CODEWITHSUNDEEP
rtime
D. Shin
D. Shin

Reputation: 335

r : calculate time interval to reference

I would like to ask your help.

tibble::tribble(
        ~Last.TP,  ~D.EndStudy,    ~EOTvisit,       ~D.END,        ~D.IP,
        "2021-02-08", "2020-08-13", "2020-08-13", "2020-08-13", "2019-07-24",
        "2021-04-19", "2021-04-26", "2021-04-26", "2021-04-26", "2020-04-06",
        "2022-01-24", "2022-02-10", "2022-02-10", "2022-02-10", "2021-01-11"
  )

Desired temporary output is below tibble. In below tibble, Days are calculated reference to D.IP

(2021-02-08 - 2019-07-24) + 1 = 566 (days)
(2021/04/19 - 2020/04/06) + 1 = 379 (days)

I would like to get below tibble, but in months.

tibble::tribble(
      ~LastTP, ~DENDSTUDY, ~EOTVISIT, ~DEND, ~DIP,
         566L,       387L,      387L,  387L,   1L,
         379L,       386L,      386L,  386L,   1L,
         379L,       396L,      396L,  396L,   1L
        )

Thank you very much!

Upvotes: 2

Views: 59

Answers (3)

jkatam
jkatam

Reputation: 3447

Alternatively, please try the below code,

code

df2 <- df %>% mutate(across(c('Last.TP','D.EndStudy','EOTvisit','D.END','D.IP'), 
~ as.numeric((as.Date(.x, '%Y-%m-%d')-as.Date(df$D.IP,'%Y-%m-%d')+1)/30.4375), 
.names = 'x{col}'))

Created on 2023-01-20 with reprex v2.0.2

output

# A tibble: 3 × 10
  Last.TP    D.EndStudy EOTvisit   D.END      D.IP       xLast.TP xD.EndStudy xEOTvisit xD.END  xD.IP
  <chr>      <chr>      <chr>      <chr>      <chr>         <dbl>       <dbl>     <dbl>  <dbl>  <dbl>
1 2021-02-08 2020-08-13 2020-08-13 2020-08-13 2019-07-24     18.6        12.7      12.7   12.7 0.0329
2 2021-04-19 2021-04-26 2021-04-26 2021-04-26 2020-04-06     12.5        12.7      12.7   12.7 0.0329
3 2022-01-24 2022-02-10 2022-02-10 2022-02-10 2021-01-11     12.5        13.0      13.0   13.0 0.0329

Upvotes: 1

Quinten
Quinten

Reputation: 41225

Another option using sapply with as.Date to make sure it is in date format. You can use the following code:

df[] <- sapply(df, \(x) {
  (as.Date(x) - as.Date("2019-07-24") + 1)/30.414
})
df
#> # A tibble: 3 × 5
#>   Last.TP D.EndStudy EOTvisit D.END    D.IP
#>     <dbl>      <dbl>    <dbl> <dbl>   <dbl>
#> 1    18.6       12.7     12.7  12.7  0.0329
#> 2    20.9       21.1     21.1  21.1  8.48  
#> 3    30.1       30.7     30.7  30.7 17.7

Created on 2023-01-20 with reprex v2.0.2

A different option like @TarJae mentioned in the comments (thanks!):

df[] <- sapply(df, \(x) {
  (as.Date(x) - as.Date(df$D.IP[1]) + 1)/30.414
})
df
#> # A tibble: 3 × 5
#>   Last.TP D.EndStudy EOTvisit D.END    D.IP
#>     <dbl>      <dbl>    <dbl> <dbl>   <dbl>
#> 1    18.6       12.7     12.7  12.7  0.0329
#> 2    20.9       21.1     21.1  21.1  8.48  
#> 3    30.1       30.7     30.7  30.7 17.7

Created on 2023-01-20 with reprex v2.0.2

Upvotes: 2

TarJae
TarJae

Reputation: 78917

Update: OP request:

library(lubridate)
library(dplyr)
df %>% 
  mutate(across(4:8, ~ . - D.IP[1])+1,
         across(4:8, ~ as.numeric(round(./30.417, digit=1)))
         )

First answer: We can do it using tripple across:

  1. First we transform character class to date class.
  2. Do the calculation with across substracting all from first row of D.IP
  3. Again across all columns we calculate the months from days:
library(dplyr)
library(lubridate)
df %>% 
  mutate(across(, ymd),
         across(, ~ . - D.IP[1])+1,
         across(, ~ as.numeric(round(./30.417, digit=1)))
         )

 Last.TP D.EndStudy EOTvisit D.END  D.IP
    <dbl>      <dbl>    <dbl> <dbl> <dbl>
1    18.6       12.7     12.7  12.7   0  
2    20.9       21.1     21.1  21.1   8.5
3    30.1       30.7     30.7  30.7  17.7

Upvotes: 2

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