I have a data.frame
which looks like the following:
id <- c("a","a","a","a","b","b","b","b")
age_from <- c(0,2,3,7,0,1,2,6)
age_to <- c(2,3,7,10,1,2,6,10)
y <- c(100,150,100,250,300,200,100,150)
df <- data.frame(id,age_from,age_to,y)
df$years <- df$age_to - df$age_from
Which gives a df
that looks like:
id age_from age_to y years
1 a 0 2 100 2
2 a 2 3 150 1
3 a 3 7 100 4
4 a 7 10 250 3
5 b 0 1 300 1
6 b 1 2 200 1
7 b 2 6 100 4
8 b 6 10 150 4
Instead of having an unequal number of years per row, I would like to have 20 rows, 10 for each id
, with each row accounting for one year. This would also involve averaging the y
column across the number of years listed in the years
column.
I believe this may have to be done using a loop 1:n
with the n
equaling a value in the years
column. Although I am not sure how to start with this.
You can use rep
to repeat the rows by the number of given years .
x <- df[rep(seq_len(nrow(df)), df$years),]
x
# id age_from age_to y years
#1 a 0 2 50.00000 2
#1.1 a 0 2 50.00000 2
#2 a 2 3 150.00000 1
#3 a 3 7 25.00000 4
#3.1 a 3 7 25.00000 4
#3.2 a 3 7 25.00000 4
#3.3 a 3 7 25.00000 4
#4 a 7 10 83.33333 3
#4.1 a 7 10 83.33333 3
#4.2 a 7 10 83.33333 3
#5 b 0 1 300.00000 1
#6 b 1 2 200.00000 1
#7 b 2 6 25.00000 4
#7.1 b 2 6 25.00000 4
#7.2 b 2 6 25.00000 4
#7.3 b 2 6 25.00000 4
#8 b 6 10 37.50000 4
#8.1 b 6 10 37.50000 4
#8.2 b 6 10 37.50000 4
#8.3 b 6 10 37.50000 4
When you mean with averaging the y column across the number of years to divide by the number of years:
x$y <- x$y / x$years
In case age_from
should go from 0
to 9
and age_to
from 1
to 10
for each id:
x$age_from <- x$age_from + ave(x$age_from, x$id, x$age_from, FUN=seq_along) - 1
#x$age_from <- ave(x$age_from, x$id, FUN=seq_along) - 1 #Alternative
x$age_to <- x$age_from + 1
Here is a solution with tidyr
and dplyr
.
First of all we complete
age_from
from 0 to 9 as you wanted, by keeping only the existing id
s.
You will have several NA
s on age_to
, y
and years
. So, we fill them by dragging down each value in order to complete the immediately following values that are NA
.
Now you can divide y
by years
(I assumed you meant this by setting the average value so to leave the sum consistent).
At that point, you only need to recalculate age_to
accordingly.
Remember to ungroup
at the end!
library(tidyr)
library(dplyr)
df %>%
complete(id, age_from = 0:9) %>%
group_by(id) %>%
fill(y, years, age_to) %>%
mutate(y = y/years) %>%
mutate(age_to = age_from + 1) %>%
ungroup()
# A tibble: 20 x 5
id age_from age_to y years
<chr> <dbl> <dbl> <dbl> <dbl>
1 a 0 1 50 2
2 a 1 2 50 2
3 a 2 3 150 1
4 a 3 4 25 4
5 a 4 5 25 4
6 a 5 6 25 4
7 a 6 7 25 4
8 a 7 8 83.3 3
9 a 8 9 83.3 3
10 a 9 10 83.3 3
11 b 0 1 300 1
12 b 1 2 200 1
13 b 2 3 25 4
14 b 3 4 25 4
15 b 4 5 25 4
16 b 5 6 25 4
17 b 6 7 37.5 4
18 b 7 8 37.5 4
19 b 8 9 37.5 4
20 b 9 10 37.5 4
A tidyverse
solution.
library(tidyverse)
df %>%
mutate(age_to = age_from + 1) %>%
group_by(id) %>%
complete(nesting(age_from = 0:9, age_to = 1:10)) %>%
fill(y, years) %>%
mutate(y = y / years)
# A tibble: 20 x 5
# Groups: id [2]
id age_from age_to y years
<chr> <dbl> <dbl> <dbl> <dbl>
1 a 0 1 50 2
2 a 1 2 50 2
3 a 2 3 150 1
4 a 3 4 25 4
5 a 4 5 25 4
6 a 5 6 25 4
7 a 6 7 25 4
8 a 7 8 83.3 3
9 a 8 9 83.3 3
10 a 9 10 83.3 3
11 b 0 1 300 1
12 b 1 2 200 1
13 b 2 3 25 4
14 b 3 4 25 4
15 b 4 5 25 4
16 b 5 6 25 4
17 b 6 7 37.5 4
18 b 7 8 37.5 4
19 b 8 9 37.5 4
20 b 9 10 37.5 4
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