def test_float_handling():
def test(e1, e2):
return len(e1.atoms(Float)) == len(e2.atoms(Float))
assert solve(x - 0.5, rational=True)[0].is_Rational
assert solve(x - 0.5, rational=False)[0].is_Float
assert solve(x - S.Half, rational=False)[0].is_Rational
assert solve(x - 0.5, rational=None)[0].is_Float
assert solve(x - S.Half, rational=None)[0].is_Rational
assert test(nfloat(1 + 2 * x), 1.0 + 2.0 * x)
for contain in [list, tuple, set]:
ans = nfloat(contain([1 + 2 * x]))
assert type(ans) is contain and test(list(ans)[0], 1.0 + 2.0 * x)
k, v = nfloat({2 * x: [1 + 2 * x]}).items()[0]
assert test(k, 2 * x) and test(v[0], 1.0 + 2.0 * x)
assert test(nfloat(cos(2 * x)), cos(2.0 * x))
assert test(nfloat(3 * x**2), 3.0 * x**2)
assert test(nfloat(3 * x**2, exponent=True), 3.0 * x**2.0)
assert test(nfloat(exp(2 * x)), exp(2.0 * x))
assert test(nfloat(x / 3), x / 3.0)
assert test(nfloat(x**4 + 2 * x + cos(S(1) / 3) + 1),
x**4 + 2.0 * x + 1.94495694631474)
# don't call nfloat if there is no solution
tot = 100 + c + z + t
assert solve(
((.7 + c) / tot - .6, (.2 + z) / tot - .3, t / tot - .1)) == []
def gaussmfAreaU(x, params1, params2, prevInter):
assert len(params1) == 2, 'sigma,c parameter must have exactly two \
elements.'
assert len(params2) == 2, 'sigma,c parameter must have exactly two \
elements.'
m1, s1 = np.r_[params1] # Zero-indexing in Python
m2, s2 = np.r_[params2] # Zero-indexing in Python
# Solving coeffiecients of quadratic poly
def solve(m1, m2, std1, std2):
a = 1 / (2 * std1**2) - 1 / (2 * std2**2)
b = m2 / (std2**2) - m1 / (std1**2)
c = m1**2 / (2 * std1**2) - m2**2 / (2 * std2**2) - np.log(
std2 / std1)
return np.roots([a, b, c])
flag = 2
result = solve(m1, m2, s1, s2)
if len(result) == 0: # Completely non-overlapping
area = 0.0
return area
if len(result) > 0: # One point of contact
for i in range(len(result)):
#print("All Points of intersection")
#print(result)
if result[i] < 0:
result[i] = 10000
flag = 0
if np.min(result) > prevInter:
r = np.min(result)
else:
r = np.max(result)
if flag == 0:
r = np.min(result)
#print("POINT OF INTERSECTION")
#print(r)
x = sympy.var("x")
func1 = sympy.exp(-(((x - params1[0]) / params1[1])**2) / 2)
func2 = sympy.exp(-(((x - params2[0]) / params2[1])**2) / 2)
int1 = sympy.integrate(func1, (x, r, 10))
int2 = sympy.integrate(func2, (x, 0, r))
area1 = nfloat(int1)
area2 = nfloat(int2)
areaTot = round(area1, 3) + round(area2, 3)
return round(areaTot, 3), r
def test_evalf_relational():
assert Eq(x / 5, y / 10).evalf() == Eq(0.2 * x, 0.1 * y)
# if this first assertion fails it should be replaced with
# one that doesn't
assert unchanged(Eq, (3 - I)**2 / 2 + I, 0)
assert Eq((3 - I)**2 / 2 + I, 0).n() is S.false
assert nfloat(Eq((3 - I)**2 + I, 0)) == S.false
def test_float_handling():
def test(e1, e2):
return len(e1.atoms(Float)) == len(e2.atoms(Float))
assert solve(x - 0.5, rational=True)[0].is_Rational
assert solve(x - 0.5, rational=False)[0].is_Float
assert solve(x - S.Half, rational=False)[0].is_Rational
assert solve(x - 0.5, rational=None)[0].is_Float
assert solve(x - S.Half, rational=None)[0].is_Rational
assert test(nfloat(1 + 2*x), 1.0 + 2.0*x)
for contain in [list, tuple, set]:
ans = nfloat(contain([1 + 2*x]))
assert type(ans) is contain and test(list(ans)[0], 1.0 + 2.0*x)
k, v = nfloat({2*x: [1 + 2*x]}).items()[0]
assert test(k, 2*x) and test(v[0], 1.0 + 2.0*x)
assert test(nfloat(cos(2*x)), cos(2.0*x))
assert test(nfloat(3*x**2), 3.0*x**2)
assert test(nfloat(3*x**2, exponent=True), 3.0*x**2.0)
assert test(nfloat(exp(2*x)), exp(2.0*x))
assert test(nfloat(x/3), x/3.0)
assert test(nfloat(x**4 + 2*x + cos(S(1)/3) + 1),
x**4 + 2.0*x + 1.94495694631474)
def gaussmfAreaC(x, params1, params2, pUnA, nUnA):
assert len(params1) == 2, 'sigma,c parameter must have exactly two \
elements.'
assert len(params2) == 2, 'sigma,c parameter must have exactly two \
elements.'
m1, s1 = np.r_[params1] # Zero-indexing in Python
m2, s2 = np.r_[params2] # Zero-indexing in Python
x = sympy.var("x")
func1 = sympy.exp(-(((x - params1[0]) / params1[1])**2) / 2)
int1 = sympy.integrate(func1, (x, 0, 10))
area1 = nfloat(int1)
areaTot = round(area1, 3) - pUnA - nUnA
return round(areaTot, 3)
def test_return_root_of():
f = x ** 5 - 15 * x ** 3 - 5 * x ** 2 + 10 * x + 20
s = list(solveset_complex(f, x))
for root in s:
assert root.func == RootOf
# if one uses solve to get the roots of a polynomial that has a RootOf
# solution, make sure that the use of nfloat during the solve process
# doesn't fail. Note: if you want numerical solutions to a polynomial
# it is *much* faster to use nroots to get them than to solve the
# equation only to get RootOf solutions which are then numerically
# evaluated. So for eq = x**5 + 3*x + 7 do Poly(eq).nroots() rather
# than [i.n() for i in solve(eq)] to get the numerical roots of eq.
assert (
nfloat(list(solveset_complex(x ** 5 + 3 * x ** 3 + 7, x))[0], exponent=False)
== RootOf(x ** 5 + 3 * x ** 3 + 7, 0).n()
)
sol = list(solveset_complex(x ** 6 - 2 * x + 2, x))
assert all(isinstance(i, RootOf) for i in sol) and len(sol) == 6
f = x ** 5 - 15 * x ** 3 - 5 * x ** 2 + 10 * x + 20
s = list(solveset_complex(f, x))
for root in s:
assert root.func == RootOf
s = x ** 5 + 4 * x ** 3 + 3 * x ** 2 + S(7) / 4
assert solveset_complex(s, x) == FiniteSet(*Poly(s * 4, domain="ZZ").all_roots())
# XXX: this comparison should work without converting the FiniteSet to list
# See #7876
eq = x * (x - 1) ** 2 * (x + 1) * (x ** 6 - x + 1)
assert list(solveset_complex(eq, x)) == list(
FiniteSet(
-1,
0,
1,
RootOf(x ** 6 - x + 1, 0),
RootOf(x ** 6 - x + 1, 1),
RootOf(x ** 6 - x + 1, 2),
RootOf(x ** 6 - x + 1, 3),
RootOf(x ** 6 - x + 1, 4),
RootOf(x ** 6 - x + 1, 5),
)
)
def test_return_root_of():
f = x**5 - 15 * x**3 - 5 * x**2 + 10 * x + 20
s = list(solveset_complex(f, x))
for root in s:
assert root.func == CRootOf
# if one uses solve to get the roots of a polynomial that has a CRootOf
# solution, make sure that the use of nfloat during the solve process
# doesn't fail. Note: if you want numerical solutions to a polynomial
# it is *much* faster to use nroots to get them than to solve the
# equation only to get CRootOf solutions which are then numerically
# evaluated. So for eq = x**5 + 3*x + 7 do Poly(eq).nroots() rather
# than [i.n() for i in solve(eq)] to get the numerical roots of eq.
assert nfloat(list(solveset_complex(x**5 + 3 * x**3 + 7, x))[0],
exponent=False) == CRootOf(x**5 + 3 * x**3 + 7, 0).n()
sol = list(solveset_complex(x**6 - 2 * x + 2, x))
assert all(isinstance(i, CRootOf) for i in sol) and len(sol) == 6
f = x**5 - 15 * x**3 - 5 * x**2 + 10 * x + 20
s = list(solveset_complex(f, x))
for root in s:
assert root.func == CRootOf
s = x**5 + 4 * x**3 + 3 * x**2 + S(7) / 4
assert solveset_complex(s, x) == \
FiniteSet(*Poly(s*4, domain='ZZ').all_roots())
# XXX: this comparison should work without converting the FiniteSet to list
# See #7876
eq = x * (x - 1)**2 * (x + 1) * (x**6 - x + 1)
assert list(solveset_complex(eq, x)) == \
list(FiniteSet(-1, 0, 1, CRootOf(x**6 - x + 1, 0),
CRootOf(x**6 - x + 1, 1),
CRootOf(x**6 - x + 1, 2),
CRootOf(x**6 - x + 1, 3),
CRootOf(x**6 - x + 1, 4),
CRootOf(x**6 - x + 1, 5)))
def simplify(expr, ratio=1.7, measure=count_ops, rational=False, inverse=False):
"""Simplifies the given expression.
Simplification is not a well defined term and the exact strategies
this function tries can change in the future versions of SymPy. If
your algorithm relies on "simplification" (whatever it is), try to
determine what you need exactly - is it powsimp()?, radsimp()?,
together()?, logcombine()?, or something else? And use this particular
function directly, because those are well defined and thus your algorithm
will be robust.
Nonetheless, especially for interactive use, or when you don't know
anything about the structure of the expression, simplify() tries to apply
intelligent heuristics to make the input expression "simpler". For
example:
>>> from sympy import simplify, cos, sin
>>> from sympy.abc import x, y
>>> a = (x + x**2)/(x*sin(y)**2 + x*cos(y)**2)
>>> a
(x**2 + x)/(x*sin(y)**2 + x*cos(y)**2)
>>> simplify(a)
x + 1
Note that we could have obtained the same result by using specific
simplification functions:
>>> from sympy import trigsimp, cancel
>>> trigsimp(a)
(x**2 + x)/x
>>> cancel(_)
x + 1
In some cases, applying :func:`simplify` may actually result in some more
complicated expression. The default ``ratio=1.7`` prevents more extreme
cases: if (result length)/(input length) > ratio, then input is returned
unmodified. The ``measure`` parameter lets you specify the function used
to determine how complex an expression is. The function should take a
single argument as an expression and return a number such that if
expression ``a`` is more complex than expression ``b``, then
``measure(a) > measure(b)``. The default measure function is
:func:`count_ops`, which returns the total number of operations in the
expression.
For example, if ``ratio=1``, ``simplify`` output can't be longer
than input.
