Reputation: 355
I have a data.frame
like this:
dat <- data.frame("ID"=c(rep(1,13),rep(2,5)), "time"=c(seq(1,13),c(seq(1,5))), "value"=c(rep(0,5), rep(1,3), 2, 0, 1, 5, 20, rep(0,2), seq(1:3)))
ID time value
1 1 1 0
2 1 2 0
3 1 3 0
4 1 4 0
5 1 5 0
6 1 6 1
7 1 7 1
8 1 8 1
9 1 9 2
10 1 10 0
11 1 11 1
12 1 12 5
13 1 13 20
14 2 1 0
15 2 2 0
16 2 3 1
17 2 4 2
18 2 5 3
My goal is to set all values to 0
, if among the remaining values there is any other 0
(for each unique ID
and sorted by time
). That means in the example data, I would like to have 0
in the rows 6:9.
I tried dat %>% group_by(ID) %>% mutate(value2 = ifelse(lead(value, order_by=time)==0, 0, value))
but I would have to run this several times, since it only changes one row at a time (i.e. row 9 first, then row 8, etc.).
dplyr
solution would be prefered but I'd take everything that works :)
Short explanation: value is the size of a tumor. If the tumor does not grow large, but actually vanishes completely at a later time, it was most likely an irrelevant encapsulation, hence should be coded as "zero tumor".
Upvotes: 3
Views: 717
Reputation: 35624
You can use accumulate(..., .dir = "backward")
in purrr
library(dplyr)
library(purrr)
dat %>%
group_by(ID) %>%
arrange(time, .by_group = T) %>%
mutate(value2 = accumulate(value, ~ if(.y == 0) 0 else .x, .dir = "backward")) %>%
ungroup()
# A tibble: 18 x 4
ID time value value2
<dbl> <int> <dbl> <dbl>
1 1 1 0 0
2 1 2 0 0
3 1 3 0 0
4 1 4 0 0
5 1 5 0 0
6 1 6 1 0
7 1 7 1 0
8 1 8 1 0
9 1 9 2 0
10 1 10 0 0
11 1 11 1 1
12 1 12 5 5
13 1 13 20 20
14 2 1 0 0
15 2 2 0 0
16 2 3 1 1
17 2 4 2 2
18 2 5 3 3
Upvotes: 0
Reputation: 28705
With data.table you can caluculate fields with the data in a certain order, without actually reordering the data frame. Useful here
library(data.table)
setDT(dat)
dat[order(-time), value := fifelse(cumsum(value == 0) > 0, 0, value), ID]
dat
# ID time value
# 1: 1 1 0
# 2: 1 2 0
# 3: 1 3 0
# 4: 1 4 0
# 5: 1 5 0
# 6: 1 6 0
# 7: 1 7 0
# 8: 1 8 0
# 9: 1 9 0
# 10: 1 10 0
# 11: 1 11 1
# 12: 1 12 5
# 13: 1 13 20
# 14: 2 1 0
# 15: 2 2 0
# 16: 2 3 1
# 17: 2 4 2
# 18: 2 5 3
Upvotes: 1
Reputation: 11981
I am not sure wether this is your desired output, but maybe it can be usefull to you
dat %>%
group_by(ID) %>%
arrange(-time) %>%
mutate(value = if_else(cumsum(value == 0) > 0, 0, value)) %>%
arrange(ID, time)
ID time value
<dbl> <int> <dbl>
1 1 1 0
2 1 2 0
3 1 3 0
4 1 4 0
5 1 5 0
6 1 6 0
7 1 7 0
8 1 8 0
9 1 9 0
10 1 10 0
11 1 11 1
12 1 12 5
13 1 13 20
14 2 1 0
15 2 2 0
16 2 3 1
17 2 4 2
18 2 5 3
Basicalyl, I first put the observations in descending order. Then I check whether there has been a zero in value (cumsum(value == 0) > 0)
). If yes, I set all remaining values to zero.
Finally, I put the observations in correct order again.
If you do not want to order and reorder the data you can use the following code, which relies on the same logic but is a bit more difficult to read:
dat %>%
group_by(ID) %>%
arrange(ID, time) %>%
mutate(value = if_else(cumsum(value == 0) < sum(value == 0), 0, value))
Or a bit more efficient without if_else
:
dat %>%
group_by(ID) %>%
arrange(ID, time) %>%
mutate(value = value * (cumsum(value == 0) >= sum(value == 0)))
Upvotes: 5
Reputation: 34751
One way could be to find the indices of the first and last occurrences of 0
and replace everything in between.
library(dplyr)
dat %>%
group_by(ID) %>%
mutate(value = replace(value, between(row_number(), which.max(value == 0), tail(which(value == 0), 1)), 0))
# A tibble: 18 x 3
# Groups: ID [2]
ID time value
<dbl> <int> <dbl>
1 1 1 0
2 1 2 0
3 1 3 0
4 1 4 0
5 1 5 0
6 1 6 0
7 1 7 0
8 1 8 0
9 1 9 0
10 1 10 0
11 1 11 1
12 1 12 5
13 1 13 20
14 2 1 0
15 2 2 0
16 2 3 1
17 2 4 2
18 2 5 3
Upvotes: 2