擅长:python、mysql、java
<p>不需要合并,您只需将<code>banana_potato_matched_df</code>转换为dict,然后在df中将<code>map</code>的值转换为<code>potato</code>。i、 e</p>
<pre><code>banana_dict = dict(zip(banana_potato_matched_df.banana, banana_potato_matched_df.potato))
</code></pre>
<p>香蕉糖:</p>
^{pr2}$
<p>映射值</p>
<pre><code>df.loc[df.potato == '','potato'] = df.loc[df.potato == '','banana'].map(banana_dict)
</code></pre>
<p>输出:</p>
<pre>
date banana potato avocado
0
33 2017-06-01 55ee4cbc 80660 b57c-2473556952a8
34 2017-06-01 391dc0f6 80686 82ff-46de03510afc
35 2017-06-01 3a1f407f 54408 a3a6-d85429eef303
36 2017-06-01 3a1f407f 54408 858d-48082acc66ed
37 2017-06-01 5222ab45 80693 ba1f-dbd387748b71
38 2017-06-01 5222ab45 80693 b085-99d58875084a
39 2017-06-01 5222ab45 80693 a570-6d4c766ff7cf
40 2017-06-01 6939ced3 55031 960c-a9ded8ed2f56
41 2017-06-01 2478913a 80610 9258-df6d26027d18
42 2017-06-01 2478913a 80610 8e53-a1d8d4e175b9
43 2017-06-01 2478913a 80610 b4b2-a9221895f8b5
</pre>
<p>希望有帮助</p>