::
>>> from sympy import sqrt, simplify, count_ops, oo
>>> root = 1/(sqrt(2)+3)
Since ``simplify(root)`` would result in a slightly longer expression,
root is returned unchanged instead::
>>> simplify(root, ratio=1) == root
True
If ``ratio=oo``, simplify will be applied anyway::
>>> count_ops(simplify(root, ratio=oo)) > count_ops(root)
True
Note that the shortest expression is not necessary the simplest, so
setting ``ratio`` to 1 may not be a good idea.
Heuristically, the default value ``ratio=1.7`` seems like a reasonable
choice.
You can easily define your own measure function based on what you feel
should represent the "size" or "complexity" of the input expression. Note
that some choices, such as ``lambda expr: len(str(expr))`` may appear to be
good metrics, but have other problems (in this case, the measure function
may slow down simplify too much for very large expressions). If you don't
know what a good metric would be, the default, ``count_ops``, is a good
one.
For example:
>>> from sympy import symbols, log
>>> a, b = symbols('a b', positive=True)
>>> g = log(a) + log(b) + log(a)*log(1/b)
>>> h = simplify(g)
>>> h
log(a*b**(-log(a) + 1))
>>> count_ops(g)
8
>>> count_ops(h)
5
So you can see that ``h`` is simpler than ``g`` using the count_ops metric.
However, we may not like how ``simplify`` (in this case, using
``logcombine``) has created the ``b**(log(1/a) + 1)`` term. A simple way
to reduce this would be to give more weight to powers as operations in
``count_ops``. We can do this by using the ``visual=True`` option:
>>> print(count_ops(g, visual=True))
2*ADD + DIV + 4*LOG + MUL
>>> print(count_ops(h, visual=True))
2*LOG + MUL + POW + SUB
>>> from sympy import Symbol, S
>>> def my_measure(expr):
... POW = Symbol('POW')
... # Discourage powers by giving POW a weight of 10
... count = count_ops(expr, visual=True).subs(POW, 10)
... # Every other operation gets a weight of 1 (the default)
... count = count.replace(Symbol, type(S.One))
... return count
>>> my_measure(g)
8
>>> my_measure(h)
14
>>> 15./8 > 1.7 # 1.7 is the default ratio
True
>>> simplify(g, measure=my_measure)
-log(a)*log(b) + log(a) + log(b)
Note that because ``simplify()`` internally tries many different
simplification strategies and then compares them using the measure
function, we get a completely different result that is still different
from the input expression by doing this.
If rational=True, Floats will be recast as Rationals before simplification.
If rational=None, Floats will be recast as Rationals but the result will
be recast as Floats. If rational=False(default) then nothing will be done
to the Floats.
If inverse=True, it will be assumed that a composition of inverse
functions, such as sin and asin, can be cancelled in any order.
For example, ``asin(sin(x))`` will yield ``x`` without checking whether
x belongs to the set where this relation is true. The default is
False.
"""
expr = sympify(expr)
try:
return expr._eval_simplify(ratio=ratio, measure=measure)
except AttributeError:
pass
original_expr = expr = signsimp(expr)
from sympy.simplify.hyperexpand import hyperexpand
from sympy.functions.special.bessel import BesselBase
from sympy import Sum, Product
if not isinstance(expr, Basic) or not expr.args: # XXX: temporary hack
return expr
if inverse and expr.has(Function):
expr = inversecombine(expr)
if not expr.args: # simplified to atomic
return expr
if not isinstance(expr, (Add, Mul, Pow, ExpBase)):
return expr.func(*[simplify(x, ratio=ratio, measure=measure, rational=rational)
for x in expr.args])
# TODO: Apply different strategies, considering expression pattern:
# is it a purely rational function? Is there any trigonometric function?...
# See also https://github.com/sympy/sympy/pull/185.
def shorter(*choices):
'''Return the choice that has the fewest ops. In case of a tie,
the expression listed first is selected.'''
