Hey Guys, i just started learning python and havent really done much programming so im pretty new

im following a book and I have to read some data from a text file

the code thats giving me trouble is

os.chdir('Users\spreston\Desktop\HeadFirst\Chapter 3')

the error says that the path does not exist, but i know it does, which leads me to believe i typed the path in wrong, any suggestions?

Guessing here, but maybe you need to give the full absolute path:
'C:\Users\spreston\Desktop\HeadFirst\Chapter 3'

assuming that is what the full absolute path is, of course

no that doesnt work either, the C: produces an error as well, a unicode error. Its the colon that throws it off it seems

If your path doesn't start with a drive letter or a (back- or forward-)slash, it's interpreted relative to the current directory. So Chophouse is right, you need to make your path absolute unless Users is a subdirectory of your current directory.

You also need to escape your backslashes or use forward-slashes instead. Otherwise \U is interpreted as the beginning of a unicode escape sequence, which is why you got a unicode error.

oh ok. well why does windows display the file paths using back slashes

You can try this too

import os
p = os.path.join('C:\\', 'Users', 'spreston', 'Desktop', 'HeadFirst', 'Chapter 3')

The best way is to always use raw string(r) when give path in windows.
There is also an option to use \\ or /.

So this do not work.
This work.


The reason why is this way,is that a singel backslashes(\) in python can(do) get read as an escape character.
@sepp2k also metion that you need escape your backslashes.
Take a look at this.

>>> s = 'C:\test'
>>> print s
C:  est
>>> print repr(s)
>>> repr(s)

Then with raw string.

>>> s = r'C:\test'
>>> print s
>>> print repr(s)
>>> repr(s)

One more \

>>> s = 'C:\\test'
>>> print s

My preferred way is to use the module I uploaded in this post

from whatever import path    
p = path("C:")/"foo"/"bar"/"qux"

Although usually I don't call it 'module whatever' but module 'kernilis.path'.