本文整理汇总了Python中sympy.functions.sign函数的典型用法代码示例。如果您正苦于以下问题:Python sign函数的具体用法?Python sign怎么用?Python sign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sign函数的20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_ccode_sign
def test_ccode_sign():
expr1, ref1 = sign(x) * y, 'y*(((x) > 0) - ((x) < 0))'
expr2, ref2 = sign(cos(x)), '(((cos(x)) > 0) - ((cos(x)) < 0))'
expr3, ref3 = sign(2 * x + x**2) * x + x**2, 'pow(x, 2) + x*(((pow(x, 2) + 2*x) > 0) - ((pow(x, 2) + 2*x) < 0))'
assert ccode(expr1) == ref1
assert ccode(expr1, 'z') == 'z = %s;' % ref1
assert ccode(expr2) == ref2
assert ccode(expr3) == ref3
开发者ID:Lenqth,项目名称:sympy,代码行数:8,代码来源:test_ccode.py
示例2: test_ccode_sign
def test_ccode_sign():
expr = sign(x) * y
assert ccode(expr) == 'y*(((x) > 0) - ((x) < 0))'
assert ccode(expr, 'z') == 'z = y*(((x) > 0) - ((x) < 0));'
assert ccode(sign(2 * x + x**2) * x + x**2) == \
'pow(x, 2) + x*(((pow(x, 2) + 2*x) > 0) - ((pow(x, 2) + 2*x) < 0))'
expr = sign(cos(x))
assert ccode(expr) == '(((cos(x)) > 0) - ((cos(x)) < 0))'
开发者ID:MCGallaspy,项目名称:sympy,代码行数:11,代码来源:test_ccode.py
示例3: test_sign
def test_sign():
expr = sign(x) * y
assert rust_code(expr) == "y*x.signum()"
assert rust_code(expr, assign_to='r') == "r = y*x.signum();"
expr = sign(x + y) + 42
assert rust_code(expr) == "(x + y).signum() + 42"
assert rust_code(expr, assign_to='r') == "r = (x + y).signum() + 42;"
expr = sign(cos(x))
assert rust_code(expr) == "x.cos().signum()"
开发者ID:abhi98khandelwal,项目名称:sympy,代码行数:11,代码来源:test_rust.py
示例4: test_rcode_sgn
def test_rcode_sgn():
expr = sign(x) * y
assert rcode(expr) == 'y*sign(x)'
p = rcode(expr, 'z')
assert p == 'z = y*sign(x);'
p = rcode(sign(2 * x + x**2) * x + x**2)
assert p == "x^2 + x*sign(x^2 + 2*x)"
expr = sign(cos(x))
p = rcode(expr)
assert p == 'sign(cos(x))'
开发者ID:chris-turner137,项目名称:sympy,代码行数:13,代码来源:test_rcode.py
示例5: test_bounded
def test_bounded():
x, y = symbols('xy')
assert ask(x, Q.bounded) == False
assert ask(x, Q.bounded, Assume(x, Q.bounded)) == True
assert ask(x, Q.bounded, Assume(y, Q.bounded)) == False
assert ask(x, Q.bounded, Assume(x, Q.complex)) == False
assert ask(x+1, Q.bounded) == False
assert ask(x+1, Q.bounded, Assume(x, Q.bounded)) == True
assert ask(x+y, Q.bounded) == None
assert ask(x+y, Q.bounded, Assume(x, Q.bounded)) == False
assert ask(x+1, Q.bounded, Assume(x, Q.bounded) & \
Assume(y, Q.bounded)) == True
assert ask(2*x, Q.bounded) == False
assert ask(2*x, Q.bounded, Assume(x, Q.bounded)) == True
assert ask(x*y, Q.bounded) == None
assert ask(x*y, Q.bounded, Assume(x, Q.bounded)) == False
assert ask(x*y, Q.bounded, Assume(x, Q.bounded) & \
Assume(y, Q.bounded)) == True
assert ask(x**2, Q.bounded) == False
assert ask(2**x, Q.bounded) == False
assert ask(2**x, Q.bounded, Assume(x, Q.bounded)) == True
assert ask(x**x, Q.bounded) == False
assert ask(Rational(1,2) ** x, Q.bounded) == True
assert ask(x ** Rational(1,2), Q.bounded) == False
# sign function
assert ask(sign(x), Q.bounded) == True
assert ask(sign(x), Q.bounded, Assume(x, Q.bounded, False)) == True
