# load the necessary libraries
library(tidyverse)
library(lubridate)Class Activity 8
Your turn 1
In the provided R code, we start with two datasets, DBP_wide and BP_wide, representing blood pressure measurements in a wide format. We then demonstrate how to transform BP_wide into a long format using pivot_longer().
DBP_wide <- tibble(id = letters[1:4],
sex = c("F", "M", "M", "F"),
v1.DBP = c(88, 84, 102, 70),
v2.DBP = c(78, 78, 96, 76),
v3.DBP = c(94, 82, 94, 74),
age=c(23, 56, 41, 38)
)
DBP_wide# A tibble: 4 × 6
id sex v1.DBP v2.DBP v3.DBP age
<chr> <chr> <dbl> <dbl> <dbl> <dbl>
1 a F 88 78 94 23
2 b M 84 78 82 56
3 c M 102 96 94 41
4 d F 70 76 74 38BP_wide <- tibble(id = letters[1:4],
sex = c("F", "M", "M", "F"),
SBP_v1 = c(130, 120, 130, 119),
SBP_v2 = c(110, 116, 136, 106),
SBP_v3 = c(112, 122, 138, 118))
BP_wide# A tibble: 4 × 5
id sex SBP_v1 SBP_v2 SBP_v3
<chr> <chr> <dbl> <dbl> <dbl>
1 a F 130 110 112
2 b M 120 116 122
3 c M 130 136 138
4 d F 119 106 118BP_long <- BP_wide %>%
pivot_longer(names_to = "visit", values_to = "SBP", SBP_v1:SBP_v3) %>%
mutate(visit = parse_number(visit))
BP_long# A tibble: 12 × 4
id sex visit SBP
<chr> <chr> <dbl> <dbl>
1 a F 1 130
2 a F 2 110
3 a F 3 112
4 b M 1 120
5 b M 2 116
6 b M 3 122
7 c M 1 130
8 c M 2 136
9 c M 3 138
10 d F 1 119
11 d F 2 106
12 d F 3 118a. Create a long dataframe from DBP_wide based on the repeated DBP columns and save it as DBP_long.
Click for answer
Answer:
DBP_long <- DBP_wide %>%
pivot_longer(names_to = "visit",
values_to = "DBP",
cols = v1.DBP:v3.DBP)
DBP_long# A tibble: 12 × 5
id sex age visit DBP
<chr> <chr> <dbl> <chr> <dbl>
1 a F 23 v1.DBP 88
2 a F 23 v2.DBP 78
3 a F 23 v3.DBP 94
4 b M 56 v1.DBP 84
5 b M 56 v2.DBP 78
6 b M 56 v3.DBP 82
7 c M 41 v1.DBP 102
8 c M 41 v2.DBP 96
9 c M 41 v3.DBP 94
10 d F 38 v1.DBP 70
11 d F 38 v2.DBP 76
12 d F 38 v3.DBP 74b. Clean up the visit column of DBP_long so that the values are 1, 2, 3, and save it as DBP_long.
Click for answer
Answer:
DBP_long <- DBP_long %>%
mutate(visit = parse_number(visit))
DBP_long# A tibble: 12 × 5
id sex age visit DBP
<chr> <chr> <dbl> <dbl> <dbl>
1 a F 23 1 88
2 a F 23 2 78
3 a F 23 3 94
4 b M 56 1 84
5 b M 56 2 78
6 b M 56 3 82
7 c M 41 1 102
8 c M 41 2 96
9 c M 41 3 94
10 d F 38 1 70
11 d F 38 2 76
12 d F 38 3 74c. Make DBP_long wide with column names visit.1, visit.2, visit.3 for the DBP values, and save it as DBP_wide2
Click for answer
Answer:
DBP_wide2 <- DBP_long %>%
pivot_wider(names_from = "visit",
values_from = "DBP",
names_prefix = "visit.")
DBP_wide2# A tibble: 4 × 6
id sex age visit.1 visit.2 visit.3
<chr> <chr> <dbl> <dbl> <dbl> <dbl>
1 a F 23 88 78 94
2 b M 56 84 78 82
3 c M 41 102 96 94
4 d F 38 70 76 74d. Join DBP_long with BP_long2 to create a single data frame with columns id, sex, visit, SBP, DBP, and age. Save this as BP_both_long.
Click for answer
Answer:
BP_both_long <- left_join(BP_long, DBP_long, by = c("id", "sex", "visit"))
BP_both_long# A tibble: 12 × 6
id sex visit SBP age DBP
<chr> <chr> <dbl> <dbl> <dbl> <dbl>
1 a F 1 130 23 88
2 a F 2 110 23 78
3 a F 3 112 23 94
4 b M 1 120 56 84
5 b M 2 116 56 78
6 b M 3 122 56 82
7 c M 1 130 41 102
8 c M 2 136 41 96
9 c M 3 138 41 94
10 d F 1 119 38 70
11 d F 2 106 38 76
12 d F 3 118 38 74e. Calculate the mean SBP and DBP for each visit and save the result as mean_BP_by_visit.
Click for answer
Answer:
mean_BP_by_visit <- BP_both_long %>%
group_by(visit) %>%
summarize(mean_SBP = mean(SBP),
mean_DBP = mean(DBP))
mean_BP_by_visit# A tibble: 3 × 3
visit mean_SBP mean_DBP
<dbl> <dbl> <dbl>
1 1 125. 86
2 2 117 82
3 3 122. 86Your turn 2
a. Parsing Complex Dates: Use dmy_hms() to parse the following date-time string: “25-Dec-2020 17:30:00”
Click for answer
Answer:
parsed_date <- dmy_hms("25-Dec-2020 17:30:00")
parsed_date[1] "2020-12-25 17:30:00 UTC"b. Advanced Date Arithmetic: Calculate the exact age in years for someone born on “1995-05-15 09:30:00”.
Click for answer
Answer:
dob <- ymd_hms("1995-05-15 09:30:00")
exact_age <- as.duration(interval(dob, now())) / dyears(1)
exact_age[1] 29.00139c. Creating Date-Time Objects: Create a date-time object for March 15, 2020, 13:30:00 using make_datetime().
Click for answer
Answer:
new_date_time <- make_datetime(2020, 3, 15, 13, 30, 0)
new_date_time[1] "2020-03-15 13:30:00 UTC"d. Extracting Components from Date-Time Objects: Extract the year, month (as a number), day, hour, and minute from “2022-07-01 14:45:00”.
Click for answer
Answer:
example_date_time <- ymd_hms("2022-07-01 14:45:00")
extracted_components <- tibble(
year = year(example_date_time),
month = month(example_date_time),
day = day(example_date_time),
hour = hour(example_date_time),
minute = minute(example_date_time)
)
extracted_components# A tibble: 1 × 5
year month day hour minute
<dbl> <dbl> <int> <int> <int>
1 2022 7 1 14 45e. Advanced Date-Time Arithmetic with Periods: Add 2 months and 15 days to “2021-08-01”.
Click for answer
Answer:
initial_date <- ymd("2021-08-01")
new_date <- initial_date + months(2) + days(15)
new_date[1] "2021-10-16"f. Duration and Time Differences: Calculate the duration in days, weeks, months, and years between “2019-04-01” and “2022-04-01”.
Click for answer
Answer:
start_date <- ymd("2019-04-01")
end_date <- ymd("2022-04-01")
time_diff <- end_date - start_date
duration_days <- as.duration(time_diff)
duration_weeks <- duration_days / dweeks(1)
duration_months <- duration_days / dmonths(1)
duration_years <- duration_days / dyears(1)
duration_results <- tibble(
days = duration_days,
weeks = duration_weeks,
months = duration_months,
years = duration_years
)
duration_results# A tibble: 1 × 4
days weeks months years
<Duration> <dbl> <dbl> <dbl>
1 94694400s (~3 years) 157. 36.0 3.00