擅长:python、mysql、java
<p>也许解决办法是:</p>
<ol>
<li><p>将month重命名为number of month</p>
<pre><code>u'january'.replace('january', 1)
</code></pre></li>
<li><p>按每月子字符串的出现次数选择</p>
<pre><code>ddrs={1:u'янв', 2:u'фев' , 3:u'мар' , 4:u'апр' , 5:u'ма' , 6:u'июн' , 7:u'июл',8:u'авг' , 9:u'сент' , 10:u'окт' , 11:u'ноя' , 12:u'дек'}
numMonth = next((x for x in ddrs if 'январь'.find(ddrs[x])>-1), None)
</code></pre></li>
</ol>
<p>第二个想法是作为类<a href="https://gist.github.com/mrbannyjo/f83b1a2ab302b0afee49d976de365aae" rel="nofollow noreferrer">https://gist.github.com/mrbannyjo/f83b1a2ab302b0afee49d976de365aae</a>实现的</p>