Python pulp.LpInteger() Examples
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code examples of pulp.LpInteger().
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Example #1
Source File: run_mskmeans.py From MinSizeKmeans with GNU General Public License v3.0 | 6 votes |
def create_model(self): def distances(assignment): return l2_distance(self.data[assignment[0]], self.centroids[assignment[1]]) clusters = list(range(self.k)) assignments = [(i, j)for i in range(self.n) for j in range(self.k)] # outflow variables for data nodes self.y = pulp.LpVariable.dicts('data-to-cluster assignments', assignments, lowBound=0, upBound=1, cat=pulp.LpInteger) # outflow variables for cluster nodes self.b = pulp.LpVariable.dicts('cluster outflows', clusters, lowBound=0, upBound=self.n-self.min_size, cat=pulp.LpContinuous) # create the model self.model = pulp.LpProblem("Model for assignment subproblem", pulp.LpMinimize) # objective function self.model += pulp.lpSum(distances(assignment) * self.y[assignment] for assignment in assignments) # flow balance constraints for data nodes for i in range(self.n): self.model += pulp.lpSum(self.y[(i, j)] for j in range(self.k)) == 1 # flow balance constraints for cluster nodes for j in range(self.k): self.model += pulp.lpSum(self.y[(i, j)] for i in range(self.n)) - self.min_size == self.b[j] # flow balance constraint for the sink node self.model += pulp.lpSum(self.b[j] for j in range(self.k)) == self.n - (self.k * self.min_size)
Example #2
Source File: minmax_kmeans.py From MinSizeKmeans with GNU General Public License v3.0 | 5 votes |
def create_model(self): def distances(assignment): return l2_distance(self.data[assignment[0]], self.centroids[assignment[1]]) clusters = list(range(self.k)) assignments = [(i, j)for i in range(self.n) for j in range(self.k)] # outflow variables for data nodes self.y = pulp.LpVariable.dicts('data-to-cluster assignments', assignments, lowBound=0, upBound=1, cat=pulp.LpInteger) # outflow variables for cluster nodes self.b = pulp.LpVariable.dicts('cluster outflows', clusters, lowBound=0, upBound=self.n-self.min_size, cat=pulp.LpContinuous) # create the model self.model = pulp.LpProblem("Model for assignment subproblem", pulp.LpMinimize) # objective function self.model += pulp.lpSum([distances(assignment) * self.y[assignment] for assignment in assignments]) # flow balance constraints for data nodes for i in range(self.n): self.model += pulp.lpSum(self.y[(i, j)] for j in range(self.k)) == 1 # flow balance constraints for cluster nodes for j in range(self.k): self.model += pulp.lpSum(self.y[(i, j)] for i in range(self.n)) - self.min_size == self.b[j] # capacity constraint on outflow of cluster nodes for j in range(self.k): self.model += self.b[j] <= self.max_size - self.min_size # flow balance constraint for the sink node self.model += pulp.lpSum(self.b[j] for j in range(self.k)) == self.n - (self.k * self.min_size)
Example #3
Source File: weighted_mm_kmeans.py From MinSizeKmeans with GNU General Public License v3.0 | 5 votes |
def create_model(self): def distances(assignment): return l2_distance(self.data[assignment[0]], self.centroids[assignment[1]]) assignments = [(i, j) for i in range(self.n) for j in range(self.k)] # assignment variables self.y = pulp.LpVariable.dicts('data-to-cluster assignments', assignments, lowBound=0, upBound=1, cat=pulp.LpInteger) # create the model self.model = pulp.LpProblem("Model for assignment subproblem", pulp.LpMinimize) # objective function self.model += pulp.lpSum([distances(assignment) * self.weights[assignment[0]] * self.y[assignment] for assignment in assignments]), 'Objective Function - sum weighted squared distances to assigned centroid' # this is also weighted, otherwise the weighted centroid computation don't make sense. # constraints on the total weights of clusters for j in range(self.k): self.model += pulp.lpSum([self.weights[i] * self.y[(i, j)] for i in range(self.n)]) >= self.min_weight, "minimum weight for cluster {}".format(j) self.model += pulp.lpSum([self.weights[i] * self.y[(i, j)] for i in range(self.n)]) <= self.max_weight, "maximum weight for cluster {}".format(j) # make sure each point is assigned at least once, and only once for i in range(self.n): self.model += pulp.lpSum([self.y[(i, j)] for j in range(self.k)]) == 1, "must assign point {}".format(i)
Example #4
Source File: upper_bound_ilp.py From acl2017-interactive_summarizer with Apache License 2.0 | 5 votes |
def solve_ilp(self, N): # build the A matrix: a_ij is 1 if j-th gram appears in the i-th sentence A = np.zeros((len(self.sentences_idx), len(self.ref_ngrams_idx))) for i in self.sentences_idx: sent = self.sentences[i].untokenized_form sngrams = list(extract_ngrams2([sent], self.stemmer, self.LANGUAGE, N)) for j in self.ref_ngrams_idx: if self.ref_ngrams[j] in sngrams: A[i][j] = 1 # Define ILP variable, x_i is 1 if sentence i is selected, z_j is 1 if gram j appears in the created summary x = pulp.LpVariable.dicts('sentences', self.sentences_idx, lowBound=0, upBound=1, cat=pulp.LpInteger) z = pulp.LpVariable.dicts('grams', self.ref_ngrams_idx, lowBound=0, upBound=1, cat=pulp.LpInteger) # Define ILP problem, maximum coverage of grams from the reference summaries prob = pulp.LpProblem("ExtractiveUpperBound", pulp.LpMaximize) prob += pulp.lpSum(z[j] for j in self.ref_ngrams_idx) # Define ILP constraints, length constraint and consistency constraint (impose that z_j is 1 if j # appears in the created summary) prob += pulp.lpSum(x[i] * self.sentences[i].length for i in self.sentences_idx) <= self.sum_length for j in self.ref_ngrams_idx: prob += pulp.lpSum(A[i][j] * x[i] for i in self.sentences_idx) >= z[j] # Solve ILP problem and post-processing to get the summary try: print('Solving using CPLEX') prob.solve(pulp.CPLEX(msg=0)) except: print('Fall back to GLPK') prob.solve(pulp.GLPK(msg=0)) summary_idx = [] for idx in self.sentences_idx: if x[idx].value() == 1.0: summary_idx.append(idx) return summary_idx
Example #5
Source File: process_path.py From Cogent with BSD 3-Clause Clear License | 5 votes |
def make_into_lp_problem(good_for, N, add_noise=False): """ Helper function for solve_with_lp_and_reduce() N --- number of isoform sequences good_for --- dict of <isoform_index> --> list of matched paths index """ prob = LpProblem("The Whiskas Problem",LpMinimize) # each good_for is (isoform_index, [list of matched paths index]) # ex: (0, [1,2,4]) # ex: (3, [2,5]) used_paths = [] for t_i, p_i_s in good_for: used_paths += p_i_s used_paths = list(set(used_paths)) variables = [LpVariable(str(i),0,1,LpInteger) for i in used_paths] #variables = [LpVariable(str(i),0,1,LpInteger) for i in xrange(N)] # objective is to minimize sum_{Xi} prob += sum(v for v in variables) already_seen = set() # constraints are for each isoform, expressed as c_i * x_i >= 1 # where c_i = 1 if x_i is matched for the isoform # ex: (0, [1,2,4]) becomes t_0 = x_1 + x_2 + x_4 >= 1 for t_i, p_i_s in good_for: #c_i_s = [1 if i in p_i_s else 0 for i in xrange(N)] #prob += sum(variables[i]*(1 if i in p_i_s else 0) for i in xrange(N)) >= 1 p_i_s.sort() pattern = ",".join(map(str,p_i_s)) #print >> sys.stderr, t_i, p_i_s, pattern if pattern not in already_seen: if add_noise: prob += sum(variables[i]*(1+random.random() if p in p_i_s else 0) for i,p in enumerate(used_paths)) >= 1 else: prob += sum(variables[i]*(1 if p in p_i_s else 0) for i,p in enumerate(used_paths)) >= 1 already_seen.add(pattern) prob.writeLP('cogent.lp') return prob
Example #6
Source File: covering.py From pyspatialopt with MIT License | 4 votes |
def create_lscp_model(coverage_dict, model_file=None, delineator="$", ): """ Creates a LSCP (Location set covering problem) using the provided coverage and parameters. Writes a .lp file which can be solved with Gurobi Church, R., & Murray, A. (2009). Coverage Business Site Selection, Location Analysis, and GIS (pp. 209-233). Hoboken, New Jersey: Wiley. :param coverage_dict: (dictionary) The coverage to use to generate the model :param model_file: (string) The model file to output :param delineator: (string) The character(s) to use to delineate the layer from the ids :return: (Pulp problem) The generated problem to solve """ validate_coverage(coverage_dict, ["coverage"], ["binary"]) if not isinstance(coverage_dict, dict): raise TypeError("coverage_dict is not a dictionary") if model_file and not (isinstance(model_file, str)): raise TypeError("model_file is not a string") if not isinstance(delineator, str): raise TypeError("delineator is not a string") # create the variables demand_vars = {} for demand_id in coverage_dict["demand"]: demand_vars[demand_id] = pulp.LpVariable("Y{}{}".format(delineator, demand_id), 0, 1, pulp.LpInteger) facility_vars = {} for facility_type in coverage_dict["facilities"]: facility_vars[facility_type] = {} for facility_id in coverage_dict["facilities"][facility_type]: facility_vars[facility_type][facility_id] = pulp.LpVariable( "{}{}{}".format(facility_type, delineator, facility_id), 0, 1, pulp.LpInteger) # create the problem prob = pulp.LpProblem("LSCP", pulp.LpMinimize) # Create objective, minimize number of facilities to_sum = [] for facility_type in coverage_dict["facilities"]: for facility_id in coverage_dict["facilities"][facility_type]: to_sum.append(facility_vars[facility_type][facility_id]) prob += pulp.lpSum(to_sum) # add coverage constraints for demand_id in coverage_dict["demand"]: to_sum = [] for facility_type in coverage_dict["demand"][demand_id]["coverage"]: for facility_id in coverage_dict["demand"][demand_id]["coverage"][facility_type]: to_sum.append(facility_vars[facility_type][facility_id]) # Hack to get model to "solve" when infeasible with GLPK. # Pulp will automatically add dummy variables when the sum is empty, since these are all the same name, # it seems that GLPK doesn't read the lp problem properly and fails if not to_sum: to_sum = [pulp.LpVariable("__dummy{}{}".format(delineator, demand_id), 0, 0, pulp.LpInteger)] prob += pulp.lpSum(to_sum) >= 1, "D{}".format(demand_id) if model_file: prob.writeLP(model_file) return prob