Say I want to make a file:
filename = "/foo/bar/baz.txt"
with open(filename, "w") as f:
f.write("FOOBAR")
This gives an IOError, since /foo/bar does not exist.
What is the most pythonic way to generate those directories automatically? Is it necessary for me explicitly call os.path.exists and os.mkdir on every single one (i.e., /foo, then /foo/bar)?
In Python 3.2+, using the APIs requested by the OP, you can elegantly do the following:
import os
filename = "/foo/bar/baz.txt"
os.makedirs(os.path.dirname(filename), exist_ok=True)
with open(filename, "w") as f:
f.write("FOOBAR")
With the Pathlib module (introduced in Python 3.4), there is an alternate syntax (thanks David258):
from pathlib import Path
output_file = Path("/foo/bar/baz.txt")
output_file.parent.mkdir(exist_ok=True, parents=True)
output_file.write_text("FOOBAR")
In older python, there is a less elegant way:
The os.makedirs function does this. Try the following:
import os
import errno
filename = "/foo/bar/baz.txt"
if not os.path.exists(os.path.dirname(filename)):
try:
os.makedirs(os.path.dirname(filename))
except OSError as exc: # Guard against race condition
if exc.errno != errno.EEXIST:
raise
with open(filename, "w") as f:
f.write("FOOBAR")
The reason to add the try-except block is to handle the case when the directory was created between the os.path.exists and the os.makedirs calls, so that to protect us from race conditions.
os.mkdir and read the documentation on one more function :)
from pathlib import Path; output_file = Path("/foo/bar/baz.txt"); output_file.parent.mkdir(exist_ok=True, parents=True); output_file.write_text("FOOBAR")