将一组数字转换为numpy，以便将每个数字转换为小于该数字的其他数字 [英] Transform a set of numbers in numpy so that each number gets converted into a number of other numbers which are less than it

问题描述

考虑一组数字:

In [8]: import numpy as np

In [9]: x = np.array([np.random.random() for i in range(10)])

In [10]: x
Out[10]: 
array([ 0.62594394,  0.03255799,  0.7768568 ,  0.03050498,  0.01951657,
        0.04767246,  0.68038553,  0.60036203,  0.3617409 ,  0.80294355])

现在，我想通过以下方式将此集合转换为另一个集合y:对于x中的每个元素i，y中的相应元素j将是其中的其他元素数小于i的x.例如，上面给出的x看起来像:

Now I want to transform this set into another set y in the following way: for every element i in x, the corresponding element j in y would be the number of other elements in x which are less than i. For example, the above given x would look like:

In [25]: y
Out[25]: array([ 6.,  2.,  8.,  1.,  0.,  3.,  7.,  5.,  4.,  9.])

现在，我可以使用简单的python循环来做到这一点:

Now, I can do this using simple python loops:

In [16]: for i in range(len(x)):
    ...:     tot = 0
    ...:     for j in range(len(x)):
    ...:         if x[i] > x[j]: tot += 1
    ...:     y[i] = int(tot)

但是，当x的长​​度很大时，代码将变得非常慢.我想知道是否可以带出任何麻木的魔法来营救.例如，如果我必须过滤小于0.5的所有元素，我将只使用布尔掩码:

However, when length of x is very large, the code becomes extremely slow. I was wondering if any numpy magic can be brought to rescue. For example, if I had to filter all the elements less than 0.5, I would have simply used a Boolean masking:

In [19]: z = x[x < 0.5]

In [20]: z
Out[20]: array([ 0.03255799,  0.03050498,  0.01951657,  0.04767246,  0.3617409 ])

可以使用类似的东西来使相同的事情更快地实现吗?

Can something like this be used so that the same thing could be achieved much faster?

推荐答案

您真正需要做的是获取数组排序顺序的 inverse :

What you actually need to do is get the inverse of the sorting order of your array:

import numpy as np
x = np.random.rand(10)
y = np.empty(x.size,dtype=np.int64)
y[x.argsort()] = np.arange(x.size)

示例运行(在ipython中):

Example run (in ipython):

In [367]: x
Out[367]: 
array([ 0.09139335,  0.29084225,  0.43560987,  0.92334644,  0.09868977,
        0.90202354,  0.80905083,  0.4801967 ,  0.99086213,  0.00933582])

In [368]: y
Out[368]: array([1, 3, 4, 8, 2, 7, 6, 5, 9, 0])

---

或者，如果要获得更大个元素的数量大于x个中每个相应元素的数量，则必须将排序从升序转换为降序.一种可能的选择是简单地交换索引的构造:

---

Alternatively, if you want to get the number of elements greater than each corresponding element in x, you have to reverse the sorting from ascending to descending. One possible option to do this is to simply swap the construction of the indexing:

y_rev = np.empty(x.size,dtype=np.int64)
y_rev[x.argsort()] = np.arange(x.size)[::-1]

另一个

another, as @unutbu suggested in a comment, is to map the original array to the new one:

y_rev = x.size - y - 1

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