) # e.g. 'Desktop'
parent_parent_directory, parent_directory_name = os.path.split(parent_directory)
print(parent_directory_name) # e.g. 'User'
This should also do the trick.

This is what I did to extract the piece of the directory:
for path in file_list:
directories = path.rsplit('\\')
directories.reverse()
line_replace_add_directory = line_replace+directories[2]
Thank you for your help.

You have to put the entire path as a parameter to os.path.split. See The docs. It doesn't work like string split.