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.
Related
I'm trying to visualize some bird data, however after grouping by month, the resulting output is out of order from the original data. It is in order for December, January, February, and March in the original, but after manipulating it results in December, February, January, March.
Any ideas how I can fix this or sort the rows?
This is the code:
BirdDataTimeClean <- BirdDataTimes %>%
group_by(Date) %>%
summarise(Gulls=sum(Gulls), Terns=sum(Terns), Sandpipers=sum(Sandpipers),
Plovers=sum(Plovers), Pelicans=sum(Pelicans), Oystercatchers=sum(Oystercatchers),
Egrets=sum(Egrets), PeregrineFalcon=sum(Peregrine_Falcon), BlackPhoebe=sum(Black_Phoebe),
Raven=sum(Common_Raven))
BirdDataTimeClean2 <- BirdDataTimeClean %>%
pivot_longer(!Date, names_to = "Species", values_to = "Count")
You haven't shared any workable data but i face this many times when reading from csv and hence all dates and data are in character.
as suggested, please convert the date data to "date" format using lubridate package or base as.Date() and then arrange() in dplyr will work or even group_by
example :toy data created
birds <- data.table(dates = c("2020-Feb-20","2020-Jan-20","2020-Dec-20","2020-Apr-20"),
species = c('Gulls','Turns','Gulls','Sandpiper'),
Counts = c(20,30,40,50)
str(birds) will show date is character (and I have not kept order)
using lubridate convert dates
birds$dates%>%lubridate::ymd() will change to date data-type
birds$dates%>%ymd()%>%str()
Date[1:4], format: "2020-02-20" "2020-01-20" "2020-12-20" "2020-04-20"
save it with birds$dates <- ymd(birds$dates) or do it in your pipeline as follows
now simply so the dplyr analysis:
birds%>%group_by(Months= ymd(dates))%>%
summarise(N=n()
,Species_Count = sum(Counts)
)%>%arrange(Months)
will give
# A tibble: 4 x 3
Months N Species_Count
<date> <int> <dbl>
1 2020-01-20 1 30
2 2020-02-20 1 20
3 2020-04-20 1 50
However, if you want Apr , Jan instead of numbers and apply as.Date() with format etc, the dates become "character" again. I woudl suggest you keep your data that way and while representing in output for others -> format it there with as.Date or if using DT or other datatables -> check the output formatting options. That way your original data remains and users see what they want.
this will make it character
birds%>%group_by(Months= as.character.Date(dates))%>%
summarise(N=n()
,Species_Count = sum(Counts)
)%>%arrange(Months)
A tibble: 4 x 3
Months N Species_Count
<chr> <int> <dbl>
1 2020-Apr-20 1 50
2 2020-Dec-20 1 40
3 2020-Feb-20 1 20
4 2020-Jan-20 1 30
This question already has answers here:
Concatenate a vector of strings/character
(8 answers)
Closed 5 years ago.
I have three variables: Year, Month, and Day. How can I merge them into one variable ("Date") so that the variable is represented as such:
yyyy-mm-dd
Thanks in advance and best regards!
How do you merge three variables into one variable?
Consider two methods:
Old school
With dplyr, lubridate, and data frames
And consider the data types. You can have:
Number or character
Date or POSIXct final type
Old School Method
The old school method is straightforward. I assume you are using vectors or lists and don't know data frames yet. Let's take your data, force it to a standardized, unambiguous format, and concatenate the data.
> y <- 2012:2015
> y
[1] 2012 2013 2014 2015
> m <- 1:4
> m
[1] 1 2 3 4
> d <- 10:13
> d
[1] 10 11 12 13
Use as.numeric if you want to be safe and convert everything to the same format before concatenation. If you get any NA values you will need to handle them with the is.na function and provide a default value.
Use paste with the sep separator value set to your delimiter, in this case, the hyphen.
