Python iterables.uniq函数代码示例

    def _eval_extract(self, rowsList, colsList):
        urow = list(uniq(rowsList))
        ucol = list(uniq(colsList))
        smat = {}
        if len(urow)*len(ucol) < len(self._smat):
            # there are fewer elements requested than there are elements in the matrix
            for i, r in enumerate(urow):
                for j, c in enumerate(ucol):
                    smat[i, j] = self._smat.get((r, c), 0)
        else:
            # most of the request will be zeros so check all of self's entries,
            # keeping only the ones that are desired
            for rk, ck in self._smat:
                if rk in urow and ck in ucol:
                    smat[(urow.index(rk), ucol.index(ck))] = self._smat[(rk, ck)]

        rv = self._new(len(urow), len(ucol), smat)
        # rv is nominally correct but there might be rows/cols
        # which require duplication
        if len(rowsList) != len(urow):
            for i, r in enumerate(rowsList):
                i_previous = rowsList.index(r)
                if i_previous != i:
                    rv = rv.row_insert(i, rv.row(i_previous))
        if len(colsList) != len(ucol):
            for i, c in enumerate(colsList):
                i_previous = colsList.index(c)
                if i_previous != i:
                    rv = rv.col_insert(i, rv.col(i_previous))
        return rv

    def __eq__(self, other):
        if not isinstance(other, Subs):
            return False
        if (len(self.point) != len(other.point) or
            self.free_symbols != other.free_symbols or
            sorted(self.point) != sorted(other.point)):
            return False

        # non-repeated point args
        selfargs = [ v[0] for v in sorted(zip(self.variables, self.point),
            key = lambda v: v[1]) if list(self.point.args).count(v[1]) == 1 ]
        otherargs = [ v[0] for v in sorted(zip(other.variables, other.point),
            key = lambda v: v[1]) if list(other.point.args).count(v[1]) == 1 ]
        # find repeated point values and subs each associated variable
        # for a single symbol
        selfrepargs = []
        otherrepargs = []
        if uniq(self.point) != self.point:
            repeated = uniq([ v for v in self.point if
                                list(self.point.args).count(v) > 1 ])
            repswap = dict(zip(repeated, [ C.Dummy() for _ in
                                            xrange(len(repeated)) ]))
            selfrepargs = [ (self.variables[i], repswap[v]) for i, v in
                            enumerate(self.point) if v in repeated ]
            otherrepargs = [ (other.variables[i], repswap[v]) for i, v in
                            enumerate(other.point) if v in repeated ]

        return self.expr.subs(selfrepargs) == other.expr.subs(
                tuple(zip(otherargs, selfargs))).subs(otherrepargs)

def bifid5_square(key):
    r"""
    5x5 Polybius square.

    Produce the Polybius square for the `5 \times 5` Bifid cipher.

    Examples
    ========

    >>> from sympy.crypto.crypto import bifid5_square
    >>> bifid5_square("gold bug")
    Matrix([
    [G, O, L, D, B],
    [U, A, C, E, F],
    [H, I, K, M, N],
    [P, Q, R, S, T],
    [V, W, X, Y, Z]])

    """
    A = alphabet_of_cipher()
    # first make sure the letters are capitalized
    # and key has no spaces or duplicates
    key = uniq(key)
    key0 = [x.capitalize() for x in key if x.isalnum()]
    # create long key
    long_key = key0 + [x for x in A if (not(x in key0) and x != "J")]
    f = lambda i, j: Symbol(long_key[5*i + j])
    M = Matrix(5, 5, f)
    return M

def bifid7_square(key):
    r"""
    7x7 Polybius square.

    Produce the Polybius square for the `7 \times 7` Bifid cipher.
    Assumes alphabet of symbols is "A", ..., "Z", "0", ..., "22".
    (Also, assumes you have some way of distinguishing "22"
    from "2", "2" juxtaposed together for deciphering...)

    Examples
    ========

    >>> from sympy.crypto.crypto import bifid7_square
    >>> bifid7_square("gold bug")
    Matrix([
    [ G,  O,  L,  D,  B,  U,  A],
    [ C,  E,  F,  H,  I,  J,  K],
    [ M,  N,  P,  Q,  R,  S,  T],
    [ V,  W,  X,  Y,  Z,  0,  1],
    [ 2,  3,  4,  5,  6,  7,  8],
    [ 9, 10, 11, 12, 13, 14, 15],
    [16, 17, 18, 19, 20, 21, 22]])

    """
    A = alphabet_of_cipher() + [str(a) for a in range(23)]
    # first make sure the letters are capitalized
    # and text has no spaces
    key = uniq(key)
    key0 = [x.capitalize() for x in key if x.isalnum()]
    # create long key
    long_key = key0 + [x for x in A if (not(x in key0))]
    f = lambda i, j: Symbol(long_key[7*i + j])
    M = Matrix(7, 7, f)
    return M

def decipher_vigenere(ct, key, symbols="ABCDEFGHIJKLMNOPQRSTUVWXYZ"):
    """
    Decode using the Vigenère cipher.

    Examples
    ========

    >>> from sympy.crypto.crypto import decipher_vigenere
    >>> key = "encrypt"
    >>> ct = "QRGK kt HRZQE BPR"
    >>> decipher_vigenere(ct, key)
    'MEETMEONMONDAY'

    """
    symbols = "".join(symbols)
    A = alphabet_of_cipher(symbols)
    N = len(A)   # normally, 26
    key0 = uniq(key)
    key0 = [x.capitalize() for x in key0 if x.isalnum()]
    K = [A.index(x) for x in key0]
    k = len(K)
    ct0 = [x.capitalize() for x in ct if x.isalnum()]
    C = [A.index(x) for x in ct0]
    n = len(C)
    #m = n//k
    P = [(-K[i % k] + C[i]) % N for i in range(n)]
    return "".join([str(A[x]) for x in P])

 def _run(coeffs):
     # find runs in coeffs such that the difference in terms (mod 1)
     # of t1, t2, ..., tn is 1/n
     u = list(uniq(coeffs))
     for i in range(len(u)):
         dj = ([((u[j] - u[i]) % 1, j) for j in range(i + 1, len(u))])
         for one, j in dj:
             if one.p == 1 and one.q != 1:
                 n = one.q
                 got = [i]
                 get = list(range(1, n))
                 for d, j in dj:
                     m = n*d
                     if m.is_Integer and m in get:
                         get.remove(m)
                         got.append(j)
                         if not get:
                             break
                 else:
                     continue
                 for i, j in enumerate(got):
                     c = u[j]
                     coeffs.remove(c)
                     got[i] = c
                 return one.q, got[0], got[1:]

    def __new__(cls, expr, variables, point, **assumptions):
        if not ordered_iter(variables, Tuple):
            variables = [variables]
        variables = Tuple(*sympify(variables))

        if uniq(variables) != variables:
            repeated = repeated = [ v for v in set(variables)
                                    if list(variables).count(v) > 1 ]
            raise ValueError('cannot substitute expressions %s more than '
                             'once.' % repeated)

        if not ordered_iter(point, Tuple):
            point = [point]
        point = Tuple(*sympify(point))

        if len(point) != len(variables):
            raise ValueError('Number of point values must be the same as '
                             'the number of variables.')

        # it's necessary to use dummy variables internally
        new_variables = Tuple(*[ arg.as_dummy() if arg.is_Symbol else
            C.Dummy(str(arg)) for arg in variables ])
        expr = sympify(expr).subs(tuple(zip(variables, new_variables)))

        if expr.is_commutative:
            assumptions['commutative'] = True

        obj = Expr.__new__(cls, expr, new_variables, point, **assumptions)
        return obj

def bifid6_square(key):
    r"""
    6x6 Polybius square.

