I have a date format like this
5170301, where it means 1st March 2017.And I have 5 attached to it
I want the format of the date to be changed.
So can anyone help me in splitting that 5 from the date?
We can use substring to read from the 2nd character onwards
v1 <- substring(df1$date, 2)
NOTE: It should work for numeric/character/factor class
Then we change it to Date class
v2 <- as.Date(v1, "%y%m%d")
and if needed change the format
format(v2, "%d %b %Y")
Or as #thelatemail mentioned, it can be mentioned in the format
as.Date(df1$date, "5%y%m%d")
You can split it quite nicely with the stringr package
Split <- stringr::str_split_fixed(string=Column_Name, pattern="5", n=2)
This will give two variables: one blank and one of your value after the "5" (170301)
Then can change it to the date as so:
Date1 <- as.Date(format="%d%m%y", x = Split)
Related
I am working a dataset which has a Date column currently stored as Character, below is the format of how the dates are:
31/12/00
01/01/01
I want to change the year part from "00" to "2000" and "01" to "2001" for the whole column and want it in date format.
How can I do it?
I tried doing the following way
if(y2k_clean$Date[7:8] == "00")
replace(y2k_clean$Date, 7, "2000")
I also tried to achieve this with gsub() but could not get the result I wanted.
Thank you.
You could first convert it to a date with the format and after that convert it to your format with %Y as 4 digit year like this:
dates <- c('31/12/00', '01/01/01')
format(as.Date(dates,format='%d/%m/%y'), "%d/%m/%Y")
#> [1] "31/12/2000" "01/01/2001"
Created on 2022-12-16 with reprex v2.0.2
You can use format:
df <- tibble(Date = c("31/12/00", "01/01/01"))
format(as.Date(df$Date, format="%d/%m/%y"))
Would like to change a column of my data.frame into the date format in R.
The problem is that the format of the column is not consistent.
Most rows are in the format "%Y-%m-%d" and I can change them easily with the as.Date() function.
Few rows are in the format "%Y/%d/%m" and can't change them with the as.Date() function but instead I get NA's.
input <- c("2019-01-22", "2019-04-17", "2019/27/05", "2019/13/05", "2019/15/06", "2019-07-30")
Input: Output:
Dates Dates
2019-01-22 2019-01-22
2019-04-17 2019-04-17
2019/27/05 2019-27-05
2019/13/05 2019-13-05
2019/15/06 2019-15-06
2019-07-30 2019-07-30
In your case, in which you have "%Y-%m-%d" and "%Y/%d/%m", you might to use as.Date including the format it has. So, for example:
input <- c("2019-10-11", "2019/27/10", "2014-12-10")
If you use:
input2 <- ifelse(grepl("/",input), format(as.Date(input,"%Y/%d/%m"),"%Y-%m-%d"), input)
then:
> input2
[1] "2019-10-11" "2019-10-27" "2014-12-10"
If you have only these two formats you can substitute all the / for -:
Example:
input <- c("2019-10-11", "2019/10/12", "2014-10-13")
as.Date(gsub("/", "-", input), format = "%Y-%m-%d")
# [1] "2019-10-11" "2019-10-12" "2014-10-13"
If you just want to replace / by - for just a few rows, I think the following code might be an efficiency way to do the replacement
output <- as.Date(input)
output[is.na(output)]<-as.Date(input[is.na(output)],format = "%Y/%d/%m")
such that
> output
[1] "2019-01-22" "2019-04-17" "2019-05-27" "2019-05-13" "2019-06-15" "2019-07-30"
We can use anydate from anytime
library(anytime)
anydate(input)
#[1] "2019-10-11" "2019-10-12" "2014-10-13"
Or using lubridate
library(lubridate)
ymd(input)
data
input <- c("2019-10-11", "2019/10/12", "2014-10-13")
I can't figure out how to turn Sys.Date() into a number in the format YYYYDDD. Where DDD is the day of the year, i.e. Jan 1 would be 2016001 Dec 31 would be 2016365
Date <- Sys.Date() ## The Variable Date is created as 2016-01-01
SomeFunction(Date) ## Returns 2016001
You can just use the format function as follows:
format(Date, '%Y%j')
which gives:
[1] "2016161" "2016162" "2016163"
If you want to format it in other ways, see ?strptime for all the possible options.
