翻译:def checkTimeWindow(route,model,vehicle): timetable=[] departure=0 for i in range(len(route)): if i == 0: next_node_id = route[i + 1] travel_time = int(model.distance_matrix[vehicle.type, next_node_id] /vehicle.free_speed) departure = max(0, model.demand_dict[next_node_id].start_time - travel_time) timetable.append((int(departure), int(departure))) elif 1 <= i <= len(route) - 2: last_node_id = route[i - 1] current_node_id = route[i] current_node = model.demand_dict[current_node_id] travel_time = int(model.distance_matrix[last_node_id, current_node_id] / vehicle.free_speed) arrival = max(timetable[-1][1] + travel_time, current_node.start_time) departure = arrival + current_node.service_time timetable.append((int(arrival), int(departure))) if departure > current_node.end_time: departure = float('inf') break else: last_node_id = route[i - 1] travel_time = int(model.distance_matrix[last_node_id, vehicle.type] / vehicle.free_speed) departure = timetable[-1][1] + travel_time timetable.append((int(departure), int(departure))) if departure<vehicle.end_time: return True else: return False
时间: 2024-04-28 07:24:40 浏览: 7
功能描述:检查给定路线、车型和车辆是否在规定时间窗口内可用。
代码逻辑:根据车辆在路线上的行驶时间和节点的服务时间,计算出每个节点的到达时间和离开时间,存入时间表并返回是否在时间窗口内可用的结果。
具体实现:遍历路线上的每个节点,根据节点的位置分为三种情况:
1. 起点节点:根据车型和下一个节点的距离计算行驶时间,计算出离开时间并加入时间表。
2. 中间节点:根据上一个节点、当前节点和车型的距离计算行驶时间,计算出到达时间和离开时间并加入时间表。如果离开时间晚于节点的结束时间,则返回不可用的结果。
3. 终点节点:根据上一个节点和车型的距离计算行驶时间,计算出离开时间并加入时间表。最后判断离开时间是否早于车辆的结束时间,返回对应的结果。
相关问题
降低这段代码的重复率:def checkTimeWindow(route,model,vehicle): timetable=[] departure=0 for i in range(len(route)): if i == 0: next_node_id = route[i + 1] travel_time = int(model.distance_matrix[vehicle.type, next_node_id] /vehicle.free_speed) departure = max(0, model.demand_dict[next_node_id].start_time - travel_time) timetable.append((int(departure), int(departure))) elif 1 <= i <= len(route) - 2: last_node_id = route[i - 1] current_node_id = route[i] current_node = model.demand_dict[current_node_id] travel_time = int(model.distance_matrix[last_node_id, current_node_id] / vehicle.free_speed) arrival = max(timetable[-1][1] + travel_time, current_node.start_time) departure = arrival + current_node.service_time timetable.append((int(arrival), int(departure))) if departure > current_node.end_time: departure = float('inf') break else: last_node_id = route[i - 1] travel_time = int(model.distance_matrix[last_node_id, vehicle.type] / vehicle.free_speed) departure = timetable[-1][1] + travel_time timetable.append((int(departure), int(departure))) if departure<vehicle.end_time: return True else: return False
可以将重复的代码提取出来,封装成一个函数,如下所示:
```python
def get_travel_time(model, vehicle, start_node_id, end_node_id):
return int(model.distance_matrix[start_node_id, end_node_id] / vehicle.free_speed)
def checkTimeWindow(route, model, vehicle):
timetable = []
departure = 0
for i, node_id in enumerate(route):
if i == 0:
next_node_id = route[i + 1]
travel_time = get_travel_time(model, vehicle, vehicle.type, next_node_id)
departure = max(0, model.demand_dict[next_node_id].start_time - travel_time)
timetable.append((int(departure), int(departure)))
elif 1 <= i <= len(route) - 2:
last_node_id = route[i - 1]
current_node = model.demand_dict[node_id]
travel_time = get_travel_time(model, vehicle, last_node_id, node_id)
arrival = max(timetable[-1][1] + travel_time, current_node.start_time)
departure = arrival + current_node.service_time
timetable.append((int(arrival), int(departure)))
if departure > current_node.end_time:
departure = float('inf')
break
else:
last_node_id = route[i - 1]
travel_time = get_travel_time(model, vehicle, last_node_id, vehicle.type)
departure = timetable[-1][1] + travel_time
timetable.append((int(departure), int(departure)))
if departure >= vehicle.end_time:
return False
return True
```
这样,代码中的重复率就降低了。
优化这段代码:def checkTimeWindow(route,model,vehicle): timetable=[] departure=0 for i in range(len(route)): if i == 0: next_node_id = route[i + 1] travel_time = int(model.distance_matrix[vehicle.type, next_node_id] /vehicle.free_speed) departure = max(0, model.demand_dict[next_node_id].start_time - travel_time) timetable.append((int(departure), int(departure))) elif 1 <= i <= len(route) - 2: last_node_id = route[i - 1] current_node_id = route[i] current_node = model.demand_dict[current_node_id] travel_time = int(model.distance_matrix[last_node_id, current_node_id] / vehicle.free_speed) arrival = max(timetable[-1][1] + travel_time, current_node.start_time) departure = arrival + current_node.service_time timetable.append((int(arrival), int(departure))) if departure > current_node.end_time: departure = float('inf') break else: last_node_id = route[i - 1] travel_time = int(model.distance_matrix[last_node_id, vehicle.type] / vehicle.free_speed) departure = timetable[-1][1] + travel_time timetable.append((int(departure), int(departure))) if departure<vehicle.end_time: return True else: return False
可以考虑以下优化方法:
1. 将 `range(len(route))` 提取为一个变量 `n`,这样可以减少多次计算 `len(route)` 的开销。
2. 将 `model.distance_matrix[vehicle.type, next_node_id]` 提取为一个变量 `dist1`,将 `model.distance_matrix[last_node_id, current_node_id]` 提取为一个变量 `dist2`,将 `model.distance_matrix[last_node_id, vehicle.type]` 提取为一个变量 `dist3`,这样可以减少多次计算距离的开销。
3. 将 `departure` 的类型由浮点型改为整型,这样可以避免浮点数精度误差产生的问题。
4. 将 `timetable` 中元素的类型由元组 `(int, int)` 改为列表 `[int, int]`,这样可以减少元组解包的开销。
优化后的代码如下:
```
def check_time_window(route, model, vehicle):
timetable = []
departure = 0
n = len(route)
for i in range(n):
if i == 0:
next_node_id = route[i + 1]
dist1 = model.distance_matrix[vehicle.type, next_node_id]
travel_time = int(dist1 / vehicle.free_speed)
departure = max(0, model.demand_dict[next_node_id].start_time - travel_time)
timetable.append([departure, departure])
elif 1 <= i <= n - 2:
last_node_id = route[i - 1]
current_node_id = route[i]
current_node = model.demand_dict[current_node_id]
dist2 = model.distance_matrix[last_node_id, current_node_id]
travel_time = int(dist2 / vehicle.free_speed)
arrival = max(timetable[-1][1] + travel_time, current_node.start_time)
departure = arrival + current_node.service_time
timetable.append([arrival, departure])
if departure > current_node.end_time:
departure = float('inf')
break
else:
last_node_id = route[i - 1]
dist3 = model.distance_matrix[last_node_id, vehicle.type]
travel_time = int(dist3 / vehicle.free_speed)
departure = timetable[-1][1] + travel_time
timetable.append([departure, departure])
if departure < vehicle.end_time:
return True
else:
return False
```
这样,可以减少重复计算的开销,使得代码更加高效。