CODEWITHSUNDEEP
rclassdate
Basil
Basil

Reputation: 1004

Function to convert character to date

I want to write a function that will accept a vector of columns that are of class character, that will convert them to date, and output the dataframe. When I run it, it outputs a list of dataframes but without the conversion taking place. Can I rewrite the below to achieve this? The context is that I might have lots of character columns that I want to convert and having a function seems the best way to achieve this.

library(tidyverse)
date1<-c("2021-01-05","2021-01-06","2021-01-07","2021-01-08","2021-01-09")
date2<-c("2021-02-05","2021-02-06","2021-02-07","2021-02-08","2021-02-09")

df<-data.frame(date1,date2,stringsAsFactors = FALSE)

#the vector of columns to convert to date format
datecols<-c("date1","date2")

datefunc<-function(a){
  df2<-df%>%
    mutate(a=as.Date(a,"%Y-%m-%d" ))
  return(df2)}

df<-lapply(datecols,datefunc)

Upvotes: 2

Views: 224

Answers (5)

Anoushiravan R
Anoushiravan R

Reputation: 21908

You can also use this.

library(dplyr)
library(lubridate)

datefunc <- function(a){
  df2 <- df %>%
    mutate(across(all_of(a), ~ ymd(.x)))
  return(df2)
  }

datefunc(datecols)

# A tibble: 5 x 2
  date1      date2     
  <date>     <date>    
1 2021-01-05 2021-02-05
2 2021-01-06 2021-02-06
3 2021-01-07 2021-02-07
4 2021-01-08 2021-02-08
5 2021-01-09 2021-02-09

Upvotes: 2

Ronak Shah
Ronak Shah

Reputation: 388817

You don't need to specify format if the date has pattern 'Y-%m-%d'. You can write a simple function which changes a single column to date and use lapply to apply it for datecols.

datefunc<-function(a) as.Date(a)
datecols<-c("date1","date2")

df[datecols] <- lapply(df[datecols], datefunc)
df

#       date1      date2
#1 2021-01-05 2021-02-05
#2 2021-01-06 2021-02-06
#3 2021-01-07 2021-02-07
#4 2021-01-08 2021-02-08
#5 2021-01-09 2021-02-09

str(df)
#'data.frame':  5 obs. of  2 variables:
# $ date1: Date, format: "2021-01-05" "2021-01-06" "2021-01-07" ...
# $ date2: Date, format: "2021-02-05" "2021-02-06" "2021-02-07" ...

Upvotes: 2

ThomasIsCoding
ThomasIsCoding

Reputation: 101064

You can try

> list2DF(Map(as.Date, df))
       date1      date2
1 2021-01-05 2021-02-05
2 2021-01-06 2021-02-06
3 2021-01-07 2021-02-07
4 2021-01-08 2021-02-08
5 2021-01-09 2021-02-09

Upvotes: 0

AnilGoyal
AnilGoyal

Reputation: 26218

df %>% mutate(across(everything(), ~as.Date(., format = "%Y-%m-%d")))

       date1      date2
1 2021-01-05 2021-02-05
2 2021-01-06 2021-02-06
3 2021-01-07 2021-02-07
4 2021-01-08 2021-02-08
5 2021-01-09 2021-02-09

Upvotes: 0

LucaCoding
LucaCoding

Reputation: 65

Could this be helpful?

 as.Date(df$date1, df$date2, format =  "%Y/%m/%d/")

Upvotes: 1

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