我有一个问题，如何将组中的值添加到组中的其余元素，然后删除该行。例如：
df <- data.frame(Year=c(1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2),
Cluster=c("a","a","a","a","a","a","a","a","a","a","a","a","a","a","a","a","a","a","a","a","c","b","b","b","b","b","b","b","b","b","b","b","b","b","b","b","b","b","b","b","b","d"),
Seed=c(1,1,1,1,1,2,2,2,2,2,3,3,3,3,3,99,99,99,99,99,99),
Day=c(1,2,3,4,5,1,2,3,4,5,1,2,3,4,5,1,2,3,4,5,1),
value=c(5,2,1,2,8,6,7,9,3,5,2,1,2,8,6,55,66,77,88,99,10))
在上面的示例中，我的数据按年，簇，种子和天分组，其中需要基于（年，簇和天）组将seed = 99值添加到上述行，然后删除该行。例如：＃16行是（Year = 1，Cluster = a，Day = 1 and Seed = 99）组的一部分，第16行的值为55应该添加到第1行（5 + 55） ，第6行（6 + 55）和第11行（2 + 55）和第16行应被删除。但是，当涉及到第21行时，它位于cluster = C中且seed = 99，应该按原样保留在数据库中，因为在year + cluster + day组合中找不到任何匹配项。
我的实际数据是100万条记录，这些记录具有10年，80个簇，500天和10 + 1（1到10和99）个种子，因此正在寻找一种有效的解决方案。
Year Cluster Seed Day value
1 1 a 1 1 60
2 1 a 1 2 68
3 1 a 1 3 78
4 1 a 1 4 90
5 1 a 1 5 107
6 1 a 2 1 61
7 1 a 2 2 73
8 1 a 2 3 86
9 1 a 2 4 91
10 1 a 2 5 104
11 1 a 3 1 57
12 1 a 3 2 67
13 1 a 3 3 79
14 1 a 3 4 96
15 1 a 3 5 105
16 1 c 99 1 10
17 2 b 1 1 60
18 2 b 1 2 68
19 2 b 1 3 78
20 2 b 1 4 90
21 2 b 1 5 107
22 2 b 2 1 61
23 2 b 2 2 73
24 2 b 2 3 86
25 2 b 2 4 91
26 2 b 2 5 104
27 2 b 3 1 57
28 2 b 3 2 67
29 2 b 3 3 79
30 2 b 3 4 96
31 2 b 3 5 105
32 2 d 99 1 10
答案 0 :(得分：0)
一种library(data.table)
df <- setDT(df)[, `:=` (value = ifelse(Seed != 99, value + value[Seed == 99], value),
flag = Seed == 99 & .N == 1), by = .(Year, Cluster, Day)][!(Seed == 99 & flag == FALSE),][, "flag" := NULL]
方法：
df[]
Year Cluster Seed Day value
1: 1 a 1 1 60
2: 1 a 1 2 68
3: 1 a 1 3 78
4: 1 a 1 4 90
5: 1 a 1 5 107
6: 1 a 2 1 61
7: 1 a 2 2 73
8: 1 a 2 3 86
9: 1 a 2 4 91
10: 1 a 2 5 104
11: 1 a 3 1 57
12: 1 a 3 2 67
13: 1 a 3 3 79
14: 1 a 3 4 96
15: 1 a 3 5 105
16: 1 c 99 1 10
17: 2 b 1 1 60
18: 2 b 1 2 68
19: 2 b 1 3 78
20: 2 b 1 4 90
21: 2 b 1 5 107
22: 2 b 2 1 61
23: 2 b 2 2 73
24: 2 b 2 3 86
25: 2 b 2 4 91
26: 2 b 2 5 104
27: 2 b 3 1 57
28: 2 b 3 2 67
29: 2 b 3 3 79
30: 2 b 3 4 96
31: 2 b 3 5 105
32: 2 d 99 1 10
输出：
{{1}}
答案 1 :(得分：0)
这是使用tidyverse的方法。如果您要寻找具有一百万行的速度，那么data.table解决方案可能会更好。
library(tidyverse)
df <- data.frame(Year=c(1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2),
Cluster=c("a","a","a","a","a","a","a","a","a","a","a","a","a","a","a","a","a","a","a","a","c","b","b","b","b","b","b","b","b","b","b","b","b","b","b","b","b","b","b","b","b","d"),
Seed=c(1,1,1,1,1,2,2,2,2,2,3,3,3,3,3,99,99,99,99,99,99),
Day=c(1,2,3,4,5,1,2,3,4,5,1,2,3,4,5,1,2,3,4,5,1),
value=c(5,2,1,2,8,6,7,9,3,5,2,1,2,8,6,55,66,77,88,99,10))
seeds <- df %>%
filter(Seed == 99)
matches <- df %>%
filter(Seed != 99) %>%
inner_join(select(seeds, -Seed), by = c("Year", "Cluster", "Day")) %>%
mutate(value = value.x + value.y) %>%
select(Year, Cluster, Seed, Day, value)
no_matches <- anti_join(seeds, matches, by = c("Year", "Cluster", "Day"))
bind_rows(matches, no_matches) %>%
arrange(Year, Cluster, Seed, Day)
#> Year Cluster Seed Day value
#> 1 1 a 1 1 60
#> 2 1 a 1 2 68
#> 3 1 a 1 3 78
#> 4 1 a 1 4 90
#> 5 1 a 1 5 107
#> 6 1 a 2 1 61
#> 7 1 a 2 2 73
#> 8 1 a 2 3 86
#> 9 1 a 2 4 91
#> 10 1 a 2 5 104
#> 11 1 a 3 1 57
#> 12 1 a 3 2 67
#> 13 1 a 3 3 79
#> 14 1 a 3 4 96
#> 15 1 a 3 5 105
#> 16 1 c 99 1 10
#> 17 2 b 1 1 60
#> 18 2 b 1 2 68
#> 19 2 b 1 3 78
#> 20 2 b 1 4 90
#> 21 2 b 1 5 107
#> 22 2 b 2 1 61
#> 23 2 b 2 2 73
#> 24 2 b 2 3 86
#> 25 2 b 2 4 91
#> 26 2 b 2 5 104
#> 27 2 b 3 1 57
#> 28 2 b 3 2 67
#> 29 2 b 3 3 79
#> 30 2 b 3 4 96
#> 31 2 b 3 5 105
#> 32 2 d 99 1 10
由reprex package（v0.2.1）于2018-11-23创建