def assignment_dlib(cost_mat):
# 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]])
cost = dlib.matrix(cost_mat)
# 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)))
return assignment
import dlib
# 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)))
# 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)))
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
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)
return out_dict
