Question

我有一个包含shop_id，纬度和经度的列表。假设我们有一定数量的积分，我想将每个商店与另一家商店相匹配，这样我们的联系就是独一无二的，我们将总距离降到最低。

我很好奇我的方法是不是太愚蠢而且更好的方法。这目前在1000分的合理时间（一小时）内工作。

想象一下，我们有一个这样的列表：

shops = [[1, '53.382072', '-2.7262165'],
[2, '52.478499', '-0.9222608'],
[3, '52.071258', '-1.2722802'],
...]

我使用PuLP使用此工作流程创建.lp文件：

prob = LpProblem("Minimise distance",LpMinimize)
# Variables
x = []
xnames = []
for one in store_connect(shops):
    xnames.append(one)
    x.append(LpVariable(one,0,1,LpInteger)) 
print("Created %d variables" % len(x))
# Objective
prob += pulp.lpSum([i[1]*x[i[0]] for i in enumerate(distances(shops))])
print("Created objective")
# Constraints
counter = 0
for constraint_rw in all_constraints(shops):
    prob += pulp.lpSum(x[xnames.index(one_cn)] for one_cn in constraint_rw) == 1
    counter += 1
    if counter % 10 == 0:
        print("Added %d contraints out of %d" % (counter,len(shops)))   
# Save "LP" file for other solvers
prob.writeLP("for_cplex.lp")

我称之为一些发电机功能（以帮助RAM ...）

def store_connect(shops):
    """
    Cross stores and return connection e.g. "v1-2" by ID
    """
    for i in range(len(shops)-1):
        for j in range(i+1,len(shops)):
            yield 'v'+str(shops[i][0])+'-'+str(shops[j][0])

def distances(shops):
    """
    Return distance in miles for crosses
    """
    for i in range(len(shops)-1):
        for j in range(i+1,len(shops)):
            yield haversine([shops[i][1],shops[i][2]],
                            [shops[j][1],shops[j][2]])

def all_constraints(shops):
    """
    Return constraint row
    """
    for a in shops:
        cons = []
        for b in shops:
            if a[0]<b[0]:
                cons.append("v"+str(a[0])+"-"+str(b[0]))
            elif a[0]>b[0]:
                cons.append("v"+str(b[0])+"-"+str(a[0]))
        yield cons

如果我在100家商店运行它，它非常快，我可以使用默认的PuLP解算器。否则，我将LP文件提交到NEOS服务器上的CPLEX。我创建了一个小帮助函数，可以显示结果匹配：

以下是我的LP文件如何查找10个商店（截断）的简单示例：

Minimize
OBJ: 131.513 v1_10 + 97.686 v1_2 + 109.107 v1_3 + 36.603 v1_4 + 11.586 v1_5
 + 109.067 v1_6 + 113.862 v1_7 + 169.371 v1_8 + 220.098 v1_9 + 101.958 v2_10
 + 31.793 v2_3 + 61.792 v2_4 + 105.822 v2_5 + 73.055 v2_6 + 32.008 v2_7
 + 81.627 v2_8 + 122.493 v2_9 + 72.945 v3_10 + 72.983 v3_4 + 114.609 v3_5
 + 46.098 v3_6 + 5.555 v3_7 + 60.305 v3_8 ...
Subject To
_C1: v1_10 + v1_2 + v1_3 + v1_4 + v1_5 + v1_6 + v1_7 + v1_8 + v1_9 = 1
_C10: v1_10 + v2_10 + v3_10 + v4_10 + v5_10 + v6_10 + v7_10 + v8_10 + v9_10
 = 1
_C2: v1_2 + v2_10 + v2_3 + v2_4 + v2_5 + v2_6 + v2_7 + v2_8 + v2_9 = 1
_C3: v1_3 + v2_3 + v3_10 + v3_4 + v3_5 + v3_6 + v3_7 + v3_8 + v3_9 = 1
_C4: v1_4 + v2_4 + v3_4 + v4_10 + v4_5 + v4_6 + v4_7 + v4_8 + v4_9 = 1
_C5: v1_5 + v2_5 + v3_5 + v4_5 + v5_10 + v5_6 + v5_7 + v5_8 + v5_9 = 1
...
Binaries
v1_10
v...
End

结果：

Status: Optimal
Total minimised distance (miles):  190.575

最后，这是一个关于1000点的更大例子：

Answer 1

Consider a pandas solution (Python's data analysis package). My elementary geometry reminds me that between two points the closest distance is a straight line which in a coordinate plane is the hypotenuse between Lat and Lng points.

Below is first runs a cartesian product between shops (all possible combination between the sets) and then calculates the hypotenuse using geographic coordinates. From there, the minimium is aggregated:

import pandas as pd

# ORIGINAL LIST
shops = [[1, 53.382072, -2.7262165],
         [2, 52.478499, -0.9222608],
         [3, 52.071258, -1.2722802]]

# CONVERTING INTO TWO DUPLICATE SHOPS DATA FRAMES    
shopsdfA = pd.DataFrame(shops, columns=['ID_A', 'Lat_A', 'Lng_A'])
shopsdfA['key']=1
shopsdfB = pd.DataFrame(shops, columns=['ID_B', 'Lat_B', 'Lng_B'])
shopsdfB['key']=1

# CROSS JOINING BOTH DATA FRAMES FOR CARTESIAN PRODUCT SET 
compareshops = pd.merge(shopsdfA, shopsdfB, on='key')[['ID_A', 'Lat_A', 'Lng_A', 
                                                       'ID_B','Lat_B', 'Lng_B']]
#  (REMOVING SHOPS COMPARED TO ITSELF)
compareshops = compareshops[compareshops['ID_A'] != compareshops['ID_B']].\
               reset_index(drop=True)
#   ID_A      Lat_A     Lng_A  ID_B      Lat_B     Lng_B
#0     1  53.382072 -2.726217     2  52.478499 -0.922261
#1     1  53.382072 -2.726217     3  52.071258 -1.272280
#2     2  52.478499 -0.922261     1  53.382072 -2.726217
#3     2  52.478499 -0.922261     3  52.071258 -1.272280
#4     3  52.071258 -1.272280     1  53.382072 -2.726217
#5     3  52.071258 -1.272280     2  52.478499 -0.922261

# CALCULATE HYPOTENUSE
compareshops['hypotenuse'] = ((compareshops['Lat_A'] - compareshops['Lat_B']) ** 2 +
                              (compareshops['Lng_A'] - compareshops['Lng_B']) ** 2) ** (1/2)
#   ID_A      Lat_A     Lng_A  ID_B      Lat_B     Lng_B  hypotenuse
#0     1  53.382072 -2.726217     2  52.478499 -0.922261    2.017598
#1     1  53.382072 -2.726217     3  52.071258 -1.272280    1.957591
#2     2  52.478499 -0.922261     1  53.382072 -2.726217    2.017598
#3     2  52.478499 -0.922261     3  52.071258 -1.272280    0.536991
#4     3  52.071258 -1.272280     1  53.382072 -2.726217    1.957591
#5     3  52.071258 -1.272280     2  52.478499 -0.922261    0.536991

# AGGREGATE FOR MINIMUM HYPOTENUSE AND MERGE CORRESPONDING PAIRED IDS
compareshops = pd.merge(compareshops[['ID_A','hypotenuse']].groupby(['ID_A']).min().reset_index(),
                        compareshops[['ID_B','hypotenuse']], on='hypotenuse')
# (REMOVING SAME STORE PAIRINGS)
compareshops = compareshops[compareshops['ID_A'] != compareshops['ID_B']].\
                     reset_index(drop=True)[['ID_A','ID_B','hypotenuse']]

#   ID_A  ID_B  hypotenuse
#0     1     3    1.957591
#1     2     3    0.536991
#2     3     2    0.536991

One foreseeable challenge is your map does not have overlapping connections, only one pairing to each store. Possibly an iterative hypotenuse calculation will need to be run to find the "next" shortest distance.

将许多点分组为最近的对 - Python + LP

1 个答案: