add values of one group into another group in R
1
I have a question on how to add the value from a group to rest of the elements in the group then delete that row. for ex:
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))
in the above example, my data is grouped by Year, Cluster, Seed and Day where seed=99 values need to be added to above rows based on (Year, Cluster and Day) group then delete this row. for ex: Row # 16, is part of (Year=1, Cluster=a,Day=1 and Seed=99) group and the value of Row #16 which is 55 should be added to Row #1 (5+55), Row # 6 (6+55) and Row # 11 (2+55) and row # 16 should be deleted. But when it comes to Row #21, which is in cluster=C with seed=99, should remain in the database as is as it cannot find any matching in year+cluster+day combination.
My actual data is of 1 million records with 10 years, 80 clusters, 500 days and 10+1 (1 to 10 and 99) seeds, so looking for so looking for an efficient solution.
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
r datatable dplyr tidyr zoo
asked Nov 23 '18 at 20:41
Ravindar GRavindar G
103
add a comment |
1
I have a question on how to add the value from a group to rest of the elements in the group then delete that row. for ex:
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))
in the above example, my data is grouped by Year, Cluster, Seed and Day where seed=99 values need to be added to above rows based on (Year, Cluster and Day) group then delete this row. for ex: Row # 16, is part of (Year=1, Cluster=a,Day=1 and Seed=99) group and the value of Row #16 which is 55 should be added to Row #1 (5+55), Row # 6 (6+55) and Row # 11 (2+55) and row # 16 should be deleted. But when it comes to Row #21, which is in cluster=C with seed=99, should remain in the database as is as it cannot find any matching in year+cluster+day combination.
My actual data is of 1 million records with 10 years, 80 clusters, 500 days and 10+1 (1 to 10 and 99) seeds, so looking for so looking for an efficient solution.
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
r datatable dplyr tidyr zoo
asked Nov 23 '18 at 20:41
Ravindar GRavindar G
103
please post expected output
– Vivek Kalyanarangan
Nov 23 '18 at 20:59
add a comment |
1
1
1
1
I have a question on how to add the value from a group to rest of the elements in the group then delete that row. for ex:
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))
in the above example, my data is grouped by Year, Cluster, Seed and Day where seed=99 values need to be added to above rows based on (Year, Cluster and Day) group then delete this row. for ex: Row # 16, is part of (Year=1, Cluster=a,Day=1 and Seed=99) group and the value of Row #16 which is 55 should be added to Row #1 (5+55), Row # 6 (6+55) and Row # 11 (2+55) and row # 16 should be deleted. But when it comes to Row #21, which is in cluster=C with seed=99, should remain in the database as is as it cannot find any matching in year+cluster+day combination.
My actual data is of 1 million records with 10 years, 80 clusters, 500 days and 10+1 (1 to 10 and 99) seeds, so looking for so looking for an efficient solution.
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
r datatable dplyr tidyr zoo
asked Nov 23 '18 at 20:41
Ravindar GRavindar G
103
I have a question on how to add the value from a group to rest of the elements in the group then delete that row. for ex:
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))
in the above example, my data is grouped by Year, Cluster, Seed and Day where seed=99 values need to be added to above rows based on (Year, Cluster and Day) group then delete this row. for ex: Row # 16, is part of (Year=1, Cluster=a,Day=1 and Seed=99) group and the value of Row #16 which is 55 should be added to Row #1 (5+55), Row # 6 (6+55) and Row # 11 (2+55) and row # 16 should be deleted. But when it comes to Row #21, which is in cluster=C with seed=99, should remain in the database as is as it cannot find any matching in year+cluster+day combination.
My actual data is of 1 million records with 10 years, 80 clusters, 500 days and 10+1 (1 to 10 and 99) seeds, so looking for so looking for an efficient solution.
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
r datatable dplyr tidyr zoo
r datatable dplyr tidyr zoo
asked Nov 23 '18 at 20:41
Ravindar GRavindar G
103
asked Nov 23 '18 at 20:41
Ravindar GRavindar G
103
edited Nov 23 '18 at 21:06
Ravindar G
asked Nov 23 '18 at 20:41
Ravindar GRavindar G
103
asked Nov 23 '18 at 20:41
Ravindar GRavindar G
103
asked Nov 23 '18 at 20:41
Ravindar GRavindar G
103
103
please post expected output
– Vivek Kalyanarangan
Nov 23 '18 at 20:59
add a comment |
please post expected output
– Vivek Kalyanarangan
Nov 23 '18 at 20:59
please post expected output
– Vivek Kalyanarangan
Nov 23 '18 at 20:59
please post expected output
– Vivek Kalyanarangan
Nov 23 '18 at 20:59
add a comment |
2 Answers
2
active
oldest
votes
0
Here's an approach using the tidyverse. If you're looking for speed with a million rows, a data.table solution will probably perform better.
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
Created on 2018-11-23 by the reprex package (v0.2.1)
answered Nov 23 '18 at 21:13
Jake KauppJake Kaupp
5,50221428
add a comment |
0
A data.table approach:
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]
Output:
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
answered Nov 23 '18 at 21:00
arg0nautarg0naut
2,239314
add a comment |
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2 Answers
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2 Answers
2
active
oldest
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active
oldest
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active
oldest
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0
Here's an approach using the tidyverse. If you're looking for speed with a million rows, a data.table solution will probably perform better.
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
Created on 2018-11-23 by the reprex package (v0.2.1)
answered Nov 23 '18 at 21:13
Jake KauppJake Kaupp
5,50221428
add a comment |
0
Here's an approach using the tidyverse. If you're looking for speed with a million rows, a data.table solution will probably perform better.
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
Created on 2018-11-23 by the reprex package (v0.2.1)
answered Nov 23 '18 at 21:13
Jake KauppJake Kaupp
5,50221428
add a comment |
0
0
0
Here's an approach using the tidyverse. If you're looking for speed with a million rows, a data.table solution will probably perform better.
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
Created on 2018-11-23 by the reprex package (v0.2.1)
answered Nov 23 '18 at 21:13
Jake KauppJake Kaupp
5,50221428
Here's an approach using the tidyverse. If you're looking for speed with a million rows, a data.table solution will probably perform better.
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
Created on 2018-11-23 by the reprex package (v0.2.1)
answered Nov 23 '18 at 21:13
Jake KauppJake Kaupp
5,50221428
answered Nov 23 '18 at 21:13
Jake KauppJake Kaupp
5,50221428
answered Nov 23 '18 at 21:13
Jake KauppJake Kaupp
5,50221428
answered Nov 23 '18 at 21:13
Jake KauppJake Kaupp
5,50221428
5,50221428
add a comment |
add a comment |
0
A data.table approach:
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]
Output:
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
answered Nov 23 '18 at 21:00
arg0nautarg0naut
2,239314
add a comment |
0
A data.table approach:
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]
Output:
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
answered Nov 23 '18 at 21:00
arg0nautarg0naut
2,239314
add a comment |
0
0
0
A data.table approach:
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]
Output:
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
answered Nov 23 '18 at 21:00
arg0nautarg0naut
2,239314
A data.table approach:
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]
Output:
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
answered Nov 23 '18 at 21:00
arg0nautarg0naut
2,239314
edited Nov 23 '18 at 21:29
answered Nov 23 '18 at 21:00
arg0nautarg0naut
2,239314
answered Nov 23 '18 at 21:00
arg0nautarg0naut
2,239314
answered Nov 23 '18 at 21:00
arg0nautarg0naut
2,239314
2,239314
add a comment |
add a comment |
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please post expected output
– Vivek Kalyanarangan
Nov 23 '18 at 20:59