numpy.argmin 用于大于阈值的元素

我感兴趣的是获取满足特定条件（在我的例子中为中等阈值）的一维 NumPy 数组中最小值的位置。例如：

import numpy as np

limit = 3
a = np.array([1, 2, 4, 5, 2, 5, 3, 6, 7, 9, 10])

我想有效地屏蔽所有数字a低于限制，这样的结果np.argmin是 6。是否有一种计算成本低的方法来屏蔽不满足条件的值，然后应用np.argmin?

您可以存储有效索引并使用它们来从中选择有效元素a并索引到argmin()在选定的元素之间进行排序以获得最终的索引输出。因此，实现看起来像这样 -

valid_idx = np.where(a >= limit)[0]
out = valid_idx[a[valid_idx].argmin()]

样本运行 -

In [32]: limit = 3
    ...: a = np.array([1, 2, 4, 5, 2, 5, 3, 6, 7, 9, 10])
    ...: 

In [33]: valid_idx = np.where(a >= limit)[0]

In [34]: valid_idx[a[valid_idx].argmin()]
Out[34]: 6

运行时测试 -

对于性能基准测试，在本节中我将比较other solution based on masked array https://stackoverflow.com/a/37973394/3293881反对regular本文前面针对各种数据大小提出了基于数组的解决方案。

def masked_argmin(a,limit): # Defining func for regular array based soln
    valid_idx = np.where(a >= limit)[0]
    return valid_idx[a[valid_idx].argmin()]

In [52]: # Inputs
    ...: a = np.random.randint(0,1000,(10000))
    ...: limit = 500
    ...: 

In [53]: %timeit np.argmin(np.ma.MaskedArray(a, a<limit))
1000 loops, best of 3: 233 µs per loop

In [54]: %timeit masked_argmin(a,limit)
10000 loops, best of 3: 101 µs per loop

In [55]: # Inputs
    ...: a = np.random.randint(0,1000,(100000))
    ...: limit = 500
    ...: 

In [56]: %timeit np.argmin(np.ma.MaskedArray(a, a<limit))
1000 loops, best of 3: 1.73 ms per loop

In [57]: %timeit masked_argmin(a,limit)
1000 loops, best of 3: 1.03 ms per loop