if not has_variety(choices):
return choices[0]
return min(choices, key=measure)
# rationalize Floats
floats = False
if rational is not False and expr.has(Float):
floats = True
expr = nsimplify(expr, rational=True)
expr = bottom_up(expr, lambda w: w.normal())
expr = Mul(*powsimp(expr).as_content_primitive())
_e = cancel(expr)
expr1 = shorter(_e, _mexpand(_e).cancel()) # issue 6829
expr2 = shorter(together(expr, deep=True), together(expr1, deep=True))
if ratio is S.Infinity:
expr = expr2
else:
expr = shorter(expr2, expr1, expr)
if not isinstance(expr, Basic): # XXX: temporary hack
return expr
expr = factor_terms(expr, sign=False)
# hyperexpand automatically only works on hypergeometric terms
expr = hyperexpand(expr)
expr = piecewise_fold(expr)
if expr.has(BesselBase):
expr = besselsimp(expr)
if expr.has(TrigonometricFunction, HyperbolicFunction):
expr = trigsimp(expr, deep=True)
if expr.has(log):
expr = shorter(expand_log(expr, deep=True), logcombine(expr))
if expr.has(CombinatorialFunction, gamma):
# expression with gamma functions or non-integer arguments is
# automatically passed to gammasimp
expr = combsimp(expr)
if expr.has(Sum):
expr = sum_simplify(expr)
if expr.has(Product):
expr = product_simplify(expr)
from sympy.physics.units import Quantity
from sympy.physics.units.util import quantity_simplify
if expr.has(Quantity):
expr = quantity_simplify(expr)
short = shorter(powsimp(expr, combine='exp', deep=True), powsimp(expr), expr)
short = shorter(short, cancel(short))
short = shorter(short, factor_terms(short), expand_power_exp(expand_mul(short)))
if short.has(TrigonometricFunction, HyperbolicFunction, ExpBase):
short = exptrigsimp(short)
# get rid of hollow 2-arg Mul factorization
hollow_mul = Transform(
lambda x: Mul(*x.args),
lambda x:
x.is_Mul and
len(x.args) == 2 and
x.args[0].is_Number and
x.args[1].is_Add and
x.is_commutative)
expr = short.xreplace(hollow_mul)
numer, denom = expr.as_numer_denom()
if denom.is_Add:
n, d = fraction(radsimp(1/denom, symbolic=False, max_terms=1))
if n is not S.One:
expr = (numer*n).expand()/d
if expr.could_extract_minus_sign():
n, d = fraction(expr)
if d != 0:
expr = signsimp(-n/(-d))
if measure(expr) > ratio*measure(original_expr):
expr = original_expr
# restore floats
if floats and rational is None:
expr = nfloat(expr, exponent=False)
return expr
def simplify(expr, ratio=1.7, measure=count_ops, rational=False):
# type: (object, object, object, object) -> object
"""
Simplifies the given expression.
Simplification is not a well defined term and the exact strategies
this function tries can change in the future versions of SymPy. If
your algorithm relies on "simplification" (whatever it is), try to
determine what you need exactly - is it powsimp()?, radsimp()?,
together()?, logcombine()?, or something else? And use this particular
function directly, because those are well defined and thus your algorithm
will be robust.
Nonetheless, especially for interactive use, or when you don't know
anything about the structure of the expression, simplify() tries to apply
intelligent heuristics to make the input expression "simpler". For
example:
>>> from sympy import simplify, cos, sin
>>> from sympy.abc import x, y
>>> a = (x + x**2)/(x*sin(y)**2 + x*cos(y)**2)
>>> a
(x**2 + x)/(x*sin(y)**2 + x*cos(y)**2)
>>> simplify(a)
x + 1
Note that we could have obtained the same result by using specific
simplification functions:
>>> from sympy import trigsimp, cancel
>>> trigsimp(a)
(x**2 + x)/x
>>> cancel(_)
x + 1
In some cases, applying :func:`simplify` may actually result in some more
complicated expression. The default ``ratio=1.7`` prevents more extreme
cases: if (result length)/(input length) > ratio, then input is returned
unmodified. The ``measure`` parameter lets you specify the function used
to determine how complex an expression is. The function should take a
single argument as an expression and return a number such that if
expression ``a`` is more complex than expression ``b``, then
``measure(a) > measure(b)``. The default measure function is
:func:`count_ops`, which returns the total number of operations in the
expression.
