scipy.optimize.linear_sum_assignment¶

scipy.optimize.
linear_sum_assignment
(cost_matrix)[source]¶ Solve the linear sum assignment problem.
The linear sum assignment problem is also known as minimum weight matching in bipartite graphs. A problem instance is described by a matrix C, where each C[i,j] is the cost of matching vertex i of the first partite set (a “worker”) and vertex j of the second set (a “job”). The goal is to find a complete assignment of workers to jobs of minimal cost.
Formally, let X be a boolean matrix where \(X[i,j] = 1\) iff row i is assigned to column j. Then the optimal assignment has cost
\[\min \sum_i \sum_j C_{i,j} X_{i,j}\]s.t. each row is assignment to at most one column, and each column to at most one row.
This function can also solve a generalization of the classic assignment problem where the cost matrix is rectangular. If it has more rows than columns, then not every row needs to be assigned to a column, and vice versa.
The method used is the Hungarian algorithm, also known as the Munkres or KuhnMunkres algorithm.
 Parameters
 cost_matrixarray
The cost matrix of the bipartite graph.
 Returns
 row_ind, col_indarray
An array of row indices and one of corresponding column indices giving the optimal assignment. The cost of the assignment can be computed as
cost_matrix[row_ind, col_ind].sum()
. The row indices will be sorted; in the case of a square cost matrix they will be equal tonumpy.arange(cost_matrix.shape[0])
.
Notes
New in version 0.17.0.
References
Harold W. Kuhn. The Hungarian Method for the assignment problem. Naval Research Logistics Quarterly, 2:8397, 1955.
Harold W. Kuhn. Variants of the Hungarian method for assignment problems. Naval Research Logistics Quarterly, 3: 253258, 1956.
Munkres, J. Algorithms for the Assignment and Transportation Problems. J. SIAM, 5(1):3238, March, 1957.
Examples
>>> cost = np.array([[4, 1, 3], [2, 0, 5], [3, 2, 2]]) >>> from scipy.optimize import linear_sum_assignment >>> row_ind, col_ind = linear_sum_assignment(cost) >>> col_ind array([1, 0, 2]) >>> cost[row_ind, col_ind].sum() 5