I have a dataframe object, and among the fields in it, I have a dates:
df$dates
I need to add a column which is 'Week Starting', i.e.
df[,'WeekStart']= manipulation
Where the week start is the date of the Monday of that week. i.e.: today is Thursday 24/09/15, would have an entry as '21-Sept'. Next thursday, 01/10/15, would be '28-Sept'.
I see that there is a weekday() function which will convert a day into a week-day, but how can I deal with this most recent monday?
A base R approach with the function strftime.
df$Week.Start <- dates-abs(1-as.numeric(strftime(df$dates, "%u")))
This can be a one-liner but we'll create a few variables to see what's happening. The %u format pattern for dates returns the day of the week as a single decimal number. We can convert that number to numeric and subtract the distance from our dates. We can then subtract that vector from our date column.
day_of_week <- as.numeric(strftime(df$dates, "%u"))
day_diff <- abs(1-day_of_week)
df$Week.Start <- dates-day_diff
# dates Week.Start
# 1 2042-10-22 2042-10-20
# 2 2026-08-14 2026-08-10
# 3 2018-11-23 2018-11-19
# 4 2017-08-21 2017-08-21
# 5 2022-05-26 2022-05-23
# 6 2037-05-27 2037-05-25
Data
set.seed(7)
all_dates <- seq(Sys.Date(), Sys.Date()+10000, by="days")
dates <- sample(all_dates, 20)
df <- data.frame(dates)
Simples:
dates <-(Sys.Date()+1:30)
week.starts <- as.Date(sapply (dates, function(d) { return (d + (-6 - as.POSIXlt(d)$wday %% -7 ))}), origin = "1970-01-01")
and running as
d <- data.frame(dataes=dates, monday=week.starts)
gives
dataes monday
1 2015-09-25 2015-09-21
2 2015-09-26 2015-09-21
3 2015-09-27 2015-09-21
4 2015-09-28 2015-09-28
5 2015-09-29 2015-09-28
6 2015-09-30 2015-09-28
7 2015-10-01 2015-09-28
8 2015-10-02 2015-09-28
9 2015-10-03 2015-09-28
10 2015-10-04 2015-09-28
11 2015-10-05 2015-10-05
12 2015-10-06 2015-10-05
13 2015-10-07 2015-10-05
14 2015-10-08 2015-10-05
15 2015-10-09 2015-10-05
16 2015-10-10 2015-10-05
17 2015-10-11 2015-10-05
18 2015-10-12 2015-10-12
19 2015-10-13 2015-10-12
20 2015-10-14 2015-10-12
21 2015-10-15 2015-10-12
22 2015-10-16 2015-10-12
23 2015-10-17 2015-10-12
24 2015-10-18 2015-10-12
25 2015-10-19 2015-10-19
26 2015-10-20 2015-10-19
27 2015-10-21 2015-10-19
28 2015-10-22 2015-10-19
29 2015-10-23 2015-10-19
30 2015-10-24 2015-10-19
Similar approach, example:
# data
d <- data.frame(date = as.Date( c("20/09/2015","24/09/2015","28/09/2015","01/10/2015"), "%d/%m/%Y"))
# get monday
d$WeekStart <- d$date - 6 - (as.POSIXlt(d$date)$wday %% -7)
d
# result
# date WeekStart
# 1 2015-09-20 2015-09-14
# 2 2015-09-24 2015-09-21
# 3 2015-09-28 2015-09-28
# 4 2015-10-01 2015-09-28
How about just subtracting from the dates the number of days required to get to the previous Monday? e.g if your data is
dates <- as.Date(c("2000-07-12", "2005-02-19", "2010-09-01"))
weekdays(dates)
# [1] "Wednesday" "Saturday" "Wednesday"
then you can compare this to a vector
wdays <- setNames(0:6, c("Monday", "Tuesday", "Wednesday",
"Thursday", "Friday", "Saturday", "Sunday"))
and subtract the required number of days from each date, ie
dates - wdays[weekdays(dates)]
# Wednesday Saturday Wednesday
#"2000-07-10" "2005-02-14" "2010-08-30"
will give the dates of the Monday preceding each date in dates. To test:
weekdays(dates - wdays[weekdays(dates)])
#Wednesday Saturday Wednesday
# "Monday" "Monday" "Monday"
Everything can be written also in one line as
dates - match(weekdays(dates), c("Monday", "Tuesday", "Wednesday",
"Thursday", "Friday", "Saturday", "Sunday")) + 1
#"2000-07-10" "2005-02-14" "2010-08-30"
a[1] <-as.Date("2016-08-20")
Finding Next day (Here "Monday")
a[1] + match("Monday",weekdays(seq(a[1]+1, a[1]+6,"days")))
"2016-08-22"
Finding Last Day (Here "Friday")
a[1] + (match("Friday",weekdays(seq(a[1]+1, a[1]+6,"days")))-7)
"2016-08-19"
A simple base-R way if your dates are properly coded as date class in R: as.Date(unclass(dates)-unclass(dates)%%7-3). You unclass it do get number of days since 1970-01-01. Then subtract remainder from division on 7 (day of the week!). Then subtract 3 because 1970-01-01 was Thursday –
Also you can group your data by week, and then create a column of "minimal date of that week". Here is how to do it in data.table package:
df=data.table(df)
df[,lastMonday:=min(dates),by=.(week(dates))]
It should work if you dont have spaces in dates.
Also, in some locales week starts with sunday, so you should be careful.
And you will need additional grouping variable, if your dates span for more than a year
If you want nearest any day and hour to the current date, use this function:
dayhour <- function(day,hour){
k <- as.Date(Sys.time())+day-as.numeric(format(strptime(Sys.time(),format="%Y-%m-%d %H:%M:%S"), format ='%u'))
dh <- format(strptime(paste(k,hour), format="%Y-%m-%d %H"), format="%A %H")
return(dh)
}
For the weekdays use 0 to 6 as day argument for sunday to saturday respectively:
> dayhour(0,17)
[1] "Sunday 17"
Related
Let's say that I have a date in R and it's formatted as follows.
date
2012-02-01
2012-02-01
2012-02-02
Is there any way in R to add another column with the day of the week associated with the date? The dataset is really large, so it would not make sense to go through manually and make the changes.
df = data.frame(date=c("2012-02-01", "2012-02-01", "2012-02-02"))
So after adding the days, it would end up looking like:
date day
2012-02-01 Wednesday
2012-02-01 Wednesday
2012-02-02 Thursday
Is this possible? Can anyone point me to a package that will allow me to do this?
Just trying to automatically generate the day by the date.
df = data.frame(date=c("2012-02-01", "2012-02-01", "2012-02-02"))
df$day <- weekdays(as.Date(df$date))
df
## date day
## 1 2012-02-01 Wednesday
## 2 2012-02-01 Wednesday
## 3 2012-02-02 Thursday
Edit: Just to show another way...
The wday component of a POSIXlt object is the numeric weekday (0-6 starting on Sunday).
as.POSIXlt(df$date)$wday
## [1] 3 3 4
which you could use to subset a character vector of weekday names
c("Sunday", "Monday", "Tuesday", "Wednesday", "Thursday",
"Friday", "Saturday")[as.POSIXlt(df$date)$wday + 1]
## [1] "Wednesday" "Wednesday" "Thursday"
Use the lubridate package and function wday:
library(lubridate)
df$date <- as.Date(df$date)
wday(df$date, label=TRUE)
[1] Wed Wed Thurs
Levels: Sun < Mon < Tues < Wed < Thurs < Fri < Sat
Look up ?strftime:
%A Full weekday name in the current locale
df$day = strftime(df$date,'%A')
Let's say you additionally want the week to begin on Monday (instead of default on Sunday), then the following is helpful:
require(lubridate)
df$day = ifelse(wday(df$time)==1,6,wday(df$time)-2)
The result is the days in the interval [0,..,6].
