從np.min獲得3個最小數（array）

Question

有沒有一種方法可以使用np.min從矩陣中獲取多於1個最小數？

x = np.array([[1,2,3],[4,5,6],[7,8,0]]) 

3個最小號碼

預期結果

>>[0,1,2] 

Answer 1

拼合並對數組進行排序並獲得前3個元素。

sorted(x.flatten())[:3] 
Out[275]: [0, 1, 2] 

或者更快的方法：

sorted(sum(x.tolist(),[]))[:3] 

%timeit sorted(sum(x.tolist(),[]))[:3] 
The slowest run took 7.38 times longer than the fastest. This could mean that an intermediate result is being cached. 
100000 loops, best of 3: 2.14 µs per loop 

%timeit sorted(x.flatten())[:3] 
The slowest run took 11.54 times longer than the fastest. This could mean that an intermediate result is being cached. 
100000 loops, best of 3: 4.97 µs per loop 

Answer 2

的一種方法是對flattend視圖中使用argsort -

x.ravel()[x.ravel().argsort()[:3]] 

更高性能一個與np.argpartition -

x.ravel()[x.ravel().argpartition(range(3))[:3]] 

或用sort事後排序呢 -

np.sort(x.ravel()[x.ravel().argpartition(3)[:3]]) 

如果你不關心這些數字進行排序，跳過sort -

x.ravel()[x.ravel().argpartition(3)[:3]] 

採樣運行 -

In [44]: x 
Out[44]: 
array([[1, 2, 3], 
     [4, 5, 6], 
     [7, 8, 0]]) 

In [45]: x.ravel()[x.ravel().argsort()[:3]] 
Out[45]: array([0, 1, 2]) 

In [48]: x.ravel()[x.ravel().argpartition(range(3))[:3]] 
Out[48]: array([0, 1, 2]) 

In [52]: np.sort(x.ravel()[x.ravel().argpartition(3)[:3]]) 
Out[52]: array([0, 1, 2]) 

In [47]: x.ravel()[x.ravel().argpartition(3)[:3]] 
Out[47]: array([0, 1, 2])