how do you work with a column of mixed date types, for example 8/2/2020,2/7/2020, and all are reflecting February,
I have tried zoo::as.Date(mixeddatescolumn,"%d/%m/%Y").The first one is right but the second is wrong.
i have tried solutions here too
Fixing mixed date formats in data frame? but the questions seems different from what i am handling.
It is really tricky to know even for a human if dates like '8/2/2020' is 8th February or 2nd August. However, we can leverage the fact that you know all these dates are in February and remove the "2" part of the date which represents the month and arrange the date in one standard format and then convert the date to an actual Date object.
x <- c('8/2/2020','2/7/2020')
lubridate::mdy(paste0('2/', sub('2/', '', x, fixed = TRUE)))
#[1] "2020-02-08" "2020-02-07"
Or same in base R :
as.Date(paste0('2/', sub('2/', '', x, fixed = TRUE)), "%m/%d/%Y")
Since we know that every month is in February search for /2/ or /02/ and if found the middle number is the month; otherwise, the first number is the month. In either case set the format appropriately and use as.Date. No packages are used.
dates <- c("8/2/2020", "2/7/2020", "2/28/2000", "28/2/2000") # test data
as.Date(dates, ifelse(grepl("/0?2/", dates), "%d/%m/%Y", "%m/%d/%Y"))
## [1] "2020-02-08" "2020-02-07" "2000-02-28" "2000-02-28"
Related
I have a data frame which is a time series of meteorological measurement with monthly resolution from 1961 till 2018. I am interested in the variable that measures the monthly average temperature since I need the multi-annual average temperature for the summers.
To do this I must filter from the "DateVaraible" column the fifth and sixth digit, which are the month.
The values in time column are formatted like this
"19610701". So I need the 07(Juli) after 1961.
I start coding for 1 month for other purposes, so I did not try anything worth to mention. I guess that .grepl could do the work, but I do not know how the "matching" operator works.
So I started with this code that works.
summersmonth<- Df[DateVariable %like% "19610101" I DateVariable %like% "19610201"]
I am expecting a code like this
summermonths <- Df[DateVariable %like% "**06**" I DateVariable%like% "**07**..]
So that all entries with month digit from 06 to 09 are saved in the new dataframe summermonths.
Thanks in advance for any reply or feedback regarding my question.
Update
Thank to your answers I got the first part, which is to convert the variable in a as.date with the format "month"(Class=char)
Now I need to select months from Juni to September .
A horrible way to get the result I wanted is to do several subset and a rbind afterward.
Sommer1<-subset(Df, MonthVar == "Mai")
Sommer2<-subset(Df, MonthVar == "Juli")
Sommer3<-subset(Df, MonthVar == "September")
SummerTotal<-rbind(Sommer1,Sommer2,Sommer3)
I would be very glad to see this written in a tidy way.
Update 2 - Solution
Here is the tidy way, as here Using multiple criteria in subset function and logical operators
Veg_Seas<-subset(Df, subset = MonthVar %in% c("Mai","Juni","Juli","August","September"))
You can convert your date variable as date (format) and take the month:
allmonths <- month(as.Date(Df$DateVariable, format="%Y%m%d"))
Note that of your column has been originally imported as factor you need to convert it to character first:
allmonths <- month(as.Date(as.character(Df$DateVariable), format="%Y%m%d"))
Then you can check whether it is a summermonth:
summersmonth <- Df[allmonths %in% 6:9, ]
Example:
as.Date("20190702", format="%Y%m%d")
[1] "2019-07-02"
month(as.Date("20190702", format="%Y%m%d"))
[1] 7
We can use anydate from anytime to convert to Date class and then extract the month
library(anytime)
month(anydate(as.character(Df$DateVariable)))
I have a column of my dataframe as
date
17-Feb
17-Mar
16-Dec
16-Nov
16-Sep
17-Feb
I am trying to convert it into a date column from string. I am using the following pieces of code:
as.Date(df$Date, format="%y-%b")
and
as.POSIXct(df$Date, format="%y-%b")
Both of them give NAs
I am getting the format from this link
The starting number is year. Sorry for the confusion.
I assume from your approach that the 17 and 16 refer to the year 2017 and 2016 respectively. You need to also specify the day of month. If you don't care about it, then set it to the 1st.
A slight modification to your code will work, by appending '-01' to the date then updating your format argument to reflect this:
df = data.frame(Date = c("17-Feb", "17-Mar", "16-Dec"))
as.Date(paste0(df$Date, "-01"), format="%y-%b-%d")
I have log files where the date is mentioned in the ordinal date format.
wikipedia page for ordinal date
i.e 14273 implies 273'rd day of 2014 so 14273 is 30-Sep-2014.
is there a function in R to convert ordinal date (14273) to (30-Sep-2014).
Tried the date package but didn come across a function that would do this.
Try as.Date with the indicated format:
as.Date(sprintf("%05d", 14273), format = "%y%j")
## [1] "2014-09-30"
Notes
For more information see ?strptime [link]
The 273 part is sometimes referred to as the day of the year (as opposed to the day of the month) or the day number or the julian day relative to the beginning of the year.
If the input were a character string of the form yyjjj (rather than numeric) then as.Date(x, format = "%y%j") will do.
Update Have updated to also handle years with one digit as per comments.
Data example
x<-as.character(c("14273", "09001", "07031", "01033"))
Data conversion
x1<-substr(x, start=0, stop=2)
x2<-substr(x, start=3, stop=5)
x3<-format(strptime(x2, format="%j"), format="%m-%d")
date<-as.Date(paste(x3, x1, sep="-"), format="%m-%d-%y")
You can use lubridate package as follows:
>library(lubridate)
# Create a template date object
>date <- as.POSIXlt("2009-02-10")
# Update the date using
> update(date, year=2014, yday=273)
[1] "2014-09-30 JST"
I have this .txt file:
http://pastebin.com/raw.php?i=0fdswDxF
First column (Date) shows date in month/day
So 0601 is the 1st of June
When I load this into R and I show the data, it removes the first 0 in the data.
So when loaded it looks like:
601
602
etc
For 1st of June, 2nd of June
For the months 10,11,12, it remains unchanged.
How do I change it back to 0601 etc.?
What I am trying to do is to change these days into the day of the year, for instance,
1st of January (0101) would be 1, and 31st of December would be 365.
There is no leap year to be considered.
I have the code to change this, if my data was shown as 0601 etc, but not as 601 etc.
copperNew$Date = as.numeric(as.POSIXct(strptime(paste0("2013",copperNew$Date), format="%Y%m%d")) -
as.POSIXct("2012-12-31"), units = "days")
Where Date of course is from the file linked above.
Please ask if you do not consider the description to be good enough.
You can use colClasses in the read.table function, then convert to POSIXlt and extract the year date. You are over complicating the process.
copperNew <- read.table("http://pastebin.com/raw.php?i=0fdswDxF", header=TRUE,
colClasses=c("character", "integer", rep("numeric", 3)))
tmp <- as.POSIXlt( copperNew$Date, format='%m%d' )
copperNew$Yday <- tmp$yday
The as.POSIXct function is able to parse a string without a year (assumes the current year) and computes the day of the year for you.
d<-as.Date("0201", format = "%m%d")
strftime(d, format="%j")
#[1] "032"
First you parse your string and obtain Date object which represents your date (notice that it will add current year, so if you want to count days for some specific year add it to your string: as.Date("1988-0201", format = "%Y-%m%d")).
Function strftime will convert your Date to POSIXlt object and return day of year. If you want the result to be a numeric value, you can do it like this: as.numeric(strftime(d, format = "%j"))(Thanks Gavin Simpson)
Convert it to POSIXlt using a year that is not a leap-year, then access the yday element and add 1 (because yday is 0 on January 1st).
strptime(paste0("2011","0201"),"%Y%m%d")$yday+1
# [1] 32
From start-to-finish:
x <- read.table("http://pastebin.com/raw.php?i=0fdswDxF",
colClasses=c("character",rep("numeric",5)), header=TRUE)
x$Date <- strptime(paste0("2011",x$Date),"%Y%m%d")$yday+1
In which language?
If it's something like C#, Java or Javascript, I'd follow these steps:
1-) parse a pair of integers from that column;
2-) create a datetime variable whose day and month are taken from the integers from step one. Set the year to some fixed value, or to the current year.
3-) create another datetime variable, whose date is the 1st of February of the same year as the one in step 2.
The number of the day is the difference in days between the datetime variables, + 1 day.
This one worked for me:
copperNew <- read.table("http://pastebin.com/raw.php?i=0fdswDxF",
header=TRUE, sep=" ", colClasses=c("character",
"integer",
rep("numeric", 3)))
copperNew$diff = difftime(as.POSIXct(strptime(paste0("2013",dat$Date),
format="%Y%m%d", tz="GMT")),
as.POSIXct("2012-12-31", tz="GMT"), units="days")
I had to specify the timezone (tz argument in as.POSIXct), otherwise I got two different timezones for the vectors I am subtracting and therefore non-integer days.
In most cases, we convert numeric time to POSIXct format using R. However, if we want to compare two time points, then we would prefer the numeric time format. For example, I have a date format like "2001-03-13 10:31:00",
begin <- "2001-03-13 10:31:00"
Using R, I want to covert this into a numeric (e.g., the Julian time), perhaps something like the passing seconds between 1970-01-01 00:00:00 and 2001-03-13 10:31:00.
Do you have any suggestions?
The Julian calendar began in 45 BC (709 AUC) as a reform of the Roman calendar by Julius Caesar. It was chosen after consultation with the astronomer Sosigenes of Alexandria and was probably designed to approximate the tropical year (known at least since Hipparchus). see http://en.wikipedia.org/wiki/Julian_calendar
If you just want to remove ":" , " ", and "-" from a character vector then this will suffice:
end <- gsub("[: -]", "" , begin, perl=TRUE)
#> end
#[1] "20010313103100"
You should read the section about 1/4 of the way down in ?regex about character classes. Since the "-" is special in that context as a range operator, it needs to be placed first or last.
After your edit then the answer is clearly what #joran wrote, except that you would need first to convert to a DateTime class:
as.numeric(as.POSIXct(begin))
#[1] 984497460
The other point to make is that comparison operators do work for Date and DateTime classed variables, so the conversion may not be necessary at all. This compares 'begin' to a time one second later and correctly reports that begin is earlier:
as.POSIXct(begin) < as.POSIXct(begin) +1
#[1] TRUE
Based on the revised question this should do what you want:
begin <- "2001-03-13 10:31:00"
as.numeric(as.POSIXct(begin))
The result is a unix timestamp, the number of seconds since epoch, assuming the timestamp is in the local time zone.
Maybe this could also work:
library(lubridate)
...
df <- '24:00:00'
as.numeric(hms(df))
hms() will convert your data from one time format into another, this will let you convert it into seconds. See full documentation.
I tried this because i had trouble with data which was in that format but over 24 hours.
The example from ?as.POSIX help gives
as.POSIXct(strptime(begin, "%Y-%m-%d %H:%M:%S"))
so for you it would be
as.numeric(as.POSIXct(strptime(begin, "%Y-%m-%d %H:%M:%S")))