> paste(y,m,d, sep = '-')
[1] "2012-1-10" "2013-2-11" "2014-3-12" "2015-4-13"
Dataframe / Dplyr / Lubridate Way
> df <- data.frame(year = y, mon = m, day = d)
> df
year mon day
1 2012 1 10
2 2013 2 11
3 2014 3 12
4 2015 4 13
Below I do the following:
Take the df object
Create a new variable name Date
Concatenate the numeric variables y, m, and d with a - separator
Convert the string literal into a Date format with ymd()
> df %>%
mutate(Date = ymd(
paste(y,m,d, sep = '-')
)
)
year mon day Date
1 2012 1 10 2012-01-10
2 2013 2 11 2013-02-11
3 2014 3 12 2014-03-12
4 2015 4 13 2015-04-13
Below we create year-month-day character strings, yyyy-mm-dd character strings (similar except one digit month and day are zero padded out to 2 digits) and Date class. The last one prints out as yyyy-mm-dd and can be manipulated in ways that character strings can't, for example adding one to a Date class object gives the next day.
First we set up some sample input:
year <- c(2017, 2015, 2014)
month <- c(3, 1, 10)
day <- c(15, 9, 25)
convert to year-month-day character string This is not quite yyyy-mm-dd since 1 digit months and days are not zero padded to 2 digits:
paste(year, month, day, sep = "-")
## [1] "2017-3-15" "2015-1-9" "2014-10-25"
convert to Date class It prints on console as yyyy-mm-dd. Two alternatives:
as.Date(paste(year, month, day, sep = "-"))
## [1] "2017-03-15" "2015-01-09" "2014-10-25"
as.Date(ISOdate(year, month, day))
## [1] "2017-03-15" "2015-01-09" "2014-10-25"
convert to character string yyyy-mm-dd In this case 1 digit month and day are zero padded out to 2 characters. Two alternatives:
as.character(as.Date(paste(year, month, day, sep = "-")))
## [1] "2017-03-15" "2015-01-09" "2014-10-25"
sprintf("%d-%02d-%02d", year, month, day)
## [1] "2017-03-15" "2015-01-09" "2014-10-25"
Here my time period range:
start_day = as.Date('1974-01-01', format = '%Y-%m-%d')
end_day = as.Date('2014-12-21', format = '%Y-%m-%d')
df = as.data.frame(seq(from = start_day, to = end_day, by = 'day'))
colnames(df) = 'date'
I need to created 10,000 data.frames with different fake years of 365days each one. This means that each of the 10,000 data.frames needs to have different start and end of year.
In total df has got 14,965 days which, divided by 365 days = 41 years. In other words, df needs to be grouped 10,000 times differently by 41 years (of 365 days each one).
The start of each year has to be random, so it can be 1974-10-03, 1974-08-30, 1976-01-03, etc... and the remaining dates at the end df need to be recycled with the starting one.
The grouped fake years need to appear in a 3rd col of the data.frames.
I would put all the data.frames into a list but I don't know how to create the function which generates 10,000 different year's start dates and subsequently group each data.frame with a 365 days window 41 times.
Can anyone help me?
#gringer gave a good answer but it solved only 90% of the problem:
dates.df <- data.frame(replicate(10000, seq(sample(df$date, 1),
length.out=365, by="day"),
simplify=FALSE))
colnames(dates.df) <- 1:10000
What I need is 10,000 columns with 14,965 rows made by dates taken from df which need to be eventually recycled when reaching the end of df.
I tried to change length.out = 14965 but R does not recycle the dates.
Another option could be to change length.out = 1 and eventually add the remaining df rows for each column by maintaining the same order:
dates.df <- data.frame(replicate(10000, seq(sample(df$date, 1),
length.out=1, by="day"),
simplify=FALSE))
colnames(dates.df) <- 1:10000
How can I add the remaining df rows to each col?
