2013-10-19 15:50:09 -03:00
|
|
|
"""
|
|
|
|
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.
|
2016-08-23 23:17:00 -03:00
|
|
|
geometric_mean Geometric mean of data.
|
2016-08-08 23:49:01 -03:00
|
|
|
harmonic_mean Harmonic mean of data.
|
2013-10-19 15:50:09 -03:00
|
|
|
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.
|
|
|
|
================== =============================================
|
|
|
|
|
|
|
|
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
|
|
|
|
|
|
|
|
|
|
|
|
Exceptions
|
|
|
|
----------
|
|
|
|
|
|
|
|
A single exception is defined: StatisticsError is a subclass of ValueError.
|
|
|
|
|
|
|
|
"""
|
|
|
|
|
|
|
|
__all__ = [ 'StatisticsError',
|
|
|
|
'pstdev', 'pvariance', 'stdev', 'variance',
|
|
|
|
'median', 'median_low', 'median_high', 'median_grouped',
|
2016-08-23 23:17:00 -03:00
|
|
|
'mean', 'mode', 'geometric_mean', 'harmonic_mean',
|
2013-10-19 15:50:09 -03:00
|
|
|
]
|
|
|
|
|
|
|
|
import collections
|
2016-08-08 23:49:01 -03:00
|
|
|
import decimal
|
2013-10-19 15:50:09 -03:00
|
|
|
import math
|
2016-08-08 23:49:01 -03:00
|
|
|
import numbers
|
2013-10-19 15:50:09 -03:00
|
|
|
|
|
|
|
from fractions import Fraction
|
|
|
|
from decimal import Decimal
|
2016-08-08 23:49:01 -03:00
|
|
|
from itertools import groupby, chain
|
2016-05-04 14:54:29 -03:00
|
|
|
from bisect import bisect_left, bisect_right
|
2015-12-01 04:59:53 -04:00
|
|
|
|
2013-10-19 15:50:09 -03:00
|
|
|
|
|
|
|
|
|
|
|
# === Exceptions ===
|
|
|
|
|
|
|
|
class StatisticsError(ValueError):
|
|
|
|
pass
|
|
|
|
|
|
|
|
|
|
|
|
# === Private utilities ===
|
|
|
|
|
|
|
|
def _sum(data, start=0):
|
2015-12-01 04:59:53 -04:00
|
|
|
"""_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.
|
2013-10-19 15:50:09 -03:00
|
|
|
|
2015-12-01 04:59:53 -04:00
|
|
|
If optional argument ``start`` is given, it is added to the total.
|
|
|
|
If ``data`` is empty, ``start`` (defaulting to 0) is returned.
|
2013-10-19 15:50:09 -03:00
|
|
|
|
|
|
|
|
|
|
|
Examples
|
|
|
|
--------
|
|
|
|
|
|
|
|
>>> _sum([3, 2.25, 4.5, -0.5, 1.0], 0.75)
|
2016-07-14 01:13:29 -03:00
|
|
|
(<class 'float'>, Fraction(11, 1), 5)
|
2013-10-19 15:50:09 -03:00
|
|
|
|
|
|
|
Some sources of round-off error will be avoided:
|
|
|
|
|
2016-08-08 23:49:01 -03:00
|
|
|
# Built-in sum returns zero.
|
|
|
|
>>> _sum([1e50, 1, -1e50] * 1000)
|
2016-07-14 01:13:29 -03:00
|
|
|
(<class 'float'>, Fraction(1000, 1), 3000)
|
2013-10-19 15:50:09 -03:00
|
|
|
|
|
|
|
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)])
|
2016-07-14 01:13:29 -03:00
|
|
|
(<class 'fractions.Fraction'>, Fraction(63, 20), 4)
|
2013-10-19 15:50:09 -03:00
|
|
|
|
|
|
|
>>> from decimal import Decimal as D
|
|
|
|
>>> data = [D("0.1375"), D("0.2108"), D("0.3061"), D("0.0419")]
|
|
|
|
>>> _sum(data)
|
2016-07-14 01:13:29 -03:00
|
|
|
(<class 'decimal.Decimal'>, Fraction(6963, 10000), 4)
|
2013-10-19 15:50:09 -03:00
|
|
|
|
2014-02-08 05:58:04 -04:00
|
|
|
Mixed types are currently treated as an error, except that int is
|
|
|
|
allowed.
