Trying to use open(filename, 'w' ) gives IOError: [Errno 2] No such file or directory if directory doesn't exist

Question:

I am trying to create and write to a text file using Python. I have searched and cannot find a solution/reason for this error.

Here’s the code that doesn’t work:

afile = 'D:\temp\test.txt'
outFile = open(afile, 'w' )
outFile.write('Test.')
outFile.close()

# Error: 2
# Traceback (most recent call last):
#   File "<maya console>", line 1, in <module>
# IOError: [Errno 2] No such file or directory: 'D:\temp\test.txt' #

Most answers I found related to the slashes in the path, so…

I tried 'D:/temp/test.txt' and got an error.
I tried r'D:temptest.txt' and got an error.

When I try to create a file at the root of D:/ I have success.

'D:/test.txt' works.
'D:\test.txt' works.
r'D:test.txt' works.

It seems that I can’t create the directory path I would like while trying to create the file. What is the correct method for creating files at a specific path with Python on Windows(7)? Am I misunderstanding what open() can do? Does it create directories if they don’t exist or do I need to explicitly create the directory path before I use open() in ‘write’ mode to create a file?

Asked By: gonzalimator

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Answers:

If you try to create a file in a directory that doesn’t exist, you will get that error.

You need to ensure the directory exists first. You can do that with os.makedirs() as per this answer.

Answered By: paxdiablo

You are correct in surmising that the parent directory for the file must exist in order for open to succeed. The simple way to deal with this is to make a call to os.makedirs.

From the documentation:

os.makedirs(path[, mode])

Recursive directory creation function. Like mkdir(), but makes all intermediate-level directories needed to contain the leaf directory.

So your code might run something like this:

filename = ...
dirname = os.path.dirname(filename)
if not os.path.exists(dirname):
    os.makedirs(dirname)
with open(filename, 'w'):
    ...
Answered By: David Heffernan

Alternately, you could check if the file exists before opening it with:

os.path.exists (afile)

Which will either say True or False, depending on whether it exists.

Answered By: aMoon