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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'

'/' 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.

Using listdir realpath abspath with symbolic links

Something that I don't understand :
In a shell :
mkdir -p /tmp/toto/titi/tutu
touch /tmp/toto/tata
ln -s /tmp/toto/tata /tmp/toto/titi/tutu/
python
Then in python:
import os
zeList = os.listdir("/tmp/toto/titi/tutu/")
print os.path.realpath(zeList[0])
>'/tata'
print os.path.abspath(zeList[0])
>'/tata'
The expected result should be : /tmp/toto/tata (or /tmp/toto/titi/tutu/tata).
Can anyone explain this result ?

os.listdir() returns base filenames, not full paths:
>>> import os
>>> os.listdir("/tmp/toto/titi/tutu/")
['tata']
Without a path, the file is then considered relative to the current working directory:
>>> os.getcwd()
'/Users/mj/Development/venvs/stackoverflow-2.7'
>>> os.path.realpath('tata')
'/Users/mj/Development/venvs/stackoverflow-2.7/tata'
Join the filename with the path first:
testdir = "/tmp/toto/titi/tutu/"
zeList = [os.path.join(testdir, fname) for fname in os.listdir(testdir)]
Now the symbolic link is properly replaced:
>>> testdir = "/tmp/toto/titi/tutu/"
>>> zeList = [os.path.join(testdir, fname) for fname in os.listdir(testdir)]
>>> print os.path.realpath(zeList[0])
/private/tmp/toto/tata
>>> print os.path.abspath(zeList[0])
/tmp/toto/titi/tutu/tata

listdir returns filenames, not paths. Therefore you are passing a relative path to realpath which is interpreted relative to your working directory /.
Use os.path.realpath(os.path.join(..., zeList[0])) for correct results.

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.

Rename multiple files in a directory in Python

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.***