|
2013-10-19 15:50:09 -03:00
|
|
|
"""
|
2015-12-01 04:59:53 -04:00
|
|
|
count = 0
|
2013-10-19 15:50:09 -03:00
|
|
|
n, d = _exact_ratio(start)
|
2015-12-01 04:59:53 -04:00
|
|
|
partials = {d: n}
|
2013-10-19 15:50:09 -03:00
|
|
|
partials_get = partials.get
|
2015-12-01 04:59:53 -04:00
|
|
|
T = _coerce(int, type(start))
|
|
|
|
for typ, values in groupby(data, type):
|
|
|
|
T = _coerce(T, typ) # or raise TypeError
|
|
|
|
for n,d in map(_exact_ratio, values):
|
|
|
|
count += 1
|
|
|
|
partials[d] = partials_get(d, 0) + n
|
2013-10-19 15:50:09 -03:00
|
|
|
if None in partials:
|
2015-12-01 04:59:53 -04:00
|
|
|
# The sum will be a NAN or INF. We can ignore all the finite
|
|
|
|
# partials, and just look at this special one.
|
|
|
|
total = partials[None]
|
|
|
|
assert not _isfinite(total)
|
|
|
|
else:
|
|
|
|
# Sum all the partial sums using builtin sum.
|
|
|
|
# FIXME is this faster if we sum them in order of the denominator?
|
|
|
|
total = sum(Fraction(n, d) for d, n in sorted(partials.items()))
|
|
|
|
return (T, total, count)
|
|
|
|
|
|
|
|
|
|
|
|
def _isfinite(x):
|
|
|
|
try:
|
|
|
|
return x.is_finite() # Likely a Decimal.
|
|
|
|
except AttributeError:
|
|
|
|
return math.isfinite(x) # Coerces to float first.
|
|
|
|
|
|
|
|
|
|
|
|
def _coerce(T, S):
|
|
|
|
"""Coerce types T and S to a common type, or raise TypeError.
|
|
|
|
|
|
|
|
Coercion rules are currently an implementation detail. See the CoerceTest
|
|
|
|
test class in test_statistics for details.
|
|
|
|
"""
|
|
|
|
# See http://bugs.python.org/issue24068.
|
|
|
|
assert T is not bool, "initial type T is bool"
|
|
|
|
# If the types are the same, no need to coerce anything. Put this
|
|
|
|
# first, so that the usual case (no coercion needed) happens as soon
|
|
|
|
# as possible.
|
|
|
|
if T is S: return T
|
|
|
|
# Mixed int & other coerce to the other type.
|
|
|
|
if S is int or S is bool: return T
|
|
|
|
if T is int: return S
|
|
|
|
# If one is a (strict) subclass of the other, coerce to the subclass.
|
|
|
|
if issubclass(S, T): return S
|
|
|
|
if issubclass(T, S): return T
|
|
|
|
# Ints coerce to the other type.
|
|
|
|
if issubclass(T, int): return S
|
|
|
|
if issubclass(S, int): return T
|
|
|
|
# Mixed fraction & float coerces to float (or float subclass).
|
|
|
|
if issubclass(T, Fraction) and issubclass(S, float):
|
|
|
|
return S
|
|
|
|
if issubclass(T, float) and issubclass(S, Fraction):
|
|
|
|
return T
|
|
|
|
# Any other combination is disallowed.
|
|
|
|
msg = "don't know how to coerce %s and %s"
|
|
|
|
raise TypeError(msg % (T.__name__, S.__name__))
|
2014-02-08 05:58:04 -04:00
|
|
|
|
|
|
|
|
2013-10-19 15:50:09 -03:00
|
|
|
def _exact_ratio(x):
|
2015-12-01 04:59:53 -04:00
|
|
|
"""Return Real number x to exact (numerator, denominator) pair.
|
2013-10-19 15:50:09 -03:00
|
|
|
|
|
|
|
>>> _exact_ratio(0.25)
|
|
|
|
(1, 4)
|
|
|
|
|
|
|
|
x is expected to be an int, Fraction, Decimal or float.
|
|
|
|
"""
|
|
|
|
try:
|
2015-12-01 04:59:53 -04:00
|
|
|
# Optimise the common case of floats. We expect that the most often
|
|
|
|
# used numeric type will be builtin floats, so try to make this as
|
|
|
|
# fast as possible.
|
2016-05-04 14:54:29 -03:00
|
|
|
if type(x) is float or type(x) is Decimal:
|
2015-12-01 04:59:53 -04:00
|
|
|
return x.as_integer_ratio()
|
2013-10-19 15:50:09 -03:00
|
|
|
try:
|
2015-12-01 04:59:53 -04:00
|
|
|
# x may be an int, Fraction, or Integral ABC.
|
2013-10-19 15:50:09 -03:00
|
|
|
return (x.numerator, x.denominator)
|
|
|
|
except AttributeError:
|
|
|
|
try:
|
2016-05-04 14:54:29 -03:00
|
|
|
# x may be a float or Decimal subclass.
|
2013-10-19 15:50:09 -03:00
|
|
|
return x.as_integer_ratio()
|
|
|
|
except AttributeError:
|
2016-05-04 14:54:29 -03:00
|
|
|
# Just give up?
|
|
|
|
pass
|
2013-10-19 15:50:09 -03:00
|
|
|
except (OverflowError, ValueError):
|
2015-12-01 04:59:53 -04:00
|
|
|
# float NAN or INF.
|
2016-05-04 14:54:29 -03:00
|
|
|
assert not _isfinite(x)
|
2013-10-19 15:50:09 -03:00
|
|
|
return (x, None)
|
2015-12-01 04:59:53 -04:00
|
|
|
msg = "can't convert type '{}' to numerator/denominator"
|
|
|
|
raise TypeError(msg.format(type(x).__name__))
|
2013-10-19 15:50:09 -03:00
|
|
|
|
|
|
|
|
2015-12-01 04:59:53 -04:00
|
|
|
def _convert(value, T):
|
|
|
|
"""Convert value to given numeric type T."""
|
|
|
|
if type(value) is T:
|
|
|
|
# This covers the cases where T is Fraction, or where value is
|
|
|
|
# a NAN or INF (Decimal or float).
|
|
|
|
return value
|
|
|
|
if issubclass(T, int) and value.denominator != 1:
|
|
|
|
T = float
|
|
|
|
try:
|
|
|
|
# FIXME: what do we do if this overflows?
|
|
|
|
return T(value)
|
|
|
|
except TypeError:
|
|
|
|
if issubclass(T, Decimal):
|
|
|
|
return T(value.numerator)/T(value.denominator)
|
|
|
|
else:
|
|
|
|
raise
|
|
|
|
|
|
|
|
|
2013-10-19 15:50:09 -03:00
|
|
|
def _counts(data):
|
|
|
|
# Generate a table of sorted (value, frequency) pairs.
|
2014-02-08 05:44:16 -04:00
|
|
|
table = collections.Counter(iter(data)).most_common()
|
2013-10-19 15:50:09 -03:00
|
|
|
if not table:
|
|
|
|
return table
|
|
|
|
# Extract the values with the highest frequency.
|
|
|
|
maxfreq = table[0][1]
|
|
|
|
for i in range(1, len(table)):
|
|
|
|
if table[i][1] != maxfreq:
|
|
|
|
table = table[:i]
|
|
|
|
break
|
|
|
|
return table
|
|
|
|
|
|
|
|
|
2016-05-04 14:54:29 -03:00
|
|
|
def _find_lteq(a, x):
|
|
|
|
'Locate the leftmost value exactly equal to x'
|
|
|
|
i = bisect_left(a, x)
|
|
|
|
if i != len(a) and a[i] == x:
|
|
|
|
return i
|
|
|
|
raise ValueError
|
|
|
|
|
|
|
|
|
|
|
|
def _find_rteq(a, l, x):
|
|
|
|
'Locate the rightmost value exactly equal to x'
|
|
|
|
i = bisect_right(a, x, lo=l)
|
|
|
|
if i != (len(a)+1) and a[i-1] == x:
|
|
|
|
return i-1
|
|
|
|
raise ValueError
|
|
|
|
|
2016-08-08 23:49:01 -03:00
|
|
|
|
|
|
|
def _fail_neg(values, errmsg='negative value'):
|
|
|
|
"""Iterate over values, failing if any are less than zero."""
|
|
|
|
for x in values:
|
|
|
|
if x < 0:
|
|
|
|
raise StatisticsError(errmsg)
|
|
|
|
yield x
|
|
|
|
|
|
|
|
|
2016-08-09 00:58:10 -03:00
|
|
|
class _nroot_NS:
|
|
|
|
"""Hands off! Don't touch!
|
|
|
|
|
|
|
|
Everything inside this namespace (class) is an even-more-private
|
|
|
|
implementation detail of the private _nth_root function.
|
|
|
|
"""
|
|
|
|
# This class exists only to be used as a namespace, for convenience
|
|
|
|
# of being able to keep the related functions together, and to
|
|
|
|
# collapse the group in an editor. If this were C# or C++, I would
|
|
|
|
# use a Namespace, but the closest Python has is a class.
|
|
|
|
#
|
|
|
|
# FIXME possibly move this out into a separate module?
|
|
|
|
# That feels like overkill, and may encourage people to treat it as
|
|
|
|
# a public feature.
|
|
|
|
def __init__(self):
|
|
|
|
raise TypeError('namespace only, do not instantiate')
|
|
|
|
|
|
|
|
def nth_root(x, n):
|
|
|
|
"""Return the positive nth root of numeric x.
