<p>有更多可能的解决方案和方法来解决这个问题。你知道吗</p>
<p>大多数人(以及其他人)都同意使用dict是正确的方法。你知道吗</p>
<p>比如这里的史蒂文。:D个</p>
<p>有人会说set()更方便、更自然,但我看到的和我自己做的大多数测试都表明,出于某种原因,使用dict()稍微快一点。至于原因,没人知道。这也可能会因Python版本的不同而有所不同。你知道吗</p>
<p>字典和集合使用哈希来访问数据,这使得它们比列表更快(O(1))。为了检查一个项目是否在一个列表中,迭代是在一个列表上执行的,在最坏的情况下,迭代次数会随着列表的增加而增加。你知道吗</p>
<p>为了进一步了解这个问题,我建议你检查一下相关的问题,特别是提到的问题。你知道吗</p>
<p>因此,我同意steveb的观点,并提出以下准则:</p>
<pre><code>chkdict = {} # A dictionary that we'll use to check for existance of an entry (whether is extension already processed or not)
setdef = chkdict.setdefault # Extracting a pointer of a method out of an instance may lead to faster access, thus improving performance a little
# Recurse through a directory:
for root, dirs, files in os.walk("ymir work"):
# Loop through all files in currently examined directory:
for file in files:
ext = path.splitext(file) # Get an extension of a file
# If file has no extension or file is named ".bashrc" or ".ds_store" for instance, then ignore it, otherwise write it to x:
if ext[0] and ext[1]: ext = ext[1].lower()
else: continue
if not ext in chkdict:
# D.setdefault(k[, d]) does: D.get(k, d), also set D[k] = d if k not in D
# You decide whether to use my method with dict.setdefault(k, k)
# Or you can write ext separately and then do: chkdict[ext] = None
# Second solution might even be faster as setdefault() will check for existance again
# But to be certain you should run the timeit test
x.write("\t\"%s\"\n" % setdef(ext, ext))
#x.write("\t\"%s\"\n" % ext)
#chkdict[ext] = None
del chkdict # If you're not inside a function, better to free the memory as soon as you can (if you don't need the data stored there any longer)
</code></pre>
<p>我在大量数据上使用了这个算法,它的性能非常好。你知道吗</p>