For example, if ``ratio=1``, ``simplify`` output can't be longer
than input.
::
>>> from sympy import sqrt, simplify, count_ops, oo
>>> root = 1/(sqrt(2)+3)
Since ``simplify(root)`` would result in a slightly longer expression,
root is returned unchanged instead::
>>> simplify(root, ratio=1) == root
True
If ``ratio=oo``, simplify will be applied anyway::
>>> count_ops(simplify(root, ratio=oo)) > count_ops(root)
True
Note that the shortest expression is not necessary the simplest, so
setting ``ratio`` to 1 may not be a good idea.
Heuristically, the default value ``ratio=1.7`` seems like a reasonable
choice.
You can easily define your own measure function based on what you feel
should represent the "size" or "complexity" of the input expression. Note
that some choices, such as ``lambda expr: len(str(expr))`` may appear to be
good metrics, but have other problems (in this case, the measure function
may slow down simplify too much for very large expressions). If you don't
know what a good metric would be, the default, ``count_ops``, is a good
one.
For example:
>>> from sympy import symbols, log
>>> a, b = symbols('a b', positive=True)
>>> g = log(a) + log(b) + log(a)*log(1/b)
>>> h = simplify(g)
>>> h
log(a*b**(-log(a) + 1))
>>> count_ops(g)
8
>>> count_ops(h)
5
So you can see that ``h`` is simpler than ``g`` using the count_ops metric.
However, we may not like how ``simplify`` (in this case, using
``logcombine``) has created the ``b**(log(1/a) + 1)`` term. A simple way
to reduce this would be to give more weight to powers as operations in
``count_ops``. We can do this by using the ``visual=True`` option:
>>> print(count_ops(g, visual=True))
2*ADD + DIV + 4*LOG + MUL
>>> print(count_ops(h, visual=True))
2*LOG + MUL + POW + SUB
>>> from sympy import Symbol, S
>>> def my_measure(expr):
... POW = Symbol('POW')
... # Discourage powers by giving POW a weight of 10
... count = count_ops(expr, visual=True).subs(POW, 10)
... # Every other operation gets a weight of 1 (the default)
... count = count.replace(Symbol, type(S.One))
... return count
>>> my_measure(g)
8
>>> my_measure(h)
14
>>> 15./8 > 1.7 # 1.7 is the default ratio
True
>>> simplify(g, measure=my_measure)
-log(a)*log(b) + log(a) + log(b)
Note that because ``simplify()`` internally tries many different
simplification strategies and then compares them using the measure
function, we get a completely different result that is still different
from the input expression by doing this.
If rational=True, Floats will be recast as Rationals before simplification.
If rational=None, Floats will be recast as Rationals but the result will
be recast as Floats. If rational=False(default) then nothing will be done
to the Floats.
"""
expr = sympify(expr)
try:
return expr._eval_simplify(ratio=ratio, measure=measure)
except AttributeError:
pass
original_expr = expr = signsimp(expr)
from sympy.simplify.hyperexpand import hyperexpand
from sympy.functions.special.bessel import BesselBase
from sympy import Sum, Product
if not isinstance(expr, Basic) or not expr.args: # XXX: temporary hack
return expr
if not isinstance(expr, (Add, Mul, Pow, ExpBase)):
if isinstance(expr, Function) and hasattr(expr, "inverse"):
if len(expr.args) == 1 and len(expr.args[0].args) == 1 and \
isinstance(expr.args[0], expr.inverse(argindex=1)):
return simplify(expr.args[0].args[0], ratio=ratio,
measure=measure, rational=rational)
return expr.func(*[simplify(x, ratio=ratio, measure=measure, rational=rational)
for x in expr.args])
# TODO: Apply different strategies, considering expression pattern:
# is it a purely rational function? Is there any trigonometric function?...
# See also https://github.com/sympy/sympy/pull/185.
def shorter(*choices):
'''Return the choice that has the fewest ops. In case of a tie,
the expression listed first is selected.'''
