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1164 lines (956 loc) · 34.9 KB
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# -*- coding: utf-8 -*-
# ------------------------------------------------------------------------------
# Name: common/numberTools.py
# Purpose: Utilities for working with numbers or number-like objects
#
# Authors: Michael Scott Asato Cuthbert
# Christopher Ariza
#
# Copyright: Copyright © 2009-2022 Michael Scott Asato Cuthbert and the music21 Project
# License: BSD, see license.txt
# ------------------------------------------------------------------------------
from functools import lru_cache
import math
import numbers
import random
import unittest
import typing as t
from fractions import Fraction
from math import isclose
from music21 import defaults
from music21.common import deprecated
from music21.common.types import OffsetQLIn, OffsetQL
__all__ = [
'ordinals', 'musicOrdinals', 'ordinalsToNumbers',
'cleanupFloat',
'numToIntOrFloat',
'opFrac', 'mixedNumeral',
'roundToHalfInteger',
'addFloatPrecision', 'strTrimFloat',
'nearestMultiple',
'dotMultiplier', 'decimalToTuplet',
'unitNormalizeProportion', 'unitBoundaryProportion',
'weightedSelection',
'euclidGCD', 'approximateGCD',
'lcm',
'contiguousList',
'groupContiguousIntegers',
'fromRoman', 'toRoman',
'ordinalAbbreviation',
]
ordinals = [
'Zeroth', 'First', 'Second', 'Third', 'Fourth', 'Fifth',
'Sixth', 'Seventh', 'Eighth', 'Ninth', 'Tenth', 'Eleventh',
'Twelfth', 'Thirteenth', 'Fourteenth', 'Fifteenth',
'Sixteenth', 'Seventeenth', 'Eighteenth', 'Nineteenth',
'Twentieth', 'Twenty-first', 'Twenty-second',
]
musicOrdinals = ordinals[:]
musicOrdinals[1] = 'Unison'
musicOrdinals[8] = 'Octave'
musicOrdinals[15] = 'Double-octave'
musicOrdinals[22] = 'Triple-octave'
@deprecated('v7.3', 'v9', 'Use common.opFrac(num) instead')
def cleanupFloat(floatNum, maxDenominator=defaults.limitOffsetDenominator): # pragma: no cover
'''
Cleans up a floating point number by converting
it to a fractions.Fraction object limited to
a denominator of maxDenominator
common.cleanupFloat(0.33333327824)
0.333333333333...
common.cleanupFloat(0.142857)
0.1428571428571...
common.cleanupFloat(1.5)
1.5
Fractions are passed through silently...
import fractions
common.cleanupFloat(fractions.Fraction(4, 3))
Fraction(4, 3)
'''
if isinstance(floatNum, Fraction):
return floatNum # do nothing to fractions
else:
f = Fraction(floatNum).limit_denominator(maxDenominator)
return float(f)
# -----------------------------------------------------------------------------
# Number methods...
def numToIntOrFloat(value: t.Union[int, float]) -> t.Union[int, float]:
'''
Given a number, return an integer if it is very close to an integer,
otherwise, return a float.
This routine is very important for conversion of
:class:`~music21.pitch.Accidental` objects' `.alter` attribute
in musicXML must be 1 (not 1.0) for sharp and -1 (not -1.0) for flat,
but allows for 0.5 for half-sharp.
>>> common.numToIntOrFloat(1.0)
1
>>> common.numToIntOrFloat(1.00003)
1.00003
>>> common.numToIntOrFloat(1.5)
1.5
>>> common.numToIntOrFloat(1.0000000005)
1
>>> common.numToIntOrFloat(0.999999999)
1
>>> sharp = pitch.Accidental('sharp')
>>> common.numToIntOrFloat(sharp.alter)
1
>>> halfFlat = pitch.Accidental('half-flat')
>>> common.numToIntOrFloat(halfFlat.alter)
-0.5
Also can take in a string representing an int or float
>>> common.numToIntOrFloat('1.0')
1
>>> common.numToIntOrFloat('1')
1
>>> common.numToIntOrFloat('1.25')
1.25
'''
try:
intVal = round(value)
except (ValueError, TypeError):
value = float(value)
intVal = round(value)
if isclose(intVal, value, abs_tol=1e-6):
return intVal
else: # source
return value
DENOM_LIMIT = defaults.limitOffsetDenominator
@lru_cache(1024)
def _preFracLimitDenominator(n: int, d: int) -> t.Tuple[int, int]:
# noinspection PyShadowingNames
'''
Used in opFrac
Copied from fractions.limit_denominator. Their method
requires creating three new Fraction instances to get one back.
This doesn't create any call before Fraction...
DENOM_LIMIT is hardcoded to defaults.limitOffsetDenominator for speed...
returns a new n, d...
>>> common.numberTools._preFracLimitDenominator(100001, 300001)
(1, 3)
>>> from fractions import Fraction
>>> Fraction(100_000_000_001, 30_0000_000_001).limit_denominator(65535)
Fraction(1, 3)
>>> Fraction(100001, 300001).limit_denominator(65535)
Fraction(1, 3)
Timing differences are huge!
t is timeit.timeit
t('Fraction(*common.numberTools._preFracLimitDenominator(*x.as_integer_ratio()))',
setup='x = 1000001/3000001.; from music21 import common;from fractions import Fraction',
number=100000)
1.0814228057861328
t('Fraction(x).limit_denominator(65535)',
setup='x = 1000001/3000001.; from fractions import Fraction',
number=100000)
7.941488981246948
Proof of working...
>>> import random
>>> myWay = lambda x: Fraction(*common.numberTools._preFracLimitDenominator(
... *x.as_integer_ratio()))
>>> theirWay = lambda x: Fraction(x).limit_denominator(65535)
>>> for _ in range(50):
... x = random.random()
... if myWay(x) != theirWay(x):
... print(f'boo: {x}, {myWay(x)}, {theirWay(x)}')
(n.b. -- nothing printed)
'''
nOrg = n
dOrg = d
if d <= DENOM_LIMIT: # faster than hard-coding 65535
return (n, d)
p0, q0, p1, q1 = 0, 1, 1, 0
while True:
a = n // d
q2 = q0 + a * q1
if q2 > DENOM_LIMIT:
break
p0, q0, p1, q1 = p1, q1, p0 + a * p1, q2
n, d = d, n - a * d
k = (DENOM_LIMIT - q0) // q1
bound1n = p0 + k * p1
bound1d = q0 + k * q1
bound2n = p1
bound2d = q1
# s = (0.0 + n)/d
bound1minusS = (abs((bound1n * dOrg) - (nOrg * bound1d)), (dOrg * bound1d))
bound2minusS = (abs((bound2n * dOrg) - (nOrg * bound2d)), (dOrg * bound2d))
difference = (bound1minusS[0] * bound2minusS[1]) - (bound2minusS[0] * bound1minusS[1])
if difference > 0:
# bound1 is farther from zero than bound2; return bound2
return (p1, q1)
else:
return (p0 + k * p1, q0 + k * q1)
# no type checking due to accessing protected attributes (for speed)
@t.no_type_check
def opFrac(num: t.Union[OffsetQLIn, None]) -> t.Union[OffsetQL, None]:
'''
opFrac -> optionally convert a number to a fraction or back.
Important music21 function for working with offsets and quarterLengths
Takes in a number (or None) and converts it to a Fraction with denominator
less than limitDenominator if it is not binary expressible; otherwise return a float.
Or if the Fraction can be converted back to a binary expressible
float then do so.
This function should be called often to ensure that values being passed around are floats
and ints wherever possible and fractions where needed.
The naming of this method violates music21's general rule of no abbreviations, but it
is important to make it short enough so that no one will be afraid of calling it often.
It also doesn't have a setting for maxDenominator so that it will expand in
Code Completion easily. That is to say, this function has been set up to be used, so please
use it.
This is a performance critical operation. Do not alter it in any way without running
many timing tests.
>>> from fractions import Fraction
>>> defaults.limitOffsetDenominator
65535
>>> common.opFrac(3)
3.0
>>> common.opFrac(1/3)
Fraction(1, 3)
>>> common.opFrac(1/4)
0.25
>>> f = Fraction(1, 3)
>>> common.opFrac(f + f + f)
1.0
>>> common.opFrac(0.123456789)
Fraction(10, 81)
>>> common.opFrac(None) is None
True
'''
# This is a performance critical operation, tuned to go as fast as possible.
# hence redundancy -- first we check for type (no inheritance) and then we
# repeat exact same test with inheritance.
# Note that the later examples are more verbose
numType = type(num)
if numType is float:
# quick test of power of whether denominator is a power
# of two, and thus representable exactly as a float: can it be
# represented w/ a denominator less than DENOM_LIMIT?
# this doesn't work:
# (denominator & (denominator-1)) != 0
# which is a nice test, but denominator here is always a power of two...
# unused_numerator, denominator = num.as_integer_ratio() # too slow
ir = num.as_integer_ratio()
if ir[1] > DENOM_LIMIT: # slightly faster[SIC!] than hard coding 65535!
return Fraction(*_preFracLimitDenominator(*ir)) # way faster!
# return Fraction(*ir).limit_denominator(DENOM_LIMIT) # *ir instead of float--can happen
# internally in Fraction constructor, but is twice as fast...
else:
return num
elif numType is int: # if vs. elif is negligible time difference.
return num + 0.0 # 8x faster than float(num)
elif numType is Fraction:
d = num._denominator # private access instead of property: 6x faster; may break later...
if (d & (d - 1)) == 0: # power of two...
return num._numerator / (d + 0.0) # 50% faster than float(num)
else:
return num # leave non-power of two fractions alone
elif num is None:
return None
# class inheritance only check AFTER ifs... this is redundant but highly optimized.
elif isinstance(num, int):
return num + 0.0
elif isinstance(num, float):
ir = num.as_integer_ratio()
if ir[1] > DENOM_LIMIT: # slightly faster than hard coding 65535!
return Fraction(*_preFracLimitDenominator(*ir)) # way faster!
else:
return num
elif isinstance(num, Fraction):
d = num._denominator # private access instead of property: 6x faster; may break later...
if (d & (d - 1)) == 0: # power of two...
return num._numerator / (d + 0.0) # 50% faster than float(num)
else:
return num # leave fraction alone
else:
raise TypeError(f'Cannot convert num: {num}')
def mixedNumeral(expr: numbers.Real,
limitDenominator=defaults.limitOffsetDenominator):
'''
Returns a string representing a mixedNumeral form of a number
>>> common.mixedNumeral(1.333333)
'1 1/3'
>>> common.mixedNumeral(0.333333)
'1/3'
>>> common.mixedNumeral(-1.333333)
'-1 1/3'
>>> common.mixedNumeral(-0.333333)
'-1/3'
>>> common.mixedNumeral(0)
'0'
>>> common.mixedNumeral(-0)
'0'
Works with Fraction objects too
>>> from fractions import Fraction
>>> common.mixedNumeral(Fraction(31, 7))
'4 3/7'
>>> common.mixedNumeral(Fraction(1, 5))
'1/5'
>>> common.mixedNumeral(Fraction(-1, 5))
'-1/5'
>>> common.mixedNumeral(Fraction(-4, 5))
'-4/5'
>>> common.mixedNumeral(Fraction(-31, 7))
'-4 3/7'
Denominator is limited by default but can be changed.
