<p>从您提供的示例df来看,元数据中的值看起来像字符串。假设它们是你在问题中提到的词典列表</p>
<pre><code>file_names = ["41456gn7L.jpg","31mndfg.jpg","dsfsdf8587eh.jpg"]
df = pd.DataFrame({'url':['http://example.com/images/41456gn7L.jpg','http://example.com/images/31mndfg.jpg','http://example.com/images/dsfsdf8587eh.jpg'],
'meta_data':[[{'id': 0, 'imageUrl': 'http://example.com/images/41dY3ASVn7L.jpg', 'score': 54.09280014038086}, {'id': 0, 'imageUrl': 'http://example.com/images/41dY3ASVn7L.jpg', 'score': 54.09280014038086}],[{'id': 0, 'imageUrl': 'http://example.com/images/31mnLrB5IHL.jpg', 'score': 99.902099609375}, {'id': 0, 'imageUrl': 'http://example.com/images/31mnLrB5IHL.jpg', 'score': 99.902099609375}],[{'id': 0, 'imageUrl': 'http://example.com/images/4189TDx0e0L.jpg' ,'score': 97.33160400390625}, {'id': 0, 'imageUrl': 'http://example.com/images/4189TDx0e0L.jpg', 'score': 97.33160400390625}]]})
</code></pre>
<p>您可以从列表的<code>first element</code>中选择文件名出现在列表文件名中的片段,并访问与键“score”关联的值</p>
<pre><code>df['score'] = df.loc[df['url'].str.rsplit('/').str[-1].isin(file_names), 'meta_data'].apply(lambda x: x[0]['score'])
url meta_data score
0 http://example.com/images/41456gn7L.jpg [{'id': 0, 'imageUrl': 'http://example.com/ima... 54.092800
1 http://example.com/images/31mndfg.jpg [{'id': 0, 'imageUrl': 'http://example.com/ima... 99.902100
2 http://example.com/images/dsfsdf8587eh.jpg [{'id': 0, 'imageUrl': 'http://example.com/ima... 97.331604
</code></pre>