Source code for music21.common.numberTools

# ------------------------------------------------------------------------------
# Name:         common/numberTools.py
# Purpose:      Utilities for working with numbers or number-like objects
#
# Authors:      Michael Scott Asato Cuthbert
#               Christopher Ariza
#
# Copyright:    Copyright © 2009-2023 Michael Scott Asato Cuthbert
# License:      BSD, see license.txt
# ------------------------------------------------------------------------------
from __future__ import annotations

from fractions import Fraction
from functools import cache
import math
from math import isclose, gcd
import numbers
import random
from typing import overload, TYPE_CHECKING
import unittest

from music21 import defaults
from music21.common.types import OffsetQLIn, OffsetQL

if TYPE_CHECKING:
    from collections.abc import Sequence, Collection


__all__ = [
    'addFloatPrecision',
    'approximateGCD',
    'contiguousList',
    'decimalToTuplet',
    'dotMultiplier',
    'fromRoman',
    'groupContiguousIntegers',
    'mixedNumeral',
    'musicOrdinals',
    'nearestMultiple',
    'numToIntOrFloat',
    'opFrac',
    'ordinalAbbreviation',
    'ordinals',
    'ordinalsToNumbers',
    'roundToHalfInteger',
    'strTrimFloat',
    'toRoman',
    'unitBoundaryProportion',
    'unitNormalizeProportion',
    'weightedSelection',
]

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'


