Earliest Date for each id in R

Question

I have a dataset where each individual (id) has an e_date, and since each individual could have more than one e_date, I\'m trying to get the earli

Answer 1

We can use data.table. Convert the 'data.frame' to 'data.table' (setDT(data_full)), grouped by 'id', we get the 1st row (head(.SD, 1L)).

library(data.table)
setDT(data_full)[order(e_date), head(.SD, 1L), by = id]

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Or using dplyr, after grouping by 'id', arrange the 'e_date' (assuming it is of Date class) and get the first row with slice.

library(dplyr)
data_full %>%
    group_by(id) %>%
    arrange(e_date) %>%
    slice(1L)

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If we need a base R option, ave can be used

data_full[with(data_full, ave(e_date, id, FUN = function(x) rank(x)==1)),]

Answer 2

Another answer that uses dplyr's filter command:

dta %>% 
  group_by(id) %>%
  filter(date == min(date))

Answer 3

I made a reproducible example, supposing that you grouped some dates by which quarter they were in.

library(lubridate)
library(dplyr)
rand_weeks <- now() + weeks(sample(100))
which_quarter <- quarter(rand_weeks)
df <- data.frame(rand_weeks, which_quarter)

df %>%
  group_by(which_quarter) %>% summarise(sort(rand_weeks)[1])

# A tibble: 4 x 2
  which_quarter sort(rand_weeks)[1]
          <dbl>              <time>
1             1 2017-01-05 05:46:32
2             2 2017-04-06 05:46:32
3             3 2016-08-18 05:46:32
4             4 2016-10-06 05:46:32

Answer 4

You may use library(sqldf) to get the minimum date as follows:

data1<-data.frame(id=c("789","123","456","123","123","456","789"),
                  e_date=c("2016-05-01","2016-07-02","2016-08-25","2015-12-11","2014-03-01","2015-07-08","2015-12-11"))  

library(sqldf)
data2 = sqldf("SELECT id,
                    min(e_date) as 'earliest_date'
                    FROM data1 GROUP BY 1", method = "name__class")    

head(data2)   

id earliest_date
123 2014-03-01
456 2015-07-08
789 2015-12-11