我有一个数据框，其中包含来自 4 个不同球队的 20 名球员(每队 5 名球员)，每个球员都从幻想选秀中获得了薪水。我希望能够创建薪水等于或小于 10000 且总分大于 x 的 8 名球员的所有组合，但不包括包含来自同一球队的 4 名或更多球员的任何组合。
这是我的数据框的样子:
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要。这是我目前所拥有的:
## 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，我能够生成所有可能的 8 名球员的阵容，然后根据各种标准(总薪水和积分数)对其进行子集化，但是在排除超过 8 名球员的阵容时我很费力同一队的 3 名球员。
我想阵容需要从 newdf 中排除，但我真的不知道从哪里开始。
这里是输出结果:
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"))
最佳答案
最好以长格式构建它，我认为:
组建团队
library(data.table)
setDT(FantasyPlayers)
xx <- combn(as.character(FantasyPlayers$Player), 8)
mxx <- setDT(melt(xx, varnames=c("jersey_no", "team_no"), value.name="Player"))
head(mxx,10)
# jersey_no team_no Player
# 1: 1 1 ExoticDeer
# 2: 2 1 Supreme
# 3: 3 1 sasu
# 4: 4 1 eL lisasH 2
# 5: 5 1 Nisha
# 6: 6 1 Swiftending
# 7: 7 1 Pajkatt
# 8: 8 1 SexyBamboe
# 9: 1 2 ExoticDeer
# 10: 2 2 Supreme
8 人一组共享 team_no并由其索引 jersey_no .看?melt.array看看这是如何工作的。 setDT只需将生成的 data.frame 转换为 data.table 以便于合并。
合并恢复 Player属性
FantasyTeams <- FantasyPlayers[mxx, on="Player"]
# Team Player K D A LH Points Salary PPS jersey_no team_no
# 1: ATN ExoticDeer 6.1 3.3 6.4 306.9 22.209 1622 1.3692 1 1
# 2: ATN Supreme 6.8 5.3 7.1 229.4 21.954 1578 1.3913 2 1
# 3: ATN sasu 3.6 6.4 11.0 95.7 19.357 1244 1.5560 3 1
# 4: ATN eL lisasH 2 2.6 6.1 7.9 29.7 12.037 998 1.2061 4 1
# 5: ATN Nisha 2.7 5.6 7.5 48.2 12.282 955 1.2861 5 1
# ---
# 1007756: GB Keyser Soze 6.0 5.0 5.6 316.0 19.120 1602 1.1935 4 125970
# 1007757: GB Madara 5.4 5.3 6.6 334.5 19.405 1577 1.2305 5 125970
# 1007758: GB SkyLark 1.8 5.3 7.0 71.8 10.218 1266 0.8071 6 125970
# 1007759: GB MNT 2.3 5.9 6.1 85.6 9.316 1007 0.9251 7 125970
# 1007760: GB SKANKS224 1.4 7.6 7.4 52.5 7.565 954 0.7930 8 125970
默认情况下，仅打印 data.table 的第一行和最后几行。要检查整个事情，请尝试 ?View或查看 ?print.data.table 的参数.
筛选出具有所选功能的一组团队
过滤那些team_no来自同一个Team的玩家不超过三名...
my_teams <- FantasyTeams[, max(table(Team)) <= 3, by=team_no][V1==TRUE]$team_no
V1是分配给构造变量的默认名称 max(table(Team)) <= 3 .这不是快如闪电，但现在您已经排除了一些团队，后面的子集步骤应该会更快:
my_new_teams <-
FantasyTeams[team_no %in% my_teams, sum(Salary) < 10000, by=team_no][V1==TRUE]$team_no
要节省几次击键和微秒，请替换为 (V1)对于 V1==TRUE .这是惯用的方式。
从一组团队中恢复花名册
要获得与每个团队关联的花名册，加入/合并 mxx
mxx[.(team_no = my_new_teams), on="team_no"]
如果您希望球员列在一行中，如 OP 中所示:
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数据，则需要使用 FantasyTeams 加入:
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
了解一下by=.EACHI正在做，需要一点背景知识。这里的合并语法是 DT[i, j, on=cols, by=.EACHI] .
- 如果
j和by被排除在外，它只是进行合并，就像在FantasyTeams的构造中一样. - 如果
by被排除在外，但是j包括在内，j在合并之后计算。 - 如果
by=.EACHI, 然后j为i中的每个值单独计算.
关于r - 我想根据同一数据框中其他列的条件从 R 数据框中的列生成 8 种名称组合，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/32855755/