(Update) Add index column to data.frame based on two columns - r
Example data.frame:
df = read.table(text = 'colA colB
2 7
2 7
2 7
2 7
1 7
1 7
1 7
89 5
89 5
89 5
88 5
88 5
70 5
70 5
70 5
69 5
69 5
44 4
44 4
44 4
43 4
42 4
42 4
41 4
41 4
120 1
100 1', header = TRUE)
I need to add an index col based on colA and colB where colB shows the exact number of rows to group but it can be duplicated. colB groups rows based on colA and colA -1.
Expected output:
colA colB index_col
2 7 1
2 7 1
2 7 1
2 7 1
1 7 1
1 7 1
1 7 1
89 5 2
89 5 2
89 5 2
88 5 2
88 5 2
70 5 3
70 5 3
70 5 3
69 5 3
69 5 3
44 4 4
44 4 4
44 4 4
43 4 4
42 4 5
42 4 5
41 4 5
41 4 5
120 1 6
100 1 7
UPDATE
How can I adapt the code that works for the above df for the same purpose but by looking at colB values grouped based on colA, colA -1 and colA -2? i.e. (instead of 2 days considering 3 days)
new_df = read.table(text = 'colA colB
3 10
3 10
3 10
2 10
2 10
2 10
2 10
1 10
1 10
1 10
90 7
90 7
89 7
89 7
89 7
88 7
88 7
71 7
71 7
70 7
70 7
70 7
69 7
69 7
44 5
44 5
44 5
43 5
42 5
41 5
41 5
41 5
40 5
40 5
120 1
100 1', header = TRUE)
Expected output:
colA colB index_col
3 10 1
3 10 1
3 10 1
2 10 1
2 10 1
2 10 1
2 10 1
1 10 1
1 10 1
1 10 1
90 7 2
90 7 2
89 7 2
89 7 2
89 7 2
88 7 2
88 7 2
71 7 3
71 7 3
70 7 3
70 7 3
70 7 3
69 7 3
69 7 3
44 5 4
44 5 4
44 5 4
43 5 4
42 5 4
41 5 5
41 5 5
41 5 5
40 5 5
40 5 5
120 1 6
100 1 7
Thanks
We can use rleid
library(data.table)
index_col <-setDT(df)[, if(colB[1L] < .N) ((seq_len(.N)-1) %/% colB[1L])+1
else as.numeric(colB), rleid(colB)][, rleid(V1)]
df[, index_col := index_col]
df
# colA colB index_col
# 1: 2 7 1
# 2: 2 7 1
# 3: 2 7 1
# 4: 2 7 1
# 5: 1 7 1
# 6: 1 7 1
# 7: 1 7 1
# 8: 70 5 2
# 9: 70 5 2
#10: 70 5 2
#11: 69 5 2
#12: 69 5 2
#13: 89 5 3
#14: 89 5 3
#15: 89 5 3
#16: 88 5 3
#17: 88 5 3
#18: 120 1 4
#19: 100 1 5
Or a one-liner would be
setDT(df)[, index_col := df[, ((seq_len(.N)-1) %/% colB[1L])+1, rleid(colB)][, as.integer(interaction(.SD, drop = TRUE, lex.order = TRUE))]]
Update
Based on the new update in the OP's post
setDT(new_df)[, index_col := cumsum(c(TRUE, abs(diff(colA))> 1))
][, colB := .N , index_col]
new_df
# colA colB index_col
# 1: 3 10 1
# 2: 3 10 1
# 3: 3 10 1
# 4: 2 10 1
# 5: 2 10 1
# 6: 2 10 1
# 7: 2 10 1
# 8: 1 10 1
# 9: 1 10 1
#10: 1 10 1
#11: 71 7 2
#12: 71 7 2
#13: 70 7 2
#14: 70 7 2
#15: 70 7 2
#16: 69 7 2
#17: 69 7 2
#18: 90 7 3
#19: 90 7 3
#20: 89 7 3
#21: 89 7 3
#22: 89 7 3
#23: 88 7 3
#24: 88 7 3
#25: 44 2 4
#26: 43 2 4
#27: 120 1 5
#28: 100 1 6
An approach in base R:
df$idxcol <- cumsum(c(1,abs(diff(df$colA)) > 1) + c(0,diff(df$colB) != 0) > 0)
which gives:
> df
colA colB idxcol
1 2 7 1
2 2 7 1
3 2 7 1
4 2 7 1
5 1 7 1
6 1 7 1
7 1 7 1
8 70 5 2
9 70 5 2
10 70 5 2
11 69 5 2
12 69 5 2
13 89 5 3
14 89 5 3
15 89 5 3
16 88 5 3
17 88 5 3
18 120 1 4
19 100 1 5
On the updated example data, you need to adapt the approach to:
n <- 1
idx1 <- cumsum(c(1, diff(df$colA) < -n) + c(0, diff(df$colB) != 0) > 0)
idx2 <- ave(df$colA, cumsum(c(1, diff(df$colA) < -n)), FUN = function(x) c(0, cumsum(diff(x)) < -n ))
idx2[idx2==1 & c(0,diff(idx2))==0] <- 0
df$idxcol <- idx1 + cumsum(idx2)
which gives:
> df
colA colB idxcol
1 2 7 1
2 2 7 1
3 2 7 1
4 2 7 1
5 1 7 1
6 1 7 1
7 1 7 1
8 89 5 2
9 89 5 2
10 89 5 2
11 88 5 2
12 88 5 2
13 70 5 3
14 70 5 3
15 70 5 3
16 69 5 3
17 69 5 3
18 44 4 4
19 44 4 4
20 44 4 4
21 43 4 4
22 42 4 5
23 42 4 5
24 41 4 5
25 41 4 5
26 120 1 6
27 100 1 7
For new_df just change n tot 2 and you will get the desired output for that as well.
Related
Recode column every nth element in R
I'm looking to recode a column, say the following: df <- data.frame(col1 = rep(3, 100), col2 = rep(NA, 100)) I want to recode col2 as 1 for rows 1:5, 2 for rows 6:10, 3 for 11:15, etc. So, every five rows I would add +1 to the assigned value. Any way to automate this process to avoid manually recoding 100 rows?
There are lot of ways to do that. Here are couple of them - Using rep : df$col2 <- rep(1:nrow(df), each = 5, length.out = nrow(df)) Using ceiling df$col2 <- ceiling(seq(nrow(df))/5)
dplyr way
df %>% mutate(col2 = ((row_number()-1) %/% 5)+1)
OR
A simple for loop
for(i in 0:((nrow(df)/5)-1)){
df[0:nrow(df) %/% 5 == i,2] <- i+1
}
> df
col1 col2
1 3 1
2 3 1
3 3 1
4 3 1
5 3 1
6 3 2
7 3 2
8 3 2
9 3 2
10 3 2
11 3 3
12 3 3
13 3 3
14 3 3
15 3 3
16 3 4
17 3 4
18 3 4
19 3 4
20 3 4
21 3 5
22 3 5
23 3 5
24 3 5
25 3 5
26 3 6
27 3 6
28 3 6
29 3 6
30 3 6
31 3 7
32 3 7
33 3 7
34 3 7
35 3 7
36 3 8
37 3 8
38 3 8
39 3 8
40 3 8
41 3 9
42 3 9
43 3 9
44 3 9
45 3 9
46 3 10
47 3 10
48 3 10
49 3 10
50 3 10
51 3 11
52 3 11
53 3 11
54 3 11
55 3 11
56 3 12
57 3 12
58 3 12
59 3 12
60 3 12
61 3 13
62 3 13
63 3 13
64 3 13
65 3 13
66 3 14
67 3 14
68 3 14
69 3 14
70 3 14
71 3 15
72 3 15
73 3 15
74 3 15
75 3 15
76 3 16
77 3 16
78 3 16
79 3 16
80 3 16
81 3 17
82 3 17
83 3 17
84 3 17
85 3 17
86 3 18
87 3 18
88 3 18
89 3 18
90 3 18
91 3 19
92 3 19
93 3 19
94 3 19
95 3 19
96 3 20
97 3 20
98 3 20
99 3 20
100 3 20