If you want the interval to be [1,..7], use the following:
df$day = ifelse(wday(df$time)==1,7,wday(df$time)-1)
... or, alternatively:
df$day = df$day + 1
This should do the trick
df = data.frame(date=c("2012-02-01", "2012-02-01", "2012-02-02"))
dow <- function(x) format(as.Date(x), "%A")
df$day <- dow(df$date)
df
#Returns:
date day
1 2012-02-01 Wednesday
2 2012-02-01 Wednesday
3 2012-02-02 Thursday
start = as.POSIXct("2017-09-01")
end = as.POSIXct("2017-09-06")
dat = data.frame(Date = seq.POSIXt(from = start,
to = end,
by = "DSTday"))
# see ?strptime for details of formats you can extract
# day of the week as numeric (Monday is 1)
dat$weekday1 = as.numeric(format(dat$Date, format = "%u"))
# abbreviated weekday name
dat$weekday2 = format(dat$Date, format = "%a")
# full weekday name
dat$weekday3 = format(dat$Date, format = "%A")
dat
# returns
Date weekday1 weekday2 weekday3
1 2017-09-01 5 Fri Friday
2 2017-09-02 6 Sat Saturday
3 2017-09-03 7 Sun Sunday
4 2017-09-04 1 Mon Monday
5 2017-09-05 2 Tue Tuesday
6 2017-09-06 3 Wed Wednesday
form comment of JStrahl format(as.Date(df$date),"%w"), we get number of current day :
as.numeric(format(as.Date("2016-05-09"),"%w"))
so i have a large data frame with a date time column of class POSIXct and a another column with price data of class numeric. the date time column has values of the form "1998-12-07 02:00:00 AEST" that are half hour observations across 20 years. a sample data set can be generated with the following code (vary the 100 to whatever number of observations are necessary):
data.frame(date.time = seq.POSIXt(as.POSIXct("1998-12-07 02:00:00 AEST"), as.POSIXct(Sys.Date()+1), by = "30 min")[1:100], price = rnorm(100))
i want to look at a typical year and typical week. so for the typical year i have the following code:
mean.year <- aggregate(df$price, by = list(format(df$date.time, "%m-%d %H:%M")), mean)
it seems to give me what i want:
Group.1 x
1 01-01 00:00 31.86200
2 01-01 00:30 34.20526
3 01-01 01:00 28.40105
4 01-01 01:30 26.01684
5 01-01 02:00 23.68895
6 01-01 02:30 23.70632
however the column "Group.1" is of class character and i would like it to be of class POSIXct. how can i do this?
for the typical week i have the following code
mean.week <- aggregate(df$price, by = list(format(df$date.time, "%wday %H:%M")), mean)
the output is as follows
Group.1 x
1 0day 00:00 33.05613
2 0day 00:30 30.92815
3 0day 01:00 29.26245
4 0day 01:30 29.47959
5 0day 02:00 29.18380
6 0day 02:30 25.99400
again, column "Group.1" is of class character and i would like POSIXct. also, i would like to have the day of the week as "Monday", "Tuesday", etc. instead of 0day. how would i do this?
Convert the datetime to a character string that can validly be converted back to POSIXct and then do so:
mean.year <- aggregate(df["price"],
by = list(time = as.POSIXct(format(df$date.time, "2000-%m-%d %H:%M"))), mean)
head(mean.year)
## time price
## 1 2000-12-07 02:00:00 -0.56047565
## 2 2000-12-07 02:30:00 -0.23017749
## 3 2000-12-07 03:00:00 1.55870831
## 4 2000-12-07 03:30:00 0.07050839
## 5 2000-12-07 04:00:00 0.12928774
## 6 2000-12-07 04:30:00 1.71506499
To get the day of the week use %a or %A -- see ?strptime for the list of percent codes.
mean.week <- aggregate(df["price"],
by = list(time = format(df$date.time, "%a %H:%M")), mean)
head(mean.week)
## time price
## 1 Mon 02:00 -0.56047565
## 2 Mon 02:30 -0.23017749
## 3 Mon 03:00 1.55870831
## 4 Mon 03:30 0.07050839
## 5 Mon 04:00 0.12928774
## 6 Mon 04:30 1.71506499
Note
The input df in reproducible form -- note that set.seed is needed to make it reproducible:
set.seed(123)
df <- data.frame(date.time = seq.POSIXt(as.POSIXct("1998-12-07 02:00:00 AEST"),
as.POSIXct(Sys.Date()+1), by = "30 min")[1:100], price = rnorm(100))
I have dataset consisting of two columns (timestamp and power) as:
str(df2)
'data.frame': 720 obs. of 2 variables:
$ timestamp: POSIXct, format: "2015-08-01 00:00:00" "2015-08-01 01:00:00" " ...