Alternatively, you could use the year and yday functions from the data.table or lubridate packages and paste them together with paste0:
library(data.table) # or: library(lubridate)
paste0(year(Date), yday(Date))
which will give you the same result.
The values that are returned by both options are of class character. Wrap the above solutions in as.numeric() to get real numbers.
Used data:
> Date <- Sys.Date() + 1:3
> Date
[1] "2016-06-09" "2016-06-10" "2016-06-11"
> class(Date)
[1] "Date"
Here's one option with lubridate:
library(lubridate)
x <- Sys.Date()
#[1] "2016-06-08"
paste0(year(x),yday(x))
#[1] "2016160"
This should work for creating a new column with the specified date format:
Date <- Sys.Date
df$Month_Yr <- format(as.Date(df$Date), "%Y%d")
But, especially when working with larger data sets, it is easier to do the following:
library(data.table)
setDT(df)[,NewDate := format(as.Date(Date), "%Y%d"
Hope this helps. May have to tinker if you only want one value and are not working with a data set.
I`m giving an input as "a <- [12/Dec/2014:05:45:10]"
a is not a time-stamp so cannot use any time and date functions
Now I want the above variable to be split down into 2 parts as:-
date --> 12/Dec/2014
time --> 05:45:10
Any help will be appreciated.
You can use gsub to create a space between the Date and Time and use that to create two columns with read.table
read.table(text=gsub('^\\[([^:]+):(.*)+\\]', '\\1 \\2', a),
sep="", col.names=c('Date', 'Time'))
# Date Time
# 1 12/Dec/2014 05:45:10
Or you can use lubridate to convert it to a 'POSIXct' class
library(lubridate)
a1 <- dmy_hms(a)
a1
#[1] "2014-12-12 05:45:10 UTC"
If we need two columns with the specified format
d1 <- data.frame(Date= format(a1, '%d/%m/%Y'), Time=format(a1, '%H:%M:%S'))
data
a <- "[12/Dec/2014:05:45:10]"
Code
a <- "[12/Dec/2014:05:45:10]"
Sys.setlocale("LC_TIME", "C") # depends on your local setting
as.POSIXlt(a, format = "[%d/%b/%Y:%H:%M:%S]")
Explanation
Depending on your local setting you need to change it such that the abbreviated month names can be read. Then you can use as.POSIXlt together with the format string to convert your string in a date.
Currently my dataframe has dates displayed in the 'Date' column as 01/01/2007 etc I would like to convert these into a week/year value i.e. 01/2007. Any ideas?
I have been trying things like this and getting no where...
enviro$Week <- strptime(enviro$Date, format= "%W/%Y")
You have to first convert to date, then you can convert back to the week of the year using format, for example:
### Converts character to date
test.date <- as.Date("10/10/2014", format="%m/%d/%Y")
### Extracts only Week of the year and year
format(test.date, format="Week number %W of %Y")
[1] "Week number 40 of 2014"
### Or if you prefer
format(date, format="%W/%Y")
[1] "40/2014"
So, in your case, you would do something like this:
enviro$Week <- format(as.Date(enviro$Date, format="%m/%d/%Y"), format= "%W/%Y")
But remember that the part as.Date(enviro$Date, format="%m/%d/%Y") is only necessary if your data is not in Date format, and you also should put the right format parameter to convert your character to Date, if that is the case.
What is the class of enviro$Date? If it is of class Date there is probably a better way of doing this, otherwise you can try
v <- strsplit(as.character(enviro$Date), split = "/")
weeks <- sapply(v, "[", 2)
years <- sapply(v, "[", 3)
enviro$Week <- paste(weeks, years, sep = "/")