/
AJI.py
234 lines (203 loc) · 8.74 KB
/
AJI.py
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
169
170
171
172
173
174
175
176
177
178
179
180
181
182
183
184
185
186
187
188
189
190
191
192
193
194
195
196
197
198
199
200
201
202
203
204
205
206
207
208
209
210
211
212
213
214
215
216
217
218
219
220
221
222
223
224
225
226
227
228
229
230
231
232
233
234
import numpy as np
from sklearn.metrics import precision_score, recall_score, f1_score
from skimage.measure import label
import torch
def get_fast_aji(true, pred):
"""AJI version distributed by MoNuSeg, has no permutation problem but suffered from
over-penalisation similar to DICE2.
Fast computation requires instance IDs are in contiguous orderding i.e [1, 2, 3, 4]
not [2, 3, 6, 10]. Please call `remap_label` before hand and `by_size` flag has no
effect on the result.
"""
true = np.copy(true) # ? do we need this
pred = np.copy(pred)
true_id_list = list(np.unique(true))
pred_id_list = list(np.unique(pred))
if len(pred_id_list)==1:
aji_score=0.
# print("111111")
return aji_score
true_masks = [
None,
]
for t in true_id_list[1:]:
t_mask = np.array(true == t, np.uint8)
true_masks.append(t_mask)
pred_masks = [
None,
]
for p in pred_id_list[1:]:
p_mask = np.array(pred == p, np.uint8)
pred_masks.append(p_mask)
# prefill with value
pairwise_inter = np.zeros(
[len(true_id_list) - 1, len(pred_id_list) - 1], dtype=np.float64
)
pairwise_union = np.zeros(
[len(true_id_list) - 1, len(pred_id_list) - 1], dtype=np.float64
)
# caching pairwise
for true_id in true_id_list[1:]: # 0-th is background
if true_id>=len(true_masks):
continue
t_mask = true_masks[true_id]
pred_true_overlap = pred[t_mask > 0]
pred_true_overlap_id = np.unique(pred_true_overlap)
pred_true_overlap_id = list(pred_true_overlap_id)
for pred_id in pred_true_overlap_id:
if pred_id == 0: # ignore
continue # overlaping background
p_mask = pred_masks[pred_id]
total = (t_mask + p_mask).sum()
inter = (t_mask * p_mask).sum()
pairwise_inter[true_id - 1, pred_id - 1] = inter
pairwise_union[true_id - 1, pred_id - 1] = total - inter
pairwise_iou = pairwise_inter / (pairwise_union + 1.0e-6)
# pair of pred that give highest iou for each true, dont care
# about reusing pred instance multiple times
# print(pairwise_iou)
paired_pred = np.argmax(pairwise_iou, axis=1)
pairwise_iou = np.max(pairwise_iou, axis=1)
# exlude those dont have intersection
paired_true = np.nonzero(pairwise_iou > 0.0)[0]
paired_pred = paired_pred[paired_true]
# print(paired_true.shape, paired_pred.shape)
overall_inter = (pairwise_inter[paired_true, paired_pred]).sum()
overall_union = (pairwise_union[paired_true, paired_pred]).sum()
paired_true = list(paired_true + 1) # index to instance ID
paired_pred = list(paired_pred + 1)
# add all unpaired GT and Prediction into the union
unpaired_true = np.array(
[idx for idx in true_id_list[1:] if idx not in paired_true]
)
unpaired_pred = np.array(
[idx for idx in pred_id_list[1:] if idx not in paired_pred]
)
for true_id in unpaired_true:
overall_union += true_masks[true_id].sum()
for pred_id in unpaired_pred:
overall_union += pred_masks[pred_id].sum()
aji_score = overall_inter / overall_union
return aji_score
def remap_label(pred, by_size=False):
"""Rename all instance id so that the id is contiguous i.e [0, 1, 2, 3]
not [0, 2, 4, 6]. The ordering of instances (which one comes first)
is preserved unless by_size=True, then the instances will be reordered
so that bigger nucler has smaller ID.
Args:
pred : the 2d array contain instances where each instances is marked
by non-zero integer
by_size : renaming with larger nuclei has smaller id (on-top)
"""
# print(pred.shape)
# pred = torch.argmax(torch.from_numpy(pred), 1)
# pred = np.expand_dims(pred.numpy(), axis=1)
pred_id = list(np.unique(pred))
pred_id.remove(0)
if len(pred_id) == 0:
return pred # no label
if by_size:
pred_size = []
for inst_id in pred_id:
size = (pred == inst_id).sum()
pred_size.append(size)
# sort the id by size in descending order
pair_list = zip(pred_id, pred_size)
pair_list = sorted(pair_list, key=lambda x: x[1], reverse=True)
pred_id, pred_size = zip(*pair_list)
new_pred = np.zeros(pred.shape, np.int32)
for idx, inst_id in enumerate(pred_id):
new_pred[pred == inst_id] = idx + 1
# print(list(np.unique(new_pred)))
return new_pred
def get_fast_pq(true, pred, match_iou=0.5):
"""
`match_iou` is the IoU threshold level to determine the pairing between
GT instances `p` and prediction instances `g`. `p` and `g` is a pair
if IoU > `match_iou`. However, pair of `p` and `g` must be unique
(1 prediction instance to 1 GT instance mapping).