# exponential functions
assert ask(log(x), Q.bounded) == False
assert ask(log(x), Q.bounded, Assume(x, Q.bounded)) == True
assert ask(exp(x), Q.bounded) == False
assert ask(exp(x), Q.bounded, Assume(x, Q.bounded)) == True
assert ask(exp(2), Q.bounded) == True
# trigonometric functions
assert ask(sin(x), Q.bounded) == True
assert ask(sin(x), Q.bounded, Assume(x, Q.bounded, False)) == True
assert ask(cos(x), Q.bounded) == True
assert ask(cos(x), Q.bounded, Assume(x, Q.bounded, False)) == True
assert ask(2*sin(x), Q.bounded) == True
assert ask(sin(x)**2, Q.bounded) == True
assert ask(cos(x)**2, Q.bounded) == True
assert ask(cos(x) + sin(x), Q.bounded) == True
开发者ID:Aang,项目名称:sympy,代码行数:48,代码来源:test_query.py
示例6: test_bounded
def test_bounded():
x, y = symbols('x,y')
assert ask(Q.bounded(x)) == False
assert ask(Q.bounded(x), Q.bounded(x)) == True
assert ask(Q.bounded(x), Q.bounded(y)) == False
assert ask(Q.bounded(x), Q.complex(x)) == False
assert ask(Q.bounded(x+1)) == False
assert ask(Q.bounded(x+1), Q.bounded(x)) == True
assert ask(Q.bounded(x+y)) == None
assert ask(Q.bounded(x+y), Q.bounded(x)) == False
assert ask(Q.bounded(x+1), Q.bounded(x) & Q.bounded(y)) == True
assert ask(Q.bounded(2*x)) == False
assert ask(Q.bounded(2*x), Q.bounded(x)) == True
assert ask(Q.bounded(x*y)) == None
assert ask(Q.bounded(x*y), Q.bounded(x)) == False
assert ask(Q.bounded(x*y), Q.bounded(x) & Q.bounded(y)) == True
assert ask(Q.bounded(x**2)) == False
assert ask(Q.bounded(2**x)) == False
assert ask(Q.bounded(2**x), Q.bounded(x)) == True
assert ask(Q.bounded(x**x)) == False
assert ask(Q.bounded(Rational(1,2) ** x)) == None
assert ask(Q.bounded(Rational(1,2) ** x), Q.positive(x)) == True
assert ask(Q.bounded(Rational(1,2) ** x), Q.negative(x)) == False
assert ask(Q.bounded(sqrt(x))) == False
# sign function
assert ask(Q.bounded(sign(x))) == True
assert ask(Q.bounded(sign(x)), ~Q.bounded(x)) == True
# exponential functions
assert ask(Q.bounded(log(x))) == False
assert ask(Q.bounded(log(x)), Q.bounded(x)) == True
assert ask(Q.bounded(exp(x))) == False
assert ask(Q.bounded(exp(x)), Q.bounded(x)) == True
assert ask(Q.bounded(exp(2))) == True
# trigonometric functions
assert ask(Q.bounded(sin(x))) == True
assert ask(Q.bounded(sin(x)), ~Q.bounded(x)) == True
assert ask(Q.bounded(cos(x))) == True
assert ask(Q.bounded(cos(x)), ~Q.bounded(x)) == True
assert ask(Q.bounded(2*sin(x))) == True
assert ask(Q.bounded(sin(x)**2)) == True
assert ask(Q.bounded(cos(x)**2)) == True
assert ask(Q.bounded(cos(x) + sin(x))) == True
开发者ID:lazovich,项目名称:sympy,代码行数:48,代码来源:test_query.py
示例7: calc_limit
def calc_limit(a, b):
"""replace x with a, using subs if possible, otherwise limit
where sign of b is considered"""
wok = inverse_mapping.subs(x, a)
if wok is S.NaN or wok.is_bounded is False and a.is_bounded:
return limit(sign(b)*inverse_mapping, x, a)
return wok
开发者ID:ValtersZ,项目名称:sympy,代码行数:7,代码来源:integrals.py
示例8: refine_Pow
def refine_Pow(expr, assumptions):
"""
Handler for instances of Pow.
