这是我的玩具数据框，真实的可能有40K-1M记录和另外五列
animal1 version1 animal2 version2 sim
53 20154620 TRUSEQ.v1 20104647 F250v1 0.3663569
854 20145687 TRUSEQ.v1 20105551 F250v1 0.5732854
3662 20154620 TRUSEQ.v1 20114509 F250v1 0.3374918
4063 20154620 TRUSEQ.v1 20114578 F250v1 0.3732692
4464 20154620 TRUSEQ.v1 20114595 F250v1 0.3772367
5262 20144516 TRUSEQ.v1 20115051 770k.v1 0.6034206
5663 20144516 TRUSEQ.v1 20115051 F250v1 0.6164795
5664 20145008 TRUSEQ.v1 20115051 F250v1 0.3146651
6064 20144516 TRUSEQ.v1 20115059 F250v1 0.3043295
6471 20165119 F250v1 20115096 F250v1 0.388435
9677 20154620 TRUSEQ.v1 20118095 F250v1 0.3079702
11281 20154620 TRUSEQ.v1 20134529 F250v1 0.3188631
12486 20165119 F250v1 20135032 F250v1 0.6091486
13282 20144516 TRUSEQ.v1 20135047 F250v1 0.3098507
14090 20165119 F250v1 20135072 F250v1 0.3025007
14892 20165119 F250v1 20135122 F250v1 0.345238
对于每只动物1，我需要所有以最高模拟量排列前3个独特动物2值的行......所以我想要的结果再现如下。
animal1 version1 animal2 version2 sim
5663 20144516 TRUSEQ.v1 20115051 F250v1 0.6164795
5262 20144516 TRUSEQ.v1 20115051 770k.v1 0.6034206
13282 20144516 TRUSEQ.v1 20135047 F250v1 0.3098507
6064 20144516 TRUSEQ.v1 20115059 F250v1 0.3043295
5664 20145008 TRUSEQ.v1 20115051 F250v1 0.3146651
854 20145687 TRUSEQ.v1 20105551 F250v1 0.5732854
4464 20154620 TRUSEQ.v1 20114595 F250v1 0.3772367
4063 20154620 TRUSEQ.v1 20114578 F250v1 0.3732692
53 20154620 TRUSEQ.v1 20104647 F250v1 0.3663569
12486 20165119 F250v1 20135032 F250v1 0.6091486
6471 20165119 F250v1 20115096 F250v1 0.388435
14892 20165119 F250v1 20135122 F250v1 0.345238
因此，在子集中，每个animal1可能有1到20个观察值，但是在这种情况下，其中n = 3的＆lt; = n唯一值为animal2。
我可以通过sim和animal1对df进行排序，就像这样
mydf <- mydf[order(-xtfrm(mydf[,"animal1"]), -mydf[,"sim"]),]
我可以像每个动物一样抓住前n个观察结果
mydf2 <- by(mydf, mydf["animal1"], head, n=1)
mydf2 <- Reduce(rbind, mydf2)
但是如何将n应用于第三列，即动物2而不是观察数量？如果这是重复的道歉，答案可能隐藏在这里，how to find the top N values by group or within category (groupwise) in an R data.frame 但我似乎无法从答案中找到解决问题的方法。
答案 0 :(得分：1)
尽管我有关于动物＆＃34; 20144516＆＃34;的问题，但这里有两个解决方案，使用dat作为您的上述样本数据（包含在底部以便重现）。我提供base-R和dplyr，但正如@Balter在评论中所建议的那样，可能还有一种简单的data.table方法。
# ordering by animal1 is not necessary, sim is priority
dat <- dat[rev(order(dat$sim)),]
dat2 <- do.call(rbind, by(dat, list(dat$animal1, dat$animal2), head, n = 1))
# ... but we need to re-sort by sim, since the ordering is lost with `by`
dat2 <- dat2[rev(order(dat2$sim)),]
head(dat2)
# animal1 version1 animal2 version2 sim
# 5663 20144516 TRUSEQ.v1 20115051 F250v1 0.6164795
# 12486 20165119 F250v1 20135032 F250v1 0.6091486
# 854 20145687 TRUSEQ.v1 20105551 F250v1 0.5732854
# 6471 20165119 F250v1 20115096 F250v1 0.3884350
# 4464 20154620 TRUSEQ.v1 20114595 F250v1 0.3772367
# 4063 20154620 TRUSEQ.v1 20114578 F250v1 0.3732692
这为我们提供了animal1和animal2的前1个配对，按sim排序（降序）。现在我们仅使用animal1有效地重复该过程：
dat3 <- do.call(rbind, by(dat, list(dat$animal1), head, n = 3))
dat3
# animal1 version1 animal2 version2 sim
# 20144516.5663 20144516 TRUSEQ.v1 20115051 F250v1 0.6164795
# 20144516.5262 20144516 TRUSEQ.v1 20115051 770k.v1 0.6034206
# 20144516.13282 20144516 TRUSEQ.v1 20135047 F250v1 0.3098507
# 20145008 20145008 TRUSEQ.v1 20115051 F250v1 0.3146651
# 20145687 20145687 TRUSEQ.v1 20105551 F250v1 0.5732854
# 20154620.4464 20154620 TRUSEQ.v1 20114595 F250v1 0.3772367
# 20154620.4063 20154620 TRUSEQ.v1 20114578 F250v1 0.3732692
# 20154620.53 20154620 TRUSEQ.v1 20104647 F250v1 0.3663569
# 20165119.12486 20165119 F250v1 20135032 F250v1 0.6091486
# 20165119.6471 20165119 F250v1 20115096 F250v1 0.3884350
# 20165119.14892 20165119 F250v1 20135122 F250v1 0.3452380
（不幸的是，这些rownames是不可思议的。如果它们有意义，我建议您将rownames(dat)放入一列并保留在那里。）