>>> common.mixedNumeral(2.0000001)
'2'
>>> common.mixedNumeral(2.0000001, limitDenominator=10000000)
'2 1/10000000'
'''
if not isinstance(expr, Fraction):
quotient, remainder = divmod(float(expr), 1.)
remainderFrac = Fraction(remainder).limit_denominator(limitDenominator)
if quotient < -1:
quotient += 1
remainderFrac = 1 - remainderFrac
elif quotient == -1:
quotient = 0.0
remainderFrac = remainderFrac - 1
else:
quotient = int(float(expr))
remainderFrac = expr - quotient
if quotient < 0:
remainderFrac *= -1
if quotient:
if remainderFrac:
return f'{int(quotient)} {remainderFrac}'
else:
return str(int(quotient))
else:
if remainderFrac != 0:
return str(remainderFrac)
return str(0)
def roundToHalfInteger(num: t.Union[float, int]) -> t.Union[float, int]:
'''
Given a floating-point number, round to the nearest half-integer. Returns int or float
>>> common.roundToHalfInteger(1.2)
1
>>> common.roundToHalfInteger(1.35)
1.5
>>> common.roundToHalfInteger(1.8)
2
>>> common.roundToHalfInteger(1.6234)
1.5
0.25 rounds up:
>>> common.roundToHalfInteger(0.25)
0.5
as does 0.75
>>> common.roundToHalfInteger(0.75)
1
unlike python round function, does the same for 1.25 and 1.75
>>> common.roundToHalfInteger(1.25)
1.5
>>> common.roundToHalfInteger(1.75)
2
negative numbers however, round up on the boundaries
>>> common.roundToHalfInteger(-0.26)
-0.5
>>> common.roundToHalfInteger(-0.25)
0
'''
intVal, floatVal = divmod(num, 1.0)
intVal = int(intVal)
if floatVal < 0.25:
floatVal = 0
elif 0.25 <= floatVal < 0.75:
floatVal = 0.5
else:
floatVal = 1
return intVal + floatVal
def addFloatPrecision(x, grain=1e-2) -> t.Union[float, Fraction]:
'''
Given a value that suggests a floating point fraction, like 0.33,
return a Fraction or float that provides greater specification, such as Fraction(1, 3)
>>> import fractions
>>> common.addFloatPrecision(0.333)
Fraction(1, 3)
>>> common.addFloatPrecision(0.33)
Fraction(1, 3)
>>> common.addFloatPrecision(0.35) == fractions.Fraction(1, 3)
False
>>> common.addFloatPrecision(0.2) == 0.2
True
>>> common.addFloatPrecision(0.125)
0.125
>>> common.addFloatPrecision(1/7) == 1/7
True
'''
if isinstance(x, str):
x = float(x)
values = (1 / 3, 2 / 3,
1 / 6, 5 / 6)
for v in values:
if isclose(x, v, abs_tol=grain):
return opFrac(v)
return x
def strTrimFloat(floatNum: float, maxNum: int = 4) -> str:
'''
returns a string from a float that is at most maxNum of
decimal digits long, but never less than 1.
>>> common.strTrimFloat(42.3333333333)
'42.3333'
>>> common.strTrimFloat(42.3333333333, 2)
'42.33'
>>> common.strTrimFloat(6.66666666666666, 2)
'6.67'
>>> common.strTrimFloat(2.0)
'2.0'
>>> common.strTrimFloat(-5)
'-5.0'
'''
# variables called 'off' because originally designed for offsets
offBuildString = r'%.' + str(maxNum) + 'f'
off = offBuildString % floatNum
offDecimal = off.index('.')
offLen = len(off)
for index in range(offLen - 1, offDecimal + 1, -1):
if off[index] != '0':
break
else:
offLen = offLen - 1
off = off[0:offLen]
return off
def nearestMultiple(n: float, unit: float) -> t.Tuple[float, float, float]:
'''
Given a positive value `n`, return the nearest multiple of the supplied `unit` as well as
the absolute difference (error) to seven significant digits and the signed difference.