>>> from sympy import Symbol, Assume, Q
>>> from sympy.assumptions.refine import refine_Pow
>>> from sympy.abc import x
>>> refine_Pow((-1)**x, Assume(x, Q.real))
>>> refine_Pow((-1)**x, Assume(x, Q.even))
1
>>> refine_Pow((-1)**x, Assume(x, Q.odd))
-1
"""
from sympy.core import Pow, Rational
from sympy.functions import sign
if ask(expr.base, Q.real, assumptions):
if expr.base.is_number:
if ask(expr.exp, Q.even, assumptions):
return abs(expr.base) ** expr.exp
if ask(expr.exp, Q.odd, assumptions):
return sign(expr.base) * abs(expr.base) ** expr.exp
if isinstance(expr.exp, Rational):
if type(expr.base) is Pow:
return abs(expr.base.base) ** (expr.base.exp * expr.exp)
开发者ID:smichr,项目名称:sympy-live,代码行数:26,代码来源:refine.py
示例9: calc_limit
def calc_limit(a, b):
"""replace x with a, using subs if possible, otherwise limit
where sign of b is considered"""
wok = inverse_mapping.subs(x, a)
if not wok is S.NaN:
return wok
return limit(sign(b)*inverse_mapping, x, a)
开发者ID:pyc111,项目名称:sympy,代码行数:7,代码来源:integrals.py
示例10: denester
def denester(nested):
"""
Denests a list of expressions that contain nested square roots.
This method should not be called directly - use 'sqrtdenest' instead.
This algorithm is based on <http://www.almaden.ibm.com/cs/people/fagin/symb85.pdf>.
It is assumed that all of the elements of 'nested' share the same
bottom-level radicand. (This is stated in the paper, on page 177, in
the paragraph immediately preceding the algorithm.)
When evaluating all of the arguments in parallel, the bottom-level
radicand only needs to be denested once. This means that calling
denester with x arguments results in a recursive invocation with x+1
arguments; hence denester has polynomial complexity.
However, if the arguments were evaluated separately, each call would
result in two recursive invocations, and the algorithm would have
exponential complexity.
This is discussed in the paper in the middle paragraph of page 179.
"""
if all((n**2).is_Number for n in nested): #If none of the arguments are nested
for f in subsets(len(nested)): #Test subset 'f' of nested
p = prod(nested[i]**2 for i in range(len(f)) if f[i]).expand()
if 1 in f and f.count(1) > 1 and f[-1]:
p = -p
if sqrt(p).is_Number:
return sqrt(p), f #If we got a perfect square, return its square root.
return nested[-1], [0]*len(nested) #Otherwise, return the radicand from the previous invocation.
else:
a, b, r, R = Wild('a'), Wild('b'), Wild('r'), None
values = [expr.match(sqrt(a + b * sqrt(r))) for expr in nested]
if any(v is None for v in values): # this pattern is not recognized
return nested[-1], [0]*len(nested) #Otherwise, return the radicand from the previous invocation.
for v in values:
if r in v: #Since if b=0, r is not defined
if R is not None:
assert R == v[r] #All the 'r's should be the same.
else:
R = v[r]
if R is None:
return nested[-1], [0]*len(nested) #return the radicand from the previous invocation.
d, f = denester([sqrt((v[a]**2).expand()-(R*v[b]**2).expand()) for v in values] + [sqrt(R)])
if all(fi == 0 for fi in f):
v = values[-1]
return sqrt(v[a] + v[b]*d), f
else:
v = prod(nested[i]**2 for i in range(len(nested)) if f[i]).expand().match(a+b*sqrt(r))
if 1 in f and f.index(1) < len(nested) - 1 and f[len(nested)-1]:
v[a] = -1 * v[a]
v[b] = -1 * v[b]
if not f[len(nested)]: #Solution denests with square roots
vad = (v[a] + d).expand()
if not vad:
return nested[-1], [0]*len(nested) #Otherwise, return the radicand from the previous invocation.
return (sqrt(vad/2) + sign(v[b])*sqrt((v[b]**2*R/(2*vad)).expand())).expand(), f
else: #Solution requires a fourth root
FR, s = (R.expand()**Rational(1,4)), sqrt((v[b]*R).expand()+d)
return (s/(sqrt(2)*FR) + v[a]*FR/(sqrt(2)*s)).expand(), f
开发者ID:AlexandruFlorescu,项目名称:sympy,代码行数:59,代码来源:sqrtdenest.py
示例11: _calc_limit_1
def _calc_limit_1(F, a, b):
"""
replace d with a, using subs if possible, otherwise limit
where sign of b is considered
"""
wok = F.subs(d, a)
if wok is S.NaN or wok.is_bounded is False and a.is_bounded:
return limit(sign(b)*F, d, a)
return wok
开发者ID:hrashk,项目名称:sympy,代码行数:9,代码来源:integrals.py
示例12: p_parameter
def p_parameter(self):
"""P is a parameter of parabola.