$ power : num 124 149 118 167 130 ..
This dataset is of entire one month duration. I want to create two subsets of it - one containing the weekend data, and other one containing weekday (Monday - Friday) data. In other words, one dataset should contain data corresponding to saturday and sunday and the other one should contain data of other days. Both of the subsets should retain both of the columns. How can I do this in R?
I tried to use the concept of aggregate and split, but I am not clear in the function parameter (FUN) of aggregate, how should I specify a divison of dataset.
You can use R base functions to do this, first use strptime to separate date data from first column and then use function weekdays.
Example:
df1<-data.frame(timestamp=c("2015-08-01 00:00:00","2015-10-13 00:00:00"),power=1:2)
df1$day<-strptime(df1[,1], "%Y-%m-%d")
df1$weekday<-weekdays(df1$day)
df1
timestamp power day weekday
2015-08-01 00:00:00 1 2015-08-01 Saturday
2015-10-13 00:00:00 2 2015-10-13 Tuesday
Building on top of #ShruS example:
df<-data.frame(timestamp=c("2015-08-01 00:00:00","2015-10-13 00:00:00", "2015-10-11 00:00:00", "2015-10-14 00:00:00"))
df$day<-strptime(df[,1], "%Y-%m-%d")
df$weekday<-weekdays(df$day)
df1 = subset(df,df$weekday == "Saturday" | df$weekday == "Sunday")
df2 = subset(df,df$weekday != "Saturday" & df$weekday != "Sunday")
> df
timestamp day weekday
1 2015-08-01 00:00:00 2015-08-01 Saturday
2 2015-10-13 00:00:00 2015-10-13 Tuesday
3 2015-10-11 00:00:00 2015-10-11 Sunday
4 2015-10-14 00:00:00 2015-10-14 Wednesday
> df1
timestamp day weekday
1 2015-08-01 00:00:00 2015-08-01 Saturday
3 2015-10-11 00:00:00 2015-10-11 Sunday
> df2
timestamp day weekday
2 2015-10-13 00:00:00 2015-10-13 Tuesday
4 2015-10-14 00:00:00 2015-10-14 Wednesday
Initially, I tried for complex approaches using extra libraries, but at the end, I came out with a basic approach using R.
#adding day column to existing set
df2$day <- weekdays(as.POSIXct(df2$timestamp))
# creating two data_subsets, i.e., week_data and weekend_data
week_data<- data.frame(timestamp=factor(), power= numeric(),day= character())
weekend_data<- data.frame(timestamp=factor(),power=numeric(),day= character())
#Specifying weekend days in vector, weekend
weekend <- c("Saturday","Sunday")
for(i in 1:nrow(df2)){
if(is.element(df2[i,3], weekend)){
weekend_data <- rbind(weekend_data, df2[i,])
} else{
week_data <- rbind(week_data, df2[i,])
}
}
The datasets created, i.e., weekend_data and week_data are my required sub datasets.
Let's say that I have a date in R and it's formatted as follows.
date
2012-02-01
2012-02-01
2012-02-02
Is there any way in R to add another column with the day of the week associated with the date? The dataset is really large, so it would not make sense to go through manually and make the changes.
df = data.frame(date=c("2012-02-01", "2012-02-01", "2012-02-02"))
So after adding the days, it would end up looking like:
date day
2012-02-01 Wednesday
2012-02-01 Wednesday
2012-02-02 Thursday
Is this possible? Can anyone point me to a package that will allow me to do this?
