Exemplo n.º 1
0
def test_matrix_from_list_as_column_vector():
    m = matrix([0, 1, 2])
    assert m.nr() == 3
    assert m.nc() == 1
    assert m.shape == (3, 1)
    assert len(m) == 3
    assert repr(m) == "< dlib.matrix containing: \n0 \n1 \n2 >"
    assert str(m) == "0 \n1 \n2"
Exemplo n.º 2
0
def test_matrix_empty_init():
    m = matrix()
    assert m.nr() == 0
    assert m.nc() == 0
    assert m.shape == (0, 0)
    assert len(m) == 0
    assert repr(m) == "< dlib.matrix containing: >"
    assert str(m) == ""
Exemplo n.º 3
0
 def test_matrix_from_object_with_2d_shape():
     m1 = numpy.array([[0, 1, 2],
                     [3, 4, 5],
                     [6, 7, 8]])
     m = matrix(m1)
     assert m.nr() == 3
     assert m.nc() == 3
     assert m.shape == (3, 3)
     assert len(m) == 3
     assert repr(m) == "< dlib.matrix containing: \n0 1 2 \n3 4 5 \n6 7 8 >"
     assert str(m) == "0 1 2 \n3 4 5 \n6 7 8"
Exemplo n.º 4
0
def test_matrix_set_size():
    m = matrix()
    m.set_size(5, 5)

    assert m.nr() == 5
    assert m.nc() == 5
    assert m.shape == (5, 5)
    assert len(m) == 5
    assert repr(m) == "< dlib.matrix containing: \n0 0 0 0 0 \n0 0 0 0 0 \n0 0 0 0 0 \n0 0 0 0 0 \n0 0 0 0 0 >"
    assert str(m) == "0 0 0 0 0 \n0 0 0 0 0 \n0 0 0 0 0 \n0 0 0 0 0 \n0 0 0 0 0"

    deser = pickle.loads(pickle.dumps(m, 2))

    for row in range(5):
        for col in range(5):
            assert m[row][col] == deser[row][col]
Exemplo n.º 5
0
def test_matrix_from_list():
    m = matrix([[0, 1, 2],
                [3, 4, 5],
                [6, 7, 8]])
    assert m.nr() == 3
    assert m.nc() == 3
    assert m.shape == (3, 3)
    assert len(m) == 3
    assert repr(m) == "< dlib.matrix containing: \n0 1 2 \n3 4 5 \n6 7 8 >"
    assert str(m) == "0 1 2 \n3 4 5 \n6 7 8"

    deser = pickle.loads(pickle.dumps(m, 2))

    for row in range(3):
        for col in range(3):
            assert m[row][col] == deser[row][col]
Exemplo n.º 6
0
        c = a
        a = b
        b = c
    d = list()
    for i in range(len(a)):
        d.append(list())
        for j in range(len(b)):
            d[-1].append(int(10 * (dist1(a[i:], b[j:]) + dist1(a[i::-1], b[j::-1]))))
    return d


# print(distM(textA, textB))

# print(textA[0] == textB[1])

cost = dlib.matrix(distM(textA, textB))
print("cost matrix ready")
before = datetime.datetime.now()
assignment = dlib.max_cost_assignment(cost)
after = datetime.datetime.now()
print(textA)
print(textB)
print(ruler)

# This prints optimal assignments:  [2, 0, 1]
# which indicates that we should assign the person from the first row of the
# cost matrix to job 2, the middle row person to job 0, and the bottom row
# person to job 1.
print("Optimal assignments: {}".format(assignment))
print(after - before)
Exemplo n.º 7
0
    print "%d source / %d target docs" \
        % (n_source, n_target)

    print datetime.now()
    if args.matching:
        print "Finding best matching"
        matching_pairs = set()

        full_matrix = np.pad(
            score_matrix,
            ((0, max(score_matrix.shape) - score_matrix.shape[0]),
             (0, max(score_matrix.shape) - score_matrix.shape[1])),
            mode='constant')

        cost = dlib.matrix(full_matrix)
        print "Searching with dlib"
        assignment = dlib.max_cost_assignment(cost)

        for sidx, tidx in enumerate(assignment):
            if sidx >= score_matrix.shape[0] or tidx >= score_matrix.shape[1]:
                continue
            matching_pairs.add((sidx, tidx))

