解释下这段代码: while len(idxs) > 0: last = len(idxs) - 1 i = idxs[last] pick.append(i) # keep top k if keep_top_k != -1: if len(pick) >= keep_top_k: break overlap_xmin = np.maximum(xmin[i], xmin[idxs[:last]]) overlap_ymin = np.maximum(ymin[i], ymin[idxs[:last]]) overlap_xmax = np.minimum(xmax[i], xmax[idxs[:last]]) overlap_ymax = np.minimum(ymax[i], ymax[idxs[:last]]) overlap_w = np.maximum(0, overlap_xmax - overlap_xmin) overlap_h = np.maximum(0, overlap_ymax - overlap_ymin) overlap_area = overlap_w * overlap_h overlap_ratio = overlap_area / (area[idxs[:last]] + area[i] - overlap_area) need_to_be_deleted_idx = np.concatenate(([last], np.where(overlap_ratio > iou_thresh)[0])) idxs = np.delete(idxs, need_to_be_deleted_idx) # if the number of final bboxes is less than keep_top_k, we need to pad it. # TODO return conf_keep_idx[pick]
时间: 2024-03-11 07:43:53 浏览: 120
这是一个非极大值抑制(Non-Maximum Suppression,简称NMS)的代码实现,用于去除重叠的边界框。输入为一组边界框(bounding boxes)及其对应的置信度(confidence scores),输出为经过NMS后剩余的边界框的索引。
具体实现过程如下:
1. 首先将所有边界框按照置信度从高到低排序。
2. 取出置信度最高的边界框,并将其加入最终输出列表中。
3. 计算当前边界框与剩余边界框的重叠部分,如果重叠部分的面积占总面积的比例大于一定阈值(iou_thresh),则将该边界框删除。
4. 重复第2步和第3步,直到所有的边界框都被处理完毕。
5. 如果最终输出的边界框数量少于指定的保留数量(keep_top_k),则需要进行填充,这里还未实现。
最终返回的是经过NMS后剩余的边界框的索引。
相关问题
def generateOwnCarRoute(service_time, model, sol): pickup_node = copy.deepcopy(model.demand_id_list[0: 16]) own_pickup_node = [] own_delivery_node = [] route = [] sol.route_list = [] depot = model.depot_dict['d1'] vehicle_number = depot.depot_capacity departure = 0 arrival = 0 for i in pickup_node: if i not in model.crowd_pickup_node: own_pickup_node.append(i) own_delivery_node.append(i+16) while vehicle_number > 0 and len(own_pickup_node) > 0: route.append(depot.depot_id) minIndex = np.argmin([model.distance_matrix[depot.depot_id, own_pickup_node[i]] for i in range(0, len(own_pickup_node))]) minnode = own_pickup_node[minIndex] route.append(minnode) arrival = model.time_matrix[depot.depot_id, minnode] departure = arrival + service_time route.append(own_delivery_node[minIndex]) arrival = departure + model.time_matrix[minnode, own_delivery_node[minIndex]] departure += arrival + service_time last_node = own_delivery_node[minIndex] own_pickup_node.remove(minnode) own_delivery_node.remove(own_delivery_node[minIndex]) for j in own_pickup_node: next_minIndex = np.argmin([model.distance_matrix[last_node, j]]) next_minnode = own_pickup_node[next_minIndex] arrival = departure + model.time_matrix[last_node, next_minnode] if arrival <= model.demand_dict[next_minnode].end_time and arrival <= depot.dend_time: route.append(next_minnode) departure = arrival + service_time route.append(own_delivery_node[next_minIndex]) arrival = departure + model.time_matrix[next_minnode, own_delivery_node[next_minIndex]] departure += arrival + service_time last_node = own_delivery_node[next_minIndex] own_pickup_node.remove(next_minnode) own_delivery_node.remove(own_delivery_node[next_minIndex]) else: continue route.append(depot.depot_id) sol.route_list.append(route) vehicle_number = vehicle_number - 1 route = [] print(sol.route_list) return sol.route_list 这段代码的问题是有可能vehicle_number为0了,但是owner_pickup_node的长度还不为0,这种情况怎么解决
这种情况下,程序会陷入无限循环,因为当所有车辆都被分配任务后,仍有剩余的pickup_node,但是vehicle_number为0,所以不会再有车辆来接这些节点。为了解决这个问题,可以在while循环条件中添加一个限制条件,即当所有车辆都被分配任务后,停止while循环。例如可以这样修改while循环的条件:
```
while vehicle_number > 0 and len(own_pickup_node) > 0 and len(sol.route_list) < model.depot_dict['d1'].depot_capacity:
```
这样,当所有车辆都被分配任务后,循环就会停止,避免了程序陷入无限循环的情况。
while len(own_pickup_node) > 0: route = [depot.depot_id] for i in range(min(len(own_pickup_node), 2)): minIndex = np.argmin([model.distance_matrix[depot.depot_id, own_pickup_node[j]] for j in range(0, len(own_pickup_node))]) minnode = own_pickup_node[minIndex] route.append(minnode) arrival = model.time_matrix[depot.depot_id, minnode] departure = arrival + service_time route.append(own_delivery_node[minIndex]) arrival = departure + model.time_matrix[minnode, own_delivery_node[minIndex]] departure += arrival + service_time last_node = own_delivery_node[minIndex] own_pickup_node.remove(minnode) own_delivery_node.remove(own_delivery_node[minIndex]) for j in own_pickup_node: next_minIndex = np.argmin([model.distance_matrix[last_node, j]]) next_minnode = own_pickup_node[next_minIndex] arrival = departure + model.time_matrix[last_node, next_minnode] if arrival <= model.demand_dict[next_minnode].end_time and arrival <= depot.dend_time: route.append(next_minnode) departure = arrival + service_time route.append(own_delivery_node[next_minIndex]) arrival = departure + model.time_matrix[next_minnode, own_delivery_node[next_minIndex]] departure += arrival + service_time last_node = own_delivery_node[next_minIndex] own_pickup_node.remove(next_minnode) own_delivery_node.remove(own_delivery_node[next_minIndex]) else: continue route.append(depot.depot_id) sol.route_list.append(route) vehicle_number = vehicle_number - 1 if vehicle_number == 0: break
这段代码实现了一个基于贪心算法的送货路径规划。具体来说,它会先将起点设为仓库,然后在所有尚未配对的取货点和配送点中,选取距离仓库最近的两个取货点进行配对,将它们的编号加入路径中。然后,它会根据这两个取货点所对应的配送点的时间窗口,计算出到达和离开的时间,并将这两个配送点的编号加入路径中。接着,它会在剩下的取货点和配送点中,选取距离上一个配送点最近的取货点进行配对,然后重复之前的操作,不断将取货点和配送点加入路径中,直到所有货物都配送完毕。
需要注意的是,这段代码中的距离矩阵、时间矩阵、时间窗口等信息都是由 model 对象提供的,而且这里采用了 NumPy 库中的 argmin 函数来查找距离最近的取货点。另外,这段代码中的 sol 对象则是一个 Solution 类的实例,它用于保存所有的路径方案。
阅读全文
相关推荐

