>>> print(common.nearestMultiple(0.25, 0.25))
(0.25, 0.0, 0.0)
>>> print(common.nearestMultiple(0.35, 0.25))
(0.25, 0.1..., 0.1...)
>>> print(common.nearestMultiple(0.20, 0.25))
(0.25, 0.05..., -0.05...)
Note that this one also has an error of 0.1, but it's a positive error off of 0.5
>>> print(common.nearestMultiple(0.4, 0.25))
(0.5, 0.1..., -0.1...)
>>> common.nearestMultiple(0.4, 0.25)[0]
0.5
>>> common.nearestMultiple(23404.001, 0.125)[0]
23404.0
>>> common.nearestMultiple(23404.134, 0.125)[0]
23404.125
Error is always positive, but signed difference can be negative.
>>> common.nearestMultiple(23404 - 0.0625, 0.125)
(23403.875, 0.0625, 0.0625)
>>> common.nearestMultiple(0.001, 0.125)[0]
0.0
>>> from math import isclose
>>> isclose(common.nearestMultiple(0.25, 1 / 3)[0], 0.33333333, abs_tol=1e-7)
True
>>> isclose(common.nearestMultiple(0.55, 1 / 3)[0], 0.66666666, abs_tol=1e-7)
True
>>> isclose(common.nearestMultiple(234.69, 1 / 3)[0], 234.6666666, abs_tol=1e-7)
True
>>> isclose(common.nearestMultiple(18.123, 1 / 6)[0], 18.16666666, abs_tol=1e-7)
True
>>> common.nearestMultiple(-0.5, 0.125)
Traceback (most recent call last):
ValueError: n (-0.5) is less than zero. Thus cannot find the nearest
multiple for a value less than the unit, 0.125
'''
if n < 0:
raise ValueError(f'n ({n}) is less than zero. '
+ 'Thus cannot find the nearest multiple for a value '
+ f'less than the unit, {unit}')
mult = math.floor(n / unit) # can start with the floor
halfUnit = unit / 2.0
matchLow = unit * mult
matchHigh = unit * (mult + 1)
# print(['mult, halfUnit, matchLow, matchHigh', mult, halfUnit, matchLow, matchHigh])
if matchLow >= n >= matchHigh:
raise Exception(f'cannot place n between multiples: {matchLow}, {matchHigh}')
if matchLow <= n <= (matchLow + halfUnit):
return matchLow, round(n - matchLow, 7), round(n - matchLow, 7)
else:
# elif n >= (matchHigh - halfUnit) and n <= matchHigh:
return matchHigh, round(matchHigh - n, 7), round(n - matchHigh, 7)
_DOT_LOOKUP = (1.0, 1.5, 1.75, 1.875, 1.9375,
1.96875, 1.984375, 1.9921875, 1.99609375)
def dotMultiplier(dots: int) -> float:
'''
dotMultiplier(dots) returns how long to multiply the note
length of a note in order to get the note length with n dots
Since dotMultiplier always returns a power of two in the denominator,
the float will be exact.
>>> from fractions import Fraction
>>> Fraction(common.dotMultiplier(1))
Fraction(3, 2)
>>> Fraction(common.dotMultiplier(2))
Fraction(7, 4)
>>> Fraction(common.dotMultiplier(3))
Fraction(15, 8)
>>> common.dotMultiplier(0)
1.0
'''
if dots < 9:
return _DOT_LOOKUP[dots]
return ((2 ** (dots + 1.0)) - 1.0) / (2 ** dots)
def decimalToTuplet(decNum: float) -> t.Tuple[int, int]:
'''
For simple decimals (usually > 1), a quick way to figure out the
fraction in lowest terms that gives a valid tuplet.
