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"
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) == ""
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"
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]
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]
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)
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:
def test_matrix_from_object_without_shape():
with raises(AttributeError):
matrix("invalid")
def test_matrix_from_object_without_2d_shape():
with raises(IndexError):
m1 = numpy.array([0, 1, 2])
matrix(m1)
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)
def test_matrix_from_list_with_invalid_rows():
with raises(ValueError):
matrix([[0, 1, 2],
[3, 4],
[5, 6, 7]])
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()
def test_matrix_from_object_without_shape():
with raises(AttributeError):
matrix("invalid")
def test_matrix_from_object_without_2d_shape():
with raises(IndexError):
m1 = numpy.array([0, 1, 2])
matrix(m1)
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
# 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)))
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.