Extract every 11 rows from data frame

Question

So I have a data frame and I want to get every 11 rows. Not just the every 11th rows but a chunk of 11 rows every time for eg:

     Subject Wt Dose  Time  conc
1        1 79.6 4.02  0.00  0.74
2        1 79.6 4.02  0.25  2.84
3        1 79.6 4.02  0.57  6.57
4        1 79.6 4.02  1.12 10.50
5        1 79.6 4.02  2.02  9.66
6        1 79.6 4.02  3.82  8.58
7        1 79.6 4.02  5.10  8.36
8        1 79.6 4.02  7.03  7.47
9        1 79.6 4.02  9.05  6.89
10       1 79.6 4.02 12.12  5.94
11       1 79.6 4.02 24.37  3.28

and then later 11 and then again the other following 11 rows.

I tried this

for (i in 1:nrow(Theoph)) {
  everyEleven = Theoph[11,i]
  everyEl
}

But it just gives me the first 11 rows and not the second chunk of 11 rows and so on

Yea I want every 11 rows, because different subjects are in every 11 rows.For eg.: subject 1 is in the first 11 and then subject 2 is in the next 11 and so on. Its because I want to make a plot that is time against concentration for every subject. — Liviosah, Apr 23 '20 at 12:42
Maybe better to assign a subject ID. `Theoph$ID <- rep (1:m, each=11)` where `m` is the number of subjects. Is it 12 by any chance? — Edward, Apr 23 '20 at 12:59
I am not sure if I understand the question completely, but I interpret you want 11 rows per group according to a grouping column. Unclear is: Which 11 rows (the first last, random...)? My first answer would use a `data.table`: `library(data.table); x <- data.table(mtcars); x[, head(.SD, 2), by = gear]` for the first 2 rows per group (column "gear") — R Yoda, Apr 23 '20 at 13:09

score 2 · Answer 1 · answered Apr 23 '20 at 12:54

2

Maybe you can try split like below

everyEleven <- split(Theoph,ceiling(seq(nrow(Theoph))/11))

answered Apr 23 '20 at 12:54

ThomasIsCoding

96,636
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Peter · Accepted Answer · 2020-04-23T13:39:52.523

Try this as adapted from [split into multiple subset of dataframes with dplyr:group_by?

library(tibble)
library(dplyr)  
library(tidyr)

Make an indicative dataframe as your data in the question is only 11 rows.

tib <- tibble(sub = rep(1:33),
              var = runif(33))

tib1 <- 
  tib %>% 
  # create a grouping variable every 11 rows , unless there is a variable in your data which does the same. 
  mutate(grp = rep(1:3, each = 11)) %>% 
  group_by(grp) %>%
  nest()%>%
  select(data) %>%
  unlist(recursive = FALSE)

Gives you:

$data1
# A tibble: 11 x 2
     sub   var
   <int> <dbl>
 1     1 0.258
 2     1 0.337
 3     1 0.463
 4     1 0.856
 5     1 0.466
 6     1 0.701
 7     1 0.548
 8     1 0.999
 9     1 0.454
10     1 0.292
11     1 0.173

$data2
# A tibble: 11 x 2
     sub   var
   <int> <dbl>
 1     2 0.148
 2     2 0.487
 3     2 0.246
 4     2 0.279
 5     2 0.130
 6     2 0.730
 7     2 0.312
 8     2 0.935
 9     2 0.968
10     2 0.745
11     2 0.485

$data3
# A tibble: 11 x 2
     sub        var
   <int>      <dbl>
 1     3 0.141     
 2     3 0.200     
 3     3 0.00000392
 4     3 0.993     
 5     3 0.644     
 6     3 0.334     
 7     3 0.567     
 8     3 0.817     
 9     3 0.0342    
10     3 0.718     
11     3 0.527

Ric S · Answer 3 · 2020-04-23T13:17:08.917

1

Since in the sample data you provided there is a column Subject, which I assume represents the subject IDs and there are only 11 rows with the same value for Subject, you can use

split(Theoph, Theoph$Subject)

edited Apr 23 '20 at 13:17

answered Apr 23 '20 at 13:09

Ric S

9,073
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score 0 · Answer 4 · answered Apr 23 '20 at 12:50

0

I will assume your data frame is 11*N rows long then

everyEleven = vector(mode = "list", length = N)

for(i in 1:N){
  start = (i - 1) * 11 + 1
  end = i * 11
  everyEleven[[i]] = Theoph[start:end, ]
}

answered Apr 23 '20 at 12:50

James Curran

1,274
7
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score 0 · Answer 5 · answered Apr 23 '20 at 19:48

0

We can use gl to create the grouping index

split(Theoph, as.integer(gl(nrow(Theoph), 11, nrow(Theoph))))

answered Apr 23 '20 at 19:48

akrun

874,273
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Extract every 11 rows from data frame

5 Answers5