Develop Reference

How to deal with multiple dots in a file name with Python pathlib?

I am having an issue with pathlib when I try to construct a file path that has a "." in its name, the pathlib module ignores it.
Here are example lines (I tried multiple versions, all resulted the same issue)
The issue is that the original file name will be coming from another application, so it is not like I can edit the name myself. I also do not want to do string replacement work arounds, if possible.
path=r"c:\temp"
1
p=Path(path).joinpath("myfile.001").with_suffix(".bat")
2
p=Path(path, "myfile.001").with_suffix(".bat")
3
p=Path(path).with_name("myfile.001").with_suffix(".bat")
All these lines will yield to
WindowsPath('C:/temp/myfile.bat')
So how do I make pathlib.Path to construct this full path properly. The final path has to be
WindowsPath('C:/temp/myfile.001.bat')
Not
WindowsPath('C:/temp/myfile.bat')
Naturally I am looking for a way to do it through pathlib itself, otherwise I can just use os.
thanks

You are telling pathlib to replace the suffix .001 with the suffix .bat. pathlib complies.
Tell pathlib to add .bat to the existing suffix.
p = Path(path, 'myfile.001')
p = p.with_suffix(p.suffix+'.001')

How to format the beginning a loop correctly?

I have a program which I designed both for myself and my colleague to use, with all the data being stored in a directories. However, I want to set up the loop so that it work both for me and him. I tried all of these:
file_location = glob.glob('/../*.nc')
file_location = glob('/../*.nc')
But none of them are picking up any files. How can I fix this?

You can get a directory relative to a user's home (called ~ in the function call) using os.path.expanduser(). In your case, the line would be
file_location = glob.glob(os.path.expanduser('~/Dropbox/Argo/Data/*.nc'))

Usually is a good practice not hardcoding paths if you're going to use your paths for other tasks which need well-formed paths (ie: subprocess, writing paths to shell scripts), I'd recommend to manage paths using the os.path module instead, for example:
import os, glob
home_path = os.path.expanduser("~")
dropbox_path = os.path.join(home_path, "Dropbox")
good_paths = glob.glob(os.path.join(dropbox_path,"Argo","Data","*.nc"))
bad_paths = glob.glob(dropbox_path+"/Argo\\Data/*.nc")
print len(good_paths)==len(bad_paths)
print all([os.path.exists(p) for p in good_paths])
print all([os.path.exists(p) for p in bad_paths])
The example shows a comparison between bad and well formed paths. Both of them will work, but good_paths will be more flexible and portable in the long term.

reading a file in python from different directory

Probably a simple query.. But basically, I have data in directory "/foo/bar/foobar.txt"
and I am working in directory "/some/path/read_foobar.py"..
Now I want to read the file "foobar.txt" but rather than giving full path, I thought of adding /foo/bar/ to the path..
So, added the following at the start of read_foobar.py
import sys
sys.path.append("/foo/bar")
But when I try to read open("foobar.txt","r"), it is not able to find the file?
how do I do this?
Thanks

You can do it like this:
import os
os.chdir('/foo/bar')
f = open('foobar.txt', 'r')

sys.path is used to set the path used to look for python modules. Short of you writing some helper function that has a list of directories to search in when opening a file, I don't believe there is a standard module that provides this functionality.

From what I gathered from here and some quick tests, appending a path to sys.path will make python search in that path when you import a file/module, but not when open-ing it. Let's say we have a file called foo.py in /foo/bar/
import sys
sys.path.append("/foo/bar/")
try:
f = open('foo.py', 'r')
except:
print('this did not work') # this will print
import foo # no problems here

Unfortunately you can't. The PATH environment variable is only used by the operating system to search for executable files, and python uses it (along with the environment variable PYTHONPATH) to search for python modules to import.
You may want to consider setting a symbolic link to that file from your current working directory
ln -s /foo/bar/foobar.txt /some/path/foobar.text

finding out absolute path to a file from python

If I have a file test.py that resides in some directory, how can I find out from test.py what directory it is in? os.path.curdir will give the current directory but not the directory where the file lives. If I invoke test.py from some directory foo, os.curdir will return foo but not the path of test.py.
thanks.

Here's how to get the directory of the current file:
import os
os.path.abspath(os.path.dirname(__file__))

the answer is to use:
__file__
which returns a relative path.
os.path.abspath(__file__)
can be used to get the full path.

The answers so far have correctly pointed you to os.path.abspath, which does exactly the job you requested. However don't forget that os.path.normpath and os.path.realpath can also be very useful in this kind of tasks (to normalize representation, and remove symbolic links, respectively) in many cases (whether your specific use case falls among these "many" is impossible to tell from the scant info we have, of course;-).

import os
dirname, filename = os.path.split(os.path.abspath(__file__))

os.path has lots of tools for dealing with paths and getting information about paths.
Particularly, you want:
os.path.abspath

How can I find path to given file?

I have a file, for example "something.exe" and I want to find path to this file
How can I do this in python?

Perhaps os.path.abspath() would do it:
import os
print os.path.abspath("something.exe")
If your something.exe is not in the current directory, you can pass any relative path and abspath() will resolve it.

use os.path.abspath to get a normalized absolutized version of the pathname
use os.walk to get it's location
import os
exe = 'something.exe'
#if the exe just in current dir
print os.path.abspath(exe)
# output
# D:\python\note\something.exe
#if we need find it first
for root, dirs, files in os.walk(r'D:\python'):
for name in files:
if name == exe:
print os.path.abspath(os.path.join(root, name))
# output
# D:\python\note\something.exe

if you absolutely do not know where it is, the only way is to find it starting from root c:\
import os
for r,d,f in os.walk("c:\\"):
for files in f:
if files == "something.exe":
print os.path.join(r,files)
else, if you know that there are only few places you store you exe, like your system32, then start finding it from there. you can also make use of os.environ["PATH"] if you always put your .exe in one of those directories in your PATH variable.
for p in os.environ["PATH"].split(";"):
for r,d,f in os.walk(p):
for files in f:
if files == "something.exe":
print os.path.join(r,files)

Just to mention, another option to achieve this task could be the subprocess module, to help us execute a command in terminal, like this:
import subprocess
command = "find"
directory = "/Possible/path/"
flag = "-iname"
file = "something.foo"
args = [command, directory, flag, file]
process = subprocess.run(args, stdout=subprocess.PIPE)
path = process.stdout.decode().strip("\n")
print(path)
With this we emulate passing the following command to the Terminal:
find /Posible/path -iname "something.foo".
After that, given that the attribute stdout is binary string, we need to decode it, and remove the trailing "\n" character.
I tested it with the %timeit magic in spyder, and the performance is 0.3 seconds slower than the os.walk() option.
I noted that you are in Windows, so you may search for a command that behaves similar to find in Unix.
Finally, if you have several files with the same name in different directories, the resulting string will contain all of them. In consequence, you need to deal with that appropriately, maybe using regular expressions.

This is really old thread, but might be useful to someone who stumbles across this. In python 3, there is a module called "glob" which takes "egrep" style search strings and returns system appropriate pathing (i.e. Unix\Linux and Windows).
https://docs.python.org/3/library/glob.html
Example usage would be:
results = glob.glob('./**/FILE_NAME')
Then you get a list of matches in the result variable.

Uh... This question is a bit unclear.
What do you mean "have"? Do you have the name of the file? Have you opened it? Is it a file object? Is it a file descriptor? What???
If it's a name, what do you mean with "find"? Do you want to search for the file in a bunch of directories? Or do you know which directory it's in?
If it is a file object, then you must have opened it, reasonably, and then you know the path already, although you can get the filename from fileob.name too.