Question

我刚刚开始学习R并遇到了一些我不确定如何处理代码的问题。

我正在创建一个data.frame，其中包含可以分配给项目的个人池。该项目需要一个BA，一个PM，两个SA，和另外一个可以是SA或BA 的人。每个人都有一个评级和相关的成本，我需要最高评级，同时保持成本低于一定的门槛。

我不确定如何实现上述方案中的粗体部分..以下代码正在运行，但并未考虑额外的BA / SA。

（这是自学..没有分配作业）

EDIT-Desired output ，其中最后一行可以是SA或BA位置。

name     position rating cost   BA PM SA  
Matt        SA     95    9500   0  0  1    
Aaron       BA     85    4700   1  0  0    
Stephanie   SA     95    9200   0  0  1    
Molly       PM     88    5500   0  1  0    
Jake        SA     74    5300   0  0  1

代码：

#load libraries
library(lpSolve)

# create data.frame
name = c("Steve", "Jeremy", "Matt", "Aaron", "Stephanie", "Molly", "Jake", "Tony", "Jay", "Katy", "Alison") 
position = c("BA", "PM", "SA", "BA", "SA", "PM", "SA", "SA", "PM", "BA", "SA")
rating = c(75, 90, 95, 85, 95, 88, 74, 81, 55, 65, 68) 
cost = c(5000, 8000, 9500, 4700, 9200, 5500, 5300, 7300, 3300, 4100, 4400)
df = data.frame(name, position, rating, cost)

# create restrictions
num_ba = 1
num_pm = 1
num_sa = 2
max_cost = 35000

# create vectors to constrain by position
df$BA = ifelse(df$position == "BA", 1, 0)
df$PM = ifelse(df$position == "PM", 1, 0)
df$SA = ifelse(df$position == "SA", 1, 0)

# vector to optimize against
objective = df$rating

# constraint directions
const_dir <- c("=", "=", "=", "<=")

# matrix
const_mat = matrix(c(df$BA, df$PM, df$SA, df$cost), 4, byrow=TRUE)
const_rhs = c(num_ba, num_pm, num_sa, max_cost)

#solve
x = lp("max", objective, const_mat, const_dir, const_rhs, all.bin=TRUE, all.int=TRUE)
print(df[which(x$solution==1), ])

Answer 1

如果我的问题是正确的，那可能会有效：

 library(lpSolve)

 # create data.frame
 name = c("Steve", "Jeremy", "Matt", "Aaron", "Stephanie", "Molly", "Jake", "Tony", "Jay", "Katy", "Alison") 
position = c("BA", "PM", "SA", "BA", "SA", "PM", "SA", "SA", "PM", "BA", "SA")
rating = c(75, 90, 95, 85, 95, 88, 74, 81, 55, 65, 68) 
cost = c(5000, 8000, 9500, 4700, 9200, 5500, 5300, 7300, 3300, 4100, 4400)

df = data.frame(name, position, rating, cost)

# create restrictions
num_pm = 1
min_num_ba = 1
min_num_sa = 2
tot_saba   = 4
max_cost = 35000

# create vectors to constrain by position
df$PM = ifelse(df$position == "PM", 1, 0)
df$minBA = ifelse(df$position == "BA", 1, 0)
df$minSA = ifelse(df$position == "SA", 1, 0)
df$SABA = ifelse(df$position %in% c("SA","BA"), 1, 0)

# vector to optimize against
objective = df$rating

# constraint directions
const_dir <- c("==", ">=", "<=", "==", "<=")

# matrix
const_mat = matrix(c(df$PM, df$minBA, df$minSA, df$SABA, df$cost), 5, byrow=TRUE)
const_rhs = c(num_pm, min_num_ba,min_num_sa, tot_saba, max_cost)

#solve
x = lp("max", objective, const_mat, const_dir, const_rhs, all.bin=TRUE, all.int=TRUE)
print(df[which(x$solution==1), ])

我正在做的是修改一些约束并添加一个新约束：BA的数量必须是> = 1. SA的数量＆gt; = 2，和 BA和SA的总和必须为4，因此您总是选择5个人。

然而，这提供了与OP所写的不同的解决方案：

       name position rating cost PM minBA minSA SABA
1     Steve       BA     75 5000  0     1     0    1
3      Matt       SA     95 9500  0     0     1    1
4     Aaron       BA     85 4700  0     1     0    1
5 Stephanie       SA     95 9200  0     0     1    1
6     Molly       PM     88 5500  1     0     0    0

然而，总结此解决方案的等级给出438，而运算结果为437，所以这应该是正确的。

HTH。

使用R找到最佳项目团队

1 个答案: