回答編集履歴
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追記
test
CHANGED
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@@ -65,3 +65,63 @@
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"""
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```
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### 参考（pandasを使わない方法）
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pandasに頼りたくない場合、pandasのgroupby相当のものがnumpyにも標準にもないので少し手間取ります。`defaultdict(list)`などを使うと比較的楽で、速度などの懸念もないかと思います。
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```python
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import numpy as np
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from operator import itemgetter
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from collections import defaultdict
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out_classes = np.array([0, 0, 3, 3, 5])
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out_scores = np.array([0.99801666, 0.7711145, 0.9962558, 0.96635705, 0.80220133])
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out_boxes = np.array(
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[[ 8.540416, 61.232857, 76.36931, 123.20946 ],
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[ 10.599434, 67.45303, 75.46433, 115.5706 ],
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[101.44424, 137.79492, 165.17267, 201.09848 ],
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[ 90.32121, 60.167862, 160.19128, 126.34792 ],
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[ 92.46553, 210.28862, 167.8457, 296.00696 ]])
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d = defaultdict(list)
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for i, c in enumerate(out_classes):
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d[c].append((i, out_scores[i]))
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idx = [max(x, key=itemgetter(1))[0] for c, x in d.items()]
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# 例示用
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print(d) # defaultdict(<class 'list'>, {0: [(0, 0.99801666), (1, 0.7711145)], 3: [(2, 0.9962558), (3, 0.96635705)], 5: [(4, 0.80220133)]})
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print(idx) # [0, 2, 4]
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# 以下はidxを用いて取り出すだけで、pandasを用いたコードと同じ
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```
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