|
|
|
|
|
|
|
|
This may be more accurate than ** or pow():
|
|
|
|
|
|
|
|
>>> math.pow(1000, 1.0/3) #doctest:+SKIP
|
|
|
|
9.999999999999998
|
|
|
|
|
|
|
|
>>> _nth_root(1000, 3)
|
|
|
|
10.0
|
|
|
|
>>> _nth_root(11**5, 5)
|
|
|
|
11.0
|
|
|
|
>>> _nth_root(2, 12)
|
|
|
|
1.0594630943592953
|
|
|
|
|
|
|
|
"""
|
|
|
|
if not isinstance(n, int):
|
|
|
|
raise TypeError('degree n must be an int')
|
|
|
|
if n < 2:
|
|
|
|
raise ValueError('degree n must be 2 or more')
|
|
|
|
if isinstance(x, decimal.Decimal):
|
|
|
|
return _nroot_NS.decimal_nroot(x, n)
|
|
|
|
elif isinstance(x, numbers.Real):
|
|
|
|
return _nroot_NS.float_nroot(x, n)
|
|
|
|
else:
|
|
|
|
raise TypeError('expected a number, got %s') % type(x).__name__
|
|
|
|
|
|
|
|
def float_nroot(x, n):
|
|
|
|
"""Handle nth root of Reals, treated as a float."""
|
|
|
|
assert isinstance(n, int) and n > 1
|
|
|
|
if x < 0:
|
2016-08-23 23:48:12 -03:00
|
|
|
raise ValueError('domain error: root of negative number')
|
2016-08-09 00:58:10 -03:00
|
|
|
elif x == 0:
|
|
|
|
return math.copysign(0.0, x)
|
|
|
|
elif x > 0:
|
|
|
|
try:
|
|
|
|
isinfinity = math.isinf(x)
|
|
|
|
except OverflowError:
|
|
|
|
return _nroot_NS.bignum_nroot(x, n)
|
|
|
|
else:
|
|
|
|
if isinfinity:
|
|
|
|
return float('inf')
|
|
|
|
else:
|
|
|
|
return _nroot_NS.nroot(x, n)
|
|
|
|
else:
|
|
|
|
assert math.isnan(x)
|
|
|
|
return float('nan')
|
|
|
|
|
|
|
|
def nroot(x, n):
|
|
|
|
"""Calculate x**(1/n), then improve the answer."""
|
|
|
|
# This uses math.pow() to calculate an initial guess for the root,
|
|
|
|
# then uses the iterated nroot algorithm to improve it.
|
|
|
|
#
|
|
|
|
# By my testing, about 8% of the time the iterated algorithm ends
|
|
|
|
# up converging to a result which is less accurate than the initial
|
|
|
|
# guess. [FIXME: is this still true?] In that case, we use the
|
|
|
|
# guess instead of the "improved" value. This way, we're never
|
|
|
|
# less accurate than math.pow().
|
|
|
|
r1 = math.pow(x, 1.0/n)
|
|
|
|
eps1 = abs(r1**n - x)
|
|
|
|
if eps1 == 0.0:
|
|
|
|
# r1 is the exact root, so we're done. By my testing, this
|
|
|
|
# occurs about 80% of the time for x < 1 and 30% of the
|
|
|
|
# time for x > 1.
|
|
|
|
return r1
|
|
|
|
else:
|
|
|
|
try:
|
|
|
|
r2 = _nroot_NS.iterated_nroot(x, n, r1)
|
|
|
|
except RuntimeError:
|
|
|
|
return r1
|
|
|
|
else:
|
|
|
|
eps2 = abs(r2**n - x)
|
|
|
|
if eps1 < eps2:
|
|
|
|
return r1
|
|
|
|
return r2
|
|
|
|
|
|
|
|
def iterated_nroot(a, n, g):
|
|
|
|
"""Return the nth root of a, starting with guess g.
|
|
|
|
|
|
|
|
This is a special case of Newton's Method.
|
|
|
|
https://en.wikipedia.org/wiki/Nth_root_algorithm
|
|
|
|
"""
|
|
|
|
np = n - 1
|
|
|
|
def iterate(r):
|
|
|
|
try:
|
|
|
|
return (np*r + a/math.pow(r, np))/n
|
|
|
|
except OverflowError:
|
|
|
|
# If r is large enough, r**np may overflow. If that
|
|
|
|
# happens, r**-np will be small, but not necessarily zero.
|
|
|
|
return (np*r + a*math.pow(r, -np))/n
|
|
|
|
# With a good guess, such as g = a**(1/n), this will converge in
|
|
|
|
# only a few iterations. However a poor guess can take thousands
|
|
|
|
# of iterations to converge, if at all. We guard against poor
|
|
|
|
# guesses by setting an upper limit to the number of iterations.
|
|
|
|
r1 = g
|
|
|
|
r2 = iterate(g)
|
|
|
|
for i in range(1000):
|
|
|
|
if r1 == r2:
|
|
|
|
break
|
|
|
|
# Use Floyd's cycle-finding algorithm to avoid being trapped
|
|
|
|
# in a cycle.
|
|
|
|
# https://en.wikipedia.org/wiki/Cycle_detection#Tortoise_and_hare
|
|
|
|
r1 = iterate(r1)
|
|
|
|
r2 = iterate(iterate(r2))
|
|
|
|
else:
|
|
|
|
# If the guess is particularly bad, the above may fail to
|
|
|
|
# converge in any reasonable time.
|
|
|
|
raise RuntimeError('nth-root failed to converge')
|
|
|
|
return r2
|
|
|
|
|
|
|
|
def decimal_nroot(x, n):
|
|
|
|
"""Handle nth root of Decimals."""
|
|
|
|
assert isinstance(x, decimal.Decimal)
|
|
|
|
assert isinstance(n, int)
|
|
|
|
if x.is_snan():
|
|
|
|
# Signalling NANs always raise.
|
|
|
|
raise decimal.InvalidOperation('nth-root of snan')
|
|
|
|
if x.is_qnan():
|
|
|
|
# Quiet NANs only raise if the context is set to raise,
|
|
|
|
# otherwise return a NAN.
|
|
|
|
ctx = decimal.getcontext()
|
|
|
|
if ctx.traps[decimal.InvalidOperation]:
|
|
|
|
raise decimal.InvalidOperation('nth-root of nan')
|
|
|
|
else:
|
|
|
|
# Preserve the input NAN.
|
|
|
|
return x
|
2016-08-23 23:48:12 -03:00
|
|
|
if x < 0:
|
|
|
|
raise ValueError('domain error: root of negative number')
|
2016-08-09 00:58:10 -03:00
|
|
|
if x.is_infinite():
|
|
|
|
return x
|
|
|
|
# FIXME this hasn't had the extensive testing of the float
|
|
|
|
# version _iterated_nroot so there's possibly some buggy
|
|
|
|
# corner cases buried in here. Can it overflow? Fail to
|
|
|
|
# converge or get trapped in a cycle? Converge to a less
|
|
|
|
# accurate root?
|
|
|
|
np = n - 1
|
|
|
|
def iterate(r):
|
|
|
|
return (np*r + x/r**np)/n
|
|
|
|
r0 = x**(decimal.Decimal(1)/n)
|
|
|
|
assert isinstance(r0, decimal.Decimal)
|
|
|
|
r1 = iterate(r0)
|
|
|
|
while True:
|
|
|
|
if r1 == r0:
|
|
|
|
return r1
|
|
|
|
r0, r1 = r1, iterate(r1)
|
|
|
|
|
|
|
|
def bignum_nroot(x, n):
|
|
|
|
"""Return the nth root of a positive huge number."""
|
|
|
|
assert x > 0
|
|
|
|
# I state without proof that ⁿ√x ≈ ⁿ√2·ⁿ√(x//2)
|
2016-08-30 14:47:49 -03:00
|
|
|
# and that for sufficiently big x the error is acceptable.
|
2016-08-09 00:58:10 -03:00
|
|
|
# We now halve x until it is small enough to get the root.
|
|
|
|
m = 0
|
|
|
|
while True:
|
|
|
|
x //= 2
|
|
|
|
m += 1
|
|
|
|
try:
|
|
|
|
y = float(x)
|
|
|
|
except OverflowError:
|
|
|
|
continue
|
|
|
|
break
|
|
|
|
a = _nroot_NS.nroot(y, n)
|
|
|
|
# At this point, we want the nth-root of 2**m, or 2**(m/n).
|
|
|
|
# We can write that as 2**(q + r/n) = 2**q * ⁿ√2**r where q = m//n.
|
|
|
|
q, r = divmod(m, n)
|
|
|
|
b = 2**q * _nroot_NS.nroot(2**r, n)
|
|
|
|
return a * b
|
|
|
|
|
|
|
|
|
|
|
|
# This is the (private) function for calculating nth roots:
|
|
|
|
_nth_root = _nroot_NS.nth_root
|
|
|
|
assert type(_nth_root) is type(lambda: None)
|
|
|
|
|
|
|
|
|
|
|
|
def _product(values):
|
|
|
|
"""Return product of values as (exponent, mantissa)."""