No it does not work really fast. No it does not return tuplets with
denominators over 100. Too bad, math geeks. This is real life. :-)
returns (numerator, denominator)
>>> common.decimalToTuplet(1.5)
(3, 2)
>>> common.decimalToTuplet(1.25)
(5, 4)
If decNum is < 1, the denominator will be greater than the numerator:
>>> common.decimalToTuplet(0.8)
(4, 5)
If decNum is <= 0, returns a ZeroDivisionError:
>>> common.decimalToTuplet(-.02)
Traceback (most recent call last):
ZeroDivisionError: number must be greater than zero
'''
def findSimpleFraction(inner_working):
'Utility function.'
for index in range(1, 1000):
for j in range(index, index * 2):
if isclose(inner_working, j / index, abs_tol=1e-7):
return (int(j), int(index))
return (0, 0)
flipNumerator = False
if decNum <= 0:
raise ZeroDivisionError('number must be greater than zero')
if decNum < 1:
flipNumerator = True
decNum = 1 / decNum
unused_remainder, multiplier = math.modf(decNum)
working = decNum / multiplier
(jy, iy) = findSimpleFraction(working)
if iy == 0:
raise Exception('No such luck')
jy *= multiplier
gcd = euclidGCD(int(jy), int(iy))
jy = jy / gcd
iy = iy / gcd
if flipNumerator is False:
return (int(jy), int(iy))
else:
return (int(iy), int(jy))
def unitNormalizeProportion(values: t.Sequence[t.Union[int, float]]) -> t.List[float]:
'''
Normalize values within the unit interval, where max is determined by the sum of the series.
>>> common.unitNormalizeProportion([0, 3, 4])
[0.0, 0.42857142857142855, 0.5714285714285714]
>>> common.unitNormalizeProportion([1, 1, 1])
[0.3333333..., 0.333333..., 0.333333...]
Works fine with a mix of ints and floats:
>>> common.unitNormalizeProportion([1.0, 1, 1.0])
[0.3333333..., 0.333333..., 0.333333...]
On 32-bit computers this number may be inexact even for small floats.
On 64-bit it works fine. This is the 32-bit output for this result.
common.unitNormalizeProportion([0.2, 0.6, 0.2])
[0.20000000000000001, 0.59999999999999998, 0.20000000000000001]
Negative values should be shifted to positive region first:
>>> common.unitNormalizeProportion([0, -2, -8])
Traceback (most recent call last):
ValueError: value members must be positive
'''
summation = 0.0
for x in values:
if x < 0.0:
raise ValueError('value members must be positive')
summation += x
unit = [] # weights on the unit interval; sum == 1
for x in values:
unit.append((x / summation))
return unit
def unitBoundaryProportion(
series: t.Sequence[t.Union[int, float]]
) -> t.List[t.Tuple[t.Union[int, float], float]]:
'''
Take a series of parts with an implied sum, and create
unit-interval boundaries proportional to the series components.
>>> common.unitBoundaryProportion([1, 1, 2])
[(0.0, 0.25), (0.25, 0.5), (0.5, 1.0)]
>>> common.unitBoundaryProportion([9, 1, 1])
[(0.0, 0.8...), (0.8..., 0.9...), (0.9..., 1.0)]
'''
unit = unitNormalizeProportion(series)
bounds = []
summation = 0.0
for index in range(len(unit)):
if index != len(unit) - 1: # not last
bounds.append((summation, summation + unit[index]))
summation += unit[index]
else: # last, avoid rounding errors
bounds.append((summation, 1.0))
return bounds
def weightedSelection(values: t.List[int],
weights: t.List[t.Union[int, float]],
randomGenerator=None) -> int:
'''
Given a list of values and an equal-sized list of weights,
return a randomly selected value using the weight.
Example: sum -1 and 1 for 100 values; should be
around 0 or at least between -50 and 50 (99.99999% of the time)
>>> -50 < sum([common.weightedSelection([-1, 1], [1, 1]) for x in range(100)]) < 50
True
'''
# See http://www.wolframalpha.com/input/?i=Probability+of+76+or+more+heads+in+100+coin+tosses
# for probability. When it was -30 to 30, failed 1 in 500 times.
if randomGenerator is not None:
q = randomGenerator() # must be in unit interval
else: # use random uniform
q = random.random()
# normalize weights w/n unit interval
boundaries = unitBoundaryProportion(weights)
index = 0
for index, (low, high) in enumerate(boundaries):
if low <= q < high: # accepts both boundaries
return values[index]
# just in case we get the high boundary
return values[index]
def euclidGCD(a: int, b: int) -> int:
'''
use Euclid's algorithm to find the GCD of a and b
>>> common.euclidGCD(2, 4)
2
>>> common.euclidGCD(20, 8)
4
>>> common.euclidGCD(20, 16)
4
'''
if b == 0:
return a
else:
return euclidGCD(b, a % b)
def approximateGCD(values: t.List[t.Union[int, float]], grain: float = 1e-4) -> float:
'''Given a list of values, find the lowest common divisor of floating point values.
>>> common.approximateGCD([2.5, 10, 0.25])
0.25
>>> common.approximateGCD([2.5, 10])
2.5
>>> common.approximateGCD([2, 10])
2.0
>>> common.approximateGCD([1.5, 5, 2, 7])
0.5
>>> common.approximateGCD([2, 5, 10])
1.0
>>> common.approximateGCD([2, 5, 10, 0.25])
0.25
>>> common.strTrimFloat(common.approximateGCD([1/3, 2/3]))
'0.3333'
>>> common.strTrimFloat(common.approximateGCD([5/3, 2/3, 4]))
'0.3333'
>>> common.strTrimFloat(common.approximateGCD([5/3, 2/3, 5]))
'0.3333'
>>> common.strTrimFloat(common.approximateGCD([5/3, 2/3, 5/6, 3/6]))
'0.1667'
'''
lowest = float(min(values))
# quick method: see if the smallest value is a common divisor of the rest
count = 0
for x in values:
# lowest is already a float
unused_int, floatingValue = divmod(x / lowest, 1.0)
# if almost an even division
if isclose(floatingValue, 0.0, abs_tol=grain):
count += 1
if count == len(values):
return lowest
# assume that one of these divisions will match
divisors = [1., 2., 3., 4., 5., 6., 7., 8., 9., 10., 11., 12., 13., 14., 15., 16.]
divisions = [] # a list of lists, one for each entry
uniqueDivisions = []
for index in values:
coll = []
for d in divisors:
v = index / d
coll.append(v) # store all divisions
if v not in uniqueDivisions:
uniqueDivisions.append(v)
divisions.append(coll)
# find a unique divisor that is found in collected divisors
commonUniqueDivisions = []
for v in uniqueDivisions:
count = 0
for coll in divisions:
for x in coll:
# grain here is set low, mostly to catch triplets
if isclose(x, v, abs_tol=grain):
count += 1
break # exit the iteration of coll; only 1 match possible
# store any division that is found in all values
if count == len(divisions):
commonUniqueDivisions.append(v)
if not commonUniqueDivisions:
raise Exception('cannot find a common divisor')
return max(commonUniqueDivisions)
def lcm(filterList: t.Iterable[int]) -> int:
'''
Find the least common multiple of a list of values
>>> common.lcm([3, 4, 5])
60
>>> common.lcm([3, 4])
12
>>> common.lcm([1, 2])
2
>>> common.lcm([3, 6])
6
Works with any iterable, like this set
>>> common.lcm({3, 5, 6})
30
'''
def _lcm(a, b):
'''find the least common multiple of a, b'''
