Question

例如，我有这个样本数据：

d=data.frame(x=c(1,1,1,2,2,3,4,4),y=c(5,6,7,8,7,5,6,5),w=c(1,2,3,4,5,6,7,8))

看起来像这样：

x和y代表datax和datay的索引。 w代表将datax[x]与datay[y]进行比较的得分。我想从w最大化总得分（或d），其中x的每个值最多匹配一个值y，反之亦然。< / p>

结果应如下所示：

所有w值的总和最大化，并且每个x和每个y在结果中只显示一次。

如何在lpSolve::lp功能中设置此问题？

Answer 1

您可以使用lpSolveAPI来解决您的问题。鉴于您的约束，您声明的解决方案并不可行。因此，让我们一起来，希望X和Y在解决方案中不再重复。

您将需要8个新的二进制变量。每个变量指定d中的行是否被选中（1）或删除（0）。

根据OP的请求进行更新

是的，lpSolveAPI代码（下面）使它看起来比实际上更复杂。这个LP公式（lpSolveAPI的输出）应该让事情更清楚：

/* Objective function */
max: +pick_1 +2 pick_2 +3 pick_3 +4 pick_4 +5 pick_5 +6 pick_6 +7 pick_7 +8 pick_8;

/* Constraints */
OneX_1: +pick_1 +pick_2 +pick_3 <= 1;
OneX_2: +pick_4 +pick_5 <= 1;
OneX_4: +pick_7 +pick_8 <= 1;
OneY_5: +pick_1 +pick_6 +pick_8 <= 1;
OneY_6: +pick_2 +pick_7 <= 1;
OneY_7: +pick_3 +pick_5 <= 1;

/* Variable bounds */
pick_1 <= 1;
pick_2 <= 1;
pick_3 <= 1;
pick_4 <= 1;
pick_5 <= 1;
pick_6 <= 1;
pick_7 <= 1;
pick_8 <= 1;

说明：第二个约束（OneX_2）简单地指出pick_4或pick_5中只有一个可以为1，因为数据框d中的第4行和第5行具有X = 2

解决方案

请注意，上面的公式会生成一个最佳解决方案，在数据框d

中选择4行

> d[c(3,4,6,7),]
  x y w
3 1 7 3
4 2 8 4
6 3 5 6
7 4 6 7

w的总和是20，这比问题中的解决方案更好。

代码

library(lpSolveAPI)
d <- data.frame(x=c(1,1,1,2,2,3,4,4),y=c(5,6,7,8,7,5,6,5),w=c(1,2,3,4,5,6,7,8))

ncol <- 8 #you have eight rows that can be picked or dropped from the solution set
lp_rowpicker <- make.lp(ncol=ncol)
set.type(lp_rowpicker, columns=1:ncol, type = c("binary"))

obj_vals <- d[, "w"]
set.objfn(lp_rowpicker, obj_vals) 
lp.control(lp_rowpicker,sense='max')

#Add constraints to limit X values from repeating
add.constraint(lp_rowpicker, xt=c(1,1,1), #xt specifies which rows of the LP
               indices=c(1,2,3), rhs=1, type="<=")
add.constraint(lp_rowpicker, xt=c(1,1), #xt specifies which rows of the LP
               indices=c(4,5), rhs=1, type="<=")
add.constraint(lp_rowpicker, xt=c(1,1), #xt specifies which rows of the LP
               indices=c(7,8), rhs=1, type="<=") #x's in dataframe rows 7 & 8 are both '4'

#Add constraints to limit Y values from repeating
add.constraint(lp_rowpicker, xt=c(1,1,1), #xt specifies which rows of the LP
               indices=c(1,6,8), rhs=1, type="<=") #Y's in df rows 1,6 & 8 are all '5'
add.constraint(lp_rowpicker, xt=c(1,1), #xt specifies which rows of the LP
               indices=c(2,7), rhs=1, type="<=") #Y's in dataframe rows 2&7 are both '6'
add.constraint(lp_rowpicker, xt=c(1,1), #xt specifies which rows of the LP
               indices=c(3,5), rhs=1, type="<=") #y's in dataframe rows 3&5 are both '7'

solve(lp_rowpicker)
get.objective(lp_rowpicker) #20
get.variables(lp_rowpicker)
#[1] 0 0 1 1 0 1 1 0
#This tells you that from d you pick rows: 3,4,6 & 7 in your optimal solution.

#If you want to look at the full formulation:
rownames1 <- paste("OneX", c(1,2,4), sep="_")
rownames2 <- paste("OneY", c(5,6,7), sep="_")
colnames<- paste("pick_",c(1:8), sep="")
dimnames(lp_rowpicker) <- list(c(rownames1, rownames2), colnames)
print(lp_rowpicker)

#write it to a text file
write.lp(lp_rowpicker,filename="max_w.lp")

希望这能让您了解如何使用lpSolveAPI来制定您的问题。

如何使用LpSolve在R中设置线性编程优化？

1 个答案:

根据OP的请求进行更新

解决方案

代码