CODEWITHSUNDEEP
rdatedatetimedata-manipulation
Venkatram Bachoti
Venkatram Bachoti

Reputation: 11

How to create a date (column) from a date-time (column) in R

I have imported a CSV containing dates in the column "Activity_Date_Minute". The date value for example is "04/12/2016 01:12:00". Now when I read the .csv into a dataframe and extract only the date this gives me date in the column as 4-12-20. Can someone help how to get the date in mm-dd-yyyy in a separate column?

Tried the below code. Was expecting to see a column with dates e.g 04/12/2016 (mm/dd/yyyy).

#Installing packages

install.packages("tidyverse")
library(tidyverse)

install.packages('ggplot2')
library(ggplot2)

install.packages("dplyr")
library(dplyr)

install.packages("lubridate")
library(lubridate)
##Installing packages

install.packages("tidyverse")
library(tidyverse)

install.packages('ggplot2')
library(ggplot2)

install.packages("dplyr")
library(dplyr)

install.packages("lubridate")
library(lubridate)

##Reading minute-wise METs into "minutewiseMET_Records" and summarizing MET per day for all the IDs

minutewiseMET_Records <- read.csv("minuteMETsNarrow_merged.csv")
str(minutewiseMET_Records)
## converting column ID to character,Activity_Date_Minute to date 
minutewiseMET_Records$Id <- as.character(minutewiseMET_Records$Id)
minutewiseMET_Records$Date <- as.Date(minutewiseMET_Records$Activity_Date_Minute)

str(minutewiseMET_Records)

The Console is as follows:

> minutewiseMET_Records <- read.csv("minuteMETsNarrow_merged.csv")
> str(minutewiseMET_Records)
'data.frame':   1048575 obs. of  3 variables:
 $ Id                  : num  1.5e+09 1.5e+09 1.5e+09 1.5e+09 1.5e+09 ...
 $ Activity_Date_Minute: chr  "04/12/2016 00:00" "04/12/2016 00:01" "04/12/2016 00:02" "04/12/2016 00:03" ...
 $ METs                : int  10 10 10 10 10 12 12 12 12 12 ...

> ## converting column ID to character,Activity_Date_Minute to date 
> minutewiseMET_Records$Id <- as.character(minutewiseMET_Records$Id)
> minutewiseMET_Records$Date <- as.Date(minutewiseMET_Records$Activity_Date_Minute)

> ## converting column ID to character,Activity_Date_Minute to date 
> minutewiseMET_Records$Id <- as.character(minutewiseMET_Records$Id)
> minutewiseMET_Records$Date <- as.Date(minutewiseMET_Records$Activity_Date_Minute)
> str(minutewiseMET_Records)
'data.frame':   1048575 obs. of  4 variables:
 $ Id                  : chr  "1503960366" "1503960366" "1503960366" "1503960366" ...
 $ Activity_Date_Minute: chr  "04/12/2016 00:00" "04/12/2016 00:01" "04/12/2016 00:02" "04/12/2016 00:03" ...
 $ METs                : int  10 10 10 10 10 12 12 12 12 12 ...
 $ Date                  : Date, format: "4-12-20" "4-12-20" ...
>

Upvotes: 0

Views: 716

Answers (2)

R18
R18

Reputation: 1560

I think this will work for you

minutewiseMET_Records$Date <- format(as.Date(minutewiseMET_Records$Activity_Date_Minute, format = "%d/%m/%Y"),"%m/%d/%Y")

Fist of all you have to tell R the format of your initial data. Then, you ask it which is the format you want for the output.

Upvotes: 1

zephryl
zephryl

Reputation: 17234

Activity_Date_Minute isn’t a datetime in your initial data, it’s a character. So you’ll have to first convert it to a datetime (e.g., using lubridate::mdy_hm()), then use as.Date().

library(dplyr)
library(lubridate)

minutewiseMET_Records %>%
  mutate(
    Activity_Date_Minute = mdy_hm(Activity_Date_Minute), 
    Activity_Date = as.Date(Activity_Date_Minute)
  )

# A tibble: 4 × 2
  Activity_Date_Minute Activity_Date
  <dttm>               <date>       
1 2016-04-12 00:00:00  2016-04-12   
2 2016-04-12 00:01:00  2016-04-12   
3 2016-04-12 00:02:00  2016-04-12   
4 2016-04-12 00:03:00  2016-04-12

Upvotes: 0

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