102

numpy.average() has a weights option, but numpy.std() does not. Does anyone have suggestions for a workaround?

Asclepius
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YGA
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    Btw, calculation of weighted std dev is actually a rather complex subject -- there's more than one way to do it. See here for a great discussion: https://www.stata.com/support/faqs/statistics/weights-and-summary-statistics/ – JohnE Nov 18 '17 at 17:09
  • http://www.ccgalberta.com/pygeostat/statistics.html#weighted-statistics – e271p314 Dec 23 '20 at 22:57

6 Answers6

162

How about the following short "manual calculation"?

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

    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))
Eric O Lebigot
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    Why not use `numpy.average` again for the variance? – user2357112 Aug 07 '13 at 01:26
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    Just wanted to point out that this will give the biased variance. For small sample sizes, you may want to re-scale the variance (before sqrt) to get the unbiased variance. See https://en.wikipedia.org/wiki/Weighted_variance#Weighted_sample_variance – Corey Mar 07 '14 at 05:17
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    Yeah, the unbiased variance estimator would be slightly different. This answer gives the standard deviation, since the question asks for a weighted version of `numpy.std()`. – Eric O Lebigot Sep 12 '14 at 09:58
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    thx for this solution... but why do you use `math.sqrt` instead of `np.sqrt` in the end? – raphael Oct 10 '18 at 08:58
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    `np.sqrt()` would work, but because `variance` is a simple (Numpy) float (and not a NumPy array), `math.sqrt()` is more explicit and appropriate (and therefore in general faster, if this matters). – Eric O Lebigot Oct 11 '18 at 10:29
45

There is a class in statsmodels that makes it easy to calculate weighted statistics: statsmodels.stats.weightstats.DescrStatsW.

Assuming this dataset and weights:

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

You initialize the class (note that you have to pass in the correction factor, the delta degrees of freedom at this point):

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

Then you can calculate:

  • .mean the weighted mean:

    >>> weighted_stats.mean      
    1.97196261682243
    
  • .std the weighted standard deviation:

    >>> weighted_stats.std       
    0.21434289609681711
    
  • .var the weighted variance:

    >>> weighted_stats.var       
    0.045942877107170932
    
  • .std_mean the standard error of weighted mean:

    >>> weighted_stats.std_mean  
    0.020818822467555047
    

    Just in case you're interested in the relation between the standard error and the standard deviation: The standard error is (for ddof == 0) calculated as the weighted standard deviation divided by the square root of the sum of the weights minus 1 (corresponding source for statsmodels version 0.9 on GitHub):

    standard_error = standard_deviation / sqrt(sum(weights) - 1)
    
MSeifert
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  • To use this approach to easily calculate the weighted coefficient of variation, see [this answer](https://stackoverflow.com/a/53748541/832230). – Asclepius Dec 12 '18 at 17:42
28

Here's one more option:

np.sqrt(np.cov(values, aweights=weights))
Leo
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6

There doesn't appear to be such a function in numpy/scipy yet, but there is a ticket proposing this added functionality. Included there you will find Statistics.py which implements weighted standard deviations.

Mad Physicist
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unutbu
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1

There is a very good example proposed by gaborous:

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

# 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
xm = xm.fillna(0) # fill NaN with 0 (because anyway a variance of 0 is 
just void, but at least it keeps the other covariance's values computed 
correctly))
sigma2 = 1./(w.sum()-1) * xm.mul(w, axis=0).T.dot(xm); # Compute the 
unbiased weighted sample covariance

Correct equation for weighted unbiased sample covariance, URL (version: 2016-06-28)

Max Bileschi
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abah
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0

A follow-up to "sample" or "unbiased" standard deviation in the "frequency weights" sense since "weighted sample standard deviation python" Google search leads to this post:

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

Or modifying the answer by @Eric as follows:

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))
Sterling
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    Thanks for this nice answer! However, for your second function `weighted_sample_avg_std()`, on the third line where you have the second part of the variance equation, the variance is not supposed to be multiplied by a ratio of the sums but by a ratio of the number of non-zeros weights (https://www.itl.nist.gov/div898/software/dataplot/refman2/ch2/weightsd.pdf). – DouglasCoenen Oct 15 '21 at 12:24
  • Hmm.. that's a good point. Would you mind suggesting an edit? I looked into this previously (but after you made the comment), but the actual change wasn't obvious to me. – Sterling Dec 09 '21 at 22:49