Calculated Column Based on Rows with Date Range - r

Question

I have a dataframe as follows:
ID
Col1
RespID
Col3
Col4
Year
Month
Day
1
blue
729Ad
3.2
A
2021
April
2
2
orange
295gS
6.5
A
2021
April
1
3
red
729Ad
8.4
B
2021
April
20
4
yellow
592Jd
2.9
A
2021
March
12
5
green
937sa
3.5
B
2021
May
13
I would like to calculate a new column, Col5, such that its value is 1 if the row has Col4 value of A and there exists another column somewhere in the dataset a row with the same RespId but a Col4 value of B. Otherwise it’s value is 0. Then I will drop all rows with Col4 value of B, to keep just those with A. I'd also like to account for the date fields (year, month, date) so that this is done in groups based on say a 30 day timeframe. So if 'B' appears within 30 days of when 'A' appears in the dataset, only then is there a 1 present (if 'B' appears within 60 days, then there is no 1. Additionally, I'd like to keep everything as data.frames.
Here is what the desired output table would look like prior to dropping rows with Col4 value of B:
ID
Col1
RespID
Col3
Col4
Col5
1
blue
729Ad
3.2
A
1
2
orange
295gS
6.5
A
0
3
red
729Ad
8.4
B
0
4
yellow
592Jd
2.9
A
0
5
green
937sa
3.5
B
0
I have found Ronak's solution in this thread (Calculated Column Based on Rows in Tidymodels Recipe) to be useful, however, would like to modify for the date range.

Answer 1

A lot of things to unpack here.
I think you're tripping up over your own feet by trying to do too many things at once. I've broken down the code into four distinct steps to make the thought process easy to follow. Obviously, for use in a production environment it should be rewritten more efficiently.
1. Generate some data
library(tidyverse)
set.seed(42)
df <- tibble(
id = c(1:10),
resp_id = c(1701, seq(2286, 2289), 1701, seq(2290, 2293)),
grouping = sample(c("A", "B"), size = 10, replace = TRUE),
date = seq.Date(as.Date("2363-10-04"), as.Date("2363-11-17"), length.out = 10)
)
Resulting data:
# A tibble: 10 × 4
id resp_id grouping date
<int> <dbl> <chr> <date>
1 1 1701 A 2363-10-04
2 2 2286 A 2363-10-08
3 3 2287 A 2363-10-13
4 4 2288 A 2363-10-18
5 5 2289 B 2363-10-23
6 6 1701 B 2363-10-28
7 7 2290 B 2363-11-02
8 8 2291 B 2363-11-07
9 9 2292 A 2363-11-12
10 10 2293 B 2363-11-17
2. Check grouping
df <- df %>%
mutate(
is_a = ifelse(grouping == "A", 1, 0),
is_b = ifelse(grouping == "B", 1, 0)
)
We have the grouping now as easy-to-use dummy variables:
> df
# A tibble: 10 × 6
id resp_id grouping date is_a is_b
<int> <dbl> <chr> <date> <dbl> <dbl>
1 1 1701 A 2363-10-04 1 0
2 2 2286 A 2363-10-08 1 0
3 3 2287 A 2363-10-13 1 0
4 4 2288 A 2363-10-18 1 0
5 5 2289 B 2363-10-23 0 1
6 6 1701 B 2363-10-28 0 1
7 7 2290 B 2363-11-02 0 1
8 8 2291 B 2363-11-07 0 1
9 9 2292 A 2363-11-12 1 0
10 10 2293 B 2363-11-17 0 1
3. Check completeness
df <- df %>%
group_by(
resp_id
) %>%
mutate(
# Check if the grouping has both "A" and "B" values
is_complete = ifelse(
sum(is_a) > 0 & sum(is_b) > 0,
1,
0
)
) %>%
ungroup()
We see that there is only one resp_id value that is complete — 1701:
> df
# A tibble: 10 × 7
id resp_id grouping date is_a is_b is_complete
<int> <dbl> <chr> <date> <dbl> <dbl> <dbl>
1 1 1701 A 2363-10-04 1 0 1
2 2 2286 A 2363-10-08 1 0 0
3 3 2287 A 2363-10-13 1 0 0
4 4 2288 A 2363-10-18 1 0 0
5 5 2289 B 2363-10-23 0 1 0
6 6 1701 B 2363-10-28 0 1 1
7 7 2290 B 2363-11-02 0 1 0
8 8 2291 B 2363-11-07 0 1 0
9 9 2292 A 2363-11-12 1 0 0
10 10 2293 B 2363-11-17 0 1 0
4. Assign target value
df <- df %>%
group_by(
resp_id
) %>%
mutate(
# Check if the "A" part of a complete grouping has a another value within 30 days
is_within_timeframe = ifelse(
is_complete == 1 & is_a == 1 & max(date) - min(date) <= 30,
1,
0
)
) %>%
ungroup()
We see that our one complete set has in fact a B value that falls within 30 days of the A observation (Caveat: This only works if there are always exactly one or two observations per grouping!). Column is_within_timeframe corresponds to your Col4:
> df
# A tibble: 10 × 8
id resp_id grouping date is_a is_b is_complete is_within_timeframe
<int> <dbl> <chr> <date> <dbl> <dbl> <dbl> <dbl>
1 1 1701 A 2363-10-04 1 0 1 1
2 2 2286 A 2363-10-08 1 0 0 0
3 3 2287 A 2363-10-13 1 0 0 0
4 4 2288 A 2363-10-18 1 0 0 0
5 5 2289 B 2363-10-23 0 1 0 0
6 6 1701 B 2363-10-28 0 1 1 0
7 7 2290 B 2363-11-02 0 1 0 0
8 8 2291 B 2363-11-07 0 1 0 0
9 9 2292 A 2363-11-12 1 0 0 0
10 10 2293 B 2363-11-17 0 1 0 0