Just trying to automatically generate the day by the date.
df = data.frame(date=c("2012-02-01", "2012-02-01", "2012-02-02"))
df$day <- weekdays(as.Date(df$date))
df
## date day
## 1 2012-02-01 Wednesday
## 2 2012-02-01 Wednesday
## 3 2012-02-02 Thursday
Edit: Just to show another way...
The wday component of a POSIXlt object is the numeric weekday (0-6 starting on Sunday).
as.POSIXlt(df$date)$wday
## [1] 3 3 4
which you could use to subset a character vector of weekday names
c("Sunday", "Monday", "Tuesday", "Wednesday", "Thursday",
"Friday", "Saturday")[as.POSIXlt(df$date)$wday + 1]
## [1] "Wednesday" "Wednesday" "Thursday"
Use the lubridate package and function wday:
library(lubridate)
df$date <- as.Date(df$date)
wday(df$date, label=TRUE)
[1] Wed Wed Thurs
Levels: Sun < Mon < Tues < Wed < Thurs < Fri < Sat
Look up ?strftime:
%A Full weekday name in the current locale
df$day = strftime(df$date,'%A')
Let's say you additionally want the week to begin on Monday (instead of default on Sunday), then the following is helpful:
require(lubridate)
df$day = ifelse(wday(df$time)==1,6,wday(df$time)-2)
The result is the days in the interval [0,..,6].
If you want the interval to be [1,..7], use the following:
df$day = ifelse(wday(df$time)==1,7,wday(df$time)-1)
... or, alternatively:
df$day = df$day + 1
This should do the trick
df = data.frame(date=c("2012-02-01", "2012-02-01", "2012-02-02"))
dow <- function(x) format(as.Date(x), "%A")
df$day <- dow(df$date)
df
#Returns:
date day
1 2012-02-01 Wednesday
2 2012-02-01 Wednesday
3 2012-02-02 Thursday
start = as.POSIXct("2017-09-01")
end = as.POSIXct("2017-09-06")
dat = data.frame(Date = seq.POSIXt(from = start,
to = end,
by = "DSTday"))
# see ?strptime for details of formats you can extract
# day of the week as numeric (Monday is 1)
dat$weekday1 = as.numeric(format(dat$Date, format = "%u"))
# abbreviated weekday name
dat$weekday2 = format(dat$Date, format = "%a")
# full weekday name
dat$weekday3 = format(dat$Date, format = "%A")
dat
# returns
Date weekday1 weekday2 weekday3
1 2017-09-01 5 Fri Friday
2 2017-09-02 6 Sat Saturday
3 2017-09-03 7 Sun Sunday
4 2017-09-04 1 Mon Monday
5 2017-09-05 2 Tue Tuesday
6 2017-09-06 3 Wed Wednesday
form comment of JStrahl format(as.Date(df$date),"%w"), we get number of current day :
as.numeric(format(as.Date("2016-05-09"),"%w"))
Looking for a function in R to convert dates into week numbers (of year) I went for week from package data.table.
However, I observed some strange behaviour:
> week("2014-03-16") # Sun, expecting 11
[1] 11
> week("2014-03-17") # Mon, expecting 12
[1] 11
> week("2014-03-18") # Tue, expecting 12
[1] 12
Why is the week number switching to 12 on tuesday, instead of monday? What am I missing? (Timezone should be irrelevant as there are just dates?!)
Other suggestions for (base) R functions are appreciated as well.
Base package Using the function strftime passing the argument %V to obtain the week of the year as decimal number (01–53) as defined in ISO 8601. (More details in the documentarion: ?strftime)
strftime(c("2014-03-16", "2014-03-17","2014-03-18", "2014-01-01"), format = "%V")
Output:
[1] "11" "12" "12" "01"
if you try with lubridate:
library(lubridate)
lubridate::week(ymd("2014-03-16", "2014-03-17","2014-03-18", '2014-01-01'))
[1] 11 11 12 1
The pattern is the same. Try isoweek
lubridate::isoweek(ymd("2014-03-16", "2014-03-17","2014-03-18", '2014-01-01'))
[1] 11 12 12 1
I understand the need for packages in certain situations, but the base language is so elegant and so proven (and debugged and optimized).