    Produces the Polybius square for the `6 \times 6` Bifid cipher.
    Assumes alphabet of symbols is "A", ..., "Z", "0", ..., "9".

    Examples
    ========

    >>> from sympy.crypto.crypto import bifid6_square
    >>> key = "encrypt"
    >>> bifid6_square(key)
    Matrix([
    [E, N, C, R, Y, P],
    [T, A, B, D, F, G],
    [H, I, J, K, L, M],
    [O, Q, S, U, V, W],
    [X, Z, 0, 1, 2, 3],
    [4, 5, 6, 7, 8, 9]])

    """
    A = alphabet_of_cipher() + [str(a) for a in range(10)]
    # first make sure the letters are capitalized
    # and text has no spaces
    key = uniq(key)
    key0 = [x.capitalize() for x in key if x.isalnum()]
    # create long key
    long_key = key0 + [x for x in A if not(x in key0)]
    f = lambda i, j: Symbol(long_key[6*i + j])
    M = Matrix(6, 6, f)
    return M

def test_uniq():
    assert list(uniq(p.copy() for p in partitions(4))) == [{4: 1}, {1: 1, 3: 1}, {2: 2}, {1: 2, 2: 1}, {1: 4}]
    assert list(uniq(x % 2 for x in range(5))) == [0, 1]
    assert list(uniq("a")) == ["a"]
    assert list(uniq("ababc")) == list("abc")
    assert list(uniq([[1], [2, 1], [1]])) == [[1], [2, 1]]
    assert list(uniq(permutations(i for i in [[1], 2, 2]))) == [([1], 2, 2), (2, [1], 2), (2, 2, [1])]
    assert list(uniq([2, 3, 2, 4, [2], [1], [2], [3], [1]])) == [2, 3, 4, [2], [1], [3]]

    def is_concyclic(self, *args):
        """Do `self` and the given sequence of points lie in a circle?

        Returns True if the set of points are concyclic and
        False otherwise. A trivial value of True is returned
        if there are fewer than 2 other points.

        Parameters
        ==========

        args : sequence of Points

        Returns
        =======

        is_concyclic : boolean


        Examples
        ========

        >>> from sympy import Point

        Define 4 points that are on the unit circle:

        >>> p1, p2, p3, p4 = Point(1, 0), (0, 1), (-1, 0), (0, -1)

        >>> p1.is_concyclic() == p1.is_concyclic(p2, p3, p4) == True
        True

        Define a point not on that circle:

        >>> p = Point(1, 1)

        >>> p.is_concyclic(p1, p2, p3)
        False

        """
        points = (self,) + args
        points = Point._normalize_dimension(*[Point(i) for i in points])
        points = list(uniq(points))
        if not Point.affine_rank(*points) <= 2:
            return False
        origin = points[0]
        points = [p - origin for p in points]
        # points are concyclic if they are coplanar and
        # there is a point c so that ||p_i-c|| == ||p_j-c|| for all
        # i and j.  Rearranging this equation gives us the following
        # condition: the matrix `mat` must not a pivot in the last
        # column.
        mat = Matrix([list(i) + [i.dot(i)] for i in points])
        rref, pivots = mat.rref()
        if len(origin) not in pivots:
            return True
        return False

    def _do_ellipse_intersection(self, o):
        """The intersection of an ellipse with another ellipse or a circle.

        Private helper method for `intersection`.

        """

        x = Dummy('x', real=True)
        y = Dummy('y', real=True)
        seq = self.equation(x, y)
        oeq = o.equation(x, y)
        result = solve([seq, oeq], [x, y])
        return [Point(*r) for r in list(uniq(result))]

    def _do_ellipse_intersection(self, o):
        """The intersection of an ellipse with another ellipse or a circle.

        Private helper method for `intersection`.

