基于多约束的2024届学生向2025组的均衡重分配技术问询

云上老周

2026-6-16

学生跨届分组优化需求及解决方案求助

我需要将2024届共105名学生（女生61人、男生44人）分配至2025年的小组中，分配目标按优先级排序如下：

同一2024组转入同一2025组的学生不超过2人；
2025组需尽可能性别均衡，理想状态为每组7-8名女生、5-6名男生；
2025组规模需尽可能均匀，理想为7个13人组+1个14人组；
尽可能随机分配。

若严格遵循前三条标准无法得到解决方案，组规模与随机性可适当灵活调整。

当前2024组的规模与性别分布较为不均：

table(df[, c("Gender", "g2024")])
#         g2024
# Gender   1 2 3 4 5 6 7 8
#   Female 7 7 7 7 9 7 9 8
#   Male   4 4 8 6 5 7 4 6

table(df[, "g2024"])
# g2024
#  1  2  3  4  5  6  7  8 
# 11 11 15 13 14 14 13 14

我曾用预设值完成了一个基础示例，但预设组规模与性别分布并不理想，且未考虑2024组学生转入同一2025组的人数限制：

# Create random vector with group ids for females per group, assign to g2025
set.seed(42)
grp_fem <- as.character(rep(1:8, sample(c(rep(7, 3), rep(8, 5)), 8)))

df$g2025 <- unlist(lapply(1:nrow(df), function(i) {  
  if (df$Gender[i] == "Female") {
    x <- sample(grp_fem, 1)
    grp_fem <<- grp_fem[-match(x, grp_fem)]
    return(x)
  } else { 
    return(NA)    
  }
}))

# Get males per group, change one group length so y sums to male count
x <- as.integer(table(df$g2025))
y <- 13 - x
z <- sample(which(y == 5), 1)
y[z] <- 6

# Create vector with group ids for males per group, assign to g2025
grp_mal <- as.character(rep(1:8, rep(y)))

df$g2025 <- unlist(lapply(1:nrow(df), function(i) {
  if (df$Gender[i] == "Male") {
    x <- sample(grp_mal, 1)
    grp_mal <<- grp_mal[-match(x, grp_mal)]
    return(x)   
  } else {
    return(df$g2025[i])  
  }
}))

# Gender distribution per group
table(df[, c("Gender", "g2025")])
#         g2025
# Gender   1 2 3 4 5 6 7 8
#   Female 7 8 8 8 7 8 7 8
#   Male   6 5 5 5 6 6 6 5

# Number of students carried from g2024 to g2025 groups
table(df[, c("g2024", "g2025")])
#      g2025
# g2024 1 2 3 4 5 6 7 8
#     1 0 2 1 2 1 2 3 0
#     2 1 3 2 2 0 0 1 2
#     3 2 2 1 2 3 2 1 2
#     4 1 1 2 1 2 2 2 2
#     5 2 1 4 0 2 3 1 1
#     6 5 3 1 2 0 0 2 1
#     7 1 1 0 1 2 2 1 5
#     8 1 0 2 3 3 3 2 0

我也曾尝试用矩阵进行组间随机分配，但仍需预设colSums()值并手动调整：

m <- structure(c(1, 1, 2, 1, 2, 2, 2, 2, 1, 1, 2, 2, 2, 2, 2, 1, 2, 
                 1, 2, 2, 2, 1, 1, 2, 1, 1, 1, 2, 2, 2, 2, 2, 2, 1, 2, 1, 2, 2, 
                 1, 2, 1, 2, 2, 2, 1, 1, 2, 2, 1, 2, 2, 2, 1, 2, 2, 1, 2, 2, 2, 
                 1, 2, 2, 1, 2), dim = c(8L, 8L))

rowSums(m)
# [1] 11 11 15 13 14 14 13 14
colSums(m)
# [1] 13 13 13 13 13 13 13 14

我猜测lpSolve包可解决此类问题，但无法掌握其使用方法，现寻求可行的实现方案。

数据：

df <- structure(list(Student = c("1", "2", "3", "4", "5", "6", "7", 
"8", "9", "10", "11", "12", "13", "14", "15", "16", "17", "18", 
"19", "20", "21", "22", "23", "24", "25", "26", "27", "28", "29", 
"30", "31", "32", "33", "34", "35", "36", "37", "38", "39", "40", 
"41", "42", "43", "44", "45", "46", "47", "48", "49", "50", "51", 
"52", "53", "54", "55", "56", "57", "58", "59", "60", "61", "62", 
"63", "64", "65", "66", "67", "68", "69", "70", "71", "72", "73", 
"74", "75", "76", "77", "78", "79", "80", "81", "82", "83", "84", 
"85", "86", "87", "88", "89", "90", "91", "92", "93", "94", "95", 
"96", "97", "98", "99", "100", "101", "102", "103", "104", "105"
), Gender = c("Female", "Male", "Male", "Male", "Female", "Female", 
"Male", "Female", "Female", "Female", "Male", "Male", "Male", 
"Female", "Male", "Female", "Male", "Female", "Female", "Female", 
"Male", "Male", "Female", "Female", "Female", "Male", "Male", 
"Male", "Female", "Female", "Male", "Female", "Female", "Female", 
"Male", "Female", "Female", "Female", "Female", "Male", "Male", 
"Female", "Female", "Female", "Female", "Male", "Female", "Male", 
"Female", "Female", "Female", "Female", "Male", "Female", "Male", 
"Female", "Male", "Male", "Male", "Female", "Female", "Female", 
"Female", "Female", "Female", "Male", "Female", "Female", "Male", 
"Male", "Female", "Female", "Male", "Female", "Male", "Female", 
"Female", "Male", "Female", "Female", "Female", "Male", "Female", 
"Male", "Female", "Female", "Male", "Female", "Male", "Male", 
"Male", "Male", "Male", "Female", "Male", "Female", "Female", 
"Male", "Female", "Male", "Female", "Female", "Male", "Male", 
"Female"), g2024 = c("4", "3", "3", "8", "2", "8", "4", "5", 
"8", "7", "2", "4", "4", "6", "6", "5", "1", "3", "7", "2", "6", 
"8", "2", "8", "1", "5", "8", "3", "3", "1", "5", "5", "1", "3", 
"8", "6", "1", "7", "5", "5", "1", "7", "4", "7", "5", "4", "8", 
"6", "3", "1", "7", "8", "7", "7", "2", "4", "8", "3", "7", "1", 
"6", "3", "6", "8", "2", "3", "3", "5", "7", "2", "2", "2", "8", 
"6", "1", "1", "4", "1", "6", "8", "2", "6", "5", "2", "5", "3", 
"3", "7", "3", "4", "3", "4", "7", "4", "6", "8", "4", "6", "6", 
"6", "5", "4", "5", "5", "7"), g2025 = c(NA, NA, NA, NA, NA, 
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, 
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, 
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, 
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, 
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, 
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA)), row.names = c(NA, -105L), class = c("tbl_df", 
"tbl", "data.frame"))

内容的提问来源于stack exchange，提问作者L Tyrone