Returns
=======
p : number or symbolic expression
Notes
=====
The absolute value of p is the focal length. The sign on p tells
which way the parabola faces. Vertical parabolas that open up
and horizontal that open right, give a positive value for p.
Vertical parabolas that open down and horizontal that open left,
give a negative value for p.
See Also
========
http://www.sparknotes.com/math/precalc/conicsections/section2.rhtml
Examples
========
>>> from sympy import Parabola, Point, Line
>>> p1 = Parabola(Point(0, 0), Line(Point(5, 8), Point(7, 8)))
>>> p1.p_parameter
-4
"""
if self.axis_of_symmetry.slope == 0:
x = self.directrix.coefficients[2]
p = sign(self.focus.args[0] + x)
else:
y = self.directrix.coefficients[2]
p = sign(self.focus.args[1] + y)
return p * self.focal_length
开发者ID:bjodah,项目名称:sympy,代码行数:39,代码来源:parabola.py
示例13: test_Function
def test_Function():
assert mcode(sin(x) ** cos(x)) == "sin(x).^cos(x)"
assert mcode(sign(x)) == "sign(x)"
assert mcode(exp(x)) == "exp(x)"
assert mcode(log(x)) == "log(x)"
assert mcode(factorial(x)) == "factorial(x)"
assert mcode(floor(x)) == "floor(x)"
assert mcode(atan2(y, x)) == "atan2(y, x)"
assert mcode(beta(x, y)) == 'beta(x, y)'
assert mcode(polylog(x, y)) == 'polylog(x, y)'
assert mcode(harmonic(x)) == 'harmonic(x)'
assert mcode(bernoulli(x)) == "bernoulli(x)"
assert mcode(bernoulli(x, y)) == "bernoulli(x, y)"
开发者ID:Lenqth,项目名称:sympy,代码行数:13,代码来源:test_octave.py
示例14: _sqrt_numeric_denest
def _sqrt_numeric_denest(a, b, r, d2):
"""Helper that denest expr = a + b*sqrt(r), with d2 = a**2 - b**2*r > 0
or returns None if not denested.
"""
from sympy.simplify.simplify import radsimp
depthr = sqrt_depth(r)
d = sqrt(d2)
vad = a + d
# sqrt_depth(res) <= sqrt_depth(vad) + 1
# sqrt_depth(expr) = depthr + 2
# there is denesting if sqrt_depth(vad)+1 < depthr + 2
# if vad**2 is Number there is a fourth root
if sqrt_depth(vad) < depthr + 1 or (vad**2).is_Rational:
vad1 = radsimp(1/vad)
return (sqrt(vad/2) + sign(b)*sqrt((b**2*r*vad1/2).expand())).expand()
开发者ID:MichaelMayorov,项目名称:sympy,代码行数:15,代码来源:sqrtdenest.py
示例15: test_PythonCodePrinter
def test_PythonCodePrinter():
prntr = PythonCodePrinter()
assert not prntr.module_imports
assert prntr.doprint(x**y) == 'x**y'
assert prntr.doprint(Mod(x, 2)) == 'x % 2'
assert prntr.doprint(And(x, y)) == 'x and y'
assert prntr.doprint(Or(x, y)) == 'x or y'
assert not prntr.module_imports
assert prntr.doprint(pi) == 'math.pi'
assert prntr.module_imports == {'math': {'pi'}}
assert prntr.doprint(acos(x)) == 'math.acos(x)'
assert prntr.doprint(Assignment(x, 2)) == 'x = 2'
assert prntr.doprint(Piecewise((1, Eq(x, 0)),
(2, x>6))) == '((1) if (x == 0) else (2) if (x > 6) else None)'
assert prntr.doprint(Piecewise((2, Le(x, 0)),
(3, Gt(x, 0)), evaluate=False)) == '((2) if (x <= 0) else'\
' (3) if (x > 0) else None)'
assert prntr.doprint(sign(x)) == '(0.0 if x == 0 else math.copysign(1, x))'
开发者ID:Lenqth,项目名称:sympy,代码行数:18,代码来源:test_pycode.py
示例16: _denester
def _denester(nested, av0, h, max_depth_level):
"""Denests a list of expressions that contain nested square roots.
