本文整理汇总了Python中sympy.expand_log函数的典型用法代码示例。如果您正苦于以下问题:Python expand_log函数的具体用法?Python expand_log怎么用?Python expand_log使用的例子?那么恭喜您, 这里精选的函数代码示例或许可以为您提供帮助。
在下文中一共展示了expand_log函数的10个代码示例,这些例子默认根据受欢迎程度排序。您可以为喜欢或者感觉有用的代码点赞,您的评价将有助于我们的系统推荐出更棒的Python代码示例。
示例1: test_log_simplify
def test_log_simplify():
x = Symbol("x", positive=True)
assert log(x**2).expand() == 2*log(x)
assert expand_log(log(x**(2 + log(2)))) == (2 + log(2))*log(x)
z = Symbol('z')
assert log(sqrt(z)).expand() == log(z)/2
assert expand_log(log(z**(log(2) - 1))) == (log(2) - 1)*log(z)
assert log(z**(-1)).expand() != -log(z)
assert log(z**(x/(x+1))).expand() == x*log(z)/(x + 1)
开发者ID:asmeurer,项目名称:sympy,代码行数:10,代码来源:test_exponential.py
示例2: test_issue_8866
def test_issue_8866():
assert simplify(log(x, 10, evaluate=False)) == simplify(log(x, 10))
assert expand_log(log(x, 10, evaluate=False)) == expand_log(log(x, 10))
y = Symbol('y', positive=True)
l1 = log(exp(y), exp(10))
b1 = log(exp(y), exp(5))
l2 = log(exp(y), exp(10), evaluate=False)
b2 = log(exp(y), exp(5), evaluate=False)
assert simplify(log(l1, b1)) == simplify(log(l2, b2))
assert expand_log(log(l1, b1)) == expand_log(log(l2, b2))
开发者ID:asmeurer,项目名称:sympy,代码行数:11,代码来源:test_exponential.py
示例3: _eval_simplify
def _eval_simplify(self, ratio, measure):
from sympy.simplify.simplify import expand_log, simplify
if (len(self.args) == 2):
return simplify(self.func(*self.args), ratio=ratio, measure=measure)
expr = self.func(simplify(self.args[0], ratio=ratio, measure=measure))
expr = expand_log(expr, deep=True)
return min([expr, self], key=measure)
开发者ID:baoqchau,项目名称:sympy,代码行数:7,代码来源:exponential.py
示例4: _eval_simplify
def _eval_simplify(self, ratio, measure, rational, inverse):
from sympy.simplify.simplify import expand_log, simplify, inversecombine
if (len(self.args) == 2):
return simplify(self.func(*self.args), ratio=ratio, measure=measure,
rational=rational, inverse=inverse)
expr = self.func(simplify(self.args[0], ratio=ratio, measure=measure,
rational=rational, inverse=inverse))
if inverse:
expr = inversecombine(expr)
expr = expand_log(expr, deep=True)
return min([expr, self], key=measure)
开发者ID:cmarqu,项目名称:sympy,代码行数:11,代码来源:exponential.py
示例5: canonise_log
def canonise_log(equation):
expanded_log = sympy.expand_log(equation, force=True)
terms = expanded_log.as_ordered_terms()
a, b = sympy.Wild('a'), sympy.Wild('b')
total_interior = 1
for term in terms:
if sympy.ask(sympy.Q.complex(term)):
term_interior *= -1
else:
term_interior = term.match(sympy.log(a) / b)[a]
if term.could_extract_minus_sign():
total_interior /= term_interior
else:
total_interior *= term_interior
if isinstance(total_interior, sympy.Add):
invert = False
elif isinstance(total_interior, sympy.Mul):
match = total_interior.together().match(x / b) # for some reason, (x/3).match(a/b) gives {a: 1/3, b: 1/x} so we have to use a workaround
if match is not None:
invert = False
else:
match = total_interior.together().match(a / b)
degree_numerator = 0 if isinstance(match[a], sympy.Rational) else match[a].as_poly().degree()
degree_denominator = 0 if isinstance(match[b], sympy.Rational) else match[b].as_poly().degree()
if degree_numerator < degree_denominator:
invert = True
else:
invert = False
elif isinstance(total_interior, sympy.Pow):
index = total_interior.as_base_exp()[1]
if index < 0:
invert = True
else:
invert = False
else: # for debugging - wtf kind of c-c-c-class is it???
print(total_interior, type(total_interior))
if invert:
return -sympy.log((1 / total_interior).together(), evaluate=False) / terms[0].as_coeff_Mul()[0].q
else:
return sympy.log(total_interior.together(), evaluate=False) / terms[0].as_coeff_Mul()[0].q
开发者ID:nebffa,项目名称:MathsExams,代码行数:51,代码来源:simplify.py
示例6: _eval_expand_log
def _eval_expand_log(self, deep=True, **hints):
from sympy import unpolarify, expand_log
from sympy.concrete import Sum, Product
force = hints.get('force', False)
if (len(self.args) == 2):
return expand_log(self.func(*self.args), deep=deep, force=force)
arg = self.args[0]
if arg.is_Integer:
# remove perfect powers
p = perfect_power(int(arg))
if p is not False:
return p[1]*self.func(p[0])
elif arg.is_Rational:
return log(arg.p) - log(arg.q)
elif arg.is_Mul:
expr = []
nonpos = []
for x in arg.args:
if force or x.is_positive or x.is_polar:
a = self.func(x)
if isinstance(a, log):
expr.append(self.func(x)._eval_expand_log(**hints))
else:
expr.append(a)
elif x.is_negative:
a = self.func(-x)
expr.append(a)
nonpos.append(S.NegativeOne)
else:
nonpos.append(x)
return Add(*expr) + log(Mul(*nonpos))
elif arg.is_Pow or isinstance(arg, exp):
if force or (arg.exp.is_real and (arg.base.is_positive or ((arg.exp+1)
.is_positive and (arg.exp-1).is_nonpositive))) or arg.base.is_polar:
b = arg.base
e = arg.exp
a = self.func(b)
if isinstance(a, log):
return unpolarify(e) * a._eval_expand_log(**hints)
else:
return unpolarify(e) * a
elif isinstance(arg, Product):
if arg.function.is_positive:
return Sum(log(arg.function), *arg.limits)
return self.func(arg)
开发者ID:cmarqu,项目名称:sympy,代码行数:46,代码来源:exponential.py
示例7: test_risch_integrate
def test_risch_integrate():
assert risch_integrate(t0*exp(x), x) == t0*exp(x)
assert risch_integrate(sin(x), x, rewrite_complex=True) == -exp(I*x)/2 - exp(-I*x)/2
# From my GSoC writeup
assert risch_integrate((1 + 2*x**2 + x**4 + 2*x**3*exp(2*x**2))/
(x**4*exp(x**2) + 2*x**2*exp(x**2) + exp(x**2)), x) == \
NonElementaryIntegral(exp(-x**2), x) + exp(x**2)/(1 + x**2)
assert risch_integrate(0, x) == 0
# also tests prde_cancel()
e1 = log(x/exp(x) + 1)
ans1 = risch_integrate(e1, x)
assert ans1 == (x*log(x*exp(-x) + 1) + NonElementaryIntegral((x**2 - x)/(x + exp(x)), x))
assert cancel(diff(ans1, x) - e1) == 0
# also tests issue #10798
e2 = (log(-1/y)/2 - log(1/y)/2)/y - (log(1 - 1/y)/2 - log(1 + 1/y)/2)/y
ans2 = risch_integrate(e2, y)
assert ans2 == log(1/y)*log(1 - 1/y)/2 - log(1/y)*log(1 + 1/y)/2 + \
NonElementaryIntegral((I*pi*y**2 - 2*y*log(1/y) - I*pi)/(2*y**3 - 2*y), y)
assert expand_log(cancel(diff(ans2, y) - e2), force=True) == 0
# These are tested here in addition to in test_DifferentialExtension above
# (symlogs) to test that backsubs works correctly. The integrals should be
# written in terms of the original logarithms in the integrands.
# XXX: Unfortunately, making backsubs work on this one is a little
# trickier, because x**x is converted to exp(x*log(x)), and so log(x**x)
# is converted to x*log(x). (x**2*log(x)).subs(x*log(x), log(x**x)) is
# smart enough, the issue is that these splits happen at different places
# in the algorithm. Maybe a heuristic is in order
assert risch_integrate(log(x**x), x) == x**2*log(x)/2 - x**2/4
assert risch_integrate(log(x**y), x) == x*log(x**y) - x*y
assert risch_integrate(log(sqrt(x)), x) == x*log(sqrt(x)) - x/2
开发者ID:Lenqth,项目名称:sympy,代码行数:38,代码来源:test_risch.py
示例8: test_log_simplify
def test_log_simplify():
x = Symbol("x", positive=True)
assert log(x**2).expand() == 2*log(x)
assert expand_log(log(x**(2+log(2)))) == (2+log(2))*log(x)
开发者ID:haz,项目名称:sympy,代码行数:4,代码来源:test_exponential.py
示例9: ReplaceOptim
new_exp_terms.append(exp_term)
else:
done = True
new_exp_terms.append(attempt)
if not done:
new_exp_terms.append(numsum)
return e.func(*chain(new_exp_terms, non_num_other))
expm1_opt = ReplaceOptim(lambda e: e.is_Add, _expm1_value)
log1p_opt = ReplaceOptim(
lambda e: isinstance(e, log),
lambda l: expand_log(l.replace(
log, lambda arg: log(arg.factor())
)).replace(log(_u+1), log1p(_u))
)
def create_expand_pow_optimization(limit):
""" Creates an instance of :class:`ReplaceOptim` for expanding ``Pow``.
The requirements for expansions are that the base needs to be a symbol
and the exponent needs to be an integer (and be less than or equal to
``limit``).
Parameters
==========
limit : int
The highest power which is expanded into multiplication.
开发者ID:asmeurer,项目名称:sympy,代码行数:31,代码来源:rewriting.py
示例10:
# <codecell>
import sympy
from sympy.abc import x, z
p=sympy.symbols('p',positive=True)
sympy.init_printing()
# <codecell>
L=p**x*(1-p)**(1-x)
J=np.prod([L.subs(x,i) for i in xs]) # objective function to maximize
J
# <codecell>
logJ=sympy.expand_log(sympy.log(J))
sol=sympy.solve(sympy.diff(logJ,p),p)[0]
x=linspace(0,1,100)
plot(x,map(sympy.lambdify(p,logJ,'numpy'),x),sol,logJ.subs(p,sol),'o',
p_true,logJ.subs(p,p_true),'s',)
xlabel('$p$',fontsize=18)
ylabel('Likelihood',fontsize=18)
title('Estimate not equal to true value',fontsize=18)
# <codecell>
L=p**x*(1-p)**(1-x)
J=np.prod([L.subs(x,i) for i in xs]) # objective function
开发者ID:gwli,项目名称:StudyNote,代码行数:29,代码来源:MLEstimation.py
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