|
|
|
|
errmsg = 'mixed Decimal and float is not supported'
|
|
|
|
prod = 1
|
|
|
|
for x in values:
|
|
|
|
if isinstance(x, float):
|
|
|
|
break
|
|
|
|
prod *= x
|
|
|
|
else:
|
|
|
|
return (0, prod)
|
|
|
|
if isinstance(prod, Decimal):
|
|
|
|
raise TypeError(errmsg)
|
|
|
|
# Since floats can overflow easily, we calculate the product as a
|
|
|
|
# sort of poor-man's BigFloat. Given that:
|
|
|
|
#
|
|
|
|
# x = 2**p * m # p == power or exponent (scale), m = mantissa
|
|
|
|
#
|
|
|
|
# we can calculate the product of two (or more) x values as:
|
|
|
|
#
|
|
|
|
# x1*x2 = 2**p1*m1 * 2**p2*m2 = 2**(p1+p2)*(m1*m2)
|
|
|
|
#
|
|
|
|
mant, scale = 1, 0 #math.frexp(prod) # FIXME
|
|
|
|
for y in chain([x], values):
|
|
|
|
if isinstance(y, Decimal):
|
|
|
|
raise TypeError(errmsg)
|
|
|
|
m1, e1 = math.frexp(y)
|
|
|
|
m2, e2 = math.frexp(mant)
|
|
|
|
scale += (e1 + e2)
|
|
|
|
mant = m1*m2
|
|
|
|
return (scale, mant)
|
|
|
|
|
|
|
|
|
2013-10-19 15:50:09 -03:00
|
|
|
# === Measures of central tendency (averages) ===
|
|
|
|
|
|
|
|
def mean(data):
|
|
|
|
"""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.
|
|
|
|
"""
|
|
|
|
if iter(data) is data:
|
|
|
|
data = list(data)
|
|
|
|
n = len(data)
|
|
|
|
if n < 1:
|
|
|
|
raise StatisticsError('mean requires at least one data point')
|
2015-12-01 04:59:53 -04:00
|
|
|
T, total, count = _sum(data)
|
|
|
|
assert count == n
|
|
|
|
return _convert(total/n, T)
|
2013-10-19 15:50:09 -03:00
|
|
|
|
|
|
|
|
2016-08-09 00:58:10 -03:00
|
|
|
def geometric_mean(data):
|
|
|
|
"""Return the geometric mean of data.
|
|
|
|
|
|
|
|
The geometric mean is appropriate when averaging quantities which
|
|
|
|
are multiplied together rather than added, for example growth rates.
|
|
|
|
Suppose an investment grows by 10% in the first year, falls by 5% in
|
|
|
|
the second, then grows by 12% in the third, what is the average rate
|
|
|
|
of growth over the three years?
|
|
|
|
|
|
|
|
>>> geometric_mean([1.10, 0.95, 1.12])
|
|
|
|
1.0538483123382172
|
|
|
|
|
|
|
|
giving an average growth of 5.385%. Using the arithmetic mean will
|
|
|
|
give approximately 5.667%, which is too high.
|
|
|
|
|
|
|
|
``StatisticsError`` will be raised if ``data`` is empty, or any
|
|
|
|
element is less than zero.
|
|
|
|
"""
|
|
|
|
if iter(data) is data:
|
|
|
|
data = list(data)
|
|
|
|
errmsg = 'geometric mean does not support negative values'
|
|
|
|
n = len(data)
|
|
|
|
if n < 1:
|
|
|
|
raise StatisticsError('geometric_mean requires at least one data point')
|
|
|
|
elif n == 1:
|
|
|
|
x = data[0]
|
|
|
|
if isinstance(g, (numbers.Real, Decimal)):
|
|
|
|
if x < 0:
|
|
|
|
raise StatisticsError(errmsg)
|
|
|
|
return x
|
|
|
|
else:
|
|
|
|
raise TypeError('unsupported type')
|
|
|
|
else:
|
|
|
|
scale, prod = _product(_fail_neg(data, errmsg))
|
|
|
|
r = _nth_root(prod, n)
|
|
|
|
if scale:
|
|
|
|
p, q = divmod(scale, n)
|
|
|
|
s = 2**p * _nth_root(2**q, n)
|
|
|
|
else:
|
|
|
|
s = 1
|
|
|
|
return s*r
|
|
|
|
|
|
|
|
|
2016-08-08 23:49:01 -03:00
|
|
|
def harmonic_mean(data):
|
|
|
|
"""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``.