Why not:
dt <- as.Date("2014-03-16")
dt2 <- as.POSIXlt(dt)
dt2$yday
[1] 74
And then your choice whether the first week of the year is zero (as in indexing in C) or 1 (as in indexing in R).
No packages to learn, update, worry about bugs in.
Actually, I think you may have discovered a bug in the week(...) function, or at least an error in the documentation. Hopefully someone will jump in and explain why I am wrong.
Looking at the code:
library(lubridate)
> week
function (x)
yday(x)%/%7 + 1
<environment: namespace:lubridate>
The documentation states:
Weeks is the number of complete seven day periods that have occured between the date and January 1st, plus one.
But since Jan 1 is the first day of the year (not the zeroth), the first "week" will be a six day period. The code should (??) be
(yday(x)-1)%/%7 + 1
NB: You are using week(...) in the data.table package, which is the same code as lubridate::week except it coerces everything to integer rather than numeric for efficiency. So this function has the same problem (??).
if you want to get the week number with the year use: "%Y-W%V":
e.g yearAndweeks <- strftime(dates, format = "%Y-W%V")
so
> strftime(c("2014-03-16", "2014-03-17","2014-03-18", "2014-01-01"), format = "%Y-W%V")
becomes:
[1] "2014-W11" "2014-W12" "2014-W12" "2014-W01"
If you want to get the week number with the year, Grant Shannon's solution using strftime works, but you need to make some corrections for the dates around january 1st. For instance, 2016-01-03 (yyyy-mm-dd) is week 53 of year 2015, not 2016. And 2018-12-31 is week 1 of 2019, not of 2018. This codes provides some examples and a solution. In column "yearweek" the years are sometimes wrong, in "yearweek2" they are corrected (rows 2 and 5).
library(dplyr)
library(lubridate)
# create a testset
test <- data.frame(matrix(data = c("2015-12-31",
"2016-01-03",
"2016-01-04",
"2018-12-30",
"2018-12-31",
"2019-01-01") , ncol=1, nrow = 6 ))
# add a colname
colnames(test) <- "date_txt"
# this codes provides correct year-week numbers
test <- test %>%
mutate(date = as.Date(date_txt, format = "%Y-%m-%d")) %>%
mutate(yearweek = as.integer(strftime(date, format = "%Y%V"))) %>%
mutate(yearweek2 = ifelse(test = day(date) > 7 & substr(yearweek, 5, 6) == '01',
yes = yearweek + 100,
no = ifelse(test = month(date) == 1 & as.integer(substr(yearweek, 5, 6)) > 51,
yes = yearweek - 100,
no = yearweek)))
# print the result
print(test)
date_txt date yearweek yearweek2
1 2015-12-31 2015-12-31 201553 201553
2 2016-01-03 2016-01-03 201653 201553
3 2016-01-04 2016-01-04 201601 201601
4 2018-12-30 2018-12-30 201852 201852
5 2018-12-31 2018-12-31 201801 201901
6 2019-01-01 2019-01-01 201901 201901
I think the problem is that the week calculation somehow uses the first day of the year. I don't understand the internal mechanics, but you can see what I mean with this example:
library(data.table)
dd <- seq(as.IDate("2013-12-20"), as.IDate("2014-01-20"), 1)
# dd <- seq(as.IDate("2013-12-01"), as.IDate("2014-03-31"), 1)
dt <- data.table(i = 1:length(dd),
day = dd,
weekday = weekdays(dd),
day_rounded = round(dd, "weeks"))
## Now let's add the weekdays for the "rounded" date
dt[ , weekday_rounded := weekdays(day_rounded)]