The seq method also works if the to argument is unspecified, so it can be used to generate a specific number of days starting at a particular date:
> seq(from=df$date[20], length.out=10, by="day")
[1] "1974-01-20" "1974-01-21" "1974-01-22" "1974-01-23" "1974-01-24"
[6] "1974-01-25" "1974-01-26" "1974-01-27" "1974-01-28" "1974-01-29"
When used in combination with replicate and sample, I think this will give what you want in a list:
> replicate(2,seq(sample(df$date, 1), length.out=10, by="day"), simplify=FALSE)
[[1]]
[1] "1985-07-24" "1985-07-25" "1985-07-26" "1985-07-27" "1985-07-28"
[6] "1985-07-29" "1985-07-30" "1985-07-31" "1985-08-01" "1985-08-02"
[[2]]
[1] "2012-10-13" "2012-10-14" "2012-10-15" "2012-10-16" "2012-10-17"
[6] "2012-10-18" "2012-10-19" "2012-10-20" "2012-10-21" "2012-10-22"
Without the simplify=FALSE argument, it produces an array of integers (i.e. R's internal representation of dates), which is a bit trickier to convert back to dates. A slightly more convoluted way to do this is and produce Date output is to use data.frame on the unsimplified replicate result. Here's an example that will produce a 10,000-column data frame with 365 dates in each column (takes about 5s to generate on my computer):
dates.df <- data.frame(replicate(10000, seq(sample(df$date, 1),
length.out=365, by="day"),
simplify=FALSE));
colnames(dates.df) <- 1:10000;
> dates.df[1:5,1:5];
1 2 3 4 5
1 1988-09-06 1996-05-30 1987-07-09 1974-01-15 1992-03-07
2 1988-09-07 1996-05-31 1987-07-10 1974-01-16 1992-03-08
3 1988-09-08 1996-06-01 1987-07-11 1974-01-17 1992-03-09
4 1988-09-09 1996-06-02 1987-07-12 1974-01-18 1992-03-10
5 1988-09-10 1996-06-03 1987-07-13 1974-01-19 1992-03-11
To get the date wraparound working, a slight modification can be made to the original data frame, pasting a copy of itself on the end:
df <- as.data.frame(c(seq(from = start_day, to = end_day, by = 'day'),
seq(from = start_day, to = end_day, by = 'day')));
colnames(df) <- "date";
This is easier to code for downstream; the alternative being a double seq for each result column with additional calculations for the start/end and if statements to deal with boundary cases.
Now instead of doing date arithmetic, the result columns subset from the original data frame (where the arithmetic is already done). Starting with one date in the first half of the frame and choosing the next 14965 values. I'm using nrow(df)/2 instead for a more generic code:
dates.df <-
as.data.frame(lapply(sample.int(nrow(df)/2, 10000),
function(startPos){
df$date[startPos:(startPos+nrow(df)/2-1)];
}));
colnames(dates.df) <- 1:10000;
>dates.df[c(1:5,(nrow(dates.df)-5):nrow(dates.df)),1:5];
1 2 3 4 5
1 1988-10-21 1999-10-18 2009-04-06 2009-01-08 1988-12-28
2 1988-10-22 1999-10-19 2009-04-07 2009-01-09 1988-12-29
3 1988-10-23 1999-10-20 2009-04-08 2009-01-10 1988-12-30
4 1988-10-24 1999-10-21 2009-04-09 2009-01-11 1988-12-31
5 1988-10-25 1999-10-22 2009-04-10 2009-01-12 1989-01-01
14960 1988-10-15 1999-10-12 2009-03-31 2009-01-02 1988-12-22
14961 1988-10-16 1999-10-13 2009-04-01 2009-01-03 1988-12-23
14962 1988-10-17 1999-10-14 2009-04-02 2009-01-04 1988-12-24
14963 1988-10-18 1999-10-15 2009-04-03 2009-01-05 1988-12-25
14964 1988-10-19 1999-10-16 2009-04-04 2009-01-06 1988-12-26
14965 1988-10-20 1999-10-17 2009-04-05 2009-01-07 1988-12-27
This takes a bit less time now, presumably because the date values have been pre-caclulated.