|
|
|
|
"""
|
|
|
|
# For a justification for using harmonic mean for P/E ratios, see
|
|
|
|
# http://fixthepitch.pellucid.com/comps-analysis-the-missing-harmony-of-summary-statistics/
|
|
|
|
# http://papers.ssrn.com/sol3/papers.cfm?abstract_id=2621087
|
|
|
|
if iter(data) is data:
|
|
|
|
data = list(data)
|
|
|
|
errmsg = 'harmonic mean does not support negative values'
|
|
|
|
n = len(data)
|
|
|
|
if n < 1:
|
|
|
|
raise StatisticsError('harmonic_mean requires at least one data point')
|
|
|
|
elif n == 1:
|
|
|
|
x = data[0]
|
|
|
|
if isinstance(x, (numbers.Real, Decimal)):
|
|
|
|
if x < 0:
|
|
|
|
raise StatisticsError(errmsg)
|
|
|
|
return x
|
|
|
|
else:
|
|
|
|
raise TypeError('unsupported type')
|
|
|
|
try:
|
|
|
|
T, total, count = _sum(1/x for x in _fail_neg(data, errmsg))
|
|
|
|
except ZeroDivisionError:
|
|
|
|
return 0
|
|
|
|
assert count == n
|
|
|
|
return _convert(n/total, T)
|
|
|
|
|
|
|
|
|
2013-10-19 15:50:09 -03:00
|
|
|
# FIXME: investigate ways to calculate medians without sorting? Quickselect?
|
|
|
|
def median(data):
|
|
|
|
"""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
|
|
|
|
|
|
|
|
"""
|
|
|
|
data = sorted(data)
|
|
|
|
n = len(data)
|
|
|
|
if n == 0:
|
|
|
|
raise StatisticsError("no median for empty data")
|
|
|
|
if n%2 == 1:
|
|
|
|
return data[n//2]
|
|
|
|
else:
|
|
|
|
i = n//2
|
|
|
|
return (data[i - 1] + data[i])/2
|
|
|
|
|
|
|
|
|
|
|
|
def median_low(data):
|
|
|
|
"""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
|
|
|
|
|
|
|
|
"""
|
|
|
|
data = sorted(data)
|
|
|
|
n = len(data)
|
|
|
|
if n == 0:
|
|
|
|
raise StatisticsError("no median for empty data")
|
|
|
|
if n%2 == 1:
|
|
|
|
return data[n//2]
|
|
|
|
else:
|
|
|
|
return data[n//2 - 1]
|
|
|
|
|
|
|
|
|
|
|
|
def median_high(data):
|
|
|
|
"""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
|
|
|
|
|
|
|
|
"""
|
|
|
|
data = sorted(data)
|
|
|
|
n = len(data)
|
|
|
|
if n == 0:
|
|
|
|
raise StatisticsError("no median for empty data")
|
|
|
|
return data[n//2]
|
|
|
|
|
|
|
|
|
|
|
|
def median_grouped(data, interval=1):
|
2015-10-28 00:00:41 -03:00
|
|
|
"""Return the 50th percentile (median) of grouped continuous data.
|
2013-10-19 15:50:09 -03:00
|
|
|
|
|
|
|
>>> 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.
|
|
|
|
"""
|
|
|
|
data = sorted(data)
|
|
|
|
n = len(data)
|
|
|
|
if n == 0:
|
|
|
|
raise StatisticsError("no median for empty data")
|
|
|
|
elif n == 1:
|
|
|
|
return data[0]
|
|
|
|
# Find the value at the midpoint. Remember this corresponds to the
|
|
|
|
# centre of the class interval.
|
|
|
|
x = data[n//2]
|
|
|
|
for obj in (x, interval):
|
|
|
|
if isinstance(obj, (str, bytes)):
|
|
|
|
raise TypeError('expected number but got %r' % obj)
|
|
|
|
try:
|
|
|
|
L = x - interval/2 # The lower limit of the median interval.
|
|
|
|
except TypeError:
|
|
|
|
# Mixed type. For now we just coerce to float.
|
|
|
|
L = float(x) - float(interval)/2
|
2016-05-04 14:54:29 -03:00
|
|
|
|
|
|
|
# Uses bisection search to search for x in data with log(n) time complexity
|
2016-05-26 03:03:33 -03:00
|
|
|
# Find the position of leftmost occurrence of x in data
|
2016-05-04 14:54:29 -03:00
|
|
|
l1 = _find_lteq(data, x)
|
2016-05-26 03:03:33 -03:00
|
|
|
# Find the position of rightmost occurrence of x in data[l1...len(data)]
|
2016-05-04 14:54:29 -03:00
|
|
|
# Assuming always l1 <= l2
|
|
|
|
l2 = _find_rteq(data, l1, x)
|
|
|
|
cf = l1
|
|
|
|
f = l2 - l1 + 1
|
2013-10-19 15:50:09 -03:00
|
|
|
return L + interval*(n/2 - cf)/f
|
|
|
|
|
|
|
|
|
|
|
|
def mode(data):
|
|
|
|
"""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.
