On Python ≥ 3.5, use pathlib.Path.mkdir
:
from pathlib import Path
Path("/my/directory").mkdir(parents=True, exist_ok=True)
For older versions of Python, I see two answers with good qualities, each with a small flaw, so I will give my take on it:
Try os.path.exists
, and consider os.makedirs
for the creation.
import os
if not os.path.exists(directory):
os.makedirs(directory)
As noted in comments and elsewhere, there's a race condition – if the directory is created between the os.path.exists
and the os.makedirs
calls, the os.makedirs
will fail with an OSError
. Unfortunately, blanket-catching OSError
and continuing is not foolproof, as it will ignore a failure to create the directory due to other factors, such as insufficient permissions, full disk, etc.
One option would be to trap the OSError
and examine the embedded error code (see Is there a cross-platform way of getting information from Python’s OSError):
import os, errno
try:
os.makedirs(directory)
except OSError as e:
if e.errno != errno.EEXIST:
raise
Alternatively, there could be a second os.path.exists
, but suppose another created the directory after the first check, then removed it before the second one – we could still be fooled.
Depending on the application, the danger of concurrent operations may be more or less than the danger posed by other factors such as file permissions. The developer would have to know more about the particular application being developed and its expected environment before choosing an implementation.
Modern versions of Python improve this code quite a bit, both by exposing FileExistsError
(in 3.3+)...
try:
os.makedirs("path/to/directory")
except FileExistsError:
# directory already exists
pass
...and by allowing a keyword argument to os.makedirs
called exist_ok
(in 3.2+).
os.makedirs("path/to/directory", exist_ok=True) # succeeds even if directory exists.
Here's a generator that yields the chunks you want:
def chunks(lst, n):
"""Yield successive n-sized chunks from lst."""
for i in range(0, len(lst), n):
yield lst[i:i + n]
import pprint
pprint.pprint(list(chunks(range(10, 75), 10)))
[[10, 11, 12, 13, 14, 15, 16, 17, 18, 19],
[20, 21, 22, 23, 24, 25, 26, 27, 28, 29],
[30, 31, 32, 33, 34, 35, 36, 37, 38, 39],
[40, 41, 42, 43, 44, 45, 46, 47, 48, 49],
[50, 51, 52, 53, 54, 55, 56, 57, 58, 59],
[60, 61, 62, 63, 64, 65, 66, 67, 68, 69],
[70, 71, 72, 73, 74]]
If you're using Python 2, you should use xrange()
instead of range()
:
def chunks(lst, n):
"""Yield successive n-sized chunks from lst."""
for i in xrange(0, len(lst), n):
yield lst[i:i + n]
Also you can simply use list comprehension instead of writing a function, though it's a good idea to encapsulate operations like this in named functions so that your code is easier to understand. Python 3:
[lst[i:i + n] for i in range(0, len(lst), n)]
Python 2 version:
[lst[i:i + n] for i in xrange(0, len(lst), n)]
Best Answer
is indeed the best you're going to get.
It's unusual to be executing a script with
exec
/execfile
; normally you should be using the module infrastructure to load scripts. If you must use these methods, I suggest setting__file__
in theglobals
you pass to the script so it can read that filename.There's no other way to get the filename in execed code: as you note, the CWD may be in a completely different place.