extract part of a date in a dataframe column - r

thanks for your help in advance. i am working with the getQuote function in the quantmod package, which returns the following data frame:
is there a way to modify all the dates in the first column to exclude the time stamp, while retaining the data frame structure? i just want the "YYYY-MM-DD" in the first column. i know that if it was a vector of dates, i would use substr(df[,1],1,10). i have also looked into the apply function, with: apply(df[,1],1,substr,1,10).

Another option not mentioned yet:
tt <- getQuote("AAPL")
trunc(tt[,1], units='days')
This returns the date in POSIXlt. You can wrap it in as.POSIXct, if you want.

using ?strptime
tt <- getQuote("AAPL")
tt[,1]
[1] "2013-01-16 02:52:00 CET"
as.POSIXct(strptime(tt[,1],format ='%Y-%m-%d')) ## as.POSIXct because strptime returns POSIXlt
[1] "2013-01-16 CET"
EDIT
You can use the format argument of POSIXct, but you need to convert the tt[,1] to character before.
as.POSIXct(as.character(tt[,1]),format ='%Y-%m-%d')
[1] "2013-01-16 CET"

I would do this with lubridate
library(plyr)
library(lubridate)
tickers <- c("AAPL","AAJX","ABR")
df <- ldply(tickers, getQuote)
rownames(df) <- tickers
df[,"Trade Time"] <- paste(year(df[,"Trade Time"]),month(df[,"Trade Time"]),day(df[,"Trade Time"]),sep="-")
There might be a more elegant way of printing the date, but this is what came to me first.

You may just use gsub. No need to convert data type.
tt <- getQuote("AAPL")
tt[, 'Trade Time']<- gsub(" [0-9]{2}:[0-9]{2}:[0-9]{2}", "", tt[, 'Trade Time'])

It can be as simple as:
tt[,1]=as.Date(tt[,1])
(where tt is tt <- getQuote("AAPL"), as shown in the alternative answers)
The blank before the comma means "do all rows" and the 1 after the comma means "operate on (just) the first column".
I prefer this solution because it gives you a Date object, which must be exactly what you want if you are trying to strip off timestamps.
agstudy's answer give you a date with a timezone, and that is going to bite you the first time you run your script in a different timezone. (Aside: I got some regressions in a unit test suite when I ran them in the U.K. while there at Christmas, due to a subtle timezone assumption in my test code.)

Related

Convert YYYYMM factor format to YYYY-MM format

I have data which have the format of YYYYMM and I wish convert it to YYYY-MM format.
exemple : 201805 should be in the format of 2018-05
How could I do it please ?
We can use as.yearmon from zoo to convert it to yearmon object and then do the format
library(zoo)
format(as.yearmon(as.character(v1), "%Y%m"), "%Y-%m")
#[1] "2018-05"
data
v1 <- 201805
I like the idea of using actual dates here. If the days component does not matter to you, then you may arbitrarily just set each of your dates to the first of the month. Then, we can leverage R's dates functions to handle the heavy lifting.
x <- "201805"
x <- paste0(x, "01")
x
y <- format(as.Date(x, format = "%Y%m%d"), "%Y-%m-%d")
substr(y, 1, 7)
[1] "20180501"
[1] "2018-05"
You could use regular expressions:
data <- "201805"
sub("(\\d{4})", "\\1-", data)
[1] "2018-05"
Another variant, using only lookarounds:
sub("(?<=\\d{4})(?=\\d{2})", "-", data, perl=TRUE)
How about following one(I am considering that OP need not to perform any checks on its variable's value here).
val="201805"
sub("(..$)","-\\1",val)
OR to perform substitution with last 2 digits only try following.
val="201805"
sub("(\\d{2}$)","-\\1",val)
[1] "2018-05"
Very similar to some of the others, but because I find the package useful I will mention it:
library(lubridate)
date <- "201805"
format(ymd(paste0(date,"01")), "%Y-%m")
Lubridate can make life easy if the formats start to vary.
Here is another option albeit a longer one:
library(tidyverse)
somestring<-"201805"
stringi::stri_sub(somestring,1,4)<-"-"
somestring1<-"201805"
somestring2<-substring(somestring1,1,4)
as.character.Date(paste0(somestring2,somestring))
Result:
"2018-05"

regex single digit

I have a question which I think is solved by regex use in R.
I have a set of dates (as chr) which I would like in a different format (as chr).
I have tried to fool around with the below examples where the first (new_dates) gives the right format for months 1-9 and wrong for 10-12 and (new_dates2) gives the right format for 10-12 but nothing for 1-9.
I see that the code in the first case matches a single digit twice for 10-12, but don't really know how to tell it to match only single digit.
The final vector of correct dates shows the result I would like.
dates <- c("1/2016", "2/2016", "3/2016", "4/2016", "5/2016", "6/2016", "7/2016", "8/2016", "9/2016", "10/2016", "11/2016", "12/2016", "1/2017")
new_dates <- sub("(\\d)[:/:](\\d{4})","\\2M0\\1", dates)
new_dates2 <- sub("(\\d{2})[:/:](\\d{4})","\\2M\\1", dates)
correctdates <- c("2016M01", "2016M02", "2016M03", "2016M04", "2016M05", "2016M06", "2016M07", "2016M08", "2016M09", "2016M10", "2016M11", "2016M12", "2017M1")
Here's a base R method that will return the desired format:
format(as.Date(paste0("1/",dates), "%d/%m/%Y"), "%YM%m")
[1] "2016M01" "2016M02" "2016M03" "2016M04" "2016M05" "2016M06" "2016M07" "2016M08" "2016M09"
[10] "2016M10" "2016M11" "2016M12" "2017M01"
The idea is to first convert to a Date object and then use the format function to create the desired character representation. I pasted on 1/ so that a day is present in each element.
As #a p o m said it might be better to look for another solution if you are manipulating dates but if you want to stick with regular expressions you can try this one.
([02-9]|1[0-2]?)[:\/](\d{4}) example
new_dates <- sub("(\\d{1,2})\\/(\\d{4})","\\2M0\\1", dates)
It's fine.

error in getting the correct date using strptime in R

I'm using strptime to extract date and the result is a wrong year
Where is the error in the below code:
strptime('8/29/2013 14:13', "%m/%d/%y")
[1] "2020-08-29 PDT"
What are the other ways to extract date and time as separate columns.
The data I have is in this format - 8/29/2013 14:13
I want to split this into two columns, one is 8/29/2013 and the other is 14:13.
You have a four digit year so you need to use %Y
strptime('8/29/2013 14:13', "%m/%d/%Y" )
[1] "2013-08-29 CEST"
Do you really want data and time in separate columns? It usually much easier to deal with a single date-time object.
Here's one possibility to separate time and date from the string.
For convenience, we could first convert the string into a POSIX object:
datetime <- '8/29/2013 14:13'
datetime.P <- as.POSIXct(datetime, format='%m/%d/%Y %H:%M')
Then we can use as.Date() to extract the date from this object and use format() to display it in the desired format:
format(as.Date(datetime.P),"%m/%d/%Y")
#[1] "08/29/2013"
To store the time separately we can use, e.g., the strftime() function:
strftime(datetime.P, '%H:%M')
#[1] "14:13"
The last function (strftime()) is not vectorized, which means that if we are dealing with a vector datetime containing several character strings with date and time in the format as described in the OP, it should be wrapped into a loop like sapply() to extract the time from each string.
Example
datetime <- c('8/29/2013 14:13', '9/15/2014 12:03')
datetime.P <- as.POSIXct(datetime, format='%m/%d/%Y %H:%M')
format(as.Date(datetime.P),"%m/%d/%Y")
#[1] "08/29/2013" "09/15/2014"
sapply(datetime.P, strftime, '%H:%M')
#[1] "14:13" "12:03"
Hope this helps.

How to lag dates in form of strings in R

The following vector of Dates is given in form of a string sequence:
d <- c("01/09/1991","01/10/1991","01/11/1991","01/12/1991")
I would like to exemplary lag this vector by 1 month, that means to produce the following structure:
d <- c("01/08/1991","01/09/1991","01/10/1991","01/11/1991")
My data is much larger and I must impose higher lags as well, but this seems to be the basis I need to know.
By doing this, I would like to have the same format in the end again:("%d/%m/%Y). How can this be done in R? I found a couple of packages (e.g. lubridate), but I always have to convert between formats (strings, dates and more) so it's a bit messy and seems prone to mistake.
edit: some more info on why I want to do this: I am using this vector as rownames of a matrix, so I would prefer a solution where the final outcome is a string vector again.
This does not use any packages. We convert to "POSIXlt" class, subtract one from the month component and convert back:
fmt <- "%d/%m/%Y"
lt <- as.POSIXlt(d, format = fmt)
lt$mon <- lt$mon - 1
format(lt, format = fmt)
## [1] "01/08/1991" "01/09/1991" "01/10/1991" "01/11/1991"
My solution uses lubridatebut it does return what you want in the specified format:
require(lubridate)
d <- c("01/09/1991","01/10/1991","01/11/1991","01/12/1991")
format(as.Date(d,format="%d/%m/%Y")-months(1),'%d/%m/%Y')
[1] "01/08/1991" "01/09/1991" "01/10/1991" "01/11/1991"
You can then change the lag and (if you want) the output (which is this part : '%d/%m/%Y') by specifying what you want.

R- date time variable loses format after ifelse [duplicate]

This question already has answers here:
How to prevent ifelse() from turning Date objects into numeric objects
(7 answers)
Closed 10 months ago.
I have a variable in the proper POSIXct format, converted with ymd_hms(DateTime) {lubridate}.
However, after a transformation the variable loses its POSIXct format:
daily$DateTime<- ifelse(daily$ID %in% "r1_1"|daily$ID %in% "r1_2",
NA,daily$DateTime)
I try to convert the variable again to POSIXct with lubridate, but it seems it does´t like the NAs, and, in addition, now the variable DateTime has a num format that lubridate does´t recognise as a date and time format (e.g. 1377419400).
Please, any help to make the required transformation to convert to NA the DateTime when ID== r1_1 and r1_2??
Thanks
The following should work:
daily <- data.frame(DateTime = seq(Sys.time(), length.out=10, by=1000), ID=rep(1:2,5))
daily$DateTime[daily$ID%in%2]<-NA
(Although the solution with is.na<- is fine too. There is just the general logic of setting is.na that doesn't make much sense - but that's no problem as long as you make sure things don't get too complicated.)
ifelse does some implicit conversions so I don't think it would ever be possible to have a date class preserved using ifelse.
The idiomatic way to set NA values is to use is.na<-, most classes (including Dates) will be dealt with appropriately
is.na(daily$DateTime) <- daily$ID %in% c('r1_1', 'r1_2')
Should do the trick.
Using the example from ?as.POSIXct
## SPSS dates (R-help 2006-02-16)
z <- c(10485849600, 10477641600, 10561104000, 10562745600)
zz <- as.POSIXct(z, origin = "1582-10-14", tz = "GMT")
is.na(zz) <- c(FALSE, TRUE, FALSE, FALSE)
zz
# [1] "1915-01-26 GMT" NA "1917-06-15 GMT" "1917-07-04 GMT"

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