|
|
|
|
"""
|
|
|
|
# Generate a table of sorted (value, frequency) pairs.
|
|
|
|
table = _counts(data)
|
|
|
|
if len(table) == 1:
|
|
|
|
return table[0][0]
|
|
|
|
elif table:
|
|
|
|
raise StatisticsError(
|
|
|
|
'no unique mode; found %d equally common values' % len(table)
|
|
|
|
)
|
|
|
|
else:
|
|
|
|
raise StatisticsError('no mode for empty data')
|
|
|
|
|
|
|
|
|
|
|
|
# === Measures of spread ===
|
|
|
|
|
|
|
|
# See http://mathworld.wolfram.com/Variance.html
|
|
|
|
# http://mathworld.wolfram.com/SampleVariance.html
|
|
|
|
# http://en.wikipedia.org/wiki/Algorithms_for_calculating_variance
|
|
|
|
#
|
|
|
|
# Under no circumstances use the so-called "computational formula for
|
|
|
|
# variance", as that is only suitable for hand calculations with a small
|
|
|
|
# amount of low-precision data. It has terrible numeric properties.
|
|
|
|
#
|
|
|
|
# See a comparison of three computational methods here:
|
|
|
|
# http://www.johndcook.com/blog/2008/09/26/comparing-three-methods-of-computing-standard-deviation/
|
|
|
|
|
|
|
|
def _ss(data, c=None):
|
|
|
|
"""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.
|
|
|
|
"""
|
|
|
|
if c is None:
|
|
|
|
c = mean(data)
|
2015-12-01 04:59:53 -04:00
|
|
|
T, total, count = _sum((x-c)**2 for x in data)
|
2013-10-19 15:50:09 -03:00
|
|
|
# The following sum should mathematically equal zero, but due to rounding
|
|
|
|
# error may not.
|
2015-12-01 04:59:53 -04:00
|
|
|
U, total2, count2 = _sum((x-c) for x in data)
|
|
|
|
assert T == U and count == count2
|
|
|
|
total -= total2**2/len(data)
|
|
|
|
assert not total < 0, 'negative sum of square deviations: %f' % total
|
|
|
|
return (T, total)
|
2013-10-19 15:50:09 -03:00
|
|
|
|
|
|
|
|
|
|
|
def variance(data, xbar=None):
|
|
|
|
"""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)
|
|
|
|
|
|
|
|
"""
|
|
|
|
if iter(data) is data:
|
|
|
|
data = list(data)
|
|
|
|
n = len(data)
|
|
|
|
if n < 2:
|
|
|
|
raise StatisticsError('variance requires at least two data points')
|
2015-12-01 04:59:53 -04:00
|
|
|
T, ss = _ss(data, xbar)
|
|
|
|
return _convert(ss/(n-1), T)
|
2013-10-19 15:50:09 -03:00
|
|
|
|
|
|
|
|
|
|
|
def pvariance(data, mu=None):
|
|
|
|
"""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)
|
|
|
|
|
|
|
|
"""
|
|
|
|
if iter(data) is data:
|
|
|
|
data = list(data)
|
|
|
|
n = len(data)
|
|
|
|
if n < 1:
|
|
|
|
raise StatisticsError('pvariance requires at least one data point')
|
2015-12-01 04:59:53 -04:00
|
|
|
T, ss = _ss(data, mu)
|
|
|
|
return _convert(ss/n, T)
|
2013-10-19 15:50:09 -03:00
|
|
|
|
|
|
|
|
|
|
|
def stdev(data, xbar=None):
|
|
|
|
"""Return 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
|
|
|
|
|
|
|
|
"""
|
|
|
|
var = variance(data, xbar)
|
|
|
|
try:
|
|
|
|
return var.sqrt()
|
|
|
|
except AttributeError:
|
|
|
|
return math.sqrt(var)
|
|
|
|
|
|
|
|
|
|
|
|
def pstdev(data, mu=None):
|
|
|
|
"""Return 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
|
|
|
|
|
|
|
|
"""
|
|
|
|
var = pvariance(data, mu)
|
|
|
|
try:
|
|
|
|
return var.sqrt()
|
|
|
|
except AttributeError:
|
|
|
|
return math.sqrt(var)
|