# -----------------------------------------------------------------------------
# Number methods


[docs] def numToIntOrFloat(value: OffsetQLIn) -> 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(2) 2 >>> common.numToIntOrFloat(-5) -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 Can also take in a string representing an int or float >>> common.numToIntOrFloat('1.0') 1 >>> common.numToIntOrFloat('1') 1 >>> common.numToIntOrFloat('1.25') 1.25 Others raise a ValueError >>> common.numToIntOrFloat('one') Traceback (most recent call last): ValueError: could not convert string to float: 'one' Fractions also become ints or floats >>> from fractions import Fraction >>> common.numToIntOrFloat(Fraction(1, 2)) 0.5 >>> common.numToIntOrFloat(Fraction(4, 3)) 1.333333333... Note: Decimal objects are not supported. ''' try: intVal = round(value) except (ValueError, TypeError): value = float(value) intVal = round(value) if isclose(intVal, value, abs_tol=1e-6): return intVal return value + 0.0 # fast opp for cast to float for fractions.Fraction
DENOM_LIMIT = defaults.limitOffsetDenominator @cache def _preFracLimitDenominator(n: int, d: int) -> 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 Nothing changed in 2023, in fact, it's faster now with the cache, and even without the cache, it's still 4x faster. 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) ''' if d <= DENOM_LIMIT: # faster than hard-coding 65535 return (n, d) nOrg = n dOrg = 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_n = abs((bound1n * dOrg) - (nOrg * bound1d)) bound1minusS_d = dOrg * bound1d bound2minusS_n = abs((bound2n * dOrg) - (nOrg * bound2d)) bound2minusS_d = dOrg * bound2d difference = (bound1minusS_n * bound2minusS_d) - (bound2minusS_n * bound1minusS_d) if difference >= 0: # bound1 is farther from zero than bound2; return bound2 return (bound2n, bound2d) else: return (bound1n, bound1d) # _KNOWN_PASSES is all values from whole to 64th notes with 0 or 1 dot # the length of this set does determine the time to search. A set with all values from maxima to # 2048th notes + 1-2 dots was half the speed _KNOWN_PASSES = frozenset([ 0.0625, 0.09375, 0.125, 0.1875, 0.25, 0.375, 0.5, 0.75, 1.0, 1.5, 2.0, 3.0, 4.0, 6.0 ]) @overload def opFrac(num: int) -> float: ... @overload def opFrac(num: float|Fraction) -> float|Fraction: ... # no type checking due to accessing protected attributes (for speed)
[docs] def opFrac(num: OffsetQLIn) -> OffsetQL: ''' opFrac -> optionally convert a number to a fraction or back. Important music21 function for working with offsets and quarterLengths Takes in a number 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(0.99999999842) 1.0 >>> common.opFrac(0.123456789) Fraction(10, 81) >>> common.opFrac(0.000001) 0.0 Please check against None before calling, but None is changed to 0.0 >>> common.opFrac(None) 0.0 * Changed in v9.3: opFrac(None) should not be called. If it is called, it now returns 0.0. ''' # 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. # # Cannot use functools's Caching mechanisms on this because then it will # return the same Fraction object for all calls, which is a problem in case # anyone sets ._numeration or ._denominator directly on that object. # if num in _KNOWN_PASSES: return num + 0.0 # need the add, because ints and Fractions satisfy the "in" # 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() # type: ignore if ir[1] > DENOM_LIMIT: # slightly faster[SIC!] than hard coding 65535! # _preFracLimitDenominator uses a cache f_out = Fraction(*_preFracLimitDenominator(*ir)) # way faster! # now, reduce denominator as Fraction -- just as below under numType == Fraction d = f_out._denominator # type: ignore if (d & (d - 1)) == 0: # power of two # 50% faster than float(num) return f_out._numerator / (d + 0.0) # type: ignore else: return f_out # leave non-power of two fractions alone # 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. # 8x faster than float(num) return num + 0.0 # type: ignore elif numType is Fraction: # private access instead of property: 6x faster; may break later d = num._denominator # type: ignore if (d & (d - 1)) == 0: # power of two # 50% faster than float(num) return num._numerator / (d + 0.0) # type: ignore else: return num # leave non-power of two fractions alone elif num is None: # undocumented -- used to be documented to return None. callers must check. return 0.0 # class inheritance only check AFTER "type is" checks this is redundant but highly optimized. elif isinstance(num, int): return num + 0.0 elif isinstance(num, float): return opFrac(float(num)) # slower for inherited floats, but simpler than duplicating elif isinstance(num, Fraction): d = num.denominator # Use properties since it is a subclass 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}')
[docs] 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), 1.0) 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)
[docs] def roundToHalfInteger(num: float|int) -> 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
[docs] def addFloatPrecision(x, grain=1e-2) -> 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
[docs] 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
[docs] def nearestMultiple(n: float, unit: float) -> 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 ValueError(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)
[docs] 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)
[docs] def decimalToTuplet(decNum: float) -> 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 ValueError('No such luck') jy *= multiplier my_gcd = gcd(int(jy), int(iy)) jy = jy / my_gcd iy = iy / my_gcd if flipNumerator is False: return (int(jy), int(iy)) else: return (int(iy), int(jy))
