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Python Mul.could_extract_minus_sign方法代码示例

本文整理汇总了Python中sympy.core.mul.Mul.could_extract_minus_sign方法的典型用法代码示例。如果您正苦于以下问题:Python Mul.could_extract_minus_sign方法的具体用法?Python Mul.could_extract_minus_sign怎么用?Python Mul.could_extract_minus_sign使用的例子?那么恭喜您, 这里精选的方法代码示例或许可以为您提供帮助。您也可以进一步了解该方法所在sympy.core.mul.Mul的用法示例。


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示例1: eval

# 需要导入模块: from sympy.core.mul import Mul [as 别名]
# 或者: from sympy.core.mul.Mul import could_extract_minus_sign [as 别名]

#.........这里部分代码省略.........
                return p
        elif isinstance(p, Add):
            # separating into modulus and non modulus
            both_l = non_mod_l, mod_l = [], []
            for arg in p.args:
                both_l[isinstance(arg, cls)].append(arg)
            # if q same for all
            if mod_l and all(inner.args[1] == q for inner in mod_l):
                net = Add(*non_mod_l) + Add(*[i.args[0] for i in mod_l])
                return cls(net, q)

        elif isinstance(p, Mul):
            # separating into modulus and non modulus
            both_l = non_mod_l, mod_l = [], []
            for arg in p.args:
                both_l[isinstance(arg, cls)].append(arg)

            if mod_l and all(inner.args[1] == q for inner in mod_l):
                # finding distributive term
                non_mod_l = [cls(x, q) for x in non_mod_l]
                mod = []
                non_mod = []
                for j in non_mod_l:
                    if isinstance(j, cls):
                        mod.append(j.args[0])
                    else:
                        non_mod.append(j)
                prod_mod = Mul(*mod)
                prod_non_mod = Mul(*non_mod)
                prod_mod1 = Mul(*[i.args[0] for i in mod_l])
                net = prod_mod1*prod_mod
                return prod_non_mod*cls(net, q)

            if q.is_Integer and q is not S.One:
                _ = []
                for i in non_mod_l:
                    if i.is_Integer and (i % q is not S.Zero):
                        _.append(i%q)
                    else:
                        _.append(i)
                non_mod_l = _

            p = Mul(*(non_mod_l + mod_l))

        # XXX other possibilities?

        # extract gcd; any further simplification should be done by the user
        G = gcd(p, q)
        if G != 1:
            p, q = [
                gcd_terms(i/G, clear=False, fraction=False) for i in (p, q)]
        pwas, qwas = p, q

        # simplify terms
        # (x + y + 2) % x -> Mod(y + 2, x)
        if p.is_Add:
            args = []
            for i in p.args:
                a = cls(i, q)
                if a.count(cls) > i.count(cls):
                    args.append(i)
                else:
                    args.append(a)
            if args != list(p.args):
                p = Add(*args)

        else:
            # handle coefficients if they are not Rational
            # since those are not handled by factor_terms
            # e.g. Mod(.6*x, .3*y) -> 0.3*Mod(2*x, y)
            cp, p = p.as_coeff_Mul()
            cq, q = q.as_coeff_Mul()
            ok = False
            if not cp.is_Rational or not cq.is_Rational:
                r = cp % cq
                if r == 0:
                    G *= cq
                    p *= int(cp/cq)
                    ok = True
            if not ok:
                p = cp*p
                q = cq*q

        # simple -1 extraction
        if p.could_extract_minus_sign() and q.could_extract_minus_sign():
            G, p, q = [-i for i in (G, p, q)]

        # check again to see if p and q can now be handled as numbers
        rv = doit(p, q)
        if rv is not None:
            return rv*G

        # put 1.0 from G on inside
        if G.is_Float and G == 1:
            p *= G
            return cls(p, q, evaluate=False)
        elif G.is_Mul and G.args[0].is_Float and G.args[0] == 1:
            p = G.args[0]*p
            G = Mul._from_args(G.args[1:])
        return G*cls(p, q, evaluate=(p, q) != (pwas, qwas))
开发者ID:bjodah,项目名称:sympy,代码行数:104,代码来源:mod.py


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