Where can I find mad (mean absolute deviation) in scipy?

Question:

It seems scipy once provided a function mad to calculate the mean absolute deviation for a set of numbers:

http://projects.scipy.org/scipy/browser/trunk/scipy/stats/models/utils.py?rev=3473

However, I can not find it anywhere in current versions of scipy. Of course it is possible to just copy the old code from repository but I prefer to use scipy’s version. Where can I find it, or has it been replaced or removed?

Asked By: Ton van den Heuvel

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Answers:

It looks like scipy.stats.models was removed in august 2008 due to insufficient baking. Development has migrated to statsmodels.

Answered By: matt

It’s not the scipy version, but here’s an implementation of the MAD using masked arrays to ignore bad values:
http://code.google.com/p/agpy/source/browse/trunk/agpy/mad.py

Edit: A more recent version is available here.

Edit 2: There’s also a version in astropy here.

Answered By: keflavich

[EDIT] Since this keeps on getting downvoted: I know that median absolute deviation is a more commonly-used statistic, but the questioner asked for mean absolute deviation, and here’s how to do it:

from numpy import mean, absolute

def mad(data, axis=None):
    return mean(absolute(data - mean(data, axis)), axis)
Answered By: mhsmith

I’m using:

from math import fabs

a = [1, 1, 2, 2, 4, 6, 9]

median = sorted(a)[len(a)//2]

for b in a:
    mad = fabs(b - median)
    print b,mad
Answered By: duhaime

For what its worth, I use this for MAD:

def mad(arr):
    """ Median Absolute Deviation: a "Robust" version of standard deviation.
        Indices variabililty of the sample.
        https://en.wikipedia.org/wiki/Median_absolute_deviation 
    """
    arr = np.ma.array(arr).compressed() # should be faster to not use masked arrays.
    med = np.median(arr)
    return np.median(np.abs(arr - med))
Answered By: Lee

I’m just learning Python and Numpy, but here is the code I wrote to check my 7th grader’s math homework which wanted the M(ean)AD of 2 sets of numbers:

Data in Numpy matrix rows:

import numpy as np

>>> a = np.matrix( [ [ 80, 76, 77, 78, 79, 81, 76, 77, 79, 84, 75, 79, 76, 78 ], \    
... [ 66, 69, 76, 72, 79, 77, 74, 77, 71, 79, 74, 66, 67, 73 ] ], dtype=float )    
>>> matMad = np.mean( np.abs( np.tile( np.mean( a, axis=1 ), ( 1, a.shape[1] ) ) - a ), axis=1 )    
>>> matMad    
matrix([[ 1.81632653],
        [ 3.73469388]])

Data in Numpy 1D arrays:

>>> a1 = np.array( [ 80, 76, 77, 78, 79, 81, 76, 77, 79, 84, 75, 79, 76, 78 ], dtype=float )    
>>> a2 = np.array( [ 66, 69, 76, 72, 79, 77, 74, 77, 71, 79, 74, 66, 67, 73 ], dtype=float )    
>>> madA1 = np.mean( np.abs( np.tile( np.mean( a1 ), ( 1, len( a1 ) ) ) - a1 ) )    
>>> madA2 = np.mean( np.abs( np.tile( np.mean( a2 ), ( 1, len( a2 ) ) ) - a2 ) )    
>>> madA1, madA2    
(1.816326530612244, 3.7346938775510199)
Answered By: RickC

The current version of statsmodels has mad in statsmodels.robust:

>>> import numpy as np
>>> from statsmodels import robust
>>> a = np.matrix( [
...     [ 80, 76, 77, 78, 79, 81, 76, 77, 79, 84, 75, 79, 76, 78 ],
...     [ 66, 69, 76, 72, 79, 77, 74, 77, 71, 79, 74, 66, 67, 73 ]
...  ], dtype=float )
>>> robust.mad(a, axis=1)
array([ 2.22390333,  5.18910776])

Note that by default this computes the robust estimate of the standard deviation assuming a normal distribution by scaling the result a scaling factor; from help:

Signature: robust.mad(a, 
                      c=0.67448975019608171, 
                      axis=0, 
                      center=<function median at 0x10ba6e5f0>)

The version in R makes a similar normalization. If you don’t want this, obviously just set c=1.

(An earlier comment mentioned this being in statsmodels.robust.scale. The implementation is in statsmodels/robust/scale.py (see github) but the robust package does not export scale, rather it exports the public functions in scale.py explicitly.)

Answered By: sfjac

If you enjoy working in Pandas (like I do), it has a useful function for the mean absolute deviation:

import pandas as pd
df = pd.DataFrame()
df['a'] = [1, 1, 2, 2, 4, 6, 9]
df['a'].mad()

Output: 2.3673469387755106

Answered By: Sam Perry

Using numpy only:

def meanDeviation(numpyArray):
    mean = np.mean(numpyArray)
    f = lambda x: abs(x - mean)
    vf = np.vectorize(f)
    return (np.add.reduce(vf(numpyArray))) / len(numpyArray)
Answered By: tonix
Answered By: Frank
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