if not has_variety(choices):
return choices[0]
return min(choices, key=measure)
# rationalize Floats
floats = False
if rational is not False and expr.has(Float):
floats = True
expr = nsimplify(expr, rational=True)
expr = bottom_up(expr, lambda w: w.normal())
expr = Mul(*powsimp(expr).as_content_primitive())
_e = cancel(expr)
expr1 = shorter(_e, _mexpand(_e).cancel()) # issue 6829
expr2 = shorter(together(expr, deep=True), together(expr1, deep=True))
if ratio is S.Infinity:
expr = expr2
else:
expr = shorter(expr2, expr1, expr)
if not isinstance(expr, Basic): # XXX: temporary hack
return expr
expr = factor_terms(expr, sign=False)
# hyperexpand automatically only works on hypergeometric terms
expr = hyperexpand(expr)
expr = piecewise_fold(expr)
if expr.has(BesselBase):
expr = besselsimp(expr)
if expr.has(TrigonometricFunction, HyperbolicFunction):
expr = trigsimp(expr, deep=True)
if expr.has(log):
expr = shorter(expand_log(expr, deep=True), logcombine(expr))
if expr.has(CombinatorialFunction, gamma):
# expression with gamma functions or non-integer arguments is
# automatically passed to gammasimp
expr = combsimp(expr)
if expr.has(Sum):
expr = sum_simplify(expr)
if expr.has(Product):
expr = product_simplify(expr)
short = shorter(powsimp(expr, combine='exp', deep=True), powsimp(expr), expr)
short = shorter(short, cancel(short))
short = shorter(short, factor_terms(short), expand_power_exp(expand_mul(short)))
if short.has(TrigonometricFunction, HyperbolicFunction, ExpBase):
short = exptrigsimp(short)
# get rid of hollow 2-arg Mul factorization
hollow_mul = Transform(
lambda x: Mul(*x.args),
lambda x:
x.is_Mul and
len(x.args) == 2 and
x.args[0].is_Number and
x.args[1].is_Add and
x.is_commutative)
expr = short.xreplace(hollow_mul)
numer, denom = expr.as_numer_denom()
if denom.is_Add:
n, d = fraction(radsimp(1/denom, symbolic=False, max_terms=1))
if n is not S.One:
expr = (numer*n).expand()/d
if expr.could_extract_minus_sign():
n, d = fraction(expr)
if d != 0:
expr = signsimp(-n/(-d))
if measure(expr) > ratio*measure(original_expr):
expr = original_expr
# restore floats
if floats and rational is None:
expr = nfloat(expr, exponent=False)
return expr
def test_issue_10395():
eq = x * Max(0, y)
assert nfloat(eq) == eq
eq = x * Max(y, -1.1)
assert nfloat(eq) == eq
assert Max(y, 4).n() == Max(4.0, y)
gaussUA[i - 1], gaussUA[i]))
#print("AreaCA {} : {} ".format(i+1, gaussCA[i]))
gaussTotCA = gaussTotCA + gaussCA[i]
#print("Total Certainity Area: {}".format(gaussTotCA))
# GAUSSIAN MF'S - FuzR MEASSURE
FuzRgaussOrig = (gaussTotCA) / (gaussTotUA)
print("FuzR Meassure - Gaussian: {}".format(FuzRgaussOrig))
# Test Something
x = sympy.var("x")
totAreaAgg = 0
for i in range(l):
func1 = sympy.exp(-(((x - gaussParam[i][0]) / gaussParam[i][1])**2) / 2)
