numpy.argmin用于大于阈值的元素 [英] numpy.argmin for elements greater than a threshold

问题描述

我有兴趣获取满足特定条件(在我的情况下为中等阈值)的一维NumPy数组中最小值的位置.例如:

I'm interested in getting the location of the minimum value in an 1-d NumPy array that meets a certain condition (in my case, a medium threshold). For example:

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a中所有在限制范围内的数字，以使np.argmin的结果为6.是否有一种计算便宜的方法来屏蔽不满足条件的值然后应用np.argmin?

I'd like to effectively mask all numbers in a that are under the limit, such that the result of np.argmin would be 6. Is there a computationally cheap way to mask values that don't meet a condition and then apply np.argmin?

推荐答案

您可以存储有效索引，并使用这些索引从a中选择有效元素，并在选定的元素之间使用argmin()进行索引.获得最终的索引输出.因此，实现看起来像这样-

You could store the valid indices and use those for both selecting the valid elements from a and also indexing into with the argmin() among the selected elements to get the final index output. Thus, the implementation would look something like this -

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

样品运行-

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 other solution based on masked array 与常规进行比较>本文前面针对各种数据大小提出的基于数组的解决方案.

For performance benchmarking, in this section I am comparing the other solution based on masked array against a regular array based solution as proposed earlier in this post for various datasizes.

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

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