I have to do a difference between dates and return the number of days.
the format of the dates is as follows:
12/9/2011 12:00:00 AM
Does anyone know how to perform the difference without using lubridate?
We can use asPOSIXct to convert to DateTime
v1 <- as.POSIXct("12/9/2011 12:00:00 AM", format = "%d/%m/%Y %I:%M:%S %p")
If we need only Date
as.Date(v1)
Convert datetime column in date format
df$DATETIME<-as.Date(df$DATETIME)
Related
For my analysis, I need to add days of the week to the dates in Rstudio. My data starts from the first day of January ( 2019-01-01 00:00:00) and time steps are 5 minutes therefore, the second term is "2019-01-01 00:05:00 to the last day of the year. Unfortunately, some rows are missing and for example, in one case next reading after 2019-05-01 10:05:00 can be 2019-05-17 23:05:00. How can I assign days of the week to my dates?
How's this work? First, we convert the date into the magic R format, and then extract it back in the format we want.
posix <- strptime(x = "2019-01-01 00:00:00",
format = "%Y-%m-%d %H:%M:%S")
strftime(posix, format = "%Y-%m-%d %H:%M:%S %A")
Here, we're essentially exporting it in the exact same format, plus the %A operator that corresponds to the weekday.
I have a data frame with two text columns which are essentially time-stamps columns namely BEFORE and AFTER and the format is 12/29/2016 4:29:00 PM. I want to compare if the gap between the BEFORE and AFTER time for each row is more than 5 minutes or not. Which package in R will allow subtraction between time stamp?
No need for extra packages, using base R you can achieve the date time comparison.
First coerce your character to a valid date time structure. Here I am using POSIXct:
d1= as.POSIXct("12/29/2016 4:29:00 PM",format="%m/%d/%Y %H:%M:%S")
d2= as.POSIXct("12/30/2016 5:29:00 PM",format="%m/%d/%Y %H:%M:%S")
Then to get the difference in minutes you can use difftime:
difftime(d1,d2,units="mins")
Note the all those functions are vectorized So the same code will work with vectors.
manage PM/AM
to take care of PM/AM we should add %p to the format that should be used in conjonction with %I not %H:
d1= as.POSIXct("12/28/2016 11:53:00 AM",
format="%m/%d/%Y %I:%M:%S %p")
d2= as.POSIXct("12/28/2016 12:03:00 PM",
format="%m/%d/%Y %I:%M:%S %p")
difftime(d2,d1,units="mins")
## Time difference of 10 mins
for more help about date/time format you can read ?strptime.
My data format is "Posted Today 10:12 AM". I want to change it to 2017-05-12 10:12:00
How can I change it using R? Thank you in advance for your help.
Try this, your question is incomplete but I'm guessing the format
#assuming this is how your date looks like
a<-data.frame("5/14/17", "10:12 AM")
#giving name
colnames(a)<-c("date", "time")
#convert to your expected format
as.POSIXlt(paste(a$date, a$time), format = "%m/%d/%y %H:%M")
The date July, 1, 2016 1:15pm and 43 seconds is given to me as the string 160701131543.
I have an entire column in my data frame of this date time. How should I go about parsing this column into usable data.
You can use the as.POSIXct function and specify the format, in your case the format is year, month, day, hour, minute, second. Read more about formatting date and time data on the ?strptime help page.
as.POSIXct("160701131543", format = "%y%m%d%H%M%S")
[1] "2016-07-01 13:15:43 EDT"
The timezone can be changed with the 'tz' parameter.
Here is another option with lubridate. The default tz is "UTC". It can be changed by specifying tz
library(lubridate)
ymd_hms("160701131543")
#[1] "2016-07-01 13:15:43 UTC"
I am trying to convert the date as factor to date using “as.date” function in R. I have the date in the following format
2008-01-01 02:30
I tried to use the following command :
as.Date(mydata$Date, format="%y-%m-%d %h:%mm")
Can somebody help me with this ? I was able to convert the format with no hour but getting difficulty with hour included.
Thank you.
Your format string is incorrect :
R> strptime("2008-01-01 02:30", format="%Y-%m-%d %H:%M")
# [1] "2008-01-01 02:30:00"
See ?strptime for the detailed values you can use to define a format.
Also note that as your string is in a standard format, you can also use directly as.POSIXlt :
R> as.POSIXlt("2008-01-01 02:30")
# [1] "2008-01-01 02:30:00"