__  __    __   __  _____      _            _          _____ _          _ _ 
 |  \/  |   \ \ / / |  __ \    (_)          | |        / ____| |        | | |
 | \  / |_ __\ V /  | |__) | __ ___   ____ _| |_ ___  | (___ | |__   ___| | |
 | |\/| | '__|> <   |  ___/ '__| \ \ / / _` | __/ _ \  \___ \| '_ \ / _ \ | |
 | |  | | |_ / . \  | |   | |  | |\ V / (_| | ||  __/  ____) | | | |  __/ | |
 |_|  |_|_(_)_/ \_\ |_|   |_|  |_| \_/ \__,_|\__\___| |_____/|_| |_|\___V 2.1
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F�ri3����UdZgd�ZddlZddlZddlZddlZddlmZddlm	Z	ddl
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lmZddlmZm Z m!Z!ed�Z"Gd
�de#�Z$d�Z%dCd�Z&d�Z'd�Z(d�Z)d�Z*dDd�Z+ddddd�de,e-fd�Z.de/de/de/fd�Z0dejbjdzd zZ3e/e4d!<de/de/de-fd"�Z5de/de/de	fd#�Z6d$�Z7dCd%�Z8d&�Z9dCd'�Z:d(�Z;d)�Z<d*�Z=dEd+�Z>d,�Z?d-�[email protected]/d0�d1�ZAdCd2�ZBdCd3�ZCdCd4�ZDdCd5�ZEd6�ZFd7�ZGd8d9�d:�ZHe d;d<�ZIdd=�d>�ZJd?�ZK	dd@lLmKZKGdA�dB�ZNy#eM$rY�wxYw)Fa�

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.

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    Return a high-precision sum of the given numeric data as a fraction,
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    (<class 'float'>, Fraction(19, 2), 5)

    Some sources of round-off error will be avoided:

    # Built-in sum returns zero.
    >>> _sum([1e50, 1, -1e50] * 1000)
    (<class 'float'>, 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)])
    (<class 'fractions.Fraction'>, 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)
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|z}||z�tt|t�}||�|fS)a3Return the exact mean and sum of square deviations of sequence data.

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    Nc3�2�K�|]}|�z
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        >>> 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']

    rxzUnknown tie resolution method: )rzr6r)ry�)	rorC�sorted�zipr�lenrr)�list)rJryrzr{r|�val_pos�i�result�_�g�group�size�rankrs�orig_poss               r4�_rankr�gs���J�y���:�4�(�C�D�D�
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%�F�B���1�u��2���z��A��B���B���� 2�2�2����Kr3c�^�t|�\}}}|dkrtd��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.
    r6z%mean requires at least one data point)rSrrt)rJrRrQr<s    r4r
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    rc3�@�K�t|d��D]	\�}|���y�w)Nr6�r|)�	enumerate)�iterablerVr<s  �r4rzfmean.<locals>.count�s&�����%�h�a�8��D�A�q��G��s�z&fmean requires at least one data pointz(data and weights must be the same lengthzsum of weights must be non-zero)	r�rir&r�
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    zGgeometric mean requires a non-empty dataset containing positive numbersN)r"rrCr%ror)rJs r4rrsE��G��5��S�$��(�)�)���G��<�=�BF�	G�G�s�!$�:c�x�t|�|urt|�}d}t|�}|dkrtd��|dk(rD|�B|d}t	|t
jtf�r|dkrt|��|Std��|�td|�}|}nQt|�|urt|�}t|�|k7rtd��td�t||�D��\}}}	t||�}td�t||�D��\}}}	|dkrtd	��t||z|�S#t$rYywxYw)
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    The harmonic mean is the reciprocal of the arithmetic mean of the
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    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 valuesr6z.harmonic_mean requires at least one data pointrzunsupported typez*Number of weights does not match data sizec3� K�|]}|���y�wr8r2)r:�ws  r4r=z harmonic_mean.<locals>.<genexpr>Ns���� G�q�� G�s�c3�4K�|]\}}|r||znd���y�w)rNr2)r:r�rVs   r4r=z harmonic_mean.<locals>.<genexpr>Qs����P�T�Q���q�1�u�q�0�P�s�zWeighted sum must be positive)�iterr�r�rr��numbers�RealrrirrSrwr��ZeroDivisionErrorrt)
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��z��=�>�>��K�%�'��+�+��	����s�",D-�-	D9�8D9c��t|�}t|�}|dk(rtd��|dzdk(r||dzS|dz}||dz
||zdzS)aBReturn 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
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    >>> median([1, 3, 5])
    3
    >>> median([1, 3, 5, 7])
    4.0

    r�no median for empty datarr6�r�r�r)rJr<r�s   r4rrYsg���$�<�D��D�	�A��A�v��8�9�9��1�u��z��A��F�|��
��F���Q��U��d�1�g�%��*�*r3c��t|�}t|�}|dk(rtd��|dzdk(r||dzS||dzdz
S)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

    rr�rr6r��rJr<s  r4rrqsU���$�<�D��D�	�A��A�v��8�9�9��1�u��z��A��F�|���A��F�Q�J��r3c�^�t|�}t|�}|dk(rtd��||dzS)aReturn 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

    rr�rr�r�s  r4r
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�s7���$�<�D��D�	�A��A�v��8�9�9���Q��<�r3c�*�t|�}t|�}|std��||dz}t||�}t	|||��}	t|�}t|�}||dzz
}|}||z
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z|zzS#t$rtd��wxYw)a�Estimates the median for numeric data binned around the midpoints
    of consecutive, fixed-width intervals.

    The *data* can be any iterable of numeric data with each value being
    exactly the midpoint of a bin.  At least one value must be present.

    The *interval* is width of each bin.

    For example, demographic information may have been summarized into
    consecutive ten-year age groups with each group being represented
    by the 5-year midpoints of the intervals:

        >>> 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.

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