Algorithm based on <http://www.almaden.ibm.com/cs/people/fagin/symb85.pdf>.
It is assumed that all of the elements of 'nested' share the same
bottom-level radicand. (This is stated in the paper, on page 177, in
the paragraph immediately preceding the algorithm.)
When evaluating all of the arguments in parallel, the bottom-level
radicand only needs to be denested once. This means that calling
_denester with x arguments results in a recursive invocation with x+1
arguments; hence _denester has polynomial complexity.
However, if the arguments were evaluated separately, each call would
result in two recursive invocations, and the algorithm would have
exponential complexity.
This is discussed in the paper in the middle paragraph of page 179.
"""
from sympy.simplify.simplify import radsimp
if h > max_depth_level:
return None, None
if av0[1] is None:
return None, None
if (av0[0] is None and
all(n.is_Number for n in nested)): # no arguments are nested
for f in subsets(len(nested)): # test subset 'f' of nested
p = _mexpand(Mul(*[nested[i] for i in range(len(f)) if f[i]]))
if f.count(1) > 1 and f[-1]:
p = -p
sqp = sqrt(p)
if sqp.is_Rational:
return sqp, f # got a perfect square so return its square root.
# Otherwise, return the radicand from the previous invocation.
return sqrt(nested[-1]), [0]*len(nested)
else:
R = None
if av0[0] is not None:
values = [av0[:2]]
R = av0[2]
nested2 = [av0[3], R]
av0[0] = None
else:
values = filter(None, [_sqrt_match(expr) for expr in nested])
for v in values:
if v[2]: #Since if b=0, r is not defined
if R is not None:
if R != v[2]:
av0[1] = None
return None, None
else:
R = v[2]
if R is None:
# return the radicand from the previous invocation
return sqrt(nested[-1]), [0]*len(nested)
nested2 = [_mexpand(v[0]**2) -
_mexpand(R*v[1]**2) for v in values] + [R]
d, f = _denester(nested2, av0, h + 1, max_depth_level)
if not f:
return None, None
if not any(f[i] for i in range(len(nested))):
v = values[-1]
return sqrt(v[0] + v[1]*d), f
else:
p = Mul(*[nested[i] for i in range(len(nested)) if f[i]])
v = _sqrt_match(p)
if 1 in f and f.index(1) < len(nested) - 1 and f[len(nested) - 1]:
v[0] = -v[0]
v[1] = -v[1]
if not f[len(nested)]: #Solution denests with square roots
vad = _mexpand(v[0] + d)
if vad <= 0:
# return the radicand from the previous invocation.
return sqrt(nested[-1]), [0]*len(nested)
if not(sqrt_depth(vad) < sqrt_depth(R) + 1 or
(vad**2).is_Number):
av0[1] = None
return None, None
vad1 = radsimp(1/vad)
return _mexpand(sqrt(vad/2) +
sign(v[1])*sqrt(_mexpand(v[1]**2*R*vad1/2))), f
else: #Solution requires a fourth root
s2 = _mexpand(v[1]*R) + d
if s2 <= 0:
return sqrt(nested[-1]), [0]*len(nested)
FR, s = root(_mexpand(R), 4), sqrt(s2)
return _mexpand(s/(sqrt(2)*FR) + v[0]*FR/(sqrt(2)*s)), f
开发者ID:MichaelMayorov,项目名称:sympy,代码行数:89,代码来源:sqrtdenest.py
示例17: Jump
def Jump(x,x_i,Xi):
return 1./2.*(sign(x-Xi)-sign(x_i-Xi))
开发者ID:axelvonderheide,项目名称:scratch,代码行数:2,代码来源:function.py
示例18: refine_Pow
def refine_Pow(expr, assumptions):
"""
Handler for instances of Pow.
