Question

我开发了一种车辆路线选择（VRP）模型，该模型可以通过为正确的车辆分配正确的交付位置来优化路线规划。我模型的输入是input.csv，可以将其作为如下所示的数据框加载，其中每个位置都分配有唯一的ID，GPS坐标和交货需求。起点是配送中心。

  ID  Location_name Location_Lat Location_Lon  Date    Demand
    0       DC      51.4457678  -0.45613      3/4/2016  0
    1   Customer_1  51.799089   -0.29716      3/4/2016  472
    2   Customer_2  52.497911   -1.903832     3/4/2016  1488
    3   Customer_3  53.74589    -2.379778     3/4/2016  5652
    4   Customer_4  52.332395   -1.346753     3/4/2016  2348
    5   Customer_5  52.276323   -1.579845     3/4/2016  12
    6   Customer_6  52.254674   -0.265726     3/4/2016  2940
    7   Customer_7  51.480381   0.332106      3/4/2016  2533
    8   Customer_8  51.466844   -0.452952     3/4/2016  1142
    9   Customer_9  51.714144   -0.462949     3/4/2016  115
    10  Customer_10 50.918412   -0.14951      3/4/2016  325
    11  Customer_11 52.042318   0.311799      3/4/2016  886
    12  Customer_12 52.084242   0.457063      3/4/2016  412
    13  Customer_13 52.126166   0.602327      3/4/2016  58
    14  Customer_14 52.16809    0.747591      3/4/2016  161
    15  Customer_15 52.210014   0.892855      3/4/2016  29

然后我运行我的模型，我将得到一个输出（一个变量作为列表的列表）： [[0,8,3,6,11,13,0]，[0,12,9,5,0]，[0,4,7,10,15,1,14,2,0]]。这意味着我必须使用3辆车：

Routes for vehicle 0: [0,8,3,6,11,13,0] #always start and back to DC
Routes for vehicle 1: [0,12,9,5,0] 
Routes for vehicle 2: [0,4,7,10,15,1,14,2,0]

现在我想添加一个新的输出函数，该函数允许模型以如下表格式重新组织上述输出结果：（请注意前三行）

ID  Location_name Location_Lat  Location_Lon    Date    demand  Vehicle_ID  Pos_in_routes
0       DC        51.4457678    -0.45613      3/4/2016  0       0           0
0       DC        51.4457678    -0.45613      3/4/2016  0       1           0
0       DC        51.4457678    -0.45613      3/4/2016  0       2           0
1   Customer_1    51.799089     -0.29716      3/4/2016  472     2           5
2   Customer_2    52.497911     -1.903832     3/4/2016  1488    2           7
3   Customer_3    53.74589      -2.379778     3/4/2016  5652    0           2
4   Customer_4    52.332395     -1.346753     3/4/2016  2348    2           1
5   Customer_5    52.276323     -1.579845     3/4/2016  12      1           3
6   Customer_6    52.254674     -0.265726     3/4/2016  2940    0           3
7   Customer_7    51.480381      0.332106     3/4/2016  2533    2           2
8   Customer_8    51.466844     -0.452952     3/4/2016  1142    0           1
9   Customer_9    51.714144     -0.462949     3/4/2016  115     1           2
10  Customer_10   50.918412     -0.14951      3/4/2016  325     2           3
11  Customer_11   52.042318      0.311799     3/4/2016  886     0           4
12  Customer_12   52.084242      0.457063     3/4/2016  412     1           1
13  Customer_13   52.126166      0.602327     3/4/2016  58      0           5
14  Customer_14   52.16809       0.747591     3/4/2016  161     2           6
15  Customer_15   52.210014      0.892855     3/4/2016  29      2           4
0       DC        51.4457678    -0.45613      3/4/2016  0       0           6
0       DC        51.4457678    -0.45613      3/4/2016  0       1           4
0       DC        51.4457678    -0.45613      3/4/2016  0       2           8

Answer 1

也许有一种更优雅的方法，但是目前可以。

构造新列：

df['Vehicle_ID'] = 0
df['Pos_in_route'] = 0

创建一个空的数据框：df1 = pd.DataFrame(columns = df.columns)

假设您的车辆分配输出为：

veh =  [[0,8,3,6,11,13,0],[0,12,9,5,0],[0,4,7,10,15,1,14,2,0]]

要创建Vehicle_ID，您想将行按其索引映射到车辆ID。以下循环将查看每个车辆分配列表，并创建除最后一个条目（在终端站结束）以外的所有条目。

for i in range(len(veh)):
    veh_id_dict = dict.fromkeys(veh[i][:-1],i)
    veh_pos_dict = dict(zip(veh[i][:-1],range(len(veh[i])-1)))
    df.loc[veh[i][:-1],'Vehicle_ID'] = df.loc[veh[i][:-1]].ID.map(veh_id_dict)
    df.loc[veh[i][:-1],'Pos_in_route'] = df.loc[veh[i][:-1]].ID.map(veh_pos_dict)
    df1 = df1.append(df.loc[veh[i][:-1]])
df1.sort_values(['ID','Vehicle_ID'],inplace =True)

然后我们可以构造最后几行：

for i in range(len(veh)):
    df.loc[veh[i][-1],'Vehicle_ID'] = i
    df.loc[veh[i][-1],'Pos_in_route'] = len(veh[i]-1)
    df1 = df1.append(df.loc[veh[i][-1]])

使用Python Pandas将列表中的数据重新组织到表中

1 个答案: