使用numpy.average的加权平均值 [英] Weighted average using numpy.average
问题描述
我有一个数组:
In [37]: bias_2e13 # our array
Out[37]:
[1.7277990734072355,
1.9718263893212737,
2.469657573252167,
2.869022991373125,
3.314720313010104,
4.232269039271717]
数组中每个值的错误是:
The error on each value in the array is:
In [38]: bias_error_2e13 # the error on each value
Out[38]:
array([ 0.13271387, 0.06842465, 0.06937965, 0.23886647, 0.30458249,
0.57906816])
现在,我将每个值的错误除以2:
Now I divide the error on each value by 2:
In [39]: error_half # error divided by 2
Out[39]:
array([ 0.06635694, 0.03421232, 0.03468982, 0.11943323, 0.15229124,
0.28953408])
现在,我使用numpy.average
计算数组的平均值,但使用errors
作为weights
.
Now I calculate the average of the array using numpy.average
, but using the errors
as weights
.
首先,我在值上使用完整错误,然后在一半上使用 错误,即错误除以2.
First I am using the full error on the values, then I am using half the error, i.e. the error divided by 2.
In [40]: test = np.average(bias_2e13,weights=bias_error_2e13)
In [41]: test_2 = np.average(bias_2e13,weights=error_half)
当一个数组的错误是另一个数组的一半时,两个平均值如何给我相同的结果?
In [42]: test
Out[42]: 3.3604746813456936
In [43]: test_2
Out[43]: 3.3604746813456936
推荐答案
因为所有错误均具有相同的相对权重.提供weight
参数不会更改您平均的实际值,它仅表示每个值对平均值贡献的权重.换句话说,在将每个传递的值乘以其相应的权重后,np.average
除以所提供的权重之和.
Because all of the errors have the same relative weight. Supplying a weight
parameter does not change the actual values you are averaging, it just indicates the weight with which each value value contributes to the average. In other words, after multiplying each value passed by its corresponding weight, np.average
divides by the sum of the weights provided.
>>> import numpy as np
>>> np.average([1, 2, 3], weights=[0.2, 0.2, 0.2])
2.0
>>> np.average([1, 2, 3])
2.0
有效地,n
维数组状容器的平均公式为
Effectively, the average formula for an n
-dimensional array-like container is
where each weight is assumed to be equal to 1 when not provided to numpy.average
.
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