As there is a pattern (each 5th row) you can use rep(row_number()) length.out = n() takes into account the length of column. Learned here dplyr: Mutate a new column with sequential repeated integers of n time in a dataframe from Ronak!!! Thanks to Ronak! df %>% mutate(col2 = rep(row_number(), each=5, length.out = n()))
Gen variable conditional on other variables from a different dataframe
I Have DF1: c01 p01 c02 p02 c03 p03 c04 p04 1 0 1 20 1 33 1 49 2 3 2 21 2 34 2 50 3 4 3 21 3 38 3 50 4 6 4 23 4 40 4 51 5 7 5 24 5 41 5 53 6 9 6 27 6 41 6 54 7 11 7 29 7 41 7 55 8 15 8 31 8 43 8 57 9 15 9 33 9 47 9 57 10 16 10 33 10 49 10 60 And i Have DF2: type round A 1 B 1 A 2 B 2 A 3 B 3 A 4 B 4 What i want is to generate a new variable in DF2 that goes like: DF2$g1<- if(DF2$round==1, 0) DF2$g2<- if(c01==4 & round==1,DF2$p01) DF2$g3<- if(c01==4 & round==2,DF2$p02) DF2$g4<- if(c01==4 & round==3,DF2$p03) DF2$g5<- if(c01==4 & round==4,DF2$p04) DF2$g6<- if(c01==4 & round==5,DF2$p05) So DF2 becomes: type round g A 1 6 B 1 6 A 2 23 B 2 23 A 3 40 B 3 40 A 4 50 B 4 50 Is there a way that i can loop this? In the original dataframe, i have 40 rounds, e C01 to C40 and P01 to P40
add values of one group into another group in R
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
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
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)
Group rows and add sum column of unique values
Here an example of my data.frame: df = read.table(text='colA colB colC 10 11 7 10 34 7 10 89 7 10 21 7 2 23 5 2 21 5 2 56 5 22 14 3 22 19 3 22 90 3 11 19 2 11 45 2 1 45 0 1 23 0 9 8 0 9 11 0 9 21 0', header = TRUE) I need to group the rows by colA and colC and add a new column which states the sum of unique values based on colB. In steps here what I need to do for this specific data.frame: group rows with colA = 10 and 9, colA = 2 and 1, colA = 22 and colA = 11; find the unique values of colB per each group; add the unique values in a new col (newcolD). Note that colC states the total number of observations for colA = 10 and 9, colA = 2 and 1, colA = 22 and colA = 11. The data.frame needs to remain ordered decreasingly by colC. My expected output is: colA colB colC newcolD 10 11 7 5 10 34 7 5 10 89 7 5 10 21 7 5 9 8 0 5 9 11 0 5 9 21 0 5 2 23 5 4 2 21 5 4 2 56 5 4 1 45 0 4 1 23 0 4 22 14 3 3 22 19 3 3 22 90 3 3 11 19 2 2 11 45 2 2 To note that in df the colB duplicated values are: 11 and 21 for group 10 and 9, and 23 for group 2 and 1.
You can do that with dplyr. The trick is to create a new grouping column which groups consecutive values in colA. This is done with cumsum(c(1, diff(colA) < -1) in the example below. df1 = read.table(text='colA colB colC 10 11 7 10 34 7 10 89 7 10 21 7 2 23 5 2 21 5 2 56 5 22 14 3 22 19 3 22 90 3 1 45 0 1 23 0 9 8 0 9 11 0 9 21 0', header = TRUE,stringsAsFactors=FALSE) library(dplyr) df1 %>% arrange(desc(colA)) %>% group_by(group_sequential = cumsum(c(1, diff(colA) < -1))) %>% mutate(newcolD=n_distinct(colB)) colA colB colC group_sequential newcolD <int> <int> <int> <dbl> <int> 1 22 14 3 1 3 2 22 19 3 1 3 3 22 90 3 1 3 4 10 11 7 2 5 5 10 34 7 2 5 6 10 89 7 2 5 7 10 21 7 2 5 8 9 8 0 2 5 9 9 11 0 2 5 10 9 21 0 2 5 11 2 23 5 3 4 12 2 21 5 3 4 13 2 56 5 3 4 14 1 45 0 3 4 15 1 23 0 3 4 EDIT FOR NEW DATA With the data you added, we need to create a custom grouping. I use case_when in the example below. This matches the order you show in the desired output column. In the text, you wrote that you wanted the table to be sorted by colC. To do so, change the last line to arrange(desc(colC)) df1 = read.table(text='colA colB colC 10 11 7 10 34 7 10 89 7 10 21 7 2 23 5 2 21 5 2 56 5 22 14 3 22 19 3 22 90 3 11 19 2 11 45 2 1 45 0 1 23 0 9 8 0 9 11 0 9 21 0', header = TRUE,stringsAsFactors=FALSE) library(dplyr) df1 %>% group_by(group_sequential = case_when(.$colA==10|.$colA==9~1, .$colA==2|.$colA==1~2, .$colA==22~3, .$colA==11~4)) %>% mutate(newcolD=n_distinct(colB)) %>% arrange(desc(newcolD)) colA colB colC group_sequential newcolD <int> <int> <int> <dbl> <int> 1 10 11 7 1 5 2 10 34 7 1 5 3 10 89 7 1 5 4 10 21 7 1 5 5 9 8 0 1 5 6 9 11 0 1 5 7 9 21 0 1 5 8 2 23 5 2 4 9 2 21 5 2 4 10 2 56 5 2 4 11 1 45 0 2 4 12 1 23 0 2 4 13 22 14 3 3 3 14 22 19 3 3 3 15 22 90 3 3 3 16 11 19 2 4 2 17 11 45 2 4 2
You're really not making it easy for us, reposting slight variations of the same question instead of updating the old one and presenting conditions that are vague and inconsistent with what the desired output implies. Anyhow, here is my attempt. This is more an answer to the second question you posted, as that was a bit more general in form.
It's a bit messy, it's pretty much a direct translation of your conditions into a for loop with some if statements. I chose to focus on your written conditions rather than the expected output as that was the easier one to understand. If you want a better answer, please consider cleaning up you question(s) considerably.
df1 <- read.table(text="
colA colB colC
10 11 7
10 34 7
10 89 7
10 21 7
2 23 5
2 21 5
2 56 5
22 14 3
22 19 3
22 90 3
11 19 2
11 45 2
1 45 0
1 23 0
9 8 0
9 11 0
9 21 0", header=TRUE)
df2 <- read.table(text="
colA colB colC
10 11 7
10 34 7
10 89 7
10 21 7
2 23 5
2 21 5
2 56 5
33 24 3
33 78 3
22 14 3
22 19 3
22 90 3
11 19 2
11 45 2
1 45 0
1 23 0
9 8 0
9 11 0
9 21 0
32 11 0", header=TRUE)
df <- df1
for (i in 1:nrow(df)) {
df$colD[i] <- ifelse(df$colC[i] == 0,
0,
length(unique(df$colA[1:i])))
if (any(df$colA[i]-1 == df$colA[1:i]) & df$colC[i] != 0) {
df$colD[i] <- df$colD[which(df$colA[i]-1 == df$colA[1:i])][1]
}
}
# colA colB colC colD
# 10 11 7 1
# 10 34 7 1
# 10 89 7 1
# 10 21 7 1
# 2 23 5 2
# 2 21 5 2
# 2 56 5 2
# 22 14 3 3
# 22 19 3 3
# 22 90 3 3
# 11 19 2 1
# 11 45 2 1
# 1 45 0 0
# 1 23 0 0
# 9 8 0 0
# 9 11 0 0
# 9 21 0 0
df <- df2
for (i in 1:nrow(df)) {
df$colD[i] <- ifelse(df$colC[i] == 0,
0,
length(unique(df$colA[1:i])))
if (any(df$colA[i]-1 == df$colA[1:i]) & df$colC[i] != 0) {
df$colD[i] <- df$colD[which(df$colA[i]-1 == df$colA[1:i])][1]
}
}
df
# colA colB colC colD
# 10 11 7 1
# 10 34 7 1
# 10 89 7 1
# 10 21 7 1
# 2 23 5 2
# 2 21 5 2
# 2 56 5 2
# 33 24 3 3
# 33 78 3 3
# 22 14 3 4
# 22 19 3 4
# 22 90 3 4
# 11 19 2 1
# 11 45 2 1
# 1 45 0 0
# 1 23 0 0
# 9 8 0 0
# 9 11 0 0
# 9 21 0 0
# 32 11 0 0
To also group the rows where colC is zero, it's sufficient to adjust the conditionals like this:
for (i in 1:nrow(df)) {
df$colD[i] <- length(unique(df$colA[1:i]))
if (any(df$colA[i]-1 == df$colA[1:i])) {
df$colD[i] <- df$colD[which(df$colA[i]-1 == df$colA[1:i])][1]
}
}
How do I use plyr to number rows?
Basically I want an autoincremented id column based on my cohorts - in this case .(kmer, cvCut) > myDataFrame size kmer cvCut cumsum 1 8132 23 10 8132 10000 778 23 10 13789274 30000 324 23 10 23658740 50000 182 23 10 28534840 100000 65 23 10 33943283 200000 25 23 10 37954383 250000 584 23 12 16546507 300000 110 23 12 29435303 400000 28 23 12 34697860 600000 127 23 2 47124443 600001 127 23 2 47124570 I want a column added that has new row names based on the kmer/cvCut group > myDataFrame size kmer cvCut cumsum newID 1 8132 23 10 8132 1 10000 778 23 10 13789274 2 30000 324 23 10 23658740 3 50000 182 23 10 28534840 4 100000 65 23 10 33943283 5 200000 25 23 10 37954383 6 250000 584 23 12 16546507 1 300000 110 23 12 29435303 2 400000 28 23 12 34697860 3 600000 127 23 2 47124443 1 600001 127 23 2 47124570 2
I'd do it like this:
library(plyr)
ddply(df, c("kmer", "cvCut"), transform, newID = seq_along(kmer))
Just add a new column each time plyr calls you: R> DF <- data.frame(kmer=sample(1:3, 50, replace=TRUE), \ cvCut=sample(LETTERS[1:3], 50, replace=TRUE)) R> library(plyr) R> ddply(DF, .(kmer, cvCut), function(X) data.frame(X, newId=1:nrow(X))) kmer cvCut newId 1 1 A 1 2 1 A 2 3 1 A 3 4 1 A 4 5 1 A 5 6 1 A 6 7 1 A 7 8 1 A 8 9 1 A 9 10 1 A 10 11 1 A 11 12 1 B 1 13 1 B 2 14 1 B 3 15 1 B 4 16 1 B 5 17 1 B 6 18 1 C 1 19 1 C 2 20 1 C 3 21 2 A 1 22 2 A 2 23 2 A 3 24 2 A 4 25 2 A 5 26 2 B 1 27 2 B 2 28 2 B 3 29 2 B 4 30 2 B 5 31 2 B 6 32 2 B 7 33 2 C 1 34 2 C 2 35 2 C 3 36 2 C 4 37 3 A 1 38 3 A 2 39 3 A 3 40 3 A 4 41 3 B 1 42 3 B 2 43 3 B 3 44 3 B 4 45 3 C 1 46 3 C 2 47 3 C 3 48 3 C 4 49 3 C 5 50 3 C 6 R>
I think that this is what you want: Load the data: x <- read.table(textConnection( "id size kmer cvCut cumsum 1 8132 23 10 8132 10000 778 23 10 13789274 30000 324 23 10 23658740 50000 182 23 10 28534840 100000 65 23 10 33943283 200000 25 23 10 37954383 250000 584 23 12 16546507 300000 110 23 12 29435303 400000 28 23 12 34697860 600000 127 23 2 47124443 600001 127 23 2 47124570"), header=TRUE) Use ddply: library(plyr) ddply(x, .(kmer, cvCut), function(x) cbind(x, 1:nrow(x)))