我有一个数据框，有来自4个不同球队的20名球员（每队5名球员），每人都分配了幻想选秀的薪水。我希望能够创建8名工资的所有组合，其薪水等于或小于10000＆amp;其总分大于x，但不包括同一团队中包含4名或更多玩家的任何组合。
以下是我的数据框：
Team Player K D A LH Points Salary PPS
4 ATN ExoticDeer 6.1 3.3 6.4 306.9 22.209 1622 1.3692
2 ATN Supreme 6.8 5.3 7.1 229.4 21.954 1578 1.3913
1 ATN sasu 3.6 6.4 11.0 95.7 19.357 1244 1.5560
3 ATN eL lisasH 2 2.6 6.1 7.9 29.7 12.037 998 1.2061
5 ATN Nisha 2.7 5.6 7.5 48.2 12.282 955 1.2861
11 CL Swiftending 6.0 5.8 7.8 360.5 22.285 1606 1.3876
13 CL Pajkatt 13.3 7.5 9.3 326.8 37.248 1489 2.5015
15 CL SexyBamboe 6.3 8.5 9.3 168.0 20.660 1256 1.6449
14 CL EGM 2.8 6.0 13.5 78.8 21.988 989 2.2233
12 CL Saksa 2.5 6.5 10.5 59.8 15.898 967 1.6441
51 DBEARS Ace 7.0 3.4 6.9 195.6 23.596 1578 1.4953
31 DBEARS HesteJoe 5.4 5.4 6.1 176.7 16.927 1512 1.1195
61 DBEARS Miggel 2.8 6.8 11.0 141.8 17.818 1212 1.4701
21 DBEARS Noia 3.0 6.0 8.0 36.1 13.161 970 1.3568
41 DBEARS Ryze 2.7 4.7 6.7 74.6 12.166 937 1.2984
8 GB Keyser Soze 6.0 5.0 5.6 316.0 19.120 1602 1.1935
9 GB Madara 5.4 5.3 6.6 334.5 19.405 1577 1.2305
10 GB SkyLark 1.8 5.3 7.0 71.8 10.218 1266 0.8071
7 GB MNT 2.3 5.9 6.1 85.6 9.316 1007 0.9251
6 GB SKANKS224 1.4 7.6 7.4 52.5 7.565 954 0.7930
调整代码以满足我的需求。这就是我到目前为止所做的：
## make a list of all combinations of 8 of Player, Points and Salary
xx <- with(FantasyPlayers, lapply(list(as.character(Player), Points, Salary), combn, 8))
## convert the names to a string,
## find the column sums of the others,
## set the names
yy <- setNames(
lapply(xx, function(x) {
if(typeof(x) == "character") apply(x, 2, toString) else colSums(x)
}),
names(FantasyPlayers)[c(2, 7, 8)]
)
## coerce to data.frame
newdf <- as.data.frame(yy)
使用上面的代码，我能够生成8个玩家的所有可能的阵容，然后通过各种标准（总薪水和积分数）对其进行分组，但是当我要排除超过阵容的阵容时我很挣扎来自同一支球队的3名球员。
我认为阵容需要从newdf中排除，但我真的不知道从哪里开始。
以下是输入结果：
structure(list(Team = c("ATN", "ATN", "ATN", "ATN", "ATN", "CL",
"CL", "CL", "CL", "CL", "DBEARS", "DBEARS", "DBEARS", "DBEARS",
"DBEARS", "GB", "GB", "GB", "GB", "GB"), Player = structure(c(2L,
5L, 4L, 1L, 3L, 15L, 12L, 14L, 11L, 13L, 16L, 18L, 19L, 20L,
21L, 6L, 7L, 10L, 8L, 9L), .Label = c("eL lisasH 2", "ExoticDeer",
"Nisha", "sasu", "Supreme", "Keyser Soze", "Madara", "MNT", "SKANKS224",
"SkyLark", "EGM", "Pajkatt", "Saksa", "SexyBamboe", "Swiftending",
"Ace", "DruidzOzoneShoc", "HesteJoe", "Miggel", "Noia", "Ryze"
), class = "factor"), K = c(6.1, 6.8, 3.6, 2.6, 2.7, 6, 13.3,
6.3, 2.8, 2.5, 7, 5.4, 2.8, 3, 2.7, 6, 5.4, 1.8, 2.3, 1.4), D = c(3.3,
5.3, 6.4, 6.1, 5.6, 5.8, 7.5, 8.5, 6, 6.5, 3.4, 5.4, 6.8, 6,
4.7, 5, 5.3, 5.3, 5.9, 7.6), A = c(6.4, 7.1, 11, 7.9, 7.5, 7.8,
9.3, 9.3, 13.5, 10.5, 6.9, 6.1, 11, 8, 6.7, 5.6, 6.6, 7, 6.1,
7.4), LH = c(306.9, 229.4, 95.7, 29.7, 48.2, 360.5, 326.8, 168,
78.8, 59.8, 195.6, 176.7, 141.8, 36.1, 74.6, 316, 334.5, 71.8,
85.6, 52.5), Points = c(22.209, 21.954, 19.357, 12.037, 12.282,
22.285, 37.248, 20.66, 21.988, 15.898, 23.596, 16.927, 17.818,
13.161, 12.166, 19.12, 19.405, 10.218, 9.316, 7.565), Salary = c(1622,
1578, 1244, 998, 955, 1606, 1489, 1256, 989, 967, 1578, 1512,
1212, 970, 937, 1602, 1577, 1266, 1007, 954), PPS = c(1.3692,
1.3913, 1.556, 1.2061, 1.2861, 1.3876, 2.5015, 1.6449, 2.2233,
1.6441, 1.4953, 1.1195, 1.4701, 1.3568, 1.2984, 1.1935, 1.2305,
0.8071, 0.9251, 0.793)), .Names = c("Team", "Player", "K", "D",
"A", "LH", "Points", "Salary", "PPS"), class = "data.frame", row.names = c("4",
"2", "1", "3", "5", "11", "13", "15", "14", "12", "51", "31",
"61", "21", "41", "8", "9", "10", "7", "6"))
答案 0 :(得分：4)
最好以长篇形式构建这个，我想：
构建团队
library(data.table)
setDT(FantasyPlayers)
xx <- combn(as.character(FantasyPlayers$Player), 8)
mxx <- setDT(melt(xx, varnames=c("jersey_no", "team_no"), value.name="Player"))
head(mxx,10)
# jersey_no team_no Player
# 1: 1 1 ExoticDeer
# 2: 2 1 Supreme
# 3: 3 1 sasu
# 4: 4 1 eL lisasH 2
# 5: 5 1 Nisha
# 6: 6 1 Swiftending
# 7: 7 1 Pajkatt
# 8: 8 1 SexyBamboe
# 9: 1 2 ExoticDeer
# 10: 2 2 Supreme
8位玩家的小组共享team_no，并按其jersey_no编入索引。查看?melt.array以了解其工作原理。 setDT只是将生成的data.frame转换为data.table，以便于合并。
合并以恢复Player属性
FantasyTeams <- FantasyPlayers[mxx, on="Player"]
# Team Player K D A LH Points Salary PPS jersey_no team_no
# 1: ATN ExoticDeer 6.1 3.3 6.4 306.9 22.209 1622 1.3692 1 1
# 2: ATN Supreme 6.8 5.3 7.1 229.4 21.954 1578 1.3913 2 1
# 3: ATN sasu 3.6 6.4 11.0 95.7 19.357 1244 1.5560 3 1
# 4: ATN eL lisasH 2 2.6 6.1 7.9 29.7 12.037 998 1.2061 4 1
# 5: ATN Nisha 2.7 5.6 7.5 48.2 12.282 955 1.2861 5 1
# ---
# 1007756: GB Keyser Soze 6.0 5.0 5.6 316.0 19.120 1602 1.1935 4 125970
# 1007757: GB Madara 5.4 5.3 6.6 334.5 19.405 1577 1.2305 5 125970
# 1007758: GB SkyLark 1.8 5.3 7.0 71.8 10.218 1266 0.8071 6 125970
# 1007759: GB MNT 2.3 5.9 6.1 85.6 9.316 1007 0.9251 7 125970
# 1007760: GB SKANKS224 1.4 7.6 7.4 52.5 7.565 954 0.7930 8 125970
默认情况下，只打印data.table的第一行和最后几行。要检查整个事情，请尝试?View或查看?print.data.table的参数。
过滤到一组具有所选功能的团队
过滤那些来自同一team_no的玩家不超过三名的Team ...
