自动创建文件输出目录 [英] Automatically creating directories with file output
问题描述
可能存在重复:
python中的mkdir -p功能
假设我想创建一个文件:
filename =/foo/bar/baz.txt
with open (文件名w)作为f:
f.write(FOOBAR)
由于 / foo / bar
不存在,所以给出 IOError
。
自动生成这些目录的最好方法是什么?是否有必要在每一个显式调用 os.path.exists
和 os.mkdir
(即/ foo,那么/ foo / bar)?
os.makedirs
这个功能。请尝试以下操作:
import os
import errno
filename =/ foo / bar / baz.txt
如果不是os.path.exists(os.path.dirname(filename)):
try:
os.makedirs(os.path.dirname(filename) )
,除了OSError,如exc:#防止竞争条件
如果exc.errno!= errno.EEXIST:
打开(文件名,w )作为f:
f.write(FOOBAR)
try-except
block用于处理在 os.path.exists
和 os.makedirs
调用,这样可以保护我们免受竞争的影响。
<在Python 3.2+中,有一个更优雅的方式这样可以避免上面的争夺情况:
filename =/foo/bar/baz.txt\"¨
os。 makedirs(os.path.dirname(文件名), (开头的文件名为w)作为f:
f.write(FOOBAR)
Possible Duplicate:
mkdir -p functionality in python
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)?
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.
In Python 3.2+, there is a more elegant way that avoids the race condition above:
filename = "/foo/bar/baz.txt"¨
os.makedirs(os.path.dirname(filename), exist_ok=True)
with open(filename, "w") as f:
f.write("FOOBAR")
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