blocks|key|1841522|text|独立于所使用的语言，您可以将目录中文件的所有引用读取到一个类似于数组的数据结构中(类似于“listFiles”)，从而获得数组的长度。计算'0‘到'arrayLength-1’范围内的一个随机数，并访问某个索引处的文件。这应该是可行的，不仅仅是在python中。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1841523|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

Independant from the language used, you can read all references to the files in a directory into a datastructure like an array (something like 'listFiles'), get the length of the array. calculate a random number in the range of '0' to 'arrayLength-1' and access the file at the certain index. This should work, not only in python.

blocks|key|2055811|text|import+os,+random
random.choice(os.listdir("C:\\"))+#change+dir+name+to+whatever|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|2055812|unstyled|2055813|关于你编辑的问题:首先，我假设你知道使用dircache的风险，以及它是deprecated+since+2.6,+and+removed+in+3.0的事实。|offset|length|style|CODE|2055814|其次，我看不到这里存在任何竞争条件。您的dircache对象基本上是不可变的(在缓存目录列表之后，它永远不会被再次读取)，因此从它进行并发读取没有什么坏处。|2055815|除此之外，我不明白为什么您会看到这个解决方案有任何问题。这很好。|2055816|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.python.org/library/dircache.html^0|0|0|K|8|10|14|0|0|K|8|0|0^^$0|@$1|2|3|4|5|6|7|X|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|Y|8|@]|9|@]|A|$]]|$1|F|3|G|5|E|7|Z|8|@$H|10|I|11|J|K]]|9|@$H|12|I|13|1|14]]|A|$]]|$1|L|3|M|5|E|7|15|8|@$H|16|I|17|J|K]]|9|@]|A|$]]|$1|N|3|O|5|E|7|18|8|@]|9|@]|A|$]]|$1|P|3|-4|5|E|7|19|8|@]|9|@]|A|$]]]|Q|$R|$5|S|T|U|A|$V|W]]]]

<pre><code>import os, random
random.choice(os.listdir("C:\\")) #change dir name to whatever
</code></pre>

<hr>

Regarding your edited question: first, I assume you know the risks of using a <code>dircache</code>, as well as the fact that it is <a href="http://docs.python.org/library/dircache.html" rel="noreferrer">deprecated since 2.6, and removed in 3.0</a>.

Second of all, I don't see where any race condition exists here. Your <code>dircache</code> object is basically immutable (after directory listing is cached, it is never read again), so no harm in concurrent reads from it.

Other than that, I do not understand why you see any problem with this solution. It is fine.

blocks|key|1841743|text|语言不可知的解决方案：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1841744|1)获取总数量指定目录中的文件数。|1841745|2)从0中选择一个随机数，总数为0。文件数-+1。|1841746|3)获得文件名列表作为适当索引的集合或类似物。|1841747|4)选择第n个元素，其中n是随机数。|1841748|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|L|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|M|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|N|8|@]|9|@]|A|$]]|$1|F|3|G|5|6|7|O|8|@]|9|@]|A|$]]|$1|H|3|I|5|6|7|P|8|@]|9|@]|A|$]]|$1|J|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|K|$]]

Language agnostic solution:

1) Get the total no. of files in specified directory.

2) Pick a random number from 0 to [total no. of files - 1].

3) Get the list of filenames as a suitably indexed collection or such.

4) Pick the nth element, where n is the random number.

blocks|key|1841791|text|如果你想包含目录，Yuval+A的答案是。否则：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1841792|import+os,+random

random.choice([x+for+x+in+os.listdir("C:\\")+if+os.path.isfile(os.path.join("C:\\",+x))])|code-block|syntax|javascript|1841793|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

If you want directories included, Yuval A's answer. Otherwise:

<pre><code>import os, random

random.choice([x for x in os.listdir("C:\\") if os.path.isfile(os.path.join("C:\\", x))])
</code></pre>

blocks|key|1841846|text|给出的大多数解决方案的问题是，您将所有输入加载到内存中，这可能会成为大型输入/层次结构的问题。这里有一个改编自Tom+Christiansen和Nat+Torkington的The+Perl+Cookbook的解决方案。要获取目录下任意位置的随机文件，请执行以下操作：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1841847|#!+/usr/bin/env+python
import+os,+random
n=0
random.seed();
for+root,+dirs,+files+in+os.walk('/tmp/foo'):
++for+name+in+files:
++++n+%2B=+1
++++if+random.uniform(0,+n)+<+1:
++++++++rfile=os.path.join(root,+name)
print+rfile|code-block|syntax|javascript|1841848|稍微泛化一下，就会得到一个方便的脚本：|1841849|$+cat+/tmp/randy.py
#!+/usr/bin/env+python
import+sys,+random
random.seed()
n+=+1
for+line+in+sys.stdin:
++if+random.uniform(0,+n)+<+1:
++++++rline=line
++n+%2B=+1
sys.stdout.write(rline)

