This question already has answers here:
Calculate ages in R
(8 answers)
Closed 5 years ago.
Sample
df <- data.frame(
Birth_Date = c("1952-03-21", "1963-12-20", "1956-02-25", "1974-08-04", "1963-06-13", "1956-11-20", "1974-03-07", "1963-10-23", "1952-11-24", "1974-12-16"),
Items_Amount = c(68,189,69,19,299,79,149,149,29,189)
)
df
I'm trying to analyse a data-set, which has column Item_Amount(in $) and customer's birth-date spread across 90 years. Goal is to compare the sales percentage based on suitable age groups.
The main data frame contains date "BirthDate" column from "1902-02-13" to "1991-12-11" as dates not string
'data.frame': 350241 obs. of 1 variable:
$ BirthDate: Date, format: "1964-06-08" "1964-06-08" "1964-06-08" "1964-06-08" ...
> min(Trans_Cust$Birth_Date)
[1] "1902-02-13"
> difftime(max(Trans_Cust$Birth_Date),min(Trans_Cust$Birth_Date),units = "auto")
Time difference of 32808 days
> max(Trans_Cust$Birth_Date)
[1] "1991-12-11"
How Do I find the present ages based on "Birth_Date" column, store it to new column "Present_ages" and then proceed with calculating sum(Items_Amount) grouped by present_ages.
I am assuming that your birth dates are just strings, so you need to convert them to some form of date. I am using POSIXct. Once converted, you can just set up the decade boundaries and use cut to break the dates into groups.
BirthDate = c("1964-06-08", "1964-06-08", "1964-06-08", "1964-06-08",
"1902-02-13", "1991-12-11", "1944-06-06", "1929-10-24")
StartDecade = seq(as.POSIXct("1900-01-01"), as.POSIXct("2000-01-01"), by="10 years")
cut(as.POSIXct(BirthDate), breaks=StartDecade)
[1] 1960-01-01 1960-01-01 1960-01-01 1960-01-01 1900-01-01 1990-01-01 1940-01-01 1920-01-01
It may be prettier to simplify the names
as.numeric(cut(as.POSIXct(BirthDate), breaks=StartDecade)) - 1
[1] 6 6 6 6 0 9 4 2
This will return a numeric value "rounded" to the decade:
BirthDate = as.Date(c("1964-06-08", "1964-06-08", "1964-06-08", "1964-06-08", "1902-02-13", "1991-12-11", "1944-06-06", "1929-10-24"))
BDdecade <- round( as.numeric( format(BirthDate, "%Y"))-5, -1)
BDdecade
#[1] 1960 1960 1960 1960 1900 1990 1940 1920
Needed to extract the year, convert to numeric and subtract 5, since the floor function does not have the same capacity for rounding to tens and hundreds as does round.
It wasn't clear what your desired starting point for the "decades was supposed to be. This would split on the basis of the minimum date.
> BDdecade2 <- cut(BirthDate, breaks= seq( min(BirthDate), max(BirthDate), by= "10 years"))
> BDdecade2
[1] 1962-02-13 1962-02-13 1962-02-13 1962-02-13 1902-02-13 <NA> 1942-02-13
[8] 1922-02-13
8 Levels: 1902-02-13 1912-02-13 1922-02-13 1932-02-13 1942-02-13 ... 1972-02-13
The NA suggest you might need to add +365 (or perhaps even more) to the max date.
Related
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 this data with 4000 observations, so this is head(both):
kön gdk age fbkurs pers stterm
1 man FALSE 69 FALSE 1941-12-23 2011-01-19
2 man NA 70 FALSE 1942-02-11 2012-01-19
3 kvinna NA 65 FALSE 1942-06-04 2007-09-01
4 kvinna TRUE 68 FALSE 1943-04-04 2011-09-01
5 kvinna NA 65 FALSE 1943-10-30 2008-09-01
6 man FALSE 70 TRUE 1944-01-27 2013-09-01
I I want to create a new column based on the column named 'stterm'.
In stterm I have different dates that I would rather name for example. VT10, VT11, etc. I like to call the new column regyear.
I have tried to enter:
regyear <- factor(both$stterm, levels = c("2007-09-01"="HT07" "2008-09-01"="HT09" "2009-01-19"="VT09" "2009-09-01"="HT09" "2010-01-19"="VT10" "2010-09-01"="HT10" "2011-01-19"="VT11"
"2011-09-01"="HT11" "2012-01-19"="VT12" "2012-09-01"="HT12" "2013-01-19"="VT13" "2013-09-01"="HT13" "2014-01-19"="VT14"))
but when I do, I get the following error message:
Error: unexpected string constant in "regyear<- factor(both$stterm, levels = c("2007-09-01"='HT07' "2008-09-01""
What should I do to make them right?
Your code relies on quite a bit of hard-coding, which may be prone to mistakes and will be tedious if you have many dates which you wish to map to periods.
Here are some alternatives, where your dates first are converted to class Date using as.Date. This makes it easier to extract and map months to the periods "VT" or "HT", and to extract the year.
In the first example, I use cut which "divides the range of x into intervals and codes the values in x according to which interval they fall.":
# some dates which are converted to proper R dates
dates <- as.Date(c("2006-09-01", "2007-02-01", "2008-09-01", "2009-01-19"))
# extract month
month <- as.integer(format(dates, "%m"))
# extract year
year <- format(dates, "%y")
# cut the months into intervals and label the levels
term <- cut(x = month, breaks = c(0, 8, 12), labels = c("VT", "HT"))
# paste 'term' and 'year' together
paste0(term, year)
# [1] "HT06" "VT07" "HT08" "VT09"
In the second example, findInterval is used to create a numerical vector of interval indices. This vector is used to extract elements from a 'period' vector. The periods are then pasted with year as above.
paste0(c("VT", "HT")[findInterval(x = month, vec = c(1, 9))], year)
# [1] "HT06" "VT07" "HT08" "VT09"
Finally, a similar, more 'manual' method, which is less convenient if you have many 'breaks' and intervals to which you wish to map your dates:
paste0(c("VT", "HT")[as.integer(month > 8) + 1], year)
# [1] "HT06" "VT07" "HT08" "VT09"
Another relevant Q&A here.
You could do it like this:
both$regyear<- factor(both$stterm, labels = c("2007-09-01"="HT07","2008-09-01"="HT09",
"2011-01-19"="VT11","2011-09-01"="HT11",
"2012-01-19"="VT12","2013-09-01"="HT13"))
There are several problems in your original code:
It did not create a new variable in your dataframe: regyear<- factor(both$stterm, ... should be both$regyear<- factor(both$stterm, ...
You had no comma's between the levels/labels.
You had to many levels for the given example dataset (see these instructions on how to give a reproducable example).
I struggle mightily with dates in R and could do this pretty easily in SPSS, but I would love to stay within R for my project.
I have a date column in my data frame and want to remove the year completely in order to leave the month and day. Here is a peak at my original data.
> head(ds$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
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.