获取最大或最小正元素出numpy的阵列？ （preferably不平坦的） [英] Get max or min n-elements out of numpy array? (preferably not flattened)

问题描述

我知道，我能得到min或以最高值：

I know that I can get min or max values with:

max(matrix)
min(matrix)

出numpy的矩阵/矢量。对于那些瓦莱斯的索引将返回：

out of a numpy matrix/vector. The indices for those vales are returned by:

argmax(matrix)
argmin(matrix)

所以，例如当我有一个5x5的矩阵：

So e.g. when I have a 5x5 matrix:

a = np.arange(5*5).reshape(5, 5) + 10

# array([[10, 11, 12, 13, 14],
#        [15, 16, 17, 18, 19],
#        [20, 21, 22, 23, 24],
#        [25, 26, 27, 28, 29],
#        [30, 31, 32, 33, 34]])

我可以通过获得最大值：

I could get the max value via:

In [86]: np.max(a) # getting the max-value out of a
Out[86]: 34

In [87]: np.argmax(a) # index of max-value 34 is 24 if array a were flattened
Out[87]: 24

...但什么是最有效的方式来获得最大或最小正元素？

...but what is the most efficient way to get the max or min n-elements?

所以我们说出来的在的我想有5最高和最低的5要素。这将返回我的 [30，31，32，33，34] 分别为5最高值 [20，21，22，23，24 ] 为他们的指数。同样 [10，11，12，13，14] 的5最低值和 [0，1，2，3，4] 的5最低元素的索引。

So let's say out of a I want to have the 5 highest and 5 lowest elements. This should return me [30, 31, 32, 33, 34] for the 5 highest values respectively [20, 21, 22, 23, 24] for their indices. Likewise [10, 11, 12, 13, 14] for the 5 lowest values and [0, 1, 2, 3, 4] for the indices of the 5 lowest elements.

什么是一个高效，合理的解决方案，这？

What would be an efficient, reasonable solution for this?

我的第一个想法为扁平化和分选阵列，并采取姓氏和5的值。后来我通过这些值的指数原来的二维矩阵搜索。 虽然此过程适用于扁平化+分拣效率不高......没有人知道一个更快的解决方案？

My first idea was flattening and sorting the array and taking the last and first 5 values. Afterwards I search through the original 2D matrix for the indices of those values. Although this procedure works flattening + sorting isn't very efficient...does anyone know a faster solution?

此外，我想有原来的二维数组，而不是扁平之一的指标。因此，而不是 24 按 np.argmax返回（一）我想有（ 4,4）。

Additionally I would like to have the indices of the original 2D array and not the flattening one. So instead of 24 returned by np.argmax(a) I would like to have (4, 4).

推荐答案

要获得最大或最小值的索引数组中的标准方法是使用的 np.argpartition 。此函数使用introselect算法和线性复杂运行 - 这比执行完全分拣更大的阵列更好（通常为O（n log n）的）

The standard way to get the indices of the largest or smallest values in an array is to use np.argpartition. This function uses an introselect algorithm and runs with linear complexity - this performs better than fully sorting for larger arrays (which is typically O(n log n)).

默认该函数沿着阵列的最后一个轴起作用。要考虑整个数组，你需要使用 拉威尔（） 。例如，这里有一个随机序列 A ：

By default this function works along the last axis of the array. To consider an entire array, you need to use ravel(). For example, here's a random array a:

>>> a = np.random.randint(0, 100, size=(5, 5))
>>> a
array([[60, 68, 86, 66,  9],
       [66, 26, 83, 87, 50],
       [41, 26,  0, 55,  9],
       [57, 80, 71, 50, 22],
       [94, 30, 95, 99, 76]])

然后得到最大的五个值的指数的（扁平化）二维数组中，使用：

Then to get the indices of the five largest values in the (flattened) 2D array, use:

>>> i = np.argpartition(a.ravel(), -5)[-5:] # argpartition(a.ravel(), 5)[:5] for smallest
>>> i
array([ 2,  8, 22, 23, 20])

要取回这些位置的对应2D指数在 A ，使用的 unravel_index ：

To get back the corresponding 2D indices of these positions in a, use unravel_index:

>>> i2d = np.unravel_index(i, a.shape)
>>> i2d
(array([0, 1, 4, 4, 4]), array([2, 3, 2, 3, 0]))

然后索引 A 与 I2D 还给五大价值观：

>>> a[i2d]
array([86, 87, 95, 99, 94])

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