open() in Python does not create a file if it doesn't exist?

Question

What is the best way to open a file as read/write if it exists, or if it does not, then create it and open it as read/write? From what I read, `file = open('myfile.dat', 'rw')` should do this, right? It is not working for me (Python 2.6.2) and I'm wondering if it is a version problem, or not supposed to work like that or what. The bottom line is, I just need a solution for the problem. I am curious about the other stuff, but all I need is a nice way to do the opening part. UPDATE: the enclosing directory was writeable by user and group, not other (I'm on a Linux system... so permissions 775 in other words), and the exact error was: > IOError: no such file or directory.

Answer 1

You need to open your file in w+ mode.

file = open('file.dat', 'w+')

Answer 2

In python2, the flag for opening a file in mode 'rw' is 'r+'

$ python2

Python 2.7.15 (default, May 1 2018, 16:44:37)

[GCC 4.2.1 Compatible Apple LLVM 9.0.0 (clang-900.0.39.2)] on darwin

Type "help", "copyright", "credits" or "license" for more information.

>>> file = open ('file_for_rw.txt','r+')

>>> file.write ('hiya\n')

>>> file.close()

>>> ^D

$ cat file_for_rw.txt

hiya

In python3, the flag for opening a file is 'x' -- because who needs consistency?

$ python3

Python 3.6.3 |Anaconda custom (64-bit)| (default, Oct 6 2017, 12:04:38)

[GCC 4.2.1 Compatible Clang 4.0.1 (tags/RELEASE_401/final)] on darwin

Type "help", "copyright", "credits" or "license" for more information.

>>> file = open ('file_for_x.txt','x')

>>> file.write('r+ for python2, x for python3')

29

>>> file.close()

>>>

$ cat file_for_x.txt

r+ for python2, x for python3