# // forces integer style division (no remainder)
return abs(a * b) // euclidGCD(a, b)
# derived from
# http://www.oreillynet.com/cs/user/view/cs_msg/41022
lcmVal = 1
for flValue in filterList:
lcmVal = _lcm(lcmVal, flValue)
return lcmVal
def contiguousList(inputListOrTuple) -> bool:
'''
returns bool True or False if a list containing ints
contains only contiguous (increasing) values
requires the list to be sorted first
>>> l = [3, 4, 5, 6]
>>> common.contiguousList(l)
True
>>> l.append(8)
>>> common.contiguousList(l)
False
Sorting matters
>>> l.append(7)
>>> common.contiguousList(l)
False
>>> common.contiguousList(sorted(l))
True
'''
currentMaxVal = inputListOrTuple[0]
for index in range(1, len(inputListOrTuple)):
newVal = inputListOrTuple[index]
if newVal != currentMaxVal + 1:
return False
currentMaxVal += 1
return True
def groupContiguousIntegers(src: t.List[int]) -> t.List[t.List[int]]:
'''
Given a list of integers, group contiguous values into sub lists
>>> common.groupContiguousIntegers([3, 5, 6])
[[3], [5, 6]]
>>> common.groupContiguousIntegers([3, 4, 6])
[[3, 4], [6]]
>>> common.groupContiguousIntegers([3, 4, 6, 7])
[[3, 4], [6, 7]]
>>> common.groupContiguousIntegers([3, 4, 6, 7, 20])
[[3, 4], [6, 7], [20]]
>>> common.groupContiguousIntegers([3, 4, 5, 6, 7])
[[3, 4, 5, 6, 7]]
>>> common.groupContiguousIntegers([3])
[[3]]
>>> common.groupContiguousIntegers([3, 200])
[[3], [200]]
'''
if len(src) <= 1:
return [src]
post = []
group = []
src.sort()
i = 0
while i < (len(src) - 1):
e = src[i]
group.append(e)
eNext = src[i + 1]
# if next is contiguous, add to group
if eNext != e + 1:
# if not contiguous
post.append(group)
group = []
# second to last elements; handle separately
if i == len(src) - 2:
# need to handle next elements
group.append(eNext)
post.append(group)
i += 1
return post
# noinspection SpellCheckingInspection
def fromRoman(num: str, *, strictModern=False) -> int:
'''
Convert a Roman numeral (upper or lower) to an int
https://code.activestate.com/recipes/81611-roman-numerals/
>>> common.fromRoman('ii')
2
>>> common.fromRoman('vii')
7
Works with both IIII and IV forms:
>>> common.fromRoman('MCCCCLXXXIX')
1489
>>> common.fromRoman('MCDLXXXIX')
1489
Some people consider this an error, but you see it in medieval and ancient roman documents:
>>> common.fromRoman('ic')
99
unless strictModern is True
>>> common.fromRoman('ic', strictModern=True)
Traceback (most recent call last):
ValueError: input contains an invalid subtraction element (modern interpretation): ic
But things like this are never seen, and thus cause an error:
>>> common.fromRoman('vx')
Traceback (most recent call last):
ValueError: input contains an invalid subtraction element: vx
'''
inputRoman = num.upper()
subtractionValues = (1, 10, 100)
nums = ('M', 'D', 'C', 'L', 'X', 'V', 'I')
ints = (1000, 500, 100, 50, 10, 5, 1)
places = []
for c in inputRoman:
if c not in nums:
raise ValueError(f'value is not a valid roman numeral: {inputRoman}')
for i in range(len(inputRoman)):
c = inputRoman[i]
value = ints[nums.index(c)]
# If the next place holds a larger number, this value is negative.
try:
nextValue = ints[nums.index(inputRoman[i + 1])]
if nextValue > value and value in subtractionValues:
if strictModern and nextValue >= value * 10:
raise ValueError(
'input contains an invalid subtraction element '
+ f'(modern interpretation): {num}')
value *= -1
elif nextValue > value: