Question

有没有办法使用np.min从矩阵中获得超过1个最小数？

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

3个最小数字的预期结果

>>[0,1,2]

Answer 1

一种方法是在展平视图上使用argsort -

public function __construct ($x, ... $ignore) { ... }

效率更高的np.argpartition -

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

或者x.ravel()[x.ravel().argpartition(range(3))[:3]]之后对其进行排序 -

sort

如果您不关心正在排序的数字，请跳过np.sort(x.ravel()[x.ravel().argpartition(3)[:3]]) -

sort

示例运行 -

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

Answer 2

对数组进行展平和排序，并获得前3个元素。

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

或更快的方法：

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

从np.min获取3个最小数字（数组）

2 个答案: