# scikit-learn/scikit-learn

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 """ Solve the unique lowest-cost assignment problem using the Hungarian algorithm (also known as Munkres algorithm). """ # Based on original code by Brain Clapper, adapted to NumPy by Gael Varoquaux. # Heavily refactored by Lars Buitinck. # # TODO: a version of this algorithm has been incorporated in SciPy; use that # when SciPy 0.17 is released. # Copyright (c) 2008 Brian M. Clapper , Gael Varoquaux # Author: Brian M. Clapper, Gael Varoquaux # LICENSE: BSD import numpy as np def linear_assignment(X): """Solve the linear assignment problem using the Hungarian algorithm. The problem is also known as maximum weight matching in bipartite graphs. The method is also known as the Munkres or Kuhn-Munkres algorithm. Parameters ---------- X : array The cost matrix of the bipartite graph Returns ------- indices : array The pairs of (row, col) indices in the original array giving the original ordering. References ---------- 1. http://www.public.iastate.edu/~ddoty/HungarianAlgorithm.html 2. Harold W. Kuhn. The Hungarian Method for the assignment problem. *Naval Research Logistics Quarterly*, 2:83-97, 1955. 3. Harold W. Kuhn. Variants of the Hungarian method for assignment problems. *Naval Research Logistics Quarterly*, 3: 253-258, 1956. 4. Munkres, J. Algorithms for the Assignment and Transportation Problems. *Journal of the Society of Industrial and Applied Mathematics*, 5(1):32-38, March, 1957. 5. https://en.wikipedia.org/wiki/Hungarian_algorithm """ indices = _hungarian(X).tolist() indices.sort() # Re-force dtype to ints in case of empty list indices = np.array(indices, dtype=int) # Make sure the array is 2D with 2 columns. # This is needed when dealing with an empty list indices.shape = (-1, 2) return indices class _HungarianState(object): """State of one execution of the Hungarian algorithm. Parameters ---------- cost_matrix : 2D matrix The cost matrix. Does not have to be square. """ def __init__(self, cost_matrix): cost_matrix = np.atleast_2d(cost_matrix) # If there are more rows (n) than columns (m), then the algorithm # will not be able to work correctly. Therefore, we # transpose the cost function when needed. Just have to # remember to swap the result columns back later. transposed = (cost_matrix.shape[1] < cost_matrix.shape[0]) if transposed: self.C = (cost_matrix.T).copy() else: self.C = cost_matrix.copy() self.transposed = transposed # At this point, m >= n. n, m = self.C.shape self.row_uncovered = np.ones(n, dtype=np.bool) self.col_uncovered = np.ones(m, dtype=np.bool) self.Z0_r = 0 self.Z0_c = 0 self.path = np.zeros((n + m, 2), dtype=int) self.marked = np.zeros((n, m), dtype=int) def _clear_covers(self): """Clear all covered matrix cells""" self.row_uncovered[:] = True self.col_uncovered[:] = True def _hungarian(cost_matrix): """The Hungarian algorithm. Calculate the Munkres solution to the classical assignment problem and return the indices for the lowest-cost pairings. Parameters ---------- cost_matrix : 2D matrix The cost matrix. Does not have to be square. Returns ------- indices : 2D array of indices The pairs of (row, col) indices in the original array giving the original ordering. """ state = _HungarianState(cost_matrix) # No need to bother with assignments if one of the dimensions # of the cost matrix is zero-length. step = None if 0 in cost_matrix.shape else _step1 while step is not None: step = step(state) # Look for the starred columns results = np.array(np.where(state.marked == 1)).T # We need to swap the columns because we originally # did a transpose on the input cost matrix. if state.transposed: results = results[:, ::-1] return results # Individual steps of the algorithm follow, as a state machine: they return # the next step to be taken (function to be called), if any. def _step1(state): """Steps 1 and 2 in the Wikipedia page.""" # Step1: For each row of the matrix, find the smallest element and # subtract it from every element in its row. state.C -= state.C.min(axis=1)[:, np.newaxis] # Step2: Find a zero (Z) in the resulting matrix. If there is no # starred zero in its row or column, star Z. Repeat for each element # in the matrix. for i, j in zip(*np.where(state.C == 0)): if state.col_uncovered[j] and state.row_uncovered[i]: state.marked[i, j] = 1 state.col_uncovered[j] = False state.row_uncovered[i] = False state._clear_covers() return _step3 def _step3(state): """ Cover each column containing a starred zero. If n columns are covered, the starred zeros describe a complete set of unique assignments. In this case, Go to DONE, otherwise, Go to Step 4. """ marked = (state.marked == 1) state.col_uncovered[np.any(marked, axis=0)] = False if marked.sum() < state.C.shape[0]: return _step4 def _step4(state): """ Find a noncovered zero and prime it. If there is no starred zero in the row containing this primed zero, Go to Step 5. Otherwise, cover this row and uncover the column containing the starred zero. Continue in this manner until there are no uncovered zeros left. Save the smallest uncovered value and Go to Step 6. """ # We convert to int as numpy operations are faster on int C = (state.C == 0).astype(np.int) covered_C = C * state.row_uncovered[:, np.newaxis] covered_C *= state.col_uncovered.astype(dtype=np.int, copy=False) n = state.C.shape[0] m = state.C.shape[1] while True: # Find an uncovered zero row, col = np.unravel_index(np.argmax(covered_C), (n, m)) if covered_C[row, col] == 0: return _step6 else: state.marked[row, col] = 2 # Find the first starred element in the row star_col = np.argmax(state.marked[row] == 1) if not state.marked[row, star_col] == 1: # Could not find one state.Z0_r = row state.Z0_c = col return _step5 else: col = star_col state.row_uncovered[row] = False state.col_uncovered[col] = True covered_C[:, col] = C[:, col] * ( state.row_uncovered.astype(dtype=np.int, copy=False)) covered_C[row] = 0 def _step5(state): """ Construct a series of alternating primed and starred zeros as follows. Let Z0 represent the uncovered primed zero found in Step 4. Let Z1 denote the starred zero in the column of Z0 (if any). Let Z2 denote the primed zero in the row of Z1 (there will always be one). Continue until the series terminates at a primed zero that has no starred zero in its column. Unstar each starred zero of the series, star each primed zero of the series, erase all primes and uncover every line in the matrix. Return to Step 3 """ count = 0 path = state.path path[count, 0] = state.Z0_r path[count, 1] = state.Z0_c while True: # Find the first starred element in the col defined by # the path. row = np.argmax(state.marked[:, path[count, 1]] == 1) if not state.marked[row, path[count, 1]] == 1: # Could not find one break else: count += 1 path[count, 0] = row path[count, 1] = path[count - 1, 1] # Find the first prime element in the row defined by the # first path step col = np.argmax(state.marked[path[count, 0]] == 2) if state.marked[row, col] != 2: col = -1 count += 1 path[count, 0] = path[count - 1, 0] path[count, 1] = col # Convert paths for i in range(count + 1): if state.marked[path[i, 0], path[i, 1]] == 1: state.marked[path[i, 0], path[i, 1]] = 0 else: state.marked[path[i, 0], path[i, 1]] = 1 state._clear_covers() # Erase all prime markings state.marked[state.marked == 2] = 0 return _step3 def _step6(state): """ Add the value found in Step 4 to every element of each covered row, and subtract it from every element of each uncovered column. Return to Step 4 without altering any stars, primes, or covered lines. """ # the smallest uncovered value in the matrix if np.any(state.row_uncovered) and np.any(state.col_uncovered): minval = np.min(state.C[state.row_uncovered], axis=0) minval = np.min(minval[state.col_uncovered]) state.C[np.logical_not(state.row_uncovered)] += minval state.C[:, state.col_uncovered] -= minval return _step4