将多个值添加到一个numpy数组索引 [英] Add multiple values to one numpy array index

问题描述

简单版本: 如果我这样做:

import numpy as np
a = np.zeros(2)
a[[1, 1]] += np.array([1, 1])

我得到[0, 1]作为输出.但我想要[0, 2].使用隐式numpy循环而不是自己遍历它，是否可能以某种方式实现?

I get [0, 1] as an output. but I would like [0, 2]. Is that possible somehow, using implicit numpy looping instead of looping over it myself?

我实际上需要做什么的版本:

我有一个结构化的数组，其中包含一个索引，一个值和一些布尔值.我想根据布尔值对那些索引处的值求和.显然，这可以通过一个简单的循环来完成，但是似乎可以通过聪明的numpy索引来实现(如上所述).

I have a structured array that contains an index, a value, and some boolean value. I would like to sum those values at those indices, based on the boolean. Clearly that can be done with a simple loop, but it seems like it should be possible with clever numpy indexing (as above).

例如，我有一个包含5个元素的数组，我希望使用值，索引和条件从该数组中填充该元素:

For example, I have an array with 5 elements that I want to populate from the array with values, indices, and conditions:

import numpy as np
size = 5
nvalues = 10
np.random.seed(1)
a = np.zeros(nvalues, dtype=[('val', float), ('ix', int), ('cond', bool)])
a = np.rec.array(a)
a.val = np.random.rand(nvalues)
a.cond = (np.random.rand(nvalues) > 0.3)
a.ix = np.random.randint(size, size=nvalues)

# obvious solution
obvssum = np.zeros(size)
for i in a:
    if i.cond:
        obvssum[i.ix] += i.val

# is something this possible?
doesntwork = np.zeros(size)
doesntwork[a[a.cond].ix] += a[a.cond].val

print(doesntwork)
print(obvssum)

输出:

[ 0.          0.          0.61927097  0.02592623  0.29965467]
[ 0.          0.          1.05459336  0.02592623  1.27063303]

我认为这里发生的是，如果保证a[a.cond].ix是唯一的，那么我的方法就可以正常工作，如简单示例中所述.

I think what's happening here is if a[a.cond].ix were guaranteed to be unique, my method would work just fine, as noted in the simple example.

推荐答案

这是

This is what the at method of NumPy ufuncs is for:

output = numpy.zeros(size)
numpy.add.at(output, a[a.cond].ix, a[a.cond].val)

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