NumPy 中的加权标准差

Question

numpy.average() 有一个权重选项，但 numpy.std() 没有。有人对解决方法有建议吗？

Answer 1

下面这个简短的“手动计算”怎么样？

def weighted_avg_and_std(values, weights):
    """
    Return the weighted average and standard deviation.

    They weights are in effect first normalized so that they 
    sum to 1 (and so they must not all be 0).

    values, weights -- NumPy ndarrays with the same shape.
    """
    average = numpy.average(values, weights=weights)
    # Fast and numerically precise:
    variance = numpy.average((values-average)**2, weights=weights)
    return (average, math.sqrt(variance))

    

Answer 2

statsmodels中有一个类可以方便计算加权统计：statsmodels.stats.weightstats.DecrStatsW。

假设此数据集和权重：

import numpy as np
from statsmodels.stats.weightstats import DescrStatsW

array = np.array([1,2,1,2,1,2,1,3])
weights = np.ones_like(array)
weights[3] = 100

您初始化类（请注意，您必须传入校正因子，增量

weighted_stats = DescrStatsW(array, weights=weights, ddof=0)

然后您可以计算：

• .mean加权平均值：

  <前><代码>>>>加权统计平均值
  1.97196261682243

• .std加权标准差< /强>：

  <前><代码>>>>加权统计.std
  0.21434289609681711
  

• .var加权方差< /强>：

  <前><代码>>>>加权统计.var
  0.045942877107170932
  

• .std_mean 加权平均值的标准误差：

  <前><代码>>>>加权统计数据.std_mean
  0.020818822467555047

  以防万一您对标准误差和标准差之间的关系感兴趣：标准误差（对于 ddof == 0）计算为加权标准差除以平方权重总和减 1 的根 (GitHub 上 statsmodels 版本 0.9 的相应源代码）：

  标准误差 = 标准偏差 / sqrt(sum(权重) - 1)
  

Answer 3

这是另一种选择：

np.sqrt(np.cov(values, aweights=weights))

Answer 4

numpy/scipy 中似乎还没有这样的函数，但是有一个 ticket< /a> 提议添加此功能。在那里你会发现 Statistics.py 它实现了加权标准偏差。

Answer 5

gaborous 提出了一个非常好的例子：

import pandas as pd
import numpy as np
# X is the dataset, as a Pandas' DataFrame
# Compute the weighted sample mean (fast, efficient and precise)
mean = np.ma.average(X, axis=0, weights=weights) 

# Convert to a Pandas' Series (it's just aesthetic and more 
# ergonomic; no difference in computed values)
mean = pd.Series(mean, index=list(X.keys())) 
xm = X-mean # xm = X diff to mean
# fill NaN with 0 
# a variance of 0 is just void, but at least it keeps the other
# covariance's values computed correctly))
xm = xm.fillna(0) 
# Compute the unbiased weighted sample covariance
sigma2 = 1./(w.sum()-1) * xm.mul(w, axis=0).T.dot(xm); 

加权无偏样本协方差的正确方程，URL（版本：2016-06-28）

Answer 6

“频率权重”自从“加权样本标准差Python”以来，谷歌搜索导致了这篇文章：

def frequency_sample_std_dev(X, n):
    """
    Sample standard deviation for X and n,
    where X[i] is the quantity each person in group i has,
    and n[i] is the number of people in group i.
    See Equation 6.4 of:
    Montgomery, Douglas, C. and George C. Runger. Applied Statistics 
     and Probability for Engineers, Enhanced eText. Available from: 
      WileyPLUS, (7th Edition). Wiley Global Education US, 2018.
    """
    n_groups = len(n)
    n_people = sum(n)
    lhs_numerator = sum([ni*Xi**2 for Xi, ni in zip(X, n)])
    rhs_numerator = sum([Xi*ni for Xi, ni in zip(X,n)])**2/n_people
    denominator = n_people-1
    var = (lhs_numerator - rhs_numerator) / denominator
    std = sqrt(var)
    return std

或者修改@Eric的答案如下：

def weighted_sample_avg_std(values, weights):
    """
    Return the weighted average and weighted sample standard deviation.

    values, weights -- Numpy ndarrays with the same shape.
    
    Assumes that weights contains only integers (e.g. how many samples in each group).
    
    See also https://en.wikipedia.org/wiki/Weighted_arithmetic_mean#Frequency_weights
    """
    average = np.average(values, weights=weights)
    variance = np.average((values-average)**2, weights=weights)
    variance = variance*sum(weights)/(sum(weights)-1)
    return (average, sqrt(variance))

print(weighted_sample_avg_std(X, n))

Answer 7

我只是在寻找与 numpy np.std 函数等效的 API，它也允许设置 axis 参数：（

我只是用二维测试了它，所以感觉如果有问题，可免费进行改进。）

def std(values, weights=None, axis=None):
    """
    Return the weighted standard deviation.
    axis -- the axis for std calculation
    values, weights -- Numpy ndarrays with the same shape on the according axis.
    """
    average = np.expand_dims(np.average(values, weights=weights, axis=axis), axis=axis)
    # Fast and numerically precise:
    variance = np.average((values-average)**2, weights=weights, axis=axis)
    return np.sqrt(variance)

感谢 Eric O Lebigot 的原始答案。