If `match_iou` < 0.5, Munkres assignment (solving minimum weight matching
in bipartite graphs) is caculated to find the maximal amount of unique pairing.
If `match_iou` >= 0.5, all IoU(p,g) > 0.5 pairing is proven to be unique and
the number of pairs is also maximal.
Fast computation requires instance IDs are in contiguous orderding
i.e [1, 2, 3, 4] not [2, 3, 6, 10]. Please call `remap_label` beforehand
and `by_size` flag has no effect on the result.
Returns:
[dq, sq, pq]: measurement statistic
[paired_true, paired_pred, unpaired_true, unpaired_pred]:
pairing information to perform measurement
"""
assert match_iou >= 0.0, "Cant' be negative"
# true = label(true, background=0)
# pred = label(pred, background=0)
true = np.copy(true)
pred = np.copy(pred)
true_id_list = list(np.unique(true))
pred_id_list = list(np.unique(pred))
if len(pred_id_list)==1:
PQ=0.
# print("111111")
return PQ
true_masks = [None,]
for t in true_id_list[1:]:
t_mask = np.array(true == t, np.uint8)
true_masks.append(t_mask)
pred_masks = [None,]
for p in pred_id_list[1:]:
p_mask = np.array(pred == p, np.uint8)
pred_masks.append(p_mask)
# prefill with value
pairwise_iou = np.zeros([len(true_id_list) -1,
len(pred_id_list) -1], dtype=np.float64)
# caching pairwise iou
for true_id in true_id_list[1:]: # 0-th is background
t_mask = true_masks[true_id]
pred_true_overlap = pred[t_mask > 0]
pred_true_overlap_id = np.unique(pred_true_overlap)
pred_true_overlap_id = list(pred_true_overlap_id)
for pred_id in pred_true_overlap_id:
if pred_id == 0: # ignore
continue # overlaping background
p_mask = pred_masks[pred_id]
total = (t_mask + p_mask).sum()
inter = (t_mask * p_mask).sum()
iou = inter / (total - inter)
pairwise_iou[true_id-1, pred_id-1] = iou
#
if match_iou >= 0.5:
paired_iou = pairwise_iou[pairwise_iou > match_iou]
pairwise_iou[pairwise_iou <= match_iou] = 0.0
paired_true, paired_pred = np.nonzero(pairwise_iou)
paired_iou = pairwise_iou[paired_true, paired_pred]
paired_true += 1 # index is instance id - 1
paired_pred += 1 # hence return back to original
else: # * Exhaustive maximal unique pairing
#### Munkres pairing with scipy library
# the algorithm return (row indices, matched column indices)
# if there is multiple same cost in a row, index of first occurence
# is return, thus the unique pairing is ensure
# inverse pair to get high IoU as minimum
paired_true, paired_pred = linear_sum_assignment(-pairwise_iou)
### extract the paired cost and remove invalid pair
paired_iou = pairwise_iou[paired_true, paired_pred]
# now select those above threshold level
# paired with iou = 0.0 i.e no intersection => FP or FN
paired_true = list(paired_true[paired_iou > match_iou] + 1)
paired_pred = list(paired_pred[paired_iou > match_iou] + 1)
paired_iou = paired_iou[paired_iou > match_iou]
# get the actual FP and FN
unpaired_true = [idx for idx in true_id_list[1:] if idx not in paired_true]
unpaired_pred = [idx for idx in pred_id_list[1:] if idx not in paired_pred]
# print(paired_iou.shape, paired_true.shape, len(unpaired_true), len(unpaired_pred))
#
tp = len(paired_true)
fp = len(unpaired_pred)
fn = len(unpaired_true)
# get the F1-score i.e DQ
dq = tp / (tp + 0.5 * fp + 0.5 * fn+ 1.0e-6)
# get the SQ, no paired has 0 iou so not impact
sq = paired_iou.sum() / (tp + 1.0e-6)
# print(dq * sq)
# return [dq, sq, dq * sq], [paired_true, paired_pred, unpaired_true, unpaired_pred]
return dq * sq