<p>人们在投票前测试过mgilson的test_re()函数吗?<a href="https://docs.python.org/3.4/library/re.html#re.sub" rel="nofollow">re.sub()</a>的参数是相反的,因此它在空字符串中进行替换,并始终返回空字符串。</p>
<p>我在Python3.4中工作;string.translate()只接受一个参数dict,因为构建这个dict有开销,所以我将它从函数中移出。公平地说,我还将regex编译从函数中移出(这没有明显的区别)。</p>
<pre><code>import re
import string
regex=re.compile('[^atgc]')
chars_to_remove = string.printable.translate({ ord('a'): None, ord('c'): None, ord('g'): None, ord('t'): None })
cmap = {}
for c in chars_to_remove:
cmap[ord(c)] = None
def test_re(s):
return regex.sub('',s)
def test_join1(s,chars_keep=set('atgc')):
return ''.join(c for c in s if c in chars_keep)
def test_join2(s,chars_keep=set('atgc')):
""" list-comp is faster, but less 'idiomatic' """
return ''.join([c for c in s if c in chars_keep])
def translate(s):
return s.translate(cmap)
import timeit
s = 'ag ct oso gcota'
for func in "test_re","test_join1","test_join2","translate":
print(func,timeit.timeit('{0}(s)'.format(func),'from __main__ import s,{0}'.format(func)))
</code></pre>
<p>以下是时间安排:</p>
<pre><code>test_re 3.3141989699797705
test_join1 2.4452173250028864
test_join2 2.081048655003542
translate 1.9390292020107154
</code></pre>
<p>字符串太糟糕了。translate()没有选项来控制如何处理不在映射中的字符。当前的实现是保留它们,但我们也可以选择删除它们,以防我们要保留的字符远少于要删除的字符(哦,hello,unicode)。</p>