擅长:python、mysql、java
<pre><code># An alternative vectorized solution.
df.options = df.options.apply(lambda x: [{k:v for k,v in e.items() if k in['a','c']} for e in x])
Out[398]:
id options
0 0 [{'a': 1, 'c': 3}, {'a': 5, 'c': 7}]
1 1 [{'a': 9, 'c': 11}, {'a': 13, 'c': 15}]
2 2 [{'a': 9, 'c': 11}, {'a': 17, 'c': 19}]
</code></pre>