I have around 10k records. There is a variable named date_time with data in the following format 2013-01-07 10:17:08.
I need to split the column and arrive at a derived variable to identify year and month separately.
We can use lubridate package to convert the 'date_time' column to POSIXct class and extract the year and month from the output
library(lubridate)
v1 <- ymd_hms(df1$date_time)
transform(df1, Year= year(v1), Month = month(v1))
# date_time Year Month
#1 2013-01-07 10:17:08 2013 1
#2 2013-01-08 10:18:08 2013 1
data
df1 <- data.frame(date_time=c('2013-01-07 10:17:08',
'2013-01-08 10:18:08'), stringsAsFactors=FALSE)
Related
I have a data frame (df) like the following:
Date Arrivals
2014-07 100
2014-08 150
2014-09 200
I know that I can convert the yearmon dates to the first date of each month as follows:
df$Date <- as.POSIXct(paste0(as.character(df[,1]),"-01"), format = "%Y-%m-%d")
However, given that my data is not available until the end of the month I want to index it to the end rather than the beginning, and I cannot figure it out. Any help appreciated.
If the Date variable is an actual yearmon class vector, from the zoo package, the as.Date.yearmon method can do what you want via its argument frac.
Using your data, and assuming that the Date was originally a character vector
library("zoo")
df <- data.frame(Date = c("2014-07", "2014-08", "2014-09"),
Arrivals = c(100, 150, 200))
I convert this to a yearmon vector:
df <- transform(df, Date2 = as.yearmon(Date))
Assuming this is what you have, then you can achieve what you want using as.Date() with frac = 1:
df <- transform(df, Date3 = as.Date(Date2, frac = 1))
which gives:
> df
Date Arrivals Date2 Date3
1 2014-07 100 Jul 2014 2014-07-31
2 2014-08 150 Aug 2014 2014-08-31
3 2014-09 200 Sep 2014 2014-09-30
That shows the individual steps. If you only want the final Date this is a one-liner
## assuming `Date` is a `yearmon` object
df <- transform(df, Date = as.Date(Date, frac = 1))
## or if not a `yearmon`
df <- transform(df, Date = as.Date(as.yearmon(Date), frac = 1))
The argument frac in the fraction of the month to assign to the resulting dates when converting from yearmon objects to Date objects. Hence, to get the first day of the month, rather than convert to a character and paste on "-01" as your Question showed, it's better to coerce to a Date object with frac = 0.
If the Date in your df is not a yearmon class object, then you can solve your problem by converting it to one and then using the as.Date() method as described above.
Here is a way to do it using the zoo package.
R code:
library(zoo)
df
# Date Arrivals
# 1 2014-07 100
# 2 2014-08 150
# 3 2014-09 200
df$Date <- as.Date(as.yearmon(df$Date), frac = 1)
# output
# Date Arrivals
# 1 2014-07-31 100
# 2 2014-08-31 150
# 3 2014-09-30 200
Using lubridate, you can add a month and subtract a day to get the last day of the month:
library(lubridate)
ymd(paste0(df$Date, '-01')) + months(1) - days(1)
# [1] "2014-07-31" "2014-08-31" "2014-09-30"
I have a data frame (df) like the following:
Date Arrivals
2014-07 100
2014-08 150
2014-09 200
I know that I can convert the yearmon dates to the first date of each month as follows:
df$Date <- as.POSIXct(paste0(as.character(df[,1]),"-01"), format = "%Y-%m-%d")
However, given that my data is not available until the end of the month I want to index it to the end rather than the beginning, and I cannot figure it out. Any help appreciated.
If the Date variable is an actual yearmon class vector, from the zoo package, the as.Date.yearmon method can do what you want via its argument frac.