        print "Found %d matches " % (len(matching_pairs))
        found = devset.intersection(matching_pairs)
        print "Found %d out of %d pairs = %f%%" \
            % (len(found), len(devset), 100. * len(found) / len(devset))
        print "RES:\t%s\t%s\t%d\t%d" % (args.prefix, args.matrix.name,
                                        len(found), len(devset))
    else:
Exemplo n.º 8
0
def test_matrix_from_object_without_shape():
    with raises(AttributeError):
        matrix("invalid")
Exemplo n.º 9
0
 def test_matrix_from_object_without_2d_shape():
     with raises(IndexError):
         m1 = numpy.array([0, 1, 2])
         matrix(m1)
Exemplo n.º 10
0
        c = a
        a = b
        b = c
    d = list()
    for i in range(len(a)):
        d.append(list())
        for j in range(len(b)):
            d[-1].append(int(10 * (
                dist1(a[i:], b[j:]) + dist1(a[i::-1], b[j::-1])
                )))
    return d

#print(distM(textA, textB))

#print(textA[0] == textB[1])

cost = dlib.matrix(distM(textA, textB))
print('cost matrix ready')
before = datetime.datetime.now()
assignment = dlib.max_cost_assignment(cost)
after = datetime.datetime.now()
print(textA)
print(textB)
print(ruler)

# This prints optimal assignments:  [2, 0, 1]
# which indicates that we should assign the person from the first row of the
# cost matrix to job 2, the middle row person to job 0, and the bottom row
# person to job 1.
print("Optimal assignments: {}".format(assignment))
print(after - before)
Exemplo n.º 11
0
def test_matrix_from_list_with_invalid_rows():
    with raises(ValueError):
        matrix([[0, 1, 2],
                [3, 4],
                [5, 6, 7]])
Exemplo n.º 12
0
    print datetime.now()
    matches = []
    if args.matching:
        print "Finding best matching"

        full_matrix = np.pad(
            score_matrix,
            ((0, max(score_matrix.shape) - score_matrix.shape[0]),
             (0, max(score_matrix.shape) - score_matrix.shape[1])),
            mode='constant')

        # print full_matrix.shape, np.sum(full_matrix)
        # print score_matrix.shape, np.sum(score_matrix)

        import dlib
        cost = dlib.matrix(full_matrix)
        print "Searching with dlib"
        assignment = dlib.max_cost_assignment(cost)
        # print assignment

        for sidx, tidx in enumerate(assignment):
            if sidx >= score_matrix.shape[0] or tidx >= score_matrix.shape[1]:
                continue
            matches.append((score_matrix[sidx, tidx], sidx, tidx))

        matches.sort(reverse=True)
        matches = [(sidx, tidx) for score, sidx, tidx in matches]

    else:
        print "Finding best match (greedy / restricted + argsort)"
        seen_cols = set()
Exemplo n.º 13
0
def test_matrix_from_object_without_shape():
    with raises(AttributeError):
        matrix("invalid")
Exemplo n.º 14
0
 def test_matrix_from_object_without_2d_shape():
     with raises(IndexError):
         m1 = numpy.array([0, 1, 2])
         matrix(m1)
Exemplo n.º 15
0
def test_matrix_from_list_with_invalid_rows():
    with raises(ValueError):
        matrix([[0, 1, 2],
                [3, 4],
                [5, 6, 7]])
    def match(input_matrix, allow_zero_scores=False):
        """
        Builds match list
        :param input_matrix: 2 dimensional array of scores
        :param allow_zero_scores: Should items with a score of 0 be considered to be matched?  Default is False
        :return:
        """
        out_dict = dict()
        out_dict["Matches"] = []
        out_dict["Details"] = {"Quality Scores": []}
        out_dict["Quality Score"] = 0
        out_dict["Precision"] = 0
        out_dict["Recall"] = 0
        out_dict["F1"] = 0
        out_dict["Mercury Score"] = 0
        warn_id_list = list(input_matrix.index)
        event_id_list = list(input_matrix.columns)
        score_matrix = np.array(input_matrix)

        if min(score_matrix.shape) == 0:
            pass
        else:
            if np.max(score_matrix) <= 0:
                pass
            else:
                score_matrix = score_matrix.copy(order="C")
                row_count, col_count = score_matrix.shape
                zero_matrix = np.zeros(score_matrix.shape)
                score_matrix = np.maximum(score_matrix, zero_matrix)

                k_max = max(score_matrix.shape[0], score_matrix.shape[1])
                score_matrix_square = np.zeros((k_max, k_max))
                score_matrix_square[:score_matrix.shape[0], :score_matrix.
                                    shape[1]] = score_matrix

                assign = dlib.max_cost_assignment(
                    dlib.matrix(score_matrix_square))
                assign = [(i, assign[i]) for i in range(k_max)]

                assign_scores = np.array(
                    [score_matrix_square[x[0], x[1]] for x in assign])
                assign = np.array(assign)

