<p>我们希望找到数组中没有重复的行,同时保持顺序。在</p>
<p>我使用这个<a href="https://stackoverflow.com/questions/16970982/find-unique-rows-in-numpy-array">solution</a>将<code>a</code>的每一行组合成一个元素,这样我们就可以使用<code>np.unique(,return_index=True, return_inverse= True)</code>来找到唯一的行。然后,我修改了这个<a href="https://stackoverflow.com/a/21124789/1435522">function</a>,使用索引和反转输出唯一行的计数。从那里,我可以选择具有<code>counts == 1</code>的所有唯一行。在</p>
<pre><code>a = np.array([[1, 2, 3],
[2, 3, 4],
[1, 2, 3],
[3, 2, 1],
[3, 4, 5]])
#use a flexible data type, np.void, to combine the columns of `a`
#size of np.void is the number of bytes for an element in `a` multiplied by number of columns
b = a.view(np.dtype((np.void, a.dtype.itemsize * a.shape[1])))
_, index, inv = np.unique(b, return_index = True, return_inverse = True)
def return_counts(index, inv):
count = np.zeros(len(index), np.int)
np.add.at(count, inv, 1)
return count
counts = return_counts(index, inv)
#if you want the indices to discard replace with: counts[i] > 1
index_keep = [i for i, j in enumerate(index) if counts[i] == 1]
>>>a[index_keep]
array([[2, 3, 4],
[3, 2, 1],
[3, 4, 5]])
#if you don't need the indices and just want the array returned while preserving the order
a_unique = np.vstack(a[idx] for i, idx in enumerate(index) if counts[i] == 1])
>>>a_unique
array([[2, 3, 4],
[3, 2, 1],
[3, 4, 5]])
</code></pre>
<p><strong>为了np.版本>;=1.9</strong></p>
^{pr2}$