Using your data, and assuming that the Date was originally a character vector
library("zoo")
df <- data.frame(Date = c("2014-07", "2014-08", "2014-09"),
Arrivals = c(100, 150, 200))
I convert this to a yearmon vector:
df <- transform(df, Date2 = as.yearmon(Date))
Assuming this is what you have, then you can achieve what you want using as.Date() with frac = 1:
df <- transform(df, Date3 = as.Date(Date2, frac = 1))
which gives:
> df
Date Arrivals Date2 Date3
1 2014-07 100 Jul 2014 2014-07-31
2 2014-08 150 Aug 2014 2014-08-31
3 2014-09 200 Sep 2014 2014-09-30
That shows the individual steps. If you only want the final Date this is a one-liner
## assuming `Date` is a `yearmon` object
df <- transform(df, Date = as.Date(Date, frac = 1))
## or if not a `yearmon`
df <- transform(df, Date = as.Date(as.yearmon(Date), frac = 1))
The argument frac in the fraction of the month to assign to the resulting dates when converting from yearmon objects to Date objects. Hence, to get the first day of the month, rather than convert to a character and paste on "-01" as your Question showed, it's better to coerce to a Date object with frac = 0.
If the Date in your df is not a yearmon class object, then you can solve your problem by converting it to one and then using the as.Date() method as described above.
Here is a way to do it using the zoo package.
R code:
library(zoo)
df
# Date Arrivals
# 1 2014-07 100
# 2 2014-08 150
# 3 2014-09 200
df$Date <- as.Date(as.yearmon(df$Date), frac = 1)
# output
# Date Arrivals
# 1 2014-07-31 100
# 2 2014-08-31 150
# 3 2014-09-30 200
Using lubridate, you can add a month and subtract a day to get the last day of the month:
library(lubridate)
ymd(paste0(df$Date, '-01')) + months(1) - days(1)
# [1] "2014-07-31" "2014-08-31" "2014-09-30"
I have an excel file, in the date column, it shows from 1/1/15 to 12/31/15. I want to change all 15(year) to 14, so that all Date looks like from 1/1/14 to 12/31/14. How to do that in R? Right now I just use replace function manually changed the date. But there are 150000 more records....
If you don't want to convert to 'Date' class and keep the same format, one option would be sub. Here we match the last two characters that are 14 and replace it with 15.
sub('14$', '15', v1)
#[1] "1/1/15" "12/31/15" "1/1/15"
data
v1 <- c('1/1/15', '12/31/15', '1/1/14')
You could use lubridate where you can just subtract 'x' number of years.
library(lubridate)
# some random 2015 dates
df <- data.frame(dates = mdy("01/13/2015", "02/25/2015"))
# subtract 1 year
df$dates <- with(df, dates - years(1))
df
dates
1 2014-01-13
2 2014-02-25
I have a data frame
> df
Age year sex
12 80210 F
13 9123 M
I want to convert the year 80210 as 26june1982. How can I do this that the new data frame contains year in day month year formate from Julian days.
You can convert Julian dates to dates using as.Date and specifying the appropriate origin:
as.Date(8210, origin=as.Date("1960-01-01"))
#[1] "1982-06-24"
However, 80210 needs an origin pretty long ago.
You should substract the origin from the year column.
as.Date(c(80210,9123)-80210,origin='1982-06-26')
[1] "1982-06-26" "1787-11-08"
There are some options for doing this job in the R package date.
See for example on page 4, the function date.mmddyy, which says:
Given a vector of Julian dates, this returns them in the form “10/11/89”, “28/7/54”, etc.
Try this code:
age = c(12,13)
year = c(8210,9123)
sex = c("F","M")
df = data.frame(cbind(age,year,sex))
library(date)
date = date.mmddyy(year, sep = "/")
df2 = transform(df,year=date) #hint provided by jilber
df2
age year sex
1 12 6/24/82 F
2 13 12/23/84 M
I need to convert date (m/d/y format) into 3 separate columns on which I hope to run an algorithm.(I'm trying to convert my dates into Julian Day Numbers). Saw this suggestion for another user for separating data out into multiple columns using Oracle. I'm using R and am throughly stuck about how to code this appropriately. Would A1,A2...represent my new column headings, and what would the format difference be with the "update set" section?
update <tablename> set A1 = substr(ORIG, 1, 4),
A2 = substr(ORIG, 5, 6),
A3 = substr(ORIG, 11, 6),
A4 = substr(ORIG, 17, 5);
I'm trying hard to improve my skills in R but cannot figure this one...any help is much appreciated. Thanks in advance... :)
I use the format() method for Date objects to pull apart dates in R. Using Dirk's datetext, here is how I would go about breaking up a date into its constituent parts:
datetxt <- c("2010-01-02", "2010-02-03", "2010-09-10")
datetxt <- as.Date(datetxt)
df <- data.frame(date = datetxt,
year = as.numeric(format(datetxt, format = "%Y")),
month = as.numeric(format(datetxt, format = "%m")),
day = as.numeric(format(datetxt, format = "%d")))
Which gives:
> df
date year month day
1 2010-01-02 2010 1 2
2 2010-02-03 2010 2 3
3 2010-09-10 2010 9 10
Note what several others have said; you can get the Julian dates without splitting out the various date components. I added this answer to show how you could do the breaking apart if you needed it for something else.