                #assign = assign[assign_scores > 0]

                if not allow_zero_scores:
                    assign_scores = np.array(
                        [score_matrix_square[x[0], x[1]] for x in assign])
                    assign = np.array(assign)
                    assign = assign[assign_scores > 0]
                assign = list([tuple(x) for x in assign])
                assign = [(int(x[0]), int(x[1])) for x in assign]
                assign = [(warn_id_list[a[0]], event_id_list[a[1]])
                          for a in assign]
                out_dict["Matches"] = assign
                scores_ = assign_scores[assign_scores > 0]
                out_dict["Quality Score"] = np.mean(scores_)
                out_dict["Details"] = {"Quality Scores": list(scores_)}
                prec = len(assign) / row_count
                rec = len(assign) / col_count
                out_dict["Precision"] = prec
                out_dict["Recall"] = rec
                out_dict["F1"] = Scorer.f1(prec, rec)
                out_dict["Mercury Score"] = (out_dict["Quality Score"] +
                                             out_dict["F1"]) / 2

        return out_dict
Exemplo n.º 17
0
# Let's imagine you need to assign N people to N jobs.  Additionally, each
# person will make your company a certain amount of money at each job, but each
# person has different skills so they are better at some jobs and worse at
# others.  You would like to find the best way to assign people to these jobs.
# In particular, you would like to maximize the amount of money the group makes
# as a whole.  This is an example of an assignment problem and is what is solved
# by the dlib.max_cost_assignment() routine.

# So in this example, let's imagine we have 3 people and 3 jobs. We represent
# the amount of money each person will produce at each job with a cost matrix.
# Each row corresponds to a person and each column corresponds to a job. So for
# example, below we are saying that person 0 will make $1 at job 0, $2 at job 1,
# and $6 at job 2.
cost = dlib.matrix([[1, 2, 6],
                    [5, 3, 6],
                    [4, 5, 0]])

# To find out the best assignment of people to jobs we just need to call this
# function.
assignment = dlib.max_cost_assignment(cost)

# This prints optimal assignments:  [2, 0, 1]
# which indicates that we should assign the person from the first row of the
# cost matrix to job 2, the middle row person to job 0, and the bottom row
# person to job 1.
print("Optimal assignments: {}".format(assignment))

# This prints optimal cost:  16.0
# which is correct since our optimal assignment is 6+5+5.
print("Optimal cost: {}".format(dlib.assignment_cost(cost, assignment)))
Exemplo n.º 18
0
import dlib

# Lets imagine you need to assign N people to N jobs.  Additionally, each person will make
# your company a certain amount of money at each job, but each person has different skills
# so they are better at some jobs and worse at others.  You would like to find the best way
# to assign people to these jobs.  In particular, you would like to maximize the amount of
# money the group makes as a whole.  This is an example of an assignment problem and is
# what is solved by the dlib.max_cost_assignment() routine.

# So in this example, lets imagine we have 3 people and 3 jobs.  We represent the amount of
# money each person will produce at each job with a cost matrix.  Each row corresponds to a
# person and each column corresponds to a job.  So for example, below we are saying that
# person 0 will make $1 at job 0, $2 at job 1, and $6 at job 2.  
cost = dlib.matrix([[1, 2, 6],
                    [5, 3, 6],
                    [4, 5, 0]])


# To find out the best assignment of people to jobs we just need to call this function.
assignment = dlib.max_cost_assignment(cost)


# This prints optimal assignments:  [2, 0, 1]
# which indicates that we should assign the person from the first row of the cost matrix to
# job 2, the middle row person to job 0, and the bottom row person to job 1.
print "optimal assignments: ", assignment


# This prints optimal cost:  16.0
# which is correct since our optimal assignment is 6+5+5.