        """
        x = Dummy('x', real=True)
        y = Dummy('y', real=True)
        seq = self.equation(x, y)
        oeq = o.equation(x, y)
        # TODO: Replace solve with nonlinsolve, when nonlinsolve will be able to solve in real domain
        result = solve([seq, oeq], x, y)
        return [Point(*r) for r in list(uniq(result))]

def DirectProduct(*groups):
    """
    Returns the direct product of several groups as a permutation group.

    This is implemented much like the __mul__ procedure for taking the direct
    product of two permutation groups, but the idea of shifting the
    generators is realized in the case of an arbitrary number of groups.
    A call to DirectProduct(G1, G2, ..., Gn) is generally expected to be faster
    than a call to G1*G2*...*Gn (and thus the need for this algorithm).

    Examples
    ========

    >>> from sympy.combinatorics.group_constructs import DirectProduct
    >>> from sympy.combinatorics.named_groups import CyclicGroup
    >>> C = CyclicGroup(4)
    >>> G = DirectProduct(C, C, C)
    >>> G.order()
    64

    See Also
    ========

    __mul__

    """
    degrees = []
    gens_count = []
    total_degree = 0
    total_gens = 0
    for group in groups:
        current_deg = group.degree
        current_num_gens = len(group.generators)
        degrees.append(current_deg)
        total_degree += current_deg
        gens_count.append(current_num_gens)
        total_gens += current_num_gens
    array_gens = []
    for i in range(total_gens):
        array_gens.append(list(range(total_degree)))
    current_gen = 0
    current_deg = 0
    for i in range(len(gens_count)):
        for j in range(current_gen, current_gen + gens_count[i]):
            gen = ((groups[i].generators)[j - current_gen]).array_form
            array_gens[j][current_deg:current_deg + degrees[i]] = \
                [x + current_deg for x in gen]
        current_gen += gens_count[i]
        current_deg += degrees[i]
    perm_gens = list(uniq([_af_new(list(a)) for a in array_gens]))
    return PermutationGroup(perm_gens, dups=False)

def test_uniq():
    assert list(uniq(p.copy() for p in partitions(4))) == \
        [{4: 1}, {1: 1, 3: 1}, {2: 2}, {1: 2, 2: 1}, {1: 4}]
    assert list(uniq(x % 2 for x in range(5))) == [0, 1]
    assert list(uniq('a')) == ['a']
    assert list(uniq('ababc')) == list('abc')
    assert list(uniq([[1], [2, 1], [1]])) == [[1], [2, 1], [1]]
    assert list(uniq(permutations(i for i in [[1], 2, 2]))) == \
        [([1], 2, 2), (2, [1], 2), (2, 2, [1]), (2, [1], 2), (2, 2, [1])]

    def are_concurrent(*planes):
        """Is a sequence of Planes concurrent?

        Two or more Planes are concurrent if their intersections
        are a common line.

        Parameters
        ==========

        planes: list

        Returns
        =======

        Boolean

        Examples
        ========

        >>> from sympy import Plane, Point3D
        >>> a = Plane(Point3D(5, 0, 0), normal_vector=(1, -1, 1))
        >>> b = Plane(Point3D(0, -2, 0), normal_vector=(3, 1, 1))
        >>> c = Plane(Point3D(0, -1, 0), normal_vector=(5, -1, 9))
        >>> Plane.are_concurrent(a, b)
        True
        >>> Plane.are_concurrent(a, b, c)
        False

        """
        planes = list(uniq(planes))
        for i in planes:
            if not isinstance(i, Plane):
                raise ValueError('All objects should be Planes but got %s' % i.func)
        if len(planes) < 2:
            return False
        planes = list(planes)
        first = planes.pop(0)
        sol = first.intersection(planes[0])
        if sol == []:
            return False
        else:
            line = sol[0]
            for i in planes[1:]:
                l = first.intersection(i)
                if not l or not l[0] in line:
                    return False
            return True