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Python functions.sign函数代码示例

原作者: [db:作者] 来自: [db:来源] 收藏 邀请

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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