int1 = sympy.integrate(func1, (x, 0, 10))
areaGauss = nfloat(int1)
areaTri = 0.5 * (triParam[i][2] - triParam[i][0])
totAreaAgg = totAreaAgg + areaGauss - areaTri
# Test Scaling Measure
print("Uncertainty Difference")
print(gaussTotUA - triTotUA)
print("Certainty Difference")
print(gaussTotCA - triTotCA)
FuzRgaussMod = (triTotCA + 2) / (triTotUA + 0.5)
print("\nFuzR Meassure - GaussianMod : {}".format(FuzRgaussMod))
#plt.margins(0)
plt.tight_layout()
plt.show()
def test_nfloat():
from sympy.core.basic import _aresame
from sympy.polys.rootoftools import rootof
x = Symbol("x")
eq = x**Rational(4, 3) + 4*x**(S.One/3)/3
assert _aresame(nfloat(eq), x**Rational(4, 3) + (4.0/3)*x**(S.One/3))
assert _aresame(nfloat(eq, exponent=True), x**(4.0/3) + (4.0/3)*x**(1.0/3))
eq = x**Rational(4, 3) + 4*x**(x/3)/3
assert _aresame(nfloat(eq), x**Rational(4, 3) + (4.0/3)*x**(x/3))
big = 12345678901234567890
# specify precision to match value used in nfloat
Float_big = Float(big, 15)
assert _aresame(nfloat(big), Float_big)
assert _aresame(nfloat(big*x), Float_big*x)
assert _aresame(nfloat(x**big, exponent=True), x**Float_big)
assert nfloat(cos(x + sqrt(2))) == cos(x + nfloat(sqrt(2)))
# issue 6342
f = S('x*lamda + lamda**3*(x/2 + 1/2) + lamda**2 + 1/4')
assert not any(a.free_symbols for a in solveset(f.subs(x, -0.139)))
# issue 6632
assert nfloat(-100000*sqrt(2500000001) + 5000000001) == \
9.99999999800000e-11
# issue 7122
eq = cos(3*x**4 + y)*rootof(x**5 + 3*x**3 + 1, 0)
assert str(nfloat(eq, exponent=False, n=1)) == '-0.7*cos(3.0*x**4 + y)'
# issue 10933
for ti in (dict, Dict):
d = ti({S.Half: S.Half})
n = nfloat(d)
assert isinstance(n, ti)
assert _aresame(list(n.items()).pop(), (S.Half, Float(.5)))
for ti in (dict, Dict):
d = ti({S.Half: S.Half})
n = nfloat(d, dkeys=True)
assert isinstance(n, ti)
assert _aresame(list(n.items()).pop(), (Float(.5), Float(.5)))
d = [S.Half]
n = nfloat(d)
assert type(n) is list
assert _aresame(n[0], Float(.5))
assert _aresame(nfloat(Eq(x, S.Half)).rhs, Float(.5))
assert _aresame(nfloat(S(True)), S(True))
assert _aresame(nfloat(Tuple(S.Half))[0], Float(.5))
assert nfloat(Eq((3 - I)**2/2 + I, 0)) == S.false
# pass along kwargs
assert nfloat([{S.Half: x}], dkeys=True) == [{Float(0.5): x}]
# Issue 17706
A = MutableMatrix([[1, 2], [3, 4]])
B = MutableMatrix(
[[Float('1.0', precision=53), Float('2.0', precision=53)],
[Float('3.0', precision=53), Float('4.0', precision=53)]])
assert _aresame(nfloat(A), B)
A = ImmutableMatrix([[1, 2], [3, 4]])
B = ImmutableMatrix(
[[Float('1.0', precision=53), Float('2.0', precision=53)],
[Float('3.0', precision=53), Float('4.0', precision=53)]])
assert _aresame(nfloat(A), B)
# issue 22524
f = Function('f')
assert not nfloat(f(2)).atoms(Float)