判断文件要写入的目录是否存在，如果不存在，使用Python创建目录，最优雅的方法是什么？以下是我try 过的:
import os
file_path = "/my/directory/filename.txt"
directory = os.path.dirname(file_path)
try:
os.stat(directory)
except:
os.mkdir(directory)
f = file(filename)
不知怎的，我错过了os.path.exists分(感谢kanja、Blair和Douglas).这就是我现在拥有的:
def ensure_dir(file_path):
directory = os.path.dirname(file_path)
if not os.path.exists(directory):
os.makedirs(directory)
有没有一个open()的旗帜，让这件事自动发生？
关于Python≥ 3.5，使用pathlib.Path.mkdir:
from pathlib import Path
Path("/my/directory").mkdir(parents=True, exist_ok=True)
对于较旧版本的Python，我看到了两个很好的答案，每个答案都有一个小缺陷，因此我将给出我的看法:
try os.path.exists，并考虑os.makedirs的创作.
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.
一种 Select 是捕获OSError并判断嵌入的错误代码(请参见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.
根据应用程序的不同，并发操作的危险可能或多或少比其他因素(如文件权限)带来的危险更大或更小.在 Select 实现之前，开发人员必须更多地了解正在开发的特定应用程序及其预期环境.
Python的现代版本大大改进了这段代码，它们都是通过公开FileExistsError(在3.3+中)和...
try:
os.makedirs("path/to/directory")
except FileExistsError:
# directory already exists
pass
.并允许a keyword argument to os.makedirs called exist_ok(3.2+).
os.makedirs("path/to/directory", exist_ok=True) # succeeds even if directory exists.