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Basic statistics module.
This module provides functions for calculating statistics of data, including
averages, variance, and standard deviation.
Calculating averages
--------------------
================== ==================================================
Function Description
================== ==================================================
mean Arithmetic mean (average) of data.
fmean Fast, floating-point arithmetic mean.
geometric_mean Geometric mean of data.
harmonic_mean Harmonic mean of data.
median Median (middle value) of data.
median_low Low median of data.
median_high High median of data.
median_grouped Median, or 50th percentile, of grouped data.
mode Mode (most common value) of data.
multimode List of modes (most common values of data).
quantiles Divide data into intervals with equal probability.
================== ==================================================
Calculate the arithmetic mean ("the average") of data:
>>> mean([-1.0, 2.5, 3.25, 5.75])
2.625
Calculate the standard median of discrete data:
>>> median([2, 3, 4, 5])
3.5
Calculate the median, or 50th percentile, of data grouped into class intervals
centred on the data values provided. E.g. if your data points are rounded to
the nearest whole number:
>>> median_grouped([2, 2, 3, 3, 3, 4]) #doctest: +ELLIPSIS
2.8333333333...
This should be interpreted in this way: you have two data points in the class
interval 1.5-2.5, three data points in the class interval 2.5-3.5, and one in
the class interval 3.5-4.5. The median of these data points is 2.8333...
Calculating variability or spread
---------------------------------
================== =============================================
Function Description
================== =============================================
pvariance Population variance of data.
variance Sample variance of data.
pstdev Population standard deviation of data.
stdev Sample standard deviation of data.
================== =============================================
Calculate the standard deviation of sample data:
>>> stdev([2.5, 3.25, 5.5, 11.25, 11.75]) #doctest: +ELLIPSIS
4.38961843444...
If you have previously calculated the mean, you can pass it as the optional
second argument to the four "spread" functions to avoid recalculating it:
>>> data = [1, 2, 2, 4, 4, 4, 5, 6]
>>> mu = mean(data)
>>> pvariance(data, mu)
2.5
Statistics for relations between two inputs
-------------------------------------------
================== ====================================================
Function Description
================== ====================================================
covariance Sample covariance for two variables.
correlation Pearson's correlation coefficient for two variables.
linear_regression Intercept and slope for simple linear regression.
================== ====================================================
Calculate covariance, Pearson's correlation, and simple linear regression
for two inputs:
>>> x = [1, 2, 3, 4, 5, 6, 7, 8, 9]
>>> y = [1, 2, 3, 1, 2, 3, 1, 2, 3]
>>> covariance(x, y)
0.75
>>> correlation(x, y) #doctest: +ELLIPSIS
0.31622776601...
>>> linear_regression(x, y) #doctest:
LinearRegression(slope=0.1, intercept=1.5)
Exceptions
----------
A single exception is defined: StatisticsError is a subclass of ValueError.
)
NormalDistStatisticsError correlation
covariancefmeangeometric_mean
harmonic_meanlinear_regressionmeanmedianmedian_grouped median_high
median_lowmode multimodepstdev pvariance quantilesstdevvariance NFraction)Decimal)countgroupbyrepeat) bisect_left bisect_right) hypotsqrtfabsexperftaulogfsumsumprod)reduce)
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Return a high-precision sum of the given numeric data as a fraction,
together with the type to be converted to and the count of items.
Examples
--------
>>> _sum([3, 2.25, 4.5, -0.5, 0.25])
(, Fraction(19, 2), 5)
Some sources of round-off error will be avoided:
# Built-in sum returns zero.
>>> _sum([1e50, 1, -1e50] * 1000)
(, Fraction(1000, 1), 3000)
Fractions and Decimals are also supported:
>>> from fractions import Fraction as F
>>> _sum([F(2, 3), F(7, 5), F(1, 4), F(5, 6)])
(, Fraction(63, 20), 4)
>>> from decimal import Decimal as D
>>> data = [D("0.1375"), D("0.2108"), D("0.3061"), D("0.0419")]
>>> _sum(data)
(, Fraction(6963, 10000), 4)
Mixed types are currently treated as an error, except that int is
allowed.
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sAu r3 c l | j S # t $ r t j | cY S w xY wr8 ) is_finiteAttributeErrormathisfinite)rV s r4 rE rE s1 {{} }}Q s 33c | t usJ d | |u r| S |t u s|t u r| S | t u r|S t || r|S t | | r| S t | t r|S t |t r| S t | t rt |t r|S t | t rt |t r| S d}t || j |j fz )zCoerce types T and S to a common type, or raise TypeError.
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test class in test_statistics for details.
zinitial type T is boolz"don't know how to coerce %s and %s)boolrI
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| j fS # t $ r% dt | j d}t | w xY w)zReturn Real number x to exact (numerator, denominator) pair.
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(1, 4)
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$ r9 t |t r' || j || j z cY S w xY w)z&Convert value to given numeric type T.r6 )rB rg rI rq rh ri r rp )valuerR s r4 _convertrt M sz E{a !Se / /1 4 x a !U__%%*;*;(<< < s > >B >B c # B K | D ] }|dk r t | | yw)z7Iterate over values, failing if any are less than zero.r N)r )rP errmsgrV s r4 _fail_negrw _ s'
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Rank order a dataset. The lowest value has rank 1.
Ties are averaged so that equal values receive the same rank:
>>> data = [31, 56, 31, 25, 75, 18]
>>> _rank(data)
[3.5, 5.0, 3.5, 2.0, 6.0, 1.0]
The operation is idempotent:
>>> _rank([3.5, 5.0, 3.5, 2.0, 6.0, 1.0])
[3.5, 5.0, 3.5, 2.0, 6.0, 1.0]
It is possible to rank the data in reverse order so that the
highest value has rank 1. Also, a key-function can extract
the field to be ranked:
>>> goals = [('eagles', 45), ('bears', 48), ('lions', 44)]
>>> _rank(goals, key=itemgetter(1), reverse=True)
[2.0, 1.0, 3.0]
Ranks are conventionally numbered starting from one; however,
setting *start* to zero allows the ranks to be used as array indices:
>>> prize = ['Gold', 'Silver', 'Bronze', 'Certificate']
>>> scores = [8.1, 7.3, 9.4, 8.3]
>>> [prize[int(i)] for i in _rank(scores, start=0, reverse=True)]
['Bronze', 'Certificate', 'Gold', 'Silver']
rx z Unknown tie resolution method: )rz r6 r )ry ) ro rC sortedzipr lenr r) list)rJ ry rz r{ r| val_posiresult_ggroupsizerankrs orig_poss r4 _rankr g s J y:4(CDD
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%FB1u2zABB 2 22 Kr3 c ^ t | \ }}}|dk r t d t ||z | S )a Return the sample arithmetic mean of data.
>>> mean([1, 2, 3, 4, 4])
2.8
>>> from fractions import Fraction as F
>>> mean([F(3, 7), F(1, 21), F(5, 3), F(1, 3)])
Fraction(13, 21)
>>> from decimal import Decimal as D
>>> mean([D("0.5"), D("0.75"), D("0.625"), D("0.375")])
Decimal('0.5625')
If ``data`` is empty, StatisticsError will be raised.
r6 z%mean requires at least one data point)rS r rt )rJ rR rQ r< s r4 r
r
s7 t*KAua1u EFF EAIq !!r3 c \ |) t | t | }s t d |z S t |t
t f s t | } t | | }t | }|s t d ||z S # t $ r dfd} || } Y w xY w# t $ r t d w xY w)zConvert data to floats and compute the arithmetic mean.
This runs faster than the mean() function and it always returns a float.
If the input dataset is empty, it raises a StatisticsError.
>>> fmean([3.5, 4.0, 5.25])
4.25
r c 3 @ K t | d D ] \ }| y w)Nr6 r| ) enumerate)iterablerV r< s r4 r zfmean..count s! %ha 8DAqG 9s z&fmean requires at least one data pointz(data and weights must be the same lengthz sum of weights must be non-zero) r ri r& r
isinstancer tupler' ro )rJ weightsr rQ numdenr< s @r4 r r s D A T
!"JK Kqy ge} -w-JdG$ w-C ?@@ 9+ A
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t d dw xY w)aY Convert data to floats and compute the geometric mean.
Raises a StatisticsError if the input dataset is empty,
if it contains a zero, or if it contains a negative value.
No special efforts are made to achieve exact results.
(However, this may change in the future.)
>>> round(geometric_mean([54, 24, 36]), 9)
36.0
zGgeometric mean requires a non-empty dataset containing positive numbersN)r" r rC r% ro r )rJ s r4 r r sE G5S$ ()) G < =BF GGs !$ :c x t | | u r t | } d}t | }|dk r t d |dk( rD|B| d }t |t
j t f r|dk r t | |S t d |t d| }|}nQt | |u r t | }t | |k7 r t d t d t || D \ }}} t | | } t d t || D \ }}} |dk r t d t ||z | S # t $ r Y yw xY w)
a Return the harmonic mean of data.
The harmonic mean is the reciprocal of the arithmetic mean of the
reciprocals of the data. It can be used for averaging ratios or
rates, for example speeds.
Suppose a car travels 40 km/hr for 5 km and then speeds-up to
60 km/hr for another 5 km. What is the average speed?
>>> harmonic_mean([40, 60])
48.0
Suppose a car travels 40 km/hr for 5 km, and when traffic clears,
speeds-up to 60 km/hr for the remaining 30 km of the journey. What
is the average speed?
>>> harmonic_mean([40, 60], weights=[5, 30])
56.0
If ``data`` is empty, or any element is less than zero,
``harmonic_mean`` will raise ``StatisticsError``.
z.harmonic mean does not support negative valuesr6 z.harmonic_mean requires at least one data pointr zunsupported typez*Number of weights does not match data sizec 3 K | ] }| y wr8 r2 )r: ws r4 r= z harmonic_mean..N s G,Fq,Fs c 3 4 K | ] \ }}|r||z nd yw)r Nr2 )r: r rV s r4 r= z harmonic_mean..Q s" P=OTQq1uq 0=Os z Weighted sum must be positive)iterr r r r numbersRealr ri r rS rw r ZeroDivisionErrorrt )
rJ r rv r< rV sum_weightsr rR rQ r s
r4 r r ! sM . DzTDz
=F D A1u NOO
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z =>> K%' ++ s ",D- - D98D9c t | } t | }|dk( r t d |dz dk( r| |dz S |dz }| |dz
| | z dz S )aB Return the median (middle value) of numeric data.
When the number of data points is odd, return the middle data point.
When the number of data points is even, the median is interpolated by
taking the average of the two middle values:
>>> median([1, 3, 5])
3
>>> median([1, 3, 5, 7])
4.0
r no median for empty datar r6 r r r )rJ r< r s r4 r r Y sg $>> median_low([1, 3, 5])
3
>>> median_low([1, 3, 5, 7])
3
r r r r6 r rJ r< s r4 r r q sU $>> median_high([1, 3, 5])
3
>>> median_high([1, 3, 5, 7])
5
r r r r r s r4 r
r
s7 $>> demographics = Counter({
... 25: 172, # 20 to 30 years old
... 35: 484, # 30 to 40 years old
... 45: 387, # 40 to 50 years old
... 55: 22, # 50 to 60 years old
... 65: 6, # 60 to 70 years old
... })
The 50th percentile (median) is the 536th person out of the 1071
member cohort. That person is in the 30 to 40 year old age group.
The regular median() function would assume that everyone in the
tricenarian age group was exactly 35 years old. A more tenable
assumption is that the 484 members of that age group are evenly
distributed between 30 and 40. For that, we use median_grouped().
>>> data = list(demographics.elements())
>>> median(data)
35
>>> round(median_grouped(data, interval=10), 1)
37.5
The caller is responsible for making sure the data points are separated
by exact multiples of *interval*. This is essential for getting a
correct result. The function does not check this precondition.
Inputs may be any numeric type that can be coerced to a float during
the interpolation step.
r r )loz$Value cannot be converted to a floatr- )r r r r r rh ro ri ) rJ intervalr< rV r jLcffs r4 r r s V $@@As A= =Bc t t | j d } |d d S # t $ r
t d dw xY w)ax Return the most common data point from discrete or nominal data.
``mode`` assumes discrete data, and returns a single value. This is the
standard treatment of the mode as commonly taught in schools:
>>> mode([1, 1, 2, 3, 3, 3, 3, 4])
3
This also works with nominal (non-numeric) data:
>>> mode(["red", "blue", "blue", "red", "green", "red", "red"])
'red'
If there are multiple modes with same frequency, return the first one
encountered:
>>> mode(['red', 'red', 'green', 'blue', 'blue'])
'red'
If *data* is empty, ``mode``, raises StatisticsError.
r6 r zno mode for empty dataN)r* r most_common
IndexErrorr )rJ pairss r4 r r sP .