## This seems to make internal sense with the "week" calculation
dt[ , weeknumber := week(day)]
dt
i day weekday day_rounded weekday_rounded weeknumber
1: 1 2013-12-20 Friday 2013-12-17 Tuesday 51
2: 2 2013-12-21 Saturday 2013-12-17 Tuesday 51
3: 3 2013-12-22 Sunday 2013-12-17 Tuesday 51
4: 4 2013-12-23 Monday 2013-12-24 Tuesday 52
5: 5 2013-12-24 Tuesday 2013-12-24 Tuesday 52
6: 6 2013-12-25 Wednesday 2013-12-24 Tuesday 52
7: 7 2013-12-26 Thursday 2013-12-24 Tuesday 52
8: 8 2013-12-27 Friday 2013-12-24 Tuesday 52
9: 9 2013-12-28 Saturday 2013-12-24 Tuesday 52
10: 10 2013-12-29 Sunday 2013-12-24 Tuesday 52
11: 11 2013-12-30 Monday 2013-12-31 Tuesday 53
12: 12 2013-12-31 Tuesday 2013-12-31 Tuesday 53
13: 13 2014-01-01 Wednesday 2014-01-01 Wednesday 1
14: 14 2014-01-02 Thursday 2014-01-01 Wednesday 1
15: 15 2014-01-03 Friday 2014-01-01 Wednesday 1
16: 16 2014-01-04 Saturday 2014-01-01 Wednesday 1
17: 17 2014-01-05 Sunday 2014-01-01 Wednesday 1
18: 18 2014-01-06 Monday 2014-01-01 Wednesday 1
19: 19 2014-01-07 Tuesday 2014-01-08 Wednesday 2
20: 20 2014-01-08 Wednesday 2014-01-08 Wednesday 2
21: 21 2014-01-09 Thursday 2014-01-08 Wednesday 2
22: 22 2014-01-10 Friday 2014-01-08 Wednesday 2
23: 23 2014-01-11 Saturday 2014-01-08 Wednesday 2
24: 24 2014-01-12 Sunday 2014-01-08 Wednesday 2
25: 25 2014-01-13 Monday 2014-01-08 Wednesday 2
26: 26 2014-01-14 Tuesday 2014-01-15 Wednesday 3
27: 27 2014-01-15 Wednesday 2014-01-15 Wednesday 3
28: 28 2014-01-16 Thursday 2014-01-15 Wednesday 3
29: 29 2014-01-17 Friday 2014-01-15 Wednesday 3
30: 30 2014-01-18 Saturday 2014-01-15 Wednesday 3
31: 31 2014-01-19 Sunday 2014-01-15 Wednesday 3
32: 32 2014-01-20 Monday 2014-01-15 Wednesday 3
i day weekday day_rounded weekday_rounded weeknumber
My workaround is this function:
https://github.com/geneorama/geneorama/blob/master/R/round_weeks.R
round_weeks <- function(x){
require(data.table)
dt <- data.table(i = 1:length(x),
day = x,
weekday = weekdays(x))
offset <- data.table(weekday = c('Sunday', 'Monday', 'Tuesday', 'Wednesday',
'Thursday', 'Friday', 'Saturday'),
offset = -(0:6))
dt <- merge(dt, offset, by="weekday")
dt[ , day_adj := day + offset]
setkey(dt, i)
return(dt[ , day_adj])
}
Of course, you can easily change the offset to make Monday first or whatever. The best way to do this would be to add an offset to the offset... but I haven't done that yet.
I provided a link to my simple geneorama package, but please don't rely on it too much because it's likely to change and not very documented.
Using only base, I wrote the following function.
Note:
Assumes Mon is day number 1 in the week
First week is week 1
Returns 0 if week is 52 from last year
Fine-tune to suit your needs.
findWeekNo <- function(myDate){
# Find out the start day of week 1; that is the date of first Mon in the year
weekday <- switch(weekdays(as.Date(paste(format(as.Date(myDate),"%Y"),"01-01", sep = "-"))),
"Monday"={1},
"Tuesday"={2},
"Wednesday"={3},
"Thursday"={4},
"Friday"={5},
"Saturday"={6},
"Sunday"={7}
)
firstMon <- ifelse(weekday==1,1, 9 - weekday )
weekNo <- floor((as.POSIXlt(myDate)$yday - (firstMon-1))/7)+1
return(weekNo)
}
findWeekNo("2017-01-15") # 2