dplyr 您也可以使用dplyr。
library(dplyr)
dat %>%
group_by(animal1, animal2) %>%
top_n(1, wt = sim) %>%
group_by(animal1) %>%
top_n(3, wt = sim) %>%
ungroup()
# # A tibble: 11 × 5
# animal1 version1 animal2 version2 sim
# <int> <fctr> <int> <fctr> <dbl>
# 1 20144516 TRUSEQ.v1 20115051 F250v1 0.6164795
# 2 20165119 F250v1 20135032 F250v1 0.6091486
# 3 20145687 TRUSEQ.v1 20105551 F250v1 0.5732854
# 4 20165119 F250v1 20115096 F250v1 0.3884350
# 5 20154620 TRUSEQ.v1 20114595 F250v1 0.3772367
# 6 20154620 TRUSEQ.v1 20114578 F250v1 0.3732692
# 7 20154620 TRUSEQ.v1 20104647 F250v1 0.3663569
# 8 20165119 F250v1 20135122 F250v1 0.3452380
# 9 20145008 TRUSEQ.v1 20115051 F250v1 0.3146651
# 10 20144516 TRUSEQ.v1 20135047 F250v1 0.3098507
# 11 20144516 TRUSEQ.v1 20115059 F250v1 0.3043295
PS：使用do.call(rbind, ...)替代Reduce(rbind, ...)明显加快速度：
library(microbenchmark)
x <- by(dat, list(dat$animal1, dat$animal2), head, n = 1)
microbenchmark(
docall = do.call(rbind, x),
reduce = Reduce(rbind, x)
)
# Unit: milliseconds
# expr min lq mean median uq max neval
# docall 1.418577 1.493335 1.809469 1.551136 1.731466 5.216277 100
# reduce 11.119961 11.829614 13.302388 12.727255 13.401535 26.897520 100
这种差异随着更多不同动物的增加而增加。 （这是因为Reduce方法会为每个唯一动物调用rbind一次，而do.call只调用rbind一次。）
此处使用的样本数据：
dat <- structure(list(animal1 = c(20154620L, 20145687L, 20154620L, 20154620L,
20154620L, 20144516L, 20144516L, 20145008L, 20144516L, 20165119L,
20154620L, 20154620L, 20165119L, 20144516L, 20165119L, 20165119L
), version1 = structure(c(2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L,
1L, 2L, 2L, 1L, 2L, 1L, 1L), .Label = c("F250v1", "TRUSEQ.v1"
), class = "factor"), animal2 = c(20104647L, 20105551L, 20114509L,
20114578L, 20114595L, 20115051L, 20115051L, 20115051L, 20115059L,
20115096L, 20118095L, 20134529L, 20135032L, 20135047L, 20135072L,
20135122L), version2 = structure(c(2L, 2L, 2L, 2L, 2L, 1L, 2L,
2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L), .Label = c("770k.v1", "F250v1"
), class = "factor"), sim = c(0.3663569, 0.5732854, 0.3374918,
0.3732692, 0.3772367, 0.6034206, 0.6164795, 0.3146651, 0.3043295,
0.388435, 0.3079702, 0.3188631, 0.6091486, 0.3098507, 0.3025007,
0.345238)), .Names = c("animal1", "version1", "animal2", "version2",
"sim"), class = "data.frame", row.names = c("53", "854", "3662",
"4063", "4464", "5262", "5663", "5664", "6064", "6471", "9677",
"11281", "12486", "13282", "14090", "14892"))
答案 1 :(得分：0)
以下代码将包含多个animal1 - animal2组合，仅适用于较低sim - animal1条目的animal2值＆＃34;将是＆＃34;在前3名。如果我误解了，请告诉我。
library(dplyr)
selected <- dat %>%
arrange(animal1,animal2,desc(sim)) %>%
group_by(animal1,animal2) %>%
mutate(rank=row_number()) %>%
filter(rank==1) %>% ungroup() %>%
group_by(animal1) %>%
top_n(3,sim) %>%
summarise(threshold = min(sim))
out <- dat %>%
inner_join(selected, by = c("animal1"="animal1")) %>%
filter(sim>=threshold) %>%
arrange(animal1,animal2,desc(sim)) %>%
select(-threshold)
> out
animal1 version1 animal2 version2 sim
1 20144516 TRUSEQ.v1 20115051 F250v1 0.6164795
2 20144516 TRUSEQ.v1 20115051 770k.v1 0.6034206
3 20144516 TRUSEQ.v1 20115059 F250v1 0.3043295
4 20144516 TRUSEQ.v1 20135047 F250v1 0.3098507
5 20145008 TRUSEQ.v1 20115051 F250v1 0.3146651
6 20145687 TRUSEQ.v1 20105551 F250v1 0.5732854
7 20154620 TRUSEQ.v1 20104647 F250v1 0.3663569
8 20154620 TRUSEQ.v1 20114578 F250v1 0.3732692
9 20154620 TRUSEQ.v1 20114595 F250v1 0.3772367
10 20165119 F250v1 20115096 F250v1 0.3884350
11 20165119 F250v1 20135032 F250v1 0.6091486
12 20165119 F250v1 20135122 F250v1 0.3452380