CODEWITHSUNDEEP
rdplyr
Rebel_47
Rebel_47

Reputation: 79

How to calculate Date difference between two records in R

Hello all my df has 3 columns and 100+ rows and looks like

PID  Record_date     DOB
123  25-02-2009  22-08-1944
165  20-04-2017  22-08-1944

I tried to calculate age for the PID but resulted like this

PID  Record_date     DOB        Age
123  25-02-2009   22-08-1944    65
165  20-04-2017   22-08-1944    73

I have calculated age for the PID by the below code

df$Age <- as.numeric(df$Record_date - df$DOB)/365
df$Age <- round (df$Age, digits = 0)

My expected Result and outcome

PID  Record_date        DOB     Age
123  25-02-2009    22-08-1944   64
165  20-04-2017    22-08-1944   72

Thanks in advance

Upvotes: 0

Views: 80

Answers (5)

Darren Tsai
Darren Tsai

Reputation: 35669

You can create a period object from the lubridate package and extract information of years.

library(dplyr)
library(lubridate)

df %>%
  mutate(across(c(Record_date, DOB), dmy),
         Age = as.period(DOB %--% Record_date)$year)

#   PID Record_date        DOB Age
# 1 123  2009-02-25 1944-08-22  64
# 2 165  2017-04-20 1944-08-22  72

If you remove $year, you can get detailed information of period:

#                 Age
# 64y 6m  3d 0H 0M 0S
# 72y 7m 29d 0H 0M 0S

Note: DOB %--% Record_date is the shortcut of interval(DOB, Record_date).

Upvotes: 1

Yuriy Saraykin
Yuriy Saraykin

Reputation: 8880

additional solution

   library(tidyverse)
   library(lubridate)
    df %>% 
      mutate(across(c(Record_date, DOB), dmy),
             age = (Record_date - DOB) %/% dyears(1))
    
      PID Record_date        DOB age
    1 123  2009-02-25 1944-08-22  64
    2 165  2017-04-20 1944-08-22  72

Upvotes: 0

Rui Barradas
Rui Barradas

Reputation: 76673

Here is a base R solution.

diffyear <- function(x, y){
  d <- difftime(x, y)
  z <- as.integer(d)
  years <- as.Date(z, origin = "0000-01-01")
  format(years, "%Y")
}

df1$Record_date <- as.Date(df1$Record_date, "%d-%m-%Y")
df1$DOB <- as.Date(df1$DOB, "%d-%m-%Y")
diffyear(df1[[2]], df1[[3]])
#[1] "64" "72"

Now assign the return value to the new column.

Upvotes: 0

Marcio Rodrigues
Marcio Rodrigues

Reputation: 349

I use this function:

age <- function(dob, age.day = today(), units = "years", floor = TRUE) {
  calc.age = interval(dob, age.day) / duration(num = 1, units = units)
  if (floor) return(as.integer(floor(calc.age)))
  return(calc.age)
}

Upvotes: 0

Ronak Shah
Ronak Shah

Reputation: 389325

If you round the values, the numbers which you have got is correct. Perhaps, you want to floor them?

library(dplyr)
df %>%
  mutate(across(-1, as.Date, '%d-%m-%Y'), 
         age = as.integer(floor((Record_date - DOB)/365)))

#  PID Record_date        DOB age
#1 123  2009-02-25 1944-08-22  64
#2 165  2017-04-20 1944-08-22  72

data

df <- structure(list(PID = c(123L, 165L), Record_date = c("25-02-2009", 
"20-04-2017"), DOB = c("22-08-1944", "22-08-1944")), 
class = "data.frame", row.names = c(NA, -2L))

Upvotes: 0

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