Given a text variable x, like this:
> x
[1] "10/3/2001"
then:
> as.Date(x,"%m/%d/%Y")
[1] "2001-10-03"
converts it to a date object. Then, if you need it:
> julian(as.Date(x,"%m/%d/%Y"))
[1] 11598
attr(,"origin")
[1] "1970-01-01"
gives you a Julian date (relative to 1970-01-01).
Don't try the substring thing...
See help(as.Date) for more.
Quick ones:
Julian date converters already exist in base R, see eg help(julian).
One approach may be to parse the date as a POSIXlt and to then read off the components. Other date / time classes and packages will work too but there is something to be said for base R.
Parsing dates as string is almost always a bad approach.
Here is an example:
datetxt <- c("2010-01-02", "2010-02-03", "2010-09-10")
dates <- as.Date(datetxt) ## you could examine these as well
plt <- as.POSIXlt(dates) ## now as POSIXlt types
plt[["year"]] + 1900 ## years are with offset 1900
#[1] 2010 2010 2010
plt[["mon"]] + 1 ## and months are on the 0 .. 11 intervasl
#[1] 1 2 9
plt[["mday"]]
#[1] 2 3 10
df <- data.frame(year=plt[["year"]] + 1900,
month=plt[["mon"]] + 1, day=plt[["mday"]])
df
# year month day
#1 2010 1 2
#2 2010 2 3
#3 2010 9 10
And of course
julian(dates)
#[1] 14611 14643 14862
#attr(,"origin")
#[1] "1970-01-01"
To convert date (m/d/y format) into 3 separate columns,consider the df,
df <- data.frame(date = c("01-02-18", "02-20-18", "03-23-18"))
df
date
1 01-02-18
2 02-20-18
3 03-23-18
Convert to date format
df$date <- as.Date(df$date, format="%m-%d-%y")
df
date
1 2018-01-02
2 2018-02-20
3 2018-03-23
To get three seperate columns with year, month and date,
library(lubridate)
df$year <- year(ymd(df$date))
df$month <- month(ymd(df$date))
df$day <- day(ymd(df$date))
df
date year month day
1 2018-01-02 2018 1 2
2 2018-02-20 2018 2 20
3 2018-03-23 2018 3 23
Hope this helps.
Hi Gavin: another way [using your idea] is:
The data-frame we will use is oilstocks which contains a variety of variables related to the changes over time of the oil and gas stocks.
The variables are:
colnames(stocks)
"bpV" "bpO" "bpC" "bpMN" "bpMX" "emdate" "emV" "emO" "emC"
"emMN" "emMN.1" "chdate" "chV" "cbO" "chC" "chMN" "chMX"
One of the first things to do is change the emdate field, which is an integer vector, into a date vector.
realdate<-as.Date(emdate,format="%m/%d/%Y")
Next we want to split emdate column into three separate columns representing month, day and year using the idea supplied by you.
> dfdate <- data.frame(date=realdate)
year=as.numeric (format(realdate,"%Y"))
month=as.numeric (format(realdate,"%m"))
day=as.numeric (format(realdate,"%d"))
ls() will include the individual vectors, day, month, year and dfdate.
Now merge the dfdate, day, month, year into the original data-frame [stocks].
ostocks<-cbind(dfdate,day,month,year,stocks)
colnames(ostocks)
"date" "day" "month" "year" "bpV" "bpO" "bpC" "bpMN" "bpMX" "emdate" "emV" "emO" "emC" "emMN" "emMX" "chdate" "chV"
"cbO" "chC" "chMN" "chMX"
Similar results and I also have date, day, month, year as separate vectors outside of the df.