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@@ -219,29 +219,27 @@ def remove_overlaps_min_blocks(res_list):
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)
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if overlap_box is not None:
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- res_to_remove = None
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- large_res = None
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- # 确定哪个是小块（要移除的）
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+ # 根据重叠框确定哪个是小块，哪个是大块
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if overlap_box == res_list[i]['bbox']:
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- res_to_remove = res_list[i]
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- large_res = res_list[j]
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+ small_res, large_res = res_list[i], res_list[j]
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elif overlap_box == res_list[j]['bbox']:
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- res_to_remove = res_list[j]
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- large_res = res_list[i]
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+ small_res, large_res = res_list[j], res_list[i]
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+ else:
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+ continue # 如果重叠框与任一块都不匹配，跳过处理
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- if res_to_remove['score'] < large_res['score']:
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+ if small_res['score'] <= large_res['score']:
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# 如果小块的分数低于大块，则小块为需要移除的块
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- if res_to_remove is not None and res_to_remove not in need_remove:
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+ if small_res is not None and small_res not in need_remove:
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# 更新大块的边界为两者的并集
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x1, y1, x2, y2 = large_res['bbox']
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- sx1, sy1, sx2, sy2 = res_to_remove['bbox']
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+ sx1, sy1, sx2, sy2 = small_res['bbox']
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x1 = min(x1, sx1)
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y1 = min(y1, sy1)
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x2 = max(x2, sx2)
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y2 = max(y2, sy2)
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large_res['bbox'] = [x1, y1, x2, y2]
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- need_remove.append(res_to_remove)
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+ need_remove.append(small_res)
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else:
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# 如果大块的分数低于小块，则大块为需要移除的块, 这时不需要更新小块的边界
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if large_res is not None and large_res not in need_remove:
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myhloli