I used this script
df?timestamp< as.POSIXct(df$timestamp,tz="",format="%Y%m%d%H%M%S",origin = "1970-01-01")
then i get NA's.
The number i have is 43466.10 and the result i should get is 2019-01-01 02:17:26.
What am i supposed to do?
In fact, 43466.10 is not a valid Unix epoch time. See https://en.wikipedia.org/wiki/Unix_time for more information.
This number looks like an Excel date/time value. If that truly is, then "2019-01-01 02:17:26" is actually 43466.0954398148, and the origin you should use is "1899-12-30", not "1970-01-01".
You cannot use as.POSIXct directly in this case because UNIX epoch time is in counting seconds, but an Excel date/time value is in counting days.
You can use the package openxlsx to do this conversion. Try this:
# install.packages("openxlsx")
df$timestamp <- openxlsx::convertToDateTime(df$timestamp, origin = "1899-12-30")
Related
I am trying to convert a supposedly UNIX epoch (obtained from GPS) to date object in R. I tried to use the following code, but I got an unrealistic answer as a result (shown below). Could anybody tell me what type of UNIX this is and how to convert it ?
x <- 2.01604*10^13 # Unix epoch I need to convert
as.POSIXct(as.numeric(as.character(x)), origin="1970-01-01", tz="Asia/Shanghai")
> > "640827-08-24 07:06:40 CST" # result
I have date format in following format in a data frame:
Jan-85
Apr-99
1-Nov
Feb-96
When I see the typeof(df$col) I get the answer as "integer".
Actually when I see the format in excel it is in m/d/yyyy format. I was trying to convert this to date format in R. All my efforts yielded NA.
I tried parse_date_time function. I tried as.date along with as.character. I tried as.POSIXct but everything is giving me NA.
My trials were as follows and everything was a failure:
as.Date.numeric(df$col,"m%d%Y")
transform(df$col, as.Date(as.character(df$col), "%m%d%Y"))
as.Date(df$col,"m%d%Y")
as.POSIXct.numeric(as.character(loan_new$issue_d), format="%Y%m%d")
as.POSIXct.date(as.character(df$col), format="%Y%m%d")
mdy(df$col)
parse_date_time(df$col,c("mdy"))
How can I convert this to date format? I have used lubridate package for parse_date_time and mdy package.
dput output is below
Label <- factor(c("Apr-08",
"Apr-09", "Apr-10", "Apr-11", "Aug-07", "Aug-08", "Aug-09", "Aug-10",
"Aug-11", "Dec-07", "Dec-08", "Dec-09", "Dec-10", "Dec-11", "Feb-08",
"Feb-09", "Feb-10", "Feb-11", "Jan-08", "Jan-09", "Jan-10", "Jan-11",
"Jul-07", "Jul-08", "Jul-09", "Jul-10", "Jul-11", "Jun-07", "Jun-08",
"Jun-09", "Jun-10", "Jun-11", "Mar-08", "Mar-09", "Mar-10", "Mar-11",
"May-08", "May-09", "May-10", "May-11", "Nov-07", "Nov-08", "Nov-09",
"Nov-10", "Nov-11", "Oct-07", "Oct-08", "Oct-09", "Oct-10", "Oct-11",
"Sep-07", "Sep-08", "Sep-09", "Sep-10", "Sep-11"))
NA is typically what you get when you misspecify the format. Which is what you do. That said, if your data is really looking like the first example you gave, it's impossible to simply convert this to a date. You have two different formats, one being month-year and the other day-month.
If your updated date (i.e. Dec-11) is the correct format, then you use the format argument of as.Date like this:
date <- "Dec-11"
as.Date(date, format = "%b-%d")
# [1] "2017-12-11"
Or on your example data:
as.Date(Label, format = "%b-%d")
# [1] "2017-04-08" "2017-04-09" "2017-04-10" "2017-04-11" "2017-08-07" "2017-08-08"
# [7] "2017-08-09" "2017-08-10" "2017-08-11" "2017-12-07" "2017-12-08" "2017-12-09"
If you want to convert something like Jan-85, you have to decide which day of the month that date should have. Say we just take the first of each month, then you can do:
x <- "Jan-85"
xd <- paste0("1-",x)
as.Date(xd, "%d-%b-%y")
# [1] "1985-01-01"
More information on the format codes can be found on ?strptime
Note that R will automatically add this year as the year. It has to, otherwise it can't specify the date. In case you do not have a day of the month (eg like Jan-85), conversion to a date is impossible because the underlying POSIX algorithms don't have all necessary information.
Also keep in mind that this only works when your locale is set to english. Otherwise you have a big chance your OS won't recognize the month abbreviations correctly. To do so, do eg:
Sys.setlocale(category = "LC_TIME", locale = "English_United Kingdom")
You can later set it back to the original one if you must, or restart your R session to reset the locale settings.
note: Please check carefully which locale notations are valid for your OS. The above example works on Windows, but is not guaranteed on either Linux or Mac.
Why you see integer
The fact that these string values are of integer type, is due to the fact that R automatically convert character vectors to factors when reading in a data frame. So typeof() returns integer because that's the internal representation of a factor.
I have a time-series object dat as
dat <- structure(list(timestamp = c("2015-07-01T00:00:06+05:30", "2015-07-01T00:00:36+05:30",
"2015-07-01T00:01:06+05:30", "2015-07-01T00:01:36+05:30", "2015-07-01T00:02:06+05:30",
"2015-07-01T00:02:37+05:30"), value = c(110535.421875, 110516.6484375,
110398.25, 110381.5703125, 110392.15625, 110471.609375)), .Names = c("timestamp",
"value"), row.names = c(NA, 6L), class = "data.frame")
This object is associated with a timezone offset of 05:30, which is of "Asia/Kolkata". Whenever I try to convert the timestamp, I face following issues:
with as.POSIXct(strptime(dat$timestamp,format ="%Y-%m-%dT%H:%M:%S%z")), it outputs empty strings
If I remove timezone information in strptime(), it gets automatically converted to the timezone of system, i.e., as.POSIXct(strptime(dat$timestamp,format ="%Y-%m-%dT%H:%M:%S")) . In other words, it takes timezone of the system.
How should I always force the timezone to be the same as associated with the original data object?
Format %z can be used for input or output: it is a character string, conventionally plus or minus followed by two digits for hours and two for minutes. So you need to clean up the data first to change +05:30 into +0530.
strptime(gsub("05:30", "0530",dat$timestamp), format="%Y-%m-%dT%H:%M:%S%z")
If it can contain a range of time zones in the data, assuming the data is in a standard format all the time, you can do this to remove the last semicolon from dat$timestamp:
strptime(gsub("(.*)\\:(.*)", "\\1\\2", dat$timestamp), "%Y-%m-%dT%H:%M:%S%z")
lubridate is your friend. It was written because the basic R functionality for handling dates and times is complete, but tiresome and unforgiving. lubridate is (comparatively) easy, flexible, and very forgiving of minor differences in format, tokens, etc.
install.packages("lubridate")
require(lubridate)
# the "z!*" will correctly parse any of these:
# "+0530", "+5:30", "+05:30"
dat$parsedTime <- parse_date_time(dat$timestamp, orders="ymd hms z!*")
#
# now print that
#
dat
format(dat$parsedTime, tz="Asia/Kolkata")
Internally, POSIXct objects are represented numerically in Unix time (https://en.wikipedia.org/wiki/Unix_time ) which has no time zone of its own, so you specify the time zone you want when you print. By default, it's usually displayed in the time zone of the R machine, but the default displayed time zone can be changed by setting the tzone attribute, like so:
attr(dat$parsedTime, "tzone") <- "Asia/Kolkata"
Cheers
Jason
I have a data frame containing what should be a datetime column that has been read into R. The time values are appearing as numeric time as seen in the below data example. I would like to convert these into datetime POSIXct or POSIXlt format, so that date and time can be viewed.
tdat <- c(974424L, 974430L, 974436L, 974442L, 974448L, 974454L, 974460L, 974466L, 974472L,
974478L, 974484L, 974490L, 974496L, 974502L, 974508L, 974514L, 974520L, 974526L,
974532L,974538L)
974424 should equate to 00:00:00 01/03/2011, but the do not know the origin time of the numeric values (i.e. 1970-01-01 used below does not work). I have tried using commands such as the below to achieve this and have spent time trying to get as.POXISct to work, but I haven’t found a solution (i.e. I either end up with a POSIXct object of NAs or end up with obscure datetime values).
Attempts to convert numeric time to datetime:
datetime <- as.POSIXct(strptime(time, format = "%d/%m/%Y %H:%M:%S"))
datetime <- as.POSIXct(as.numeric(time), origin='1970-01-01')
I am sure that this is a simple thing to do. Any help would be greatly received. Thanks!
Try one of these depending on which time zone you want:
t.gmt <- as.POSIXct(3600 * (tdat - 974424), origin = '2011-03-01', tz = "GMT")
t.local <- as.POSIXct(format(t.gmt))
My code:
axis.Date(1,sites$date, origin="1970-01-01")
Error:
Error in as.Date.numeric(x) : 'origin' must be supplied
Why is it asking me for the origin when I supplied it in the above code?
I suspect you meant:
axis.Date(1, as.Date(sites$date, origin = "1970-01-01"))
as the 'x' argument to as.Date() has to be of type Date.
As an aside, this would have been appropriate as a follow-up or edit of your previous question.
My R use 1970-01-01:
>as.Date(15103, origin="1970-01-01")
[1] "2011-05-09"
and this matches the calculation from
>as.numeric(as.Date(15103, origin="1970-01-01"))
So generally this has been solved, but you might get this error message because the date you use is not in the correct format.
I know this is an old post, but whenever I run this I get NA all the way down my date column. My dates are in this format 20150521 – NealC Jun 5 '15 at 16:06
If you have dates of this format just check the format of your dates with:
str(sides$date)
If the format is not a character, then convert it:
as.character(sides$date)
For as.Date, you won't need an origin any longer, because this is supplied for numeric values only. Thus you can use (assuming you have the format of NealC):
as.Date(as.character(sides$date),format="%Y%m%d")
I hope this might help some of you.
Another option is the lubridate package:
library(lubridate)
x <- 15103
as_date(x, origin = lubridate::origin)
"2011-05-09"
y <- 1442866615
as_datetime(y, origin = lubridate::origin)
"2015-09-21 20:16:55 UTC"
From the docs:
Origin is the date-time for 1970-01-01 UTC in POSIXct format. This date-time is the origin for the numbering system used by POSIXct, POSIXlt, chron, and Date classes.
If you have both date and time information in the numeric value, then use as.POSIXct. Data.table package IDateTime format is such a case. If you use fwrite to save a file, the package automatically converts date-times to idatetime format which is unix time. To convert back to normal format following can be done.
Example: Let's say you have a unix time stamp with date and time info: 1442866615
> as.POSIXct(1442866615,origin="1970-01-01")
[1] "2015-09-21 16:16:54 EDT"
by the way, the zoo package, if it is loaded, overrides the base as.Date() with its own which, by default, provides origin="1970-01-01".
(i mention this in case you find that sometimes you need to add the origin, and sometimes you don't.)