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Python polytools.factor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.polys.polytools.factor函数的典型用法代码示例。如果您正苦于以下问题:Python factor函数的具体用法?Python factor怎么用?Python factor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factor函数的8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: roots_linear

def roots_linear(f):
    """Returns a list of roots of a linear polynomial."""
    r = -f.nth(0)/f.nth(1)
    dom = f.get_domain()

    if not dom.is_Numerical:
        if dom.is_Composite:
            r = factor(r)
        else:
            r = simplify(r)

    return [r]
开发者ID:bjodah,项目名称:sympy,代码行数:12,代码来源:polyroots.py


示例2: _rational_case

    def _rational_case(cls, poly, func):
        """Handle the rational function case. """
        roots = symbols('r:%d' % poly.degree())
        var, expr = func.variables[0], func.expr

        f = sum(expr.subs(var, r) for r in roots)
        p, q = together(f).as_numer_denom()

        domain = QQ[roots]

        p = p.expand()
        q = q.expand()

        try:
            p = Poly(p, domain=domain, expand=False)
        except GeneratorsNeeded:
            p, p_coeff = None, (p,)
        else:
            p_monom, p_coeff = zip(*p.terms())

        try:
            q = Poly(q, domain=domain, expand=False)
        except GeneratorsNeeded:
            q, q_coeff = None, (q,)
        else:
            q_monom, q_coeff = zip(*q.terms())

        coeffs, mapping = symmetrize(p_coeff + q_coeff, formal=True)
        formulas, values = viete(poly, roots), []

        for (sym, _), (_, val) in zip(mapping, formulas):
            values.append((sym, val))

        for i, (coeff, _) in enumerate(coeffs):
            coeffs[i] = coeff.subs(values)

        n = len(p_coeff)

        p_coeff = coeffs[:n]
        q_coeff = coeffs[n:]

        if p is not None:
            p = Poly(dict(zip(p_monom, p_coeff)), *p.gens).as_expr()
        else:
            (p,) = p_coeff

        if q is not None:
            q = Poly(dict(zip(q_monom, q_coeff)), *q.gens).as_expr()
        else:
            (q,) = q_coeff

        return factor(p/q)
开发者ID:A-turing-machine,项目名称:sympy,代码行数:52,代码来源:rootoftools.py


示例3: _eval_sum_hyper

def _eval_sum_hyper(f, i, a):
    """ Returns (res, cond). Sums from a to oo. """
    from sympy.functions import hyper
    from sympy.simplify import hyperexpand, hypersimp, fraction, simplify
    from sympy.polys.polytools import Poly, factor
    from sympy.core.numbers import Float

    if a != 0:
        return _eval_sum_hyper(f.subs(i, i + a), i, 0)

    if f.subs(i, 0) == 0:
        if simplify(f.subs(i, Dummy('i', integer=True, positive=True))) == 0:
            return S(0), True
        return _eval_sum_hyper(f.subs(i, i + 1), i, 0)

    hs = hypersimp(f, i)
    if hs is None:
        return None

    if isinstance(hs, Float):
        from sympy.simplify.simplify import nsimplify
        hs = nsimplify(hs)

    numer, denom = fraction(factor(hs))
    top, topl = numer.as_coeff_mul(i)
    bot, botl = denom.as_coeff_mul(i)
    ab = [top, bot]
    factors = [topl, botl]
    params = [[], []]
    for k in range(2):
        for fac in factors[k]:
            mul = 1
            if fac.is_Pow:
                mul = fac.exp
                fac = fac.base
                if not mul.is_Integer:
                    return None
            p = Poly(fac, i)
            if p.degree() != 1:
                return None
            m, n = p.all_coeffs()
            ab[k] *= m**mul
            params[k] += [n/m]*mul

    # Add "1" to numerator parameters, to account for implicit n! in
    # hypergeometric series.
    ap = params[0] + [1]
    bq = params[1]
    x = ab[0]/ab[1]
    h = hyper(ap, bq, x)

    return f.subs(i, 0)*hyperexpand(h), h.convergence_statement
开发者ID:carstimon,项目名称:sympy,代码行数:52,代码来源:summations.py


示例4: roots_linear

def roots_linear(f):
    """Returns a list of roots of a linear polynomial."""
    
    add_comment('This equation is linear')

    r = -f.nth(0)/f.nth(1)
    dom = f.get_domain()

    if not dom.is_Numerical:
        if dom.is_Composite:
            r = factor(r)
        else:
            r = simplify(r)
    add_comment("The root of this equation is")
    add_eq(f.gen, r)
    return [r]
开发者ID:hrashk,项目名称:sympy,代码行数:16,代码来源:polyroots.py


示例5: roots_linear

def roots_linear(f):
    """Returns a list of roots of a linear polynomial."""
    
    add_eq(f.as_expr(), 0)
    
    tmp = str(Poly(f.as_expr()).factor_list()).split(',')[2][1:]

    r = -f.nth(0)/f.nth(1)
    dom = f.get_domain()

    if not dom.is_Numerical:
        if dom.is_Composite:
            r = factor(r)
        else:
            r = simplify(r)
    tmp = str(Poly(f.as_expr()).factor_list()).split(',')[2][1:]
    x = r
    add_eq(tmp, r)
    add_comment('')
    return [r]
开发者ID:Frifon,项目名称:sympy,代码行数:20,代码来源:polyroots.py


示例6: _simplify

 def _simplify(expr):
     if dom.is_Composite:
         return factor(expr)
     else:
         return simplify(expr)
开发者ID:bjodah,项目名称:sympy,代码行数:5,代码来源:polyroots.py


示例7: roots_quintic

def roots_quintic(f):
    """
    Calculate exact roots of a solvable quintic
    """
    result = []
    coeff_5, coeff_4, p, q, r, s = f.all_coeffs()

    # Eqn must be of the form x^5 + px^3 + qx^2 + rx + s
    if coeff_4:
        return result

    if coeff_5 != 1:
        l = [p/coeff_5, q/coeff_5, r/coeff_5, s/coeff_5]
        if not all(coeff.is_Rational for coeff in l):
            return result
        f = Poly(f/coeff_5)
    quintic = PolyQuintic(f)

    # Eqn standardized. Algo for solving starts here
    if not f.is_irreducible:
        return result

    f20 = quintic.f20
    # Check if f20 has linear factors over domain Z
    if f20.is_irreducible:
        return result

    # Now, we know that f is solvable
    for _factor in f20.factor_list()[1]:
        if _factor[0].is_linear:
            theta = _factor[0].root(0)
            break
    d = discriminant(f)
    delta = sqrt(d)
    # zeta = a fifth root of unity
    zeta1, zeta2, zeta3, zeta4 = quintic.zeta
    T = quintic.T(theta, d)
    tol = S(1e-10)
    alpha = T[1] + T[2]*delta
    alpha_bar = T[1] - T[2]*delta
    beta = T[3] + T[4]*delta
    beta_bar = T[3] - T[4]*delta

    disc = alpha**2 - 4*beta
    disc_bar = alpha_bar**2 - 4*beta_bar

    l0 = quintic.l0(theta)

    l1 = _quintic_simplify((-alpha + sqrt(disc)) / S(2))
    l4 = _quintic_simplify((-alpha - sqrt(disc)) / S(2))

    l2 = _quintic_simplify((-alpha_bar + sqrt(disc_bar)) / S(2))
    l3 = _quintic_simplify((-alpha_bar - sqrt(disc_bar)) / S(2))

    order = quintic.order(theta, d)
    test = (order*delta.n()) - ( (l1.n() - l4.n())*(l2.n() - l3.n()) )
    # Comparing floats
    if not comp(test, 0, tol):
        l2, l3 = l3, l2

    # Now we have correct order of l's
    R1 = l0 + l1*zeta1 + l2*zeta2 + l3*zeta3 + l4*zeta4
    R2 = l0 + l3*zeta1 + l1*zeta2 + l4*zeta3 + l2*zeta4
    R3 = l0 + l2*zeta1 + l4*zeta2 + l1*zeta3 + l3*zeta4
    R4 = l0 + l4*zeta1 + l3*zeta2 + l2*zeta3 + l1*zeta4

    Res = [None, [None]*5, [None]*5, [None]*5, [None]*5]
    Res_n = [None, [None]*5, [None]*5, [None]*5, [None]*5]
    sol = Symbol('sol')

    # Simplifying improves performance a lot for exact expressions
    R1 = _quintic_simplify(R1)
    R2 = _quintic_simplify(R2)
    R3 = _quintic_simplify(R3)
    R4 = _quintic_simplify(R4)

    # Solve imported here. Causing problems if imported as 'solve'
    # and hence the changed name
    from sympy.solvers.solvers import solve as _solve
    a, b = symbols('a b', cls=Dummy)
    _sol = _solve( sol**5 - a - I*b, sol)
    for i in range(5):
        _sol[i] = factor(_sol[i])
    R1 = R1.as_real_imag()
    R2 = R2.as_real_imag()
    R3 = R3.as_real_imag()
    R4 = R4.as_real_imag()

    for i, currentroot in enumerate(_sol):
        Res[1][i] = _quintic_simplify(currentroot.subs({ a: R1[0], b: R1[1] }))
        Res[2][i] = _quintic_simplify(currentroot.subs({ a: R2[0], b: R2[1] }))
        Res[3][i] = _quintic_simplify(currentroot.subs({ a: R3[0], b: R3[1] }))
        Res[4][i] = _quintic_simplify(currentroot.subs({ a: R4[0], b: R4[1] }))

    for i in range(1, 5):
        for j in range(5):
            Res_n[i][j] = Res[i][j].n()
            Res[i][j] = _quintic_simplify(Res[i][j])
    r1 = Res[1][0]
    r1_n = Res_n[1][0]
#.........这里部分代码省略.........
开发者ID:bjodah,项目名称:sympy,代码行数:101,代码来源:polyroots.py


示例8: _solve_lambert

def _solve_lambert(f, symbol, gens):
    """Return solution to ``f`` if it is a Lambert-type expression
    else raise NotImplementedError.

    The equality, ``f(x, a..f) = a*log(b*X + c) + d*X - f = 0`` has the
    solution,  `X = -c/b + (a/d)*W(d/(a*b)*exp(c*d/a/b)*exp(f/a))`. There
    are a variety of forms for `f(X, a..f)` as enumerated below:

    1a1)
      if B**B = R for R not [0, 1] then
      log(B) + log(log(B)) = log(log(R))
      X = log(B), a = 1, b = 1, c = 0, d = 1, f = log(log(R))
    1a2)
      if B*(b*log(B) + c)**a = R then
      log(B) + a*log(b*log(B) + c) = log(R)
      X = log(B); d=1, f=log(R)
    1b)
      if a*log(b*B + c) + d*B = R then
      X = B, f = R
    2a)
      if (b*B + c)*exp(d*B + g) = R then
      log(b*B + c) + d*B + g = log(R)
      a = 1, f = log(R) - g, X = B
    2b)
      if -b*B + g*exp(d*B + h) = c then
      log(g) + d*B + h - log(b*B + c) = 0
      a = -1, f = -h - log(g), X = B
    3)
      if d*p**(a*B + g) - b*B = c then
      log(d) + (a*B + g)*log(p) - log(c + b*B) = 0
      a = -1, d = a*log(p), f = -log(d) - g*log(p)
    """

    nrhs, lhs = f.as_independent(symbol, as_Add=True)
    rhs = -nrhs

    lamcheck = [tmp for tmp in gens
                if (tmp.func in [exp, log] or
                (tmp.is_Pow and symbol in tmp.exp.free_symbols))]
    if not lamcheck:
        raise NotImplementedError()

    if lhs.is_Mul:
        lhs = expand_log(log(lhs))
        rhs = log(rhs)

    lhs = factor(lhs, deep=True)
    # make sure we are inverted as completely as possible
    r = Dummy()
    i, lhs = _invert(lhs - r, symbol)
    rhs = i.xreplace({r: rhs})

    # For the first ones:
    # 1a1) B**B = R != 0 (when 0, there is only a solution if the base is 0,
    #                     but if it is, the exp is 0 and 0**0=1
    #                     comes back as B*log(B) = log(R)
    # 1a2) B*(a + b*log(B))**p = R or with monomial expanded or with whole
    #                              thing expanded comes back unchanged
    #     log(B) + p*log(a + b*log(B)) = log(R)
    #     lhs is Mul:
    #         expand log of both sides to give:
    #         log(B) + log(log(B)) = log(log(R))
    # 1b) d*log(a*B + b) + c*B = R
    #     lhs is Add:
    #         isolate c*B and expand log of both sides:
    #         log(c) + log(B) = log(R - d*log(a*B + b))

    soln = []
    if not soln:
        mainlog = _mostfunc(lhs, log, symbol)
        if mainlog:
            if lhs.is_Mul and rhs != 0:
                soln = _lambert(log(lhs) - log(rhs), symbol)
            elif lhs.is_Add:
                other = lhs.subs(mainlog, 0)
                if other and not other.is_Add and [
                        tmp for tmp in other.atoms(Pow)
                        if symbol in tmp.free_symbols]:
                    if not rhs:
                        diff = log(other) - log(other - lhs)
                    else:
                        diff = log(lhs - other) - log(rhs - other)
                    soln = _lambert(expand_log(diff), symbol)
                else:
                    #it's ready to go
                    soln = _lambert(lhs - rhs, symbol)

    # For the next two,
    #     collect on main exp
    #     2a) (b*B + c)*exp(d*B + g) = R
    #         lhs is mul:
    #             log to give
    #             log(b*B + c) + d*B = log(R) - g
    #     2b) -b*B + g*exp(d*B + h) = R
    #         lhs is add:
    #             add b*B
    #             log and rearrange
    #             log(R + b*B) - d*B = log(g) + h

    if not soln:
#.........这里部分代码省略.........
开发者ID:AALEKH,项目名称:sympy,代码行数:101,代码来源:bivariate.py



注:本文中的sympy.polys.polytools.factor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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