>>> from sympy import Symbol, Q
>>> from sympy.assumptions.refine import refine_Pow
>>> from sympy.abc import x,y,z
>>> refine_Pow((-1)**x, Q.real(x))
>>> refine_Pow((-1)**x, Q.even(x))
1
>>> refine_Pow((-1)**x, Q.odd(x))
-1
For powers of -1, even parts of the exponent can be simplified:
>>> refine_Pow((-1)**(x+y), Q.even(x))
(-1)**y
>>> refine_Pow((-1)**(x+y+z), Q.odd(x) & Q.odd(z))
(-1)**y
>>> refine_Pow((-1)**(x+y+2), Q.odd(x))
(-1)**(y + 1)
>>> refine_Pow((-1)**(x+3), True)
(-1)**(x + 1)
"""
from sympy.core import Pow, Rational
from sympy.functions.elementary.complexes import Abs
from sympy.functions import sign
if isinstance(expr.base, Abs):
if ask(Q.real(expr.base.args[0]), assumptions) and \
ask(Q.even(expr.exp), assumptions):
return expr.base.args[0] ** expr.exp
if ask(Q.real(expr.base), assumptions):
if expr.base.is_number:
if ask(Q.even(expr.exp), assumptions):
return abs(expr.base) ** expr.exp
if ask(Q.odd(expr.exp), assumptions):
return sign(expr.base) * abs(expr.base) ** expr.exp
if isinstance(expr.exp, Rational):
if type(expr.base) is Pow:
return abs(expr.base.base) ** (expr.base.exp * expr.exp)
if expr.base is S.NegativeOne:
if expr.exp.is_Add:
old = expr
# For powers of (-1) we can remove
# - even terms
# - pairs of odd terms
# - a single odd term + 1
# - A numerical constant N can be replaced with mod(N,2)
coeff, terms = expr.exp.as_coeff_add()
terms = set(terms)
even_terms = set([])
odd_terms = set([])
initial_number_of_terms = len(terms)
for t in terms:
if ask(Q.even(t), assumptions):
even_terms.add(t)
elif ask(Q.odd(t), assumptions):
odd_terms.add(t)
terms -= even_terms
if len(odd_terms) % 2:
terms -= odd_terms
new_coeff = (coeff + S.One) % 2
else:
terms -= odd_terms
new_coeff = coeff % 2
if new_coeff != coeff or len(terms) < initial_number_of_terms:
terms.add(new_coeff)
expr = expr.base**(Add(*terms))
# Handle (-1)**((-1)**n/2 + m/2)
e2 = 2*expr.exp
if ask(Q.even(e2), assumptions):
if e2.could_extract_minus_sign():
e2 *= expr.base
if e2.is_Add:
i, p = e2.as_two_terms()
if p.is_Pow and p.base is S.NegativeOne:
if ask(Q.integer(p.exp), assumptions):
i = (i + 1)/2
if ask(Q.even(i), assumptions):
return expr.base**p.exp
elif ask(Q.odd(i), assumptions):
return expr.base**(p.exp + 1)
else:
return expr.base**(p.exp + i)
if old != expr:
return expr
开发者ID:kumarkrishna,项目名称:sympy,代码行数:96,代码来源:refine.py
示例19: test_NumPyPrinter
def test_NumPyPrinter():
p = NumPyPrinter()
assert p.doprint(sign(x)) == 'numpy.sign(x)'
A = MatrixSymbol("A", 2, 2)
assert p.doprint(A**(-1)) == "numpy.linalg.inv(A)"
assert p.doprint(A**5) == "numpy.linalg.matrix_power(A, 5)"
开发者ID:asmeurer,项目名称:sympy,代码行数:6,代码来源:test_pycode.py
示例20: test_MpmathPrinter
def test_MpmathPrinter():
p = MpmathPrinter()
assert p.doprint(sign(x)) == 'mpmath.sign(x)'
assert p.doprint(Rational(1, 2)) == 'mpmath.mpf(1)/mpmath.mpf(2)'
开发者ID:Lenqth,项目名称:sympy,代码行数:4,代码来源:test_pycode.py
注:本文中的sympy.functions.sign函数示例由纯净天空整理自Github/MSDocs等源码及文档管理平台,相关代码片段筛选自各路编程大神贡献的开源项目,源码版权归原作者所有,传播和使用请参考对应项目的License;未经允许,请勿转载。 |
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