Using dplyr: Within groups, select the first value meeting a condition - r

Question

I need assistance obtaining a solution that will scan backwards in time and obtain the first value meeting a condition. I have data similar to:
set.seed(42)
df <- data.frame(
id = sample(LETTERS[1:3], 20, replace = TRUE),
time.var = sample(1:20, 20, replace = TRUE),
x = sample(c(1:10), 20, replace = TRUE)
)
df <- df[order(df$id, df$time.var),]
id time.var x
A 5 2
A 14 8
A 19 7
A 20 1
B 1 1
B 2 5
B 9 10
B 11 10
B 13 6
B 15 4
B 19 3
C 1 7
C 3 5
C 8 9
C 8 4
C 17 7
C 17 4
C 17 8
C 19 4
C 19 10
For the last member of each group defined in time order by time.var, I'd like to obtain the first value from x less than 5 by scanning in descending time order.
I have tried:
test <- df %>%
group_by(id) %>%
arrange(id, time.var) %>%
mutate(less.5 = which.max(x[x < 5]) )
What strategy can I use to obtain this type of output:
id time.var x previous.less.5
A 5 2
A 14 8
A 19 7
A 20 1 2
B 1 1
B 2 5
B 9 10
B 11 10
B 13 6
B 15 4
B 19 3 4
C 1 7
C 3 5
C 8 9
C 8 4
C 17 7
C 17 4
C 17 8
C 19 4
C 19 10 4

Answer 1

Using library(dplyr):
df %>%
arrange(id, time.var) %>%
group_by(id) %>%
mutate(previous.less.5 = tail(c(x[c((x[-n()] < 5), FALSE)]),1)) %>%
group_by(id) %>%
mutate(previous.less.5 = if_else(row_number() == n(), previous.less.5, NULL))
or
df %>%
arrange(id, time.var) %>%
group_by(id) %>%
slice(1:(n()-1)) %>%
filter(x < 5) %>%
slice(n()) %>%
select(-time.var) %>%
right_join(df, ., by="id", suffix =c("",".y")) %>%
group_by(id) %>%
mutate(previous.less.5 = if_else(row_number() == n(), x.y, NULL)) %>%
select(-x.y)
giving:
#> # A tibble: 20 x 4
#> # Groups: id [3]
#> id time.var x previous.less.5
#> <fct> <int> <int> <int>
#> 1 A 3 10 NA
#> 2 A 4 8 NA
#> 3 A 4 6 NA
#> 4 A 5 2 NA
#> 5 A 5 8 NA
#> 6 A 5 7 NA
#> 7 A 11 6 NA
#> 8 A 13 3 NA
#> 9 A 15 2 3
#> 10 B 2 1 NA
#> 11 B 4 3 NA
#> 12 B 4 6 NA
#> 13 B 8 5 NA
#> 14 B 8 4 NA
#> 15 B 20 7 4
#> 16 C 1 2 NA
#> 17 C 2 10 NA
#> 18 C 10 6 NA
#> 19 C 13 2 NA
#> 20 C 18 5 2
Update:
If there's a group with no record less than 5 (or only last record less than 5) then following works:
df %>%
arrange(id, time.var) %>%
group_by(id) %>%
mutate(previous.less.5 = if_else(row_number() == n(),
max(tail(c( x[ c( x[-n()] < 5, FALSE) ] ), 1)),
NULL)) %>%
mutate(previous.less.5 = replace(previous.less.5, is.infinite(previous.less.5), NA))
Data:
set.seed(42) # I am getting different data than what you've shown with this seed
df <- data.frame(
id = sample(LETTERS[1:3], 20, replace = TRUE),
time.var = sample(1:20, 20, replace = TRUE),
x = sample(c(1:10), 20, replace = TRUE)
)
df <- df[order(df$id, df$time.var),]

Answer 2

We can reverse value of x by id get the first number which is less than 5 using which. The last replace is to assign NA to all the values in previous.less.5 except the last one.
library(dplyr)
df %>%
#Data is already sorted by `id` and `time.var` but if your still need use
#arrange(id, time.var) %>%
group_by(id) %>%
mutate(rev_x = c(NA, rev(x)[-1]), previous.less.5 = rev_x[which(rev_x < 5)[1]],
previous.less.5 = replace(previous.less.5, row_number() != n(), NA)) %>%
select(-rev_x)
# id time.var x previous.less.5
# <fct> <int> <int> <int>
# 1 A 5 2 NA
# 2 A 14 8 NA
# 3 A 19 7 NA
# 4 A 20 1 2
# 5 B 1 1 NA
# 6 B 2 5 NA
# 7 B 9 10 NA
# 8 B 11 10 NA
# 9 B 13 6 NA
#10 B 15 4 NA
#11 B 19 3 4
#12 C 1 7 NA
#13 C 3 5 NA
#14 C 8 9 NA
#15 C 8 4 NA
#16 C 17 7 NA
#17 C 17 4 NA
#18 C 17 8 NA
#19 C 19 4 NA
#20 C 19 10 4
This should also handle the case and return NA's if there is no value less than 5 in an id.