R：提取满足多个条件的 ID 数量

Question

I want to identify those IDs in a dataset, which newly developed a disease. The dataset is in form of a diary in which people daily answer a "yes/no" question on whether they have the disease.

ID <- c(1,1,1,1,1,2,2,2,2,2,2,2,3,3,3,3,3,3,3,3,3)
Date <- c("2020-03-10","2020-03-11","2020-03-12","2020-03-13","2020-03-14","2020-03-12","2020-03-13","2020-03-14","2020-03-15","2020-03-16","2020-03-17","2020-03-18", "2020-03-12","2020-03-13","2020-03-14","2020-03-15","2020-03-16","2020-03-17","2020-03-18","2020-03-19","2020-03-20")
Disease <- c("No","No","Yes","Yes","Yes","No","No","No", "Yes","Yes","Yes","No","Yes","Yes","No","No","No","Yes","Yes","Yes","Yes")

df <- data.frame(ID, Date, Disease)

df
ID   Date         Disease
1    2020-03-10   No
1    2020-03-11   No
1    2020-03-12   Yes
1    2020-03-13   Yes
1    2020-03-14   Yes
2    2020-03-12   No
2    2020-03-13   No
2    2020-03-14   No
2    2020-03-15   Yes
2    2020-03-16   Yes
2    2020-03-17   Yes
2    2020-03-18   No
3    2020-03-12   Yes
3    2020-03-13   Yes
3    2020-03-14   No
3    2020-03-15   No
3    2020-03-16   No
3    2020-03-17   Yes
3    2020-03-18   Yes
3    2020-03-19   Yes
3    2020-03-20   Yes

However, in order to be characterized as "newly developed the disease" the person has to meet the following conditions: 1. The person has to have "yes" for at least two days in a row 2. The person must have answered "no" for at least 3 days in a row before the first "yes".

As an Output, I would like to have the number of people fulfilling these conditions. So in the extraction of the dataset above, this would be two (IDs 2+3).

Does anybody know a way how to achieve this? Thanks in advance for your time!

Answer 1

A slightly messy way of doing this would be to use the dplyr::lag() function.

 library(tidyverse)
 library(lubridate)
 df %>% 
    mutate(Date = ymd(Date)) %>%
    group_by(ID) %>% 
    mutate(day_1 = lag(Disease, 1, order_by = Date), 
           day_2 = lag(Disease, 2, order_by = Date), 
           day_3 = lag(Disease, 3, order_by = Date), 
           day_4 = lag(Disease, 4, order_by = Date)) %>% 
    filter(day_1 == "No" & day_2 == "No" & day_3 == "No" & day_4 == "Yes" &        Disease == "Yes")
    distinct(ID) %>% 
    summarise("Number of patients matching the condition" = n())

This groups the rows by ID, so all calculations are computed individually for each person. It then gets the value of Disease in the column the day before, the day before that, and so on for the last 4 days. Then, check if each row in the dataset matches the conditions. Then take unique IDs and count them.

Answer 2

Here might be a compact way to detect for patterns in Disease column. This is based on a similar answer provided here:

https://stackoverflow.com/a/41131260/3460670

Define the pattern you want (in this case, 3 "No" followed by 2 "Yes"). Filter the rows that meet this pattern; include shift from data.table as this uses a vector for Map , instead of lead from dplyr which requires length 1 for n .

library(tidyverse)
library(data.table)

pattern = c("No", "No", "No", "Yes", "Yes")

df %>%
  group_by(ID) %>%
  filter(Reduce("&", Map("==", shift(Disease, n = 0:(length(pattern) - 1), type = "lead"), pattern))) %>% 
  ungroup() %>%
  summarise(Total = n_distinct(ID))