擅长:python、mysql、java
<p>受<a href="https://stackoverflow.com/a/38145104/3293881">^{<cd1>}</a>的启发,在本例中我们也可以替换{<cd2>}操作。实现应该是这样的-</p>
<pre><code># Create numerical IDs for relevant columns and a combined one
ID1 = np.unique(df['pborderid'],return_inverse=True)[1]
ID2 = np.unique(df['to_wpadr'],return_inverse=True)[1]
ID = np.column_stack((ID1,ID2))
# Convert to linear indices
lidx = np.ravel_multi_index(ID.T,ID.max(0)+1)
# Get unique IDs for each element based on grouped uniqueness and group counts
_,ID,count = np.unique(lidx,return_inverse=True,return_counts=True)
# Look for counts>1 and collect respective IDs and thus respective rows off df
df_out = df[np.in1d(ID,np.where(count>1)[0])]
</code></pre>
<p>样本运行-</p>
^{pr2}$
<p>在我这边的运行时测试似乎没有显示出这种方法比另一个解决方案中列出的<code>groupby</code>方法有任何改进。所以,看起来<code>df.groupby</code>将是首选的方式!在</p>