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.
Given a list of lists t
,
flat_list = [item for sublist in t for item in sublist]
which means:
flat_list = []
for sublist in t:
for item in sublist:
flat_list.append(item)
is faster than the shortcuts posted so far. (t
is the list to flatten.)
Here is the corresponding function:
def flatten(t):
return [item for sublist in t for item in sublist]
As evidence, you can use the timeit
module in the standard library:
$ python -mtimeit -s't=[[1,2,3],[4,5,6], [7], [8,9]]*99' '[item for sublist in t for item in sublist]'
10000 loops, best of 3: 143 usec per loop
$ python -mtimeit -s't=[[1,2,3],[4,5,6], [7], [8,9]]*99' 'sum(t, [])'
1000 loops, best of 3: 969 usec per loop
$ python -mtimeit -s't=[[1,2,3],[4,5,6], [7], [8,9]]*99' 'reduce(lambda x,y: x+y,t)'
1000 loops, best of 3: 1.1 msec per loop
Explanation: the shortcuts based on +
(including the implied use in sum
) are, of necessity, O(T**2)
when there are T sublists -- as the intermediate result list keeps getting longer, at each step a new intermediate result list object gets allocated, and all the items in the previous intermediate result must be copied over (as well as a few new ones added at the end). So, for simplicity and without actual loss of generality, say you have T sublists of k items each: the first k items are copied back and forth T-1 times, the second k items T-2 times, and so on; total number of copies is k times the sum of x for x from 1 to T excluded, i.e., k * (T**2)/2
.
The list comprehension just generates one list, once, and copies each item over (from its original place of residence to the result list) also exactly once.
Best Answer
Using
state=DISABLED
is the correct way to do this.However, you must be putting it in the wrong place.
state
is an option ofCheckbutton
, so it needs to be used like this:Below is a sample script to demonstrate:
If you want to change a checkbutton's state programmatically, use
Tkinter.Checkbutton.config
.Below is a sample script to demonstrate: