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Python iterables.common_suffix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.utilities.iterables.common_suffix函数的典型用法代码示例。如果您正苦于以下问题:Python common_suffix函数的具体用法?Python common_suffix怎么用?Python common_suffix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common_suffix函数的2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_common_prefix_suffix

def test_common_prefix_suffix():
    assert common_prefix([], [1]) == []
    assert common_prefix(range(3)) == [0, 1, 2]
    assert common_prefix(range(3), range(4)) == [0, 1, 2]
    assert common_prefix([1, 2, 3], [1, 2, 5]) == [1, 2]
    assert common_prefix([1, 2, 3], [1, 3, 5]) == [1]

    assert common_suffix([], [1]) == []
    assert common_suffix(range(3)) == [0, 1, 2]
    assert common_suffix(range(3), range(3)) == [0, 1, 2]
    assert common_suffix(range(3), range(4)) == []
    assert common_suffix([1, 2, 3], [9, 2, 3]) == [2, 3]
    assert common_suffix([1, 2, 3], [9, 7, 3]) == [3]
开发者ID:Acebulf,项目名称:sympy,代码行数:13,代码来源:test_iterables.py


示例2: factor_nc

def factor_nc(expr):
    """Return the factored form of ``expr`` while handling non-commutative
    expressions.

    **examples**
    >>> from sympy.core.exprtools import factor_nc
    >>> from sympy import Symbol
    >>> from sympy.abc import x
    >>> A = Symbol('A', commutative=False)
    >>> B = Symbol('B', commutative=False)
    >>> factor_nc((x**2 + 2*A*x + A**2).expand())
    (x + A)**2
    >>> factor_nc(((x + A)*(x + B)).expand())
    (x + A)*(x + B)
    """
    from sympy.simplify.simplify import _mexpand
    from sympy.polys import gcd, factor

    expr = sympify(expr)
    if not isinstance(expr, Expr) or not expr.args:
        return expr
    if not expr.is_Add:
        return expr.func(*[factor_nc(a) for a in expr.args])

    expr, rep, nc_symbols = _mask_nc(expr)
    if rep:
        return factor(expr).subs(rep)
    else:
        args = [a.args_cnc() for a in Add.make_args(expr)]
        c = g = l = r = S.One
        hit = False
        # find any commutative gcd term
        for i, a in enumerate(args):
            if i == 0:
                c = Mul._from_args(a[0])
            elif a[0]:
                c = gcd(c, Mul._from_args(a[0]))
            else:
                c = S.One
        if c is not S.One:
            hit = True
            c, g = c.as_coeff_Mul()
            if g is not S.One:
                for i, (cc, _) in enumerate(args):
                    cc = list(Mul.make_args(Mul._from_args(list(cc))/g))
                    args[i][0] = cc
            else:
                for i, (cc, _) in enumerate(args):
                    cc[0] = cc[0]/c
                    args[i][0] = cc
        # find any noncommutative common prefix
        for i, a in enumerate(args):
            if i == 0:
                n = a[1][:]
            else:
                n = common_prefix(n, a[1])
            if not n:
                # is there a power that can be extracted?
                if not args[0][1]:
                    break
                b, e = args[0][1][0].as_base_exp()
                ok = False
                if e.is_Integer:
                    for t in args:
                        if not t[1]:
                            break
                        bt, et = t[1][0].as_base_exp()
                        if et.is_Integer and bt == b:
                            e = min(e, et)
                        else:
                            break
                    else:
                        ok = hit = True
                        l = b**e
                        il = b**-e
                        for i, a in enumerate(args):
                            args[i][1][0] = il*args[i][1][0]
                        break
                if not ok:
                    break
        else:
            hit = True
            lenn = len(n)
            l = Mul(*n)
            for i, a in enumerate(args):
                args[i][1] = args[i][1][lenn:]
        # find any noncommutative common suffix
        for i, a in enumerate(args):
            if i == 0:
                n = a[1][:]
            else:
                n = common_suffix(n, a[1])
            if not n:
                # is there a power that can be extracted?
                if not args[0][1]:
                    break
                b, e = args[0][1][-1].as_base_exp()
                ok = False
                if e.is_Integer:
                    for t in args:
#.........这里部分代码省略.........
开发者ID:FireJade,项目名称:sympy,代码行数:101,代码来源:exprtools.py



注:本文中的sympy.utilities.iterables.common_suffix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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Python iterables.common_prefix函数代码示例发布时间:2022-05-27
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