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How do you define a local path in python

Good morning, I can indicate how to enter a path of internal hard disk in python, currently use the statement:
file = GETfile() or 'http://**********'
I would like to put a path to a local file, but it does not work, where am I wrong?
file = GETfile() or 'D:\xxx\xxxx\playlist\playlist.m3u'

\ is a escape character. You have three options.
1) use /. This, as a bonus works for linux as well:
'D:/xxx/xxxx/playlist/playlist.m3u'
2) escape the backslash
'D:\\xxx\\xxxx\\playlist\\playlist.m3u'
3) use raw strings:
r'D:\xxx\xxxx\playlist\playlist.m3u'

A correct answer is already given, but some additional information when working with local drive paths on Windows operating system.
Personally I would go with the r'D:\dir\subdir\filename.ext' format, however the other two methods already mentioned are valid as well.
Furthermore, file operations on Windows are limited by Explorer to a 256 character limit. Longer path names will usually result in an OS error.
However there is a workaround, by pre fixing "\\?\" to a long path.
Example of a path which does not work:
D:\reallyreallyreallyreallyreallylonglonglonglongdir\reallyreallyreallyreallyreallylonglonglonglongdir\reallyreallyreallyreallyreallylonglonglonglongdir\reallyreallyreallyreallyreallylonglonglonglongdir\reallyreallyreallyreallyreallylonglonglonglongdir\reallyreallyreallyreallyreallylonglonglonglongdir\reallyreallyreallyreallyreallylonglonglonglongdir\reallyreallyreallyreallyreallylonglonglonglongdir\filename.ext
Same file path which does work:
\\?\D:\reallyreallyreallyreallyreallylonglonglonglongdir\reallyreallyreallyreallyreallylonglonglonglongdir\reallyreallyreallyreallyreallylonglonglonglongdir\reallyreallyreallyreallyreallylonglonglonglongdir\reallyreallyreallyreallyreallylonglonglonglongdir\reallyreallyreallyreallyreallylonglonglonglongdir\reallyreallyreallyreallyreallylonglonglonglongdir\reallyreallyreallyreallyreallylonglonglonglongdir\filename.ext
so the following code I use to change filenames to include the "\\?\":
import os
import platform
def full_path_windows(filepath):
if platform.system() == 'Windows':
if filepath[1:3] == ':\\':
return u'\\\\?\\' + os.path.normcase(filepath)
return os.path.normcase(filepath)
I use this for every path to file (or directories), it will return the path with a prefix. The path does not need to exist; so you can use this also before you create a file or directory, to ensure you are not running into the Windows Explorer limitations.
HTH

Python for Data Analysis, Chapter 2, first example

I'm following along with the examples in a translated version of Wes McKinney's "Python for Data Analysis" and I was blocked in first example of Chapter 2
I think my problem arose because I saved a data file in a wrong path. is that right?
I stored a file, usagov_bitly_data2012-03-16-1331923249.txt, in C:\Users\HRR
and also stored folder, pydata-book-mater, that can be downloaded from http://github.com/pydata-book in C:\Users\HRR\Anaconda2\Library\bin.

Depends.
You might change the location you save your File or eddit the path you give to your code in Line 10. Since you're yousing relativ Paths i guess your script runs in C:\Users\HRR\Anaconda2\Library\bin, which means you have to go back to C:\Users\HRR or use an absolute Path ... or move the File, but hell you don't want to move a file every time you want to open it, like moving word files into msoffice file to open it, so try to change the Path.
And allways try harder ;)

In python open() will open from the current directory down unless given a full path (in linux that starts with / and windows <C>://). In your case the command is open the folder ch02 in the directory the script is running from and then open usagov_bitly_data2012-03-16-1331923249.txt in that folder.
Since you are storing the text file in C:\Users\HRR\usagov_bitly_data2012-03-16-1331923249.txt and you did not specify the directory of the script. I recommend the following command instead open(C:\\Users\\HRR\\usagov_bitly_data2012-03-16-1331923249.txt)
Note: the double \ is to escape the characters and avoid tabs and newlines showing up in the path.

How to create page links?

I know current user's location. It can be one of the following url:
(1) http://myapp.appspot.com/something/something-else/
(2) http://myapp.appspot.com/something/something-else
(3) http://myapp.appspot.com/something/something-else/page1
(4) http://myapp.appspot.com/something/something-else/page3
(actually, addresses 1, 2 and 3 are for the same page1)
I need to display on these pages link for page2:
http://myapp.appspot.com/something/something-else/page2
The question is how to generate such link?
I've tried to use relative links: /page2 and page2 - doesn't work properly. I am not sure how to create absolute link with self.request.path - it doesn't work properly also.

/page2 will never work; the leading / makes it relative to the website root rather than the current directory.
page2 should work for everything except #2; without a trailing slash, something-else is interpreted as a file rather than the current directory.
One solution would be to link to /something/something-else/page2 so your link doesn't change based on the user's address.

import something #refers to your .py file with the template handler
...
application = webapp.WSGIApplication([
('/something/something-else/', something.SomeThingElseHandler),
('/something/something-else', something.SomeThingElseHandler),
('/something/something-else/' + '([^/]+)/', something.PageHandler),
#The above pattern will be recognized if you close the url with /
#If you want your url to end without the slash your remove it for the reg ex. like
('/something/something-else/' + '([^/]+)', something.PageHandler),
],
debug=config.DEBUG)
util.run_wsgi_app(application)
In something.py, in your class PageHandler you have to parse the key or the id you are parsing manually to render the correct content.

how to check for platform incompatible folder (file) names in python

I would like to be able to check from python if a given string could be a valid cross platform folder name - below is the concrete problem I ran into (folder name ending in .), but I'm sure there are some more special cases (e.g.: con, etc.).
Is there a library for this?
From python (3.2) I created a folder on Windows (7) with a name ending in dot ('.'), e.g. (without square brackets): [What I've done on my holidays, Part II.]
When the created folder was ftp'd (to linux, but I guess that's irrelevant), it did not have the dot in it anymore (and in return, this broke a lot of hyperlinks).
I've checked it from the command line, and it seems that the folder doesn't have the '.' in the filename
mkdir tmp.
dir
cd tmp
cd ..\tmp.
Apparently, adding a single dot at the end of the folder name is ignored, e.g.:
cd c:\Users.
works just as expected.

Nope there's sadly no way to do this. For windows you basically can use the following code to remove all illegal characters - but if someone still has a FAT filesystem you'd have to handle these too since those are stricter. Basically you'll have to read the documentation for all filesystem and come up with a complete list. Here's the NTFS one as a starting point:
ILLEGAL_NTFS_CHARS = "[<>:/\\|?*\"]|[\0-\31]"
def __removeIllegalChars(name):
# removes characters that are invalid for NTFS
return re.sub(ILLEGAL_NTFS_CHARS, "", name)
And then you need some "forbidden" name list as well to get rid of COM. Pretty much a complete mess that.. and that's ignoring linux (although there it's pretty relaxed afaik)

Do not end a file or directory name with a space or a period. Although
the underlying file system may support such names, the Windows shell
and user interface does not.
http://msdn.microsoft.com/en-us/library/aa365247.aspx#naming_conventions
That page will give you information about other illegal names too, for Windows that is. Including CON as you said your self.
If you respect those (seemingly harsh) rules, I think you'll be safe on Linux and most other systems too.

How do I use '~' (tilde) in the context of paths?

To fix a problem in code for work, I was told to "use a path relative to ~". What does ~ mean in a file path? How can I make a path that is relative to ~, and use that path to open files in Python?

it is your $HOME var in UNIX, which usually is /home/username.
"Your home" meaning the home of the user who's executing a command like cd ~/MyDocuments/ is cd /home/user_executing_cd_commnd/MyDocuments

Unless you're writing a shell script or using some other language that knows to substitute the value of $HOME for ~, tildes in file paths have no special meaning and will be treated as any other non-special character.
If you are writing a shell script, shells don't interpret tildes unless they occur as the first character in an argument. In other words, ~/file will become /path/to/users/home/directory/file, but ./~/file will be interpreted literally (i.e., "a file called file in a subdirectory of . called ~").
Used in URLs, interpretation of the tilde as a shorthand for a user's home directory (e.g., http://www.foo.org/~bob) is a convention borrowed from Unix. Implementation is entirely server-specific, so you'd need to check the documentation for your web server to see if it has any special meaning.

If you are using pathlib for filenames then you can use on both Windows and Linux (I came here for a windows answer):
from pathlib import Path
p = Path('~').expanduser()
print(p)