Try this one, using subsetting instead:
start_day = as.Date('1974-01-01', format = '%Y-%m-%d')
end_day = as.Date('2014-12-21', format = '%Y-%m-%d')
date_vec <- seq.Date(from=start_day, to=end_day, by="day")
Now, I create a vector long enough so that I can use easy subsetting later on:
date_vec2 <- rep(date_vec,2)
Now, create the random start dates for 100 instances (replace this with 10000 for your application):
random_starts <- sample(1:14965, 100)
Now, create a list of dates by simply subsetting date_vec2 with your desired length:
dates <- lapply(random_starts, function(x) date_vec2[x:(x+14964)])
date_df <- data.frame(dates)
names(date_df) <- 1:100
date_df[1:5,1:5]
1 2 3 4 5
1 1997-05-05 2011-12-10 1978-11-11 1980-09-16 1989-07-24
2 1997-05-06 2011-12-11 1978-11-12 1980-09-17 1989-07-25
3 1997-05-07 2011-12-12 1978-11-13 1980-09-18 1989-07-26
4 1997-05-08 2011-12-13 1978-11-14 1980-09-19 1989-07-27
5 1997-05-09 2011-12-14 1978-11-15 1980-09-20 1989-07-28
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 struggle mightily with dates in R and could do this pretty easily in SPSS, but I would love to stay within R for my project.
I have a date column in my data frame and want to remove the year completely in order to leave the month and day. Here is a peak at my original data.
> head(ds$date)
[1] "2003-10-09" "2003-10-11" "2003-10-13" "2003-10-15" "2003-10-18" "2003-10-20"
> class((ds$date))
[1] "Date"
I "want" it to be.
> head(ds$date)
[1] "10-09" "10-11" "10-13" "10-15" "10-18" "10-20"
> class((ds$date))
[1] "Date"
If possible, I would love to set the first date to be October 1st instead of January 1st.
Any help you can provide will be greatly appreciated.
EDIT: I felt like I should add some context. I want to plot an NHL player's performance over the course of a season which starts in October and ends in April. To add to this, I would like to facet the plots by each season which is a separate column in my data frame. Because I want to compare cumulative performance over the course of the season, I believe that I need to remove the year portion, but maybe I don't; as I indicated, I struggle with dates in R. What I am looking to accomplish is a plot that compares cumulative performance over relative dates by season and have the x-axis start in October and end in April.
> d = as.Date("2003-10-09", format="%Y-%m-%d")
> format(d, "%m-%d")
[1] "10-09"
Is this what you are looking for?
library(ggplot2)
## make up data for two seasons a and b
a = as.Date("2010/10/1")
b = as.Date("2011/10/1")
a.date <- seq(a, by='1 week', length=28)
b.date <- seq(b, by='1 week', length=28)
## make up some score data
a.score <- abs(trunc(rnorm(28, mean = 10, sd = 5)))
b.score <- abs(trunc(rnorm(28, mean = 10, sd = 5)))
## create a data frame
df <- data.frame(a.date, b.date, a.score, b.score)
df
## Since I am using ggplot I better create a "long formated" data frame
df.molt <- melt(df, measure.vars = c("a.score", "b.score"))
levels(df.molt$variable) <- c("First season", "Second season")
df.molt
Then, I am using ggplot2 for plotting the data:
## plot it
ggplot(aes(y = value, x = a.date), data = df.molt) + geom_point() +
geom_line() + facet_wrap(~variable, ncol = 1) +
scale_x_date("Date", format = "%m-%d")
If you want to modify the x-axis (e.g., display format), then you'll probably be interested in scale_date.
You have to remember Date is a numeric format, representing the number of days passed since the "origin" of the internal date counting :
> str(Date)
Class 'Date' num [1:10] 14245 14360 14475 14590 14705 ...
This is the same as in EXCEL, if you want a reference. Hence the solution with format as perfectly valid.
Now if you want to set the first date of a year as October 1st, you can construct some year index like this :
redefine.year <- function(x,start="10-1"){
year <- as.numeric(strftime(x,"%Y"))
yearstart <- as.Date(paste(year,start,sep="-"))
year + (x >= yearstart) - min(year) + 1
}
Testing code :
Start <- as.Date("2009-1-1")
Stop <- as.Date("2011-11-1")
Date <- seq(Start,Stop,length.out=10)
data.frame( Date=as.character(Date),
year=redefine.year(Date))
gives
Date year
1 2009-01-01 1
2 2009-04-25 1
3 2009-08-18 1
4 2009-12-11 2
5 2010-04-05 2
6 2010-07-29 2
7 2010-11-21 3
8 2011-03-16 3
9 2011-07-09 3
10 2011-11-01 4