DJ + +A .EBQx{ B 67TABs - Ac t t | }|sg S t |j }|j D cg c]
\ }}||k( s | c}}S c c}}w )a. Return a list of the most frequently occurring values.
Will return more than one result if there are multiple modes
or an empty list if *data* is empty.
>>> multimode('aabbbbbbbbcc')
['b']
>>> multimode('aabbbbccddddeeffffgg')
['b', 'd', 'f']
>>> multimode('')
[]
)r* r maxrP rG )rJ countsmaxcountrs r s r4 r r sS T$Z
F 6==?#H&,lln JnleU8IEn JJ Js
AAr exclusive)r< methodc ( |dk r t d t | } t | }|dk r t d |dk( rW|dz
}g }t d| D ]? }t ||z | \ }}| | ||z
z | |dz |z z |z } |j | A |S |dk( rn|dz }g }t d| D ]V }||z |z }|dk rdn||dz
kD r|dz
n|}||z ||z z
}| |dz
||z
z | | |z z |z } |j | X |S t
d| )a Divide *data* into *n* continuous intervals with equal probability.
Returns a list of (n - 1) cut points separating the intervals.
Set *n* to 4 for quartiles (the default). Set *n* to 10 for deciles.
Set *n* to 100 for percentiles which gives the 99 cuts points that
separate *data* in to 100 equal sized groups.
The *data* can be any iterable containing sample.
The cut points are linearly interpolated between data points.
If *method* is set to *inclusive*, *data* is treated as population
data. The minimum value is treated as the 0th percentile and the
maximum value is treated as the 100th percentile.
r6 zn must be at least 1r z"must have at least two data points inclusiver Unknown method: )r r r rangedivmodappendro )
rJ r< r ldr r r r delta interpolateds
r4 r r 9 sa 1u 455 $>> data = [2.75, 1.75, 1.25, 0.25, 0.5, 1.25, 3.5]
>>> variance(data)
1.3720238095238095
If you have already calculated the mean of your data, you can pass it as
the optional second argument ``xbar`` to avoid recalculating it:
>>> m = mean(data)
>>> variance(data, m)
1.3720238095238095
This function does not check that ``xbar`` is actually the mean of
``data``. Giving arbitrary values for ``xbar`` may lead to invalid or
impossible results.
Decimals and Fractions are supported:
>>> from decimal import Decimal as D
>>> variance([D("27.5"), D("30.25"), D("30.25"), D("34.5"), D("41.75")])
Decimal('31.01875')
>>> from fractions import Fraction as F
>>> variance([F(1, 6), F(1, 2), F(5, 3)])
Fraction(67, 108)
r z*variance requires at least two data pointsr6 r_ r rt )rJ xbarrR ssrW r< s r4 r r j s@ L dD/KAr1a1u JKK B!a%L! $$r3 c b t | | \ }}}}|dk r t d t ||z | S )a, Return the population variance of ``data``.
data should be a sequence or iterable of Real-valued numbers, with at least one
value. The optional argument mu, if given, should be the mean of
the data. If it is missing or None, the mean is automatically calculated.
Use this function to calculate the variance from the entire population.
To estimate the variance from a sample, the ``variance`` function is
usually a better choice.
Examples:
>>> data = [0.0, 0.25, 0.25, 1.25, 1.5, 1.75, 2.75, 3.25]
>>> pvariance(data)
1.25
If you have already calculated the mean of the data, you can pass it as
the optional second argument to avoid recalculating it:
>>> mu = mean(data)
>>> pvariance(data, mu)
1.25
Decimals and Fractions are supported:
>>> from decimal import Decimal as D
>>> pvariance([D("27.5"), D("30.25"), D("30.25"), D("34.5"), D("41.75")])
Decimal('24.815')
>>> from fractions import Fraction as F
>>> pvariance([F(1, 4), F(5, 4), F(1, 2)])
Fraction(13, 72)
r6 z*pvariance requires at least one data pointr )rJ murR r rW r< s r4 r r s< F dB-KAr1a1u JKK BFA r3 c t | | \ }}}}|dk r t d ||dz
z }t |t r t |j
|j S t |j
|j S )zReturn the square root of the sample variance.
See ``variance`` for arguments and other details.
>>> stdev([1.5, 2.5, 2.5, 2.75, 3.25, 4.75])
1.0810874155219827
r 'stdev requires at least two data pointsr6 r_ r rg r r rp rq r )rJ r rR r rW r< msss r4 r r sk dD/KAr1a1u GHH
A,C!W $S]]COODD s}}coo >>r3 c t | | \ }}}}|dk r t d ||z }t |t r t |j
|j S t |j
|j S )zReturn the square root of the population variance.
See ``pvariance`` for arguments and other details.
>>> pstdev([1.5, 2.5, 2.5, 2.75, 3.25, 4.75])
0.986893273527251
r6 z'pstdev requires at least one data pointr )rJ r rR r rW r< r s r4 r r sg dB-KAr1a1u GHH
q&C!W $S]]COODD s}}coo >>r3 c t | \ }}}}|dk r t d ||dz
z } t | t |j |j
fS # t $ r% t | t | t | z fcY S w xY w)zFIn one pass, compute the mean and sample standard deviation as floats.r r r6 )r_ r rh r rp rq rb )rJ rR r r r< r s r4 _mean_stdevr s YNAr41u GHH
A,C4T{ /
s OOO 4T{E$K%) 3334s *A +BBc t | }t | |k7 r t d |dk r t d t | |z t | |z t fd| D fd|D }||dz
z S )ap Covariance
Return the sample covariance of two inputs *x* and *y*. Covariance
is a measure of the joint variability of two inputs.
>>> x = [1, 2, 3, 4, 5, 6, 7, 8, 9]
>>> y = [1, 2, 3, 1, 2, 3, 1, 2, 3]
>>> covariance(x, y)
0.75
>>> z = [9, 8, 7, 6, 5, 4, 3, 2, 1]
>>> covariance(x, z)
-7.5
>>> covariance(z, x)
-7.5
zDcovariance requires that both inputs have same number of data pointsr z,covariance requires at least two data pointsc 3 ( K | ] }|z
y wr8 r2 )r: xir s r4 r= z covariance.. s )q29q c 3 ( K | ] }|z
y wr8 r2 r: yiybars r4 r= z covariance.. s +B"BIr r6 )r r r& r' )rV yr< sxyr r s @@r4 r r sx " AA
1v{ dee1u LMM 7Q;D 7Q;D
)q)+B+B
CC !a%=r3 linear)r c t | }t | |k7 r t d |dk r t d |dvrt d| |dk( r#|dz
dz }t | | } t || }n@t | |z }t | |z }| D cg c] }||z
} }|D cg c] }||z
}}t | | } t | | }
t || } | t
|
| z z S c c}w c c}w # t $ r t d
w xY w) al Pearson's correlation coefficient
Return the Pearson's correlation coefficient for two inputs. Pearson's
correlation coefficient *r* takes values between -1 and +1. It measures
the strength and direction of a linear relationship.
>>> x = [1, 2, 3, 4, 5, 6, 7, 8, 9]
>>> y = [9, 8, 7, 6, 5, 4, 3, 2, 1]
>>> correlation(x, x)
1.0
>>> correlation(x, y)
-1.0
If *method* is "ranked", computes Spearman's rank correlation coefficient
for two inputs. The data is replaced by ranks. Ties are averaged
so that equal values receive the same rank. The resulting coefficient
measures the strength of a monotonic relationship.
Spearman's rank correlation coefficient is appropriate for ordinal
data or for continuous data that doesn't meet the linear proportion
requirement for Pearson's correlation coefficient.
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Return the slope and intercept of simple linear regression
parameters estimated using ordinary least squares. Simple linear
regression describes relationship between an independent variable
*x* and a dependent variable *y* in terms of a linear function:
y = slope * x + intercept + noise
where *slope* and *intercept* are the regression parameters that are
estimated, and noise represents the variability of the data that was
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variable).
The parameters are returned as a named tuple.
>>> x = [1, 2, 3, 4, 5]
>>> noise = NormalDist().samples(5, seed=42)
>>> y = [3 * x[i] + 2 + noise[i] for i in range(5)]
>>> linear_regression(x, y) #doctest: +ELLIPSIS
LinearRegression(slope=3.09078914170..., intercept=1.75684970486...)
If *proportional* is true, the independent variable *x* and the
dependent variable *y* are assumed to be directly proportional.
The data is fit to a line passing through the origin.
Since the *intercept* will always be 0.0, the underlying linear
function simplifies to:
y = slope * x + noise
>>> y = [3 * x[i] + noise[i] for i in range(5)]
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LinearRegression(slope=3.02447542484..., intercept=0.0)
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0.8035050657330205
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