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@@ -119,12 +119,6 @@ jobs: | |
with: | ||
name: python-package-distributions | ||
path: dist/ | ||
- name: Sign the dists with Sigstore | ||
uses: sigstore/[email protected] | ||
with: | ||
inputs: >- | ||
./dist/*.tar.gz | ||
./dist/*.whl | ||
- name: Create GitHub Release | ||
env: | ||
GITHUB_TOKEN: ${{ github.token }} | ||
|
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Changes for this version of DIPLOMAT: | ||
- Added new memory mode hybrid and set it to the default | ||
- Added a save to disk button to the UI which saves frames to disk when in memory mode | ||
- New configuration file with fixes for the CPU conda environment | ||
- Rename CLI commands from supervised / unsupervised / track / restore to track_and_interact / track / track_with / interact | ||
- Minor UI bug fixes | ||
- sidebar radiobutton label colors now update with video label colors when the colormap is changed in settings menu. |
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## COPIED FROM https://tedboy.github.io/scipy/_modules/scipy/optimize/_hungarian.html#linear_sum_assignment | ||
## This function is defined locally, rather than by importing scipy.optimize, due to installation problems. | ||
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# Hungarian algorithm (Kuhn-Munkres) for solving the linear sum assignment | ||
# problem. Taken from scikit-learn. Based on original code by Brian Clapper, | ||
# adapted to NumPy by Gael Varoquaux. | ||
# Further improvements by Ben Root, Vlad Niculae and Lars Buitinck. | ||
# | ||
# Copyright (c) 2008 Brian M. Clapper <[email protected]>, Gael Varoquaux | ||
# Author: Brian M. Clapper, Gael Varoquaux | ||
# License: 3-clause BSD | ||
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import numpy as np | ||
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def linear_sum_assignment(cost_matrix): | ||
"""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 :math:`X[i,j] = 1` iff row i is | ||
assigned to column j. Then the optimal assignment has cost | ||
.. math:: | ||
\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 | ||
Kuhn-Munkres algorithm. | ||
Parameters | ||
---------- | ||
cost_matrix : array | ||
The cost matrix of the bipartite graph. | ||
Returns | ||
------- | ||
row_ind, col_ind : array | ||
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 to | ||
``numpy.arange(cost_matrix.shape[0])``. | ||
Notes | ||
----- | ||
.. versionadded:: 0.17.0 | ||
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 | ||
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. | ||
*J. SIAM*, 5(1):32-38, March, 1957. | ||
5. https://en.wikipedia.org/wiki/Hungarian_algorithm | ||
""" | ||
cost_matrix = np.asarray(cost_matrix) | ||
if len(cost_matrix.shape) != 2: | ||
raise ValueError("expected a matrix (2-d array), got a %r array" | ||
% (cost_matrix.shape,)) | ||
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# The algorithm expects more columns than rows in the cost matrix. | ||
if cost_matrix.shape[1] < cost_matrix.shape[0]: | ||
cost_matrix = cost_matrix.T | ||
transposed = True | ||
else: | ||
transposed = False | ||
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state = _Hungary(cost_matrix) | ||
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# 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 | ||
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while step is not None: | ||
step = step(state) | ||
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if transposed: | ||
marked = state.marked.T | ||
else: | ||
marked = state.marked | ||
return np.where(marked == 1) | ||
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class _Hungary(object): | ||
"""State of the Hungarian algorithm. | ||
Parameters | ||
---------- | ||
cost_matrix : 2D matrix | ||
The cost matrix. Must have shape[1] >= shape[0]. | ||
""" | ||
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def __init__(self, cost_matrix): | ||
self.C = cost_matrix.copy() | ||
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n, m = self.C.shape | ||
self.row_uncovered = np.ones(n, dtype=bool) | ||
self.col_uncovered = np.ones(m, dtype=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) | ||
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def _clear_covers(self): | ||
"""Clear all covered matrix cells""" | ||
self.row_uncovered[:] = True | ||
self.col_uncovered[:] = True | ||
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# Individual steps of the algorithm follow, as a state machine: they return | ||
# the next step to be taken (function to be called), if any. | ||
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def _step1(state): | ||
"""Steps 1 and 2 in the Wikipedia page.""" | ||
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# Step 1: 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] | ||
# Step 2: 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 | ||
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state._clear_covers() | ||
return _step3 | ||
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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 | ||
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if marked.sum() < state.C.shape[0]: | ||
return _step4 | ||
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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(int) | ||
covered_C = C * state.row_uncovered[:, np.newaxis] | ||
covered_C *= np.asarray(state.col_uncovered, dtype=int) | ||
n = state.C.shape[0] | ||
m = state.C.shape[1] | ||
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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 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] * ( | ||
np.asarray(state.row_uncovered, dtype=int)) | ||
covered_C[row] = 0 | ||
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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 | ||
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while True: | ||
# Find the first starred element in the col defined by | ||
# the path. | ||
row = np.argmax(state.marked[:, path[count, 1]] == 1) | ||
if 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] | ||
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# 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 | ||
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# 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 | ||
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state._clear_covers() | ||
# Erase all prime markings | ||
state.marked[state.marked == 2] = 0 | ||
return _step3 | ||
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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[~state.row_uncovered] += minval | ||
state.C[:, state.col_uncovered] -= minval | ||
return _step4 |
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