my_teams <- FantasyTeams[, max(table(Team)) <= 3, by=team_no][V1==TRUE]$team_no
V1是分配给构造变量max(table(Team)) <= 3的默认名称。这不是一件闪电般快速的事情，但现在你已经排除了一些团队，后来的子集化步骤应该更快：
my_new_teams <-
FantasyTeams[team_no %in% my_teams, sum(Salary) < 10000, by=team_no][V1==TRUE]$team_no
要保存几个击键和微秒，请将(V1)替换为V1==TRUE。这是惯用的方式。
从一组球队中恢复名单
要获得与每个团队相关联的名单，请加入/合并mxx
mxx[.(team_no = my_new_teams), on="team_no"]
如果你想让球员列在一条线上，就像OP：
mxx[.(team_no = my_new_teams), .(roster = toString(Player)), on="team_no", by=.EACHI]
如果您想要每个团队的汇总统计信息，则需要加入FantasyTeams：
FantasyTeams[.(team_no = my_new_teams), .(
roster = toString(Player),
tot_salary = sum(Salary),
tot_points = sum(Points)
), on="team_no", by=.EACHI]
# team_no roster tot_salary tot_points
# 1: 3716 ExoticDeer, Supreme, sasu, Swiftending, EGM, Saksa, Noia, Ryze 9913 149.018
# 2: 3720 ExoticDeer, Supreme, sasu, Swiftending, EGM, Saksa, Noia, MNT 9983 146.168
# 3: 3721 ExoticDeer, Supreme, sasu, Swiftending, EGM, Saksa, Noia, SKANKS224 9930 144.417
# 4: 3725 ExoticDeer, Supreme, sasu, Swiftending, EGM, Saksa, Ryze, MNT 9950 145.173
# 5: 3726 ExoticDeer, Supreme, sasu, Swiftending, EGM, Saksa, Ryze, SKANKS224 9897 143.422
# ---
# 40202: 125663 EGM, Saksa, Miggel, Noia, Ryze, Keyser Soze, MNT, SKANKS224 8638 117.032
# 40203: 125664 EGM, Saksa, Miggel, Noia, Ryze, Madara, SkyLark, MNT 8925 119.970
# 40204: 125665 EGM, Saksa, Miggel, Noia, Ryze, Madara, SkyLark, SKANKS224 8872 118.219
# 40205: 125666 EGM, Saksa, Miggel, Noia, Ryze, Madara, MNT, SKANKS224 8613 117.317
# 40206: 125667 EGM, Saksa, Miggel, Noia, Ryze, SkyLark, MNT, SKANKS224 8302 108.130
要了解by=.EACHI正在做什么，需要一些背景知识。这里的合并语法是DT[i, j, on=cols, by=.EACHI]。
j和by，则只会进行合并，就像FantasyTeams的构造一样。 by，但包含j，则合并后会计算j。by=.EACHI，则会j中的每个值单独计算i。答案 1 :(得分：3)
这是一种方式：
splt.names <- strsplit(as.character(newdf$Player), ", ")
indices <- lapply(splt.names, function(x) match(x, FantasyPlayers$Player))
exclude <- lapply(indices, function(x) any(table(FantasyPlayers$Team[x]) > 3))
newdf2 <- newdf[!unlist(exclude), ]
首先用逗号分隔Player列。然后将玩家名称与Fantasy Players玩家名称列匹配。对于那些indices，我们可以完成any(table(FantasyPlayers$Team[x]) > 3)的主要工作。这是对超过3的团队计数的检查，这将指示来自同一团队的3个或更多玩家。