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Zd(d Zd)d Z d*d Z d+d Z d7d-d Z d.d Z d8d/d0Z!d9d1dZ"d:d2dZ#d;d3dZ$d>> 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
Exceptions
----------
A single exception is defined: StatisticsError is a subclass of ValueError.
StatisticsErrorpstdev pvariancestdevvariancemedian
median_low median_highmedian_groupedmeanmode
harmonic_mean N)Fraction)Decimal)groupbychain) bisect_left bisect_rightc @ s e Zd ZdS )r N)__name__
__module__ __qualname__ r r "/usr/lib64/python3.6/statistics.pyr c s c C s d}t |\}}||i}|j}ttt|}xRt| tD ]D\}} t||}x0tt | D ]"\}}|d7 }||d| ||< qVW q8W d|kr|d }
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|fS )aC _sum(data [, start]) -> (type, sum, count)
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.
If optional argument ``start`` is given, it is added to the total.
If ``data`` is empty, ``start`` (defaulting to 0) is returned.
Examples
--------
>>> _sum([3, 2.25, 4.5, -0.5, 1.0], 0.75)
(, Fraction(11, 1), 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.
r
Nc s s | ]\}}t ||V qd S )N)r ).0dnr r r s z_sum..) _exact_ratioget_coerceinttyper map _isfiniteAssertionErrorsumsorteditems) datastartcountr r ZpartialsZ partials_getTtypvaluestotalr r r _sumi s $
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r" tj| S X d S )N)Z is_finiteAttributeErrormathZisfinite)xr r r r$ s r$ c C s | t k std| |kr | S |tks,|t kr0| S | tkr<|S t|| rJ|S t| |rX| S t| trf|S t|trt| S t| trt|tr|S t| trt|tr| S d}t|| j|jf dS )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 %sN)boolr% r!
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r` y| j S tk
rZ Y nX Y nX W n* ttfk
r t | st
| dfS X d}t |j t | j
dS )zReturn Real number x to exact (numerator, denominator) pair.
>>> _exact_ratio(0.25)
(1, 4)
x is expected to be an int, Fraction, Decimal or float.
Nz0can't convert type '{}' to numerator/denominator)r" r6 r as_integer_ratio numerator denominatorr1
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r c C sj t | |kr| S t|tr(| jdkr(t}y|| S tk
rd t|tr^|| j|| j S Y nX dS )z&Convert value to given numeric type T.r N)r" r5 r! r< r6 r7 r r; )valuer, r r r _convert s
rA c C s` t jt| j }|s|S |d d }x4tdt|D ]"}|| d |kr6|d | }P q6W |S )Nr
r ) collectionsCounteriter most_commonrangelen)r) tableZmaxfreqir r r _counts s rJ c C s. t | |}|t| kr&| | |kr&|S tdS )z,Locate the leftmost value exactly equal to xN)r rG r> )ar3 rI r r r
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rL c C s> t | ||d}|t| d kr6| |d |kr6|d S tdS )z-Locate the rightmost value exactly equal to x)lor N)r rG r> )rK lr3 rI r r r
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N)r )r. errmsgr3 r r r _fail_neg s
rR c C sT t | | krt| } t| }|dk r,tdt| \}}}||ksFtt|| |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.
r z%mean requires at least one data point)rD listrG r r0 r% rA )r) r r, r/ r+ r r r r
# s c C s t | | krt| } d}t| }|dk r2tdn<|dkrn| d }t|tjtfrf|dk rbt||S tdy"t dd t
| |D \}}}W n t k
r dS X ||kst t
|| |S )a Return the harmonic mean of data.
The harmonic mean, sometimes called the subcontrary mean, is the
reciprocal of the arithmetic mean of the reciprocals of the data,
and is often appropriate when averaging quantities which are rates
or ratios, for example speeds. Example:
Suppose an investor purchases an equal value of shares in each of
three companies, with P/E (price/earning) ratios of 2.5, 3 and 10.
What is the average P/E ratio for the investor's portfolio?
>>> harmonic_mean([2.5, 3, 10]) # For an equal investment portfolio.
3.6
Using the arithmetic mean would give an average of about 5.167, which
is too high.
If ``data`` is empty, or any element is less than zero,
``harmonic_mean`` will raise ``StatisticsError``.
z.harmonic mean does not support negative valuesr z.harmonic_mean requires at least one data pointr
zunsupported typec s s | ]}d | V qdS )r Nr )r r3 r r r r d s z harmonic_mean..)rD rS rG r
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" c C s\ t | } t| }|dkr td|d dkr8| |d S |d }| |d | | d S dS )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
zno median for empty data r N)r' rG r )r) r rI r r r r l s
c C sL t | } t| }|dkr td|d dkr8| |d S | |d d S dS )a Return the low median of numeric data.
When the number of data points is odd, the middle value is returned.
When it is even, the smaller of the two middle values is returned.
>>> median_low([1, 3, 5])
3
>>> median_low([1, 3, 5, 7])
3
r
zno median for empty datarW r N)r' rG r )r) r r r r r s c C s, t | } t| }|dkr td| |d S )a Return the high median of data.
When the number of data points is odd, the middle value is returned.
When it is even, the larger of the two middle values is returned.
>>> median_high([1, 3, 5])
3
>>> median_high([1, 3, 5, 7])
5
r
zno median for empty datarW )r' rG r )r) r r r r r s
r c
C s t | } t| }|dkr"tdn|dkr2| d S | |d }x*||fD ] }t|ttfrHtd| qHW y||d }W n( tk
r t|t|d }Y nX t| |}t | ||}|}|| d } |||d | | S )a Return the 50th percentile (median) of grouped continuous data.
>>> median_grouped([1, 2, 2, 3, 4, 4, 4, 4, 4, 5])
3.7
>>> median_grouped([52, 52, 53, 54])
52.5
This calculates the median as the 50th percentile, and should be
used when your data is continuous and grouped. In the above example,
the values 1, 2, 3, etc. actually represent the midpoint of classes
0.5-1.5, 1.5-2.5, 2.5-3.5, etc. The middle value falls somewhere in
class 3.5-4.5, and interpolation is used to estimate it.
Optional argument ``interval`` represents the class interval, and
defaults to 1. Changing the class interval naturally will change the
interpolated 50th percentile value:
>>> median_grouped([1, 3, 3, 5, 7], interval=1)
3.25
>>> median_grouped([1, 3, 3, 5, 7], interval=2)
3.5
This function does not check whether the data points are at least
``interval`` apart.
r
zno median for empty datar rW zexpected number but got %r)
r' rG r rT strbytesr7 r6 rL rO )
r) Zintervalr r3 objLl1l2Zcffr r r r s&
c C sB t | }t|dkr |d d S |r6tdt| ntddS )a 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 is not exactly one most common value, ``mode`` will raise
StatisticsError.
r r
z.no unique mode; found %d equally common valueszno mode for empty dataN)rJ rG r )r) rH r r r r s c s dkrt | t fdd| D \}}}t fdd| D \}}}||krX||ks\t||d t| 8 }|dk std| ||fS )a; Return sum of square deviations of sequence data.
If ``c`` is None, the mean is calculated in one pass, and the deviations
from the mean are calculated in a second pass. Otherwise, deviations are
calculated from ``c`` as given. Use the second case with care, as it can
lead to garbage results.
Nc 3 s | ]}| d V qdS )rW Nr )r r3 )cr r r s z_ss..c 3 s | ]}| V qd S )Nr )r r3 )r_ r r r s rW r
z%negative sum of square deviations: %f)r
r0 r% rG )r) r_ r, r/ r+ UZtotal2Zcount2r )r_ r _ss s ra c C sL t | | krt| } t| }|dk r,tdt| |\}}t||d |S )a Return the sample variance of data.
data should be an iterable of Real-valued numbers, with at least two
values. The optional argument xbar, if given, should be the mean of
the data. If it is missing or None, the mean is automatically calculated.
Use this function when your data is a sample from a population. To
calculate the variance from the entire population, see ``pvariance``.
Examples:
>>> 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)
rW z*variance requires at least two data pointsr )rD rS rG r ra rA )r) xbarr r, ssr r r r " s & c C sH t | | krt| } t| }|dk r,tdt| |\}}t|| |S )a Return the population variance of ``data``.
data should be an 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
This function does not check that ``mu`` is actually the mean of ``data``.
Giving arbitrary values for ``mu`` may lead to invalid or impossible
results.
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)
r z*pvariance requires at least one data point)rD rS rG r ra rA )r) mur r, rc r r r r Q s ' c C s2 t | |}y|j S tk
r, tj|S X dS )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
N)r sqrtr1 r2 )r) rb varr r r r s
c C s2 t | |}y|j S tk
r, tj|S X dS )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
N)r re r1 r2 )r) rd rf r r r r s
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)rP )r )N)N)N)N)N)&__doc____all__rB Zdecimalr2 rU Z fractionsr r itertoolsr r Zbisectr r r> r r0 r$ r r rA rJ rL rO rR r
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