[docs] def unitNormalizeProportion(values: Sequence[int|float]) -> 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
[docs] def unitBoundaryProportion( series: Sequence[int|float] ) -> list[tuple[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
[docs] def weightedSelection(values: list[int], weights: list[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]
[docs] def approximateGCD(values: Collection[int|float|Fraction], 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: x_adjust = x / lowest floatingValue = x_adjust - int(x_adjust) # 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 = ( 1.0, 2.0, 3.0, 4.0, 5.0, 6.0, 7.0, 8.0, 9.0, 10.0, 11.0, 12.0, 13.0, 14.0, 15.0, 16.0, ) divisions = [] # a list of lists, one for each entry uniqueDivisions = set() for index in values: coll = [] for d in divisors: v = index / d coll.append(v) # store all divisions uniqueDivisions.add(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 ValueError('cannot find a common divisor') return max(commonUniqueDivisions)
[docs] 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
[docs] def groupContiguousIntegers(src: list[int]) -> list[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
[docs] 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: raise ValueError( f'input contains an invalid subtraction element: {num}') except IndexError: # there is no next place. pass places.append(value) summation = 0 for n in places: summation += n return summation
# noinspection SpellCheckingInspection
[docs] def toRoman(num: int) -> str: ''' Convert a number from 1 to 3999 to a roman numeral >>> common.toRoman(2) 'II' >>> common.toRoman(7) 'VII' >>> common.toRoman(1999) 'MCMXCIX' >>> common.toRoman('hi') Traceback (most recent call last): TypeError: expected integer, got <class 'str'> >>> common.toRoman(0) Traceback (most recent call last): ValueError: Argument must be between 1 and 3999 ''' if not isinstance(num, int): raise TypeError(f'expected integer, got {type(num)}') if not 0 < num < 4000: raise ValueError('Argument must be between 1 and 3999') ints = (1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1) nums = ('M', 'CM', 'D', 'CD', 'C', 'XC', 'L', 'XL', 'X', 'IX', 'V', 'IV', 'I') result = '' for i in range(len(ints)): count = int(num / ints[i]) result += nums[i] * count num -= ints[i] * count return result
[docs] def ordinalAbbreviation(value: int, plural=False) -> str: ''' Return the ordinal abbreviations for integers >>> common.ordinalAbbreviation(3) 'rd' >>> common.ordinalAbbreviation(255) 'th' >>> common.ordinalAbbreviation(255, plural=True) 'ths' ''' valueHundredths = value % 100 if valueHundredths in (11, 12, 13): post = 'th' else: valueMod = value % 10 if valueMod == 1: post = 'st' elif valueMod in (0, 4, 5, 6, 7, 8, 9): post = 'th' elif valueMod == 2: post = 'nd' elif valueMod == 3: post = 'rd' else: raise ValueError('Something really weird') # pragma: no cover if post != 'st' and plural: post += 's' return post
ordinalsToNumbers = {} for ordinal_index in range(len(ordinals)): ordinalName = ordinals[ordinal_index] ordinalNameLower = ordinalName.lower() ordinalsToNumbers[ordinalName] = ordinal_index ordinalsToNumbers[ordinalNameLower] = ordinal_index ordinalsToNumbers[str(ordinal_index) + ordinalAbbreviation(ordinal_index)] = ordinal_index musicOrdinalName = musicOrdinals[ordinal_index] if musicOrdinalName != ordinalName: musicOrdinalNameLower = musicOrdinalName.lower() ordinalsToNumbers[musicOrdinalName] = ordinal_index ordinalsToNumbers[musicOrdinalNameLower] = ordinal_index del ordinal_index del ordinalName del ordinalNameLower del musicOrdinalName del musicOrdinalNameLower class Test(unittest.TestCase): ''' Tests not requiring file output. ''' def setUp(self): pass def testToRoman(self): for src, dst in [(1, 'I'), (3, 'III'), (5, 'V')]: self.assertEqual(dst, toRoman(src)) def testOrdinalsToNumbers(self): self.assertEqual(ordinalsToNumbers['unison'], 1) self.assertEqual(ordinalsToNumbers['Unison'], 1) self.assertEqual(ordinalsToNumbers['first'], 1) self.assertEqual(ordinalsToNumbers['First'], 1) self.assertEqual(ordinalsToNumbers['1st'], 1) self.assertEqual(ordinalsToNumbers['octave'], 8) self.assertEqual(ordinalsToNumbers['Octave'], 8) self.assertEqual(ordinalsToNumbers['eighth'], 8) self.assertEqual(ordinalsToNumbers['Eighth'], 8) self.assertEqual(ordinalsToNumbers['8th'], 8) def testWeightedSelection(self): # test equal selection for j in range(10): x = 0 for i in range(1000): # equal chance of -1, 1 x += weightedSelection([-1, 1], [1, 1]) # environLocal.printDebug(['weightedSelection([-1, 1], [1, 1])', x]) self.assertTrue(-250 < x < 250) # test a strongly weighed boundary for j in range(10): x = 0 for i in range(1000): # 10000 more chance of 0 than 1. x += weightedSelection([0, 1], [10000, 1]) # environLocal.printDebug(['weightedSelection([0, 1], [10000, 1])', x]) self.assertTrue(0 <= x < 20) for j in range(10): x = 0 for i in range(1000): # 10,000 times more likely 1 than 0. x += weightedSelection([0, 1], [1, 10000]) # environLocal.printDebug(['weightedSelection([0, 1], [1, 10000])', x]) self.assertTrue(900 <= x <= 1000) for unused_j in range(10): x = 0 for i in range(1000): # no chance of anything but 0. x += weightedSelection([0, 1], [1, 0]) # environLocal.printDebug(['weightedSelection([0, 1], [1, 0])', x]) self.assertEqual(x, 0) # ------------------------------------------------------------------------------ # define presented order in documentation _DOC_ORDER = [fromRoman, toRoman] if __name__ == '__main__': import music21 music21.mainTest(Test)