Question

我有一个ND数组（向量）列表，每个向量都有(1,300)形状我的目标是在列表中找到重复的向量，对它们求和，然后将它们除以列表的大小，结果值（向量）将替换重复的向量。
例如，a是ND数组列表a = [[2,3,1],[5,65,-1],[2,3,1]]，然后第一个和最后一个元素是重复的。他们的sum将是：[4,6,2]，它将除以矢量列表的大小size = 3。

输出：a = [[4/3,6/3,2/3],[5,65,-1],[4/3,6/3,2/3]]

我尝试使用Counter，但它对ndarrays不起作用。

Numpy的方式是什么？感谢。

Answer 1

If you have numpy 1.13 or higher, this is pretty simple:

def f(a):
    u, inv, c = np.unique(a, return_counts = True, return_inverse = True, axis = 0)
    p = np.where(c > 1,  c / a.shape[0], 1)[:, None]
    return (u * p)[inv]

If you don't have 1.13, you'll need some trick to convert a into a 1-d array first. I recommend @Jaime's excellent answer using np.void here

How it works:

u is the unique rows of a (usually not in their original order)
c is the number of times each row of u are repeated in a
inv is the indices to get u back to a, i.e. u[inv] = a
p is the multiplier for each row of u based on your requirements. 1 if c == 1 and c / n (where n is the number of rows in a) if c > 1. [:, None] turns it into a column vector so that it broadcasts well with u

return u * p indexed back to their original locations by [inv]

Answer 2

你可以使用numpy unique，with count return count

 elements, count = np.unique(a, axis=0, return_counts=True)

返回计数允许返回数组中每个元素的出现次数

输出就像这样，

(array([[ 2,  3,  1],
        [ 5, 65, -1]]), array([2, 1]))

然后你可以像这样繁殖它们：

(count * elements.T).T

输出：

array([[ 4,  6,  2],
       [ 5, 65, -1]])

在Python中。我有一个ND数组列表，我想计算重复数组，以计算每个重复数组值的平均值

2 个答案: