例如，我有以下示例数据:
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 y w
1 1 5 1
2 1 6 2
3 1 7 3
4 2 8 4
5 2 7 5
6 3 5 6
7 4 6 7
8 4 5 8
x和y表示datax和datay的索引。 w表示将datax[x]与datay[y]比较得到的分数。我想最大化w的总得分(或d)，其中x的每个值最多匹配y的一个值，反之亦然。结果应如下所示:
x y w
1 2 7 5
2 3 5 6
3 4 6 7
其中所有
w值的总和最大化，并且每个x和每个y在结果中仅显示一次。如何在
lpSolve::lp函数中设置此问题？ 最佳答案
您可以使用lpSolveAPI来解决您的问题。考虑到您的约束，您陈述的解决方案不太可行。因此，我们希望您不要在解决方案中重复X和Y。
您将需要8个新的二进制变量。每个变量指定d中的该行是被选择(1)还是被删除(0)。
根据OP的要求进行更新
是的，下面的lpSolveAPI代码使它看起来比实际复杂得多。这个LP公式(lpSolveAPI的输出)应该使事情更清楚:
/* 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;
pick_4或pick_5中只有一个可以为1，因为数据帧d中的第4行和第5行的X = 2解决方案
请注意，上面的公式产生了一个最佳解决方案，该解决方案在数据帧
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
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")