This probably seems straightforward, but I am pretty stumped.
I have a set of dates ~ August 1 of each year and need to sum sales by week number. The earliest date is 2008-12-08 (YYYY-MM-DD). I need to create a "week_id" field where week #1 begins on 2008-12-08. And the date 2011-09-03 is week 142. Note that this is different since the calculation of week number does not reset every year.
I am putting up a small example dataset here:
data <- data.frame(
dates = c("2008-12-08", "2009-08-10", "2010-03-31", "2011-10-16", "2008-06-03", "2009-11-14" , "2010-05-05", "2011-09-03"))
data$date = as.Date(data$date)
Any help is appreciated
data$week_id = as.numeric(data$date - as.Date("2008-12-08")) %/% 7 + 1
This would take the day difference between the two dates and find the integer number of 7 days elapsed. I add one since we want the dates where zero weeks have elapsed since the start to be week 1 instead of week 0.
dates date week_id
1 2008-12-07 2008-12-07 0 # added for testing
2 2008-12-08 2008-12-08 1
3 2008-12-09 2008-12-09 1 # added for testing
4 2008-12-14 2008-12-14 1 # added for testing
5 2008-12-15 2008-12-15 2 # added for testing
6 2009-08-10 2009-08-10 36
7 2010-03-31 2010-03-31 69
8 2011-10-16 2011-10-16 149
9 2008-06-03 2008-06-03 -26
10 2009-11-14 2009-11-14 49
11 2010-05-05 2010-05-05 74
12 2011-09-03 2011-09-03 143
I have a folder with several files where the name of each file is the respective userID. Something like this:
Time Sms
1 2012-01-01 00:00:00 10
2 2012-01-01 00:30:00 11
3 2012-01-01 01:00:00 13
4 2012-01-01 01:30:00 10
How can i aggretate by moth, week, hour and minute? Something like this:
Month DayofWeek hour min SMS
1 Mon 0 0 14 <-mean
1 Mon 0 30 12
1 Mon 1 0 17
1 Mon 1 30 21
.............................
12 Sunday 23 30 12
I had a similar issue aggregating hourly data into daily data. This is the code that worked for me.
fun <- function(s,i,j) { sum(s[i:(i+j-1)]) }
radday<-sapply(X=seq(1,24*nb_of_days,24),FUN=fun,s=your_time_series,j=24)
This sums data across a period j, which in my case since I was summing over 24 hours was 24. By changing the j value you can adjust it for your different periods of hour, day, week, month assuming that you have a constant period.
thanks for the help. I solved my problem by applying this code:
df<-aggregate(Sms~month(Time)+weekdays(Time)+hour(Time)+minute(Time),df,FUN='mean')
Edited for clarity:
I'm using R and I have a set of data that consists of order days:
> orders <- data.frame(order.num=1:4,
+ day = c("Mon", "Mon", "Mon", "Tue"))
> orders
order.num day
1 1 Mon
2 2 Mon
3 3 Mon
4 4 Tue
...
Orders typically come in on a consistent day (Monday in example above), but sometimes they come in on an alternate day (Tuesday in example above).
Here is actual data, spread into columns using dplyr::spread function
Outlet.number Sun Mon Tue Wed Thu Fri Sat
1 1 0 530 162 0 629 49 0
2 2 0 784 123 0 854 65 0
3 3 24 15 483 0 365 0 0
For Outlet 1, the "typical" order days are Monday and Thursday
For Outlet 2, the "typical" order days are Monday and Thursday
For Outlet 3, the "typical" order days are Tuesday and Thursday
I want to be able to predict if an order on an atypical day (e.g. Tue for Outlet 1) is more likely to be associated with the first typical day (Monday) or the second typical day (Thursday)
Neither of these examples have any orders on Wednesdays so I was able to hard code this small set, but for future outlets, Wednesday may either be a typical or atypical day.
Is there a way to ingest the data as shown above and then classify them?
I'm trying to define a custom week for a dataframe.
I have a dataframe with timestamps.
I've read the questions on here regarding isocalendar. While it does the job. It's not what I want.
I'm trying to define the weeks from Friday to Thrusday.
For example:
Friday 2nd Jan 2015 would be the first day of the week.
Thursday 8th Jan 2015 would be the last day of the week.
And this would be week 1.
Is there a way to set a custom weekday? so when I access the the datetime library, I get the result that I expect.
df['Week_Number'] = df['Date'].dt.week
Here's one solution - convert your dates to a Period representing weeks that end on Thursday.
In [39]: df = pd.DataFrame({'Date':pd.date_range('2015-1-1', '2015-12-31')})
In [40]: df['Period'] = df['Date'].dt.to_period('W-THU')
In [41]: df['Week_Number'] = df['Period'].dt.week
In [44]: df.head()
Out[44]:
Date Period Week_Number
0 2015-01-01 2014-12-26/2015-01-01 1
1 2015-01-02 2015-01-02/2015-01-08 2
2 2015-01-03 2015-01-02/2015-01-08 2
3 2015-01-04 2015-01-02/2015-01-08 2
4 2015-01-05 2015-01-02/2015-01-08 2
Note that it follows the same convention as datetimes, where week 1 can be incomplete, so you may have to do a little extra munging if you want 1 to be the first complete week.
Is there a good way to get a year + week number converted a date in R? I have tried the following:
> as.POSIXct("2008 41", format="%Y %U")
[1] "2008-02-21 EST"
> as.POSIXct("2008 42", format="%Y %U")
[1] "2008-02-21 EST"
According to ?strftime:
%Y Year with century. Note that whereas there was no zero in the
original Gregorian calendar, ISO 8601:2004 defines it to be valid
(interpreted as 1BC): see http://en.wikipedia.org/wiki/0_(year). Note
that the standard also says that years before 1582 in its calendar
should only be used with agreement of the parties involved.
%U Week of the year as decimal number (00–53) using Sunday as the
first day 1 of the week (and typically with the first Sunday of the
year as day 1 of week 1). The US convention.
This is kinda like another question you may have seen before. :)
The key issue is: what day should a week number specify? Is it the first day of the week? The last? That's ambiguous. I don't know if week one is the first day of the year or the 7th day of the year, or possibly the first Sunday or Monday of the year (which is a frequent interpretation). (And it's worse than that: these generally appear to be 0-indexed, rather than 1-indexed.) So, an enumerated day of the week needs to be specified.
For instance, try this:
as.POSIXlt("2008 42 1", format = "%Y %U %u")
The %u indicator specifies the day of the week.
Additional note: See ?strptime for the various options for format conversion. It's important to be careful about the enumeration of weeks, as these can be split across the end of the year, and day 1 is ambiguous: is it specified based on a Sunday or Monday, or from the first day of the year? This should all be specified and tested on the different systems where the R code will run. I'm not certain that Windows and POSIX systems sing the same tune on some of these conversions, hence I'd test and test again.
Day-of-week == zero in the POSIXlt DateTimesClasses system is Sunday. Not exactly Biblical and not in agreement with the R indexing that starts at "1" convention either, but that's what it is. Week zero is the first (partial) week in the year. Week one (but day of week zero) starts with the first Sunday. And all the other sequence types in POSIXlt have 0 as their starting point. It kind of interesting to see what coercing the list elements of POSIXlt objects do. The only way you can actually change a POSIXlt date is to alter the $year, the $mon or the $mday elements. The others seem to be epiphenomena.
today <- as.POSIXlt(Sys.Date())
today # Tuesday
#[1] "2012-02-21 UTC"
today$wday <- 0 # attempt to make it Sunday
today
# [1] "2012-02-21 UTC" The attempt fails
today$mday <- 19
today
#[1] "2012-02-19 UTC" Success
I did not come up with this myself (it's taken from a blog post by Forester), but nevertheless I thought I'd add this to the answer list because it's the first implementation of the ISO 8601 week number convention that I've seen in R.
No doubt, week numbers are a very ambiguous topic, but I prefer an ISO standard over the current implementation of week numbers via format(..., "%U") because it seems that this is what most people agreed on, at least in Germany (calendars etc.).
I've put the actual function def at the bottom to facilitate focusing on the output first. Also, I just stumbled across package ISOweek, maybe worth a try.
Approach Comparison
x.days <- c("Mon", "Tue", "Wed", "Thu", "Fri", "Sat", "Sun")
x.names <- sapply(1:length(posix), function(x) {
x.day <- as.POSIXlt(posix[x], tz="Europe/Berlin")$wday
if (x.day == 0) {
x.day <- 7
}
out <- x.days[x.day]
})
data.frame(
posix,
name=x.names,
week.r=weeknum,
week.iso=ISOweek(as.character(posix), tzone="Europe/Berlin")$weeknum
)
# Result
posix name week.r week.iso
1 2012-01-01 Sun 1 4480458
2 2012-01-02 Mon 1 1
3 2012-01-03 Tue 1 1
4 2012-01-04 Wed 1 1
5 2012-01-05 Thu 1 1
6 2012-01-06 Fri 1 1
7 2012-01-07 Sat 1 1
8 2012-01-08 Sun 2 1
9 2012-01-09 Mon 2 2
10 2012-01-10 Tue 2 2
11 2012-01-11 Wed 2 2
12 2012-01-12 Thu 2 2
13 2012-01-13 Fri 2 2
14 2012-01-14 Sat 2 2
15 2012-01-15 Sun 3 2
16 2012-01-16 Mon 3 3
17 2012-01-17 Tue 3 3
18 2012-01-18 Wed 3 3
19 2012-01-19 Thu 3 3
20 2012-01-20 Fri 3 3
21 2012-01-21 Sat 3 3
22 2012-01-22 Sun 4 3
23 2012-01-23 Mon 4 4
24 2012-01-24 Tue 4 4
25 2012-01-25 Wed 4 4
26 2012-01-26 Thu 4 4
27 2012-01-27 Fri 4 4
28 2012-01-28 Sat 4 4
29 2012-01-29 Sun 5 4
30 2012-01-30 Mon 5 5
31 2012-01-31 Tue 5 5
Function Def
It's taken directly from the blog post, I've just changed a couple of minor things. The function is still kind of sketchy (e.g. the week number of the first date is far off), but I find it to be a nice start!
ISOweek <- function(
date,
format="%Y-%m-%d",
tzone="UTC",
return.val="weekofyear"
){
##converts dates into "dayofyear" or "weekofyear", the latter providing the ISO-8601 week
##date should be a vector of class Date or a vector of formatted character strings
##format refers to the date form used if a vector of
## character strings is supplied
##convert date to POSIXt format
if(class(date)[1]%in%c("Date","character")){
date=as.POSIXlt(date,format=format, tz=tzone)
}
# if(class(date)[1]!="POSIXt"){
if (!inherits(date, "POSIXt")) {
print("Date is of wrong format.")
break
}else if(class(date)[2]=="POSIXct"){
date=as.POSIXlt(date, tz=tzone)
}
print(date)
if(return.val=="dayofyear"){
##add 1 because POSIXt is base zero
return(date$yday+1)
}else if(return.val=="weekofyear"){
##Based on the ISO8601 weekdate system,
## Monday is the first day of the week
## W01 is the week with 4 Jan in it.
year=1900+date$year
jan4=strptime(paste(year,1,4,sep="-"),format="%Y-%m-%d")
wday=jan4$wday
wday[wday==0]=7 ##convert to base 1, where Monday == 1, Sunday==7
##calculate the date of the first week of the year
weekstart=jan4-(wday-1)*86400
weeknum=ceiling(as.numeric((difftime(date,weekstart,units="days")+0.1)/7))
#########################################################################
##calculate week for days of the year occuring in the next year's week 1.
#########################################################################
mday=date$mday
wday=date$wday
wday[wday==0]=7
year=ifelse(weeknum==53 & mday-wday>=28,year+1,year)
weeknum=ifelse(weeknum==53 & mday-wday>=28,1,weeknum)
################################################################
##calculate week for days of the year occuring prior to week 1.
################################################################
##first calculate the numbe of weeks in the previous year
year.shift=year-1
jan4.shift=strptime(paste(year.shift,1,4,sep="-"),format="%Y-%m-%d")
wday=jan4.shift$wday
wday[wday==0]=7 ##convert to base 1, where Monday == 1, Sunday==7
weekstart=jan4.shift-(wday-1)*86400
weeknum.shift=ceiling(as.numeric((difftime(date,weekstart)+0.1)/7))
##update year and week
year=ifelse(weeknum==0,year.shift,year)
weeknum=ifelse(weeknum==0,weeknum.shift,weeknum)
return(list("year"=year,"weeknum"=weeknum))
}else{
print("Unknown return.val")
break
}
}