擅长:python、mysql、java
<p>你可以用熊猫,它很快。你知道吗</p>
<pre><code>import pandas as pd
array_element = [('T10', 'R1T0'), ('T20', 'R2T0'), ('T31', 'R3T1'),
('T21', 'R2T1'), ('T10', 'R1T0'), ('T20', 'R2T0')]
k = pd.Index(tuple(array_element)).value_counts()
list(zip(k.index, k))
</code></pre>
<p>出去</p>
<pre><code>[(('T10', 'R1T0'), 2),
(('T20', 'R2T0'), 2),
(('T31', 'R3T1'), 1),
(('T21', 'R2T1'), 1)]
</code></pre>
<p>或者另一个只有numpy的解决方案:</p>
<pre><code>b = np.unique(array_element,return_counts=True, axis=0)
list(zip(zip(*b[0].T.tolist()), b[1]))
</code></pre>
<p>出去</p>
<pre><code>[(('T10', 'R1T0'), 2),
(('T20', 'R2T0'), 2),
(('T21', 'R2T1'), 1),
(('T31', 'R3T1'), 1)]
</code></pre>