$+/tmp/randy.py+<+/usr/share/dict/words+
chrysochlore

$+find+/tmp/foo+-type+f+%7C+/tmp/randy.py
/tmp/foo/bar|1841850|entityMap|0|LINK|mutability|MUTABLE|url|http://docstore.mik.ua/orelly/perl/cookbook/ch08_07.htm^0|2F|H|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@$A|V|B|W|1|X]]|C|$]]|$1|D|3|E|5|F|7|Y|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|Z|8|@]|9|@]|C|$]]|$1|K|3|L|5|F|7|10|8|@]|9|@]|C|$G|H]]|$1|M|3|-4|5|6|7|11|8|@]|9|@]|C|$]]]|N|$O|$5|P|Q|R|C|$S|T]]]]

The problem with most of the solutions given is you load all your input into memory which can become a problem for large inputs/hierarchies. Here's a solution adapted from <a href="http://docstore.mik.ua/orelly/perl/cookbook/ch08_07.htm" rel="nofollow noreferrer">The Perl Cookbook</a> by Tom Christiansen and Nat Torkington. To get a random file anywhere beneath a directory:
<pre><code>#! /usr/bin/env python
import os, random
n=0
random.seed();
for root, dirs, files in os.walk('/tmp/foo'):
 for name in files:
 n += 1
 if random.uniform(0, n) &lt; 1:
 rfile=os.path.join(root, name)
print rfile
</code></pre>
Generalizing a bit makes a handy script:
<pre><code>$ cat /tmp/randy.py
#! /usr/bin/env python
import sys, random
random.seed()
n = 1
for line in sys.stdin:
 if random.uniform(0, n) &lt; 1:
 rline=line
 n += 1
sys.stdout.write(rline)

$ /tmp/randy.py &lt; /usr/share/dict/words 
chrysochlore

$ find /tmp/foo -type f | /tmp/randy.py
/tmp/foo/bar
</code></pre>

blocks|key|2056054|text|Python3有pathlib模块，可以用来以更面向对象的方式推断文件和目录：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|2056055|from+random+import+choice
from+pathlib+import+Path

path:+Path+=+Path()
#+The+Path.iterdir+method+returns+a+generator,+so+we+must+convert+it+to+a+list
#+before+passing+it+to+random.choice,+which+expects+an+iterable.
random_path+=+choice(list(path.iterdir()))|code-block|syntax|javascript|2056056|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.python.org/3/library/pathlib.html^0|8|7|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@$A|R|B|S|1|T]]|C|$]]|$1|D|3|E|5|F|7|U|8|@]|9|@]|C|$G|H]]|$1|I|3|-4|5|6|7|V|8|@]|9|@]|C|$]]]|J|$K|$5|L|M|N|C|$O|P]]]]

Python 3 has the <a href="https://docs.python.org/3/library/pathlib.html" rel="nofollow noreferrer">pathlib</a> module, which can be used to reason about files and directories in a more object oriented fashion:
<pre class="lang-py prettyprint-override"><code>from random import choice
from pathlib import Path

path: Path = Path()
# The Path.iterdir method returns a generator, so we must convert it to a list
# before passing it to random.choice, which expects an iterable.
random_path = choice(list(path.iterdir()))
</code></pre>

What is the best way to choose a random file from a directory in Python?

Edit: Here is what I am doing:

<pre><code>import os
import random
import dircache

dir = 'some/directory'
filename = random.choice(dircache.listdir(dir))
path = os.path.join(dir, filename)
</code></pre>

Is this particularly bad, or is there a particularly better way?

Best way to choose a random file from a directory

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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在Python中从目录中选择随机文件的最佳方法是什么？编辑:这是我正在做的事情：import osimport randomimport dircachedir = 'some/directory'filename = random.choice(dircache.listdir(dir))path = os.path...

问从目录中选择随机文件的最佳方法
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问从目录中选择随机文件的最佳方法EN