[R] summing values by week - based on daily dates - but with some dates missing

Henrique Dallazuanna wwwhsd at gmail.com
Wed Mar 30 23:25:55 CEST 2011


You're right:

wk <- as.numeric(format(myframe$dates, "%Y.%W"))
is.na(wk) <- wk %% 1 == 0
solution<-aggregate(value ~ group + na.locf(wk), myframe, FUN = sum)


On Wed, Mar 30, 2011 at 6:10 PM, Dimitri Liakhovitski
<dimitri.liakhovitski at gmail.com> wrote:
> Yes, zoo! That's what I forgot. It's great.
> Henrique, thanks a lot! One question:
>
> if the data are as I originally posted - then week numbered 52 is
> actually the very first week (it straddles 2008-2009).
> What if the data much longer (like in the code below - same as before,
> but more dates) so that we have more than 1 year to deal with.
> It looks like this code is lumping everything into 52 weeks. And my
> goal is to keep each week independent. If I have 2 years, then it
> should be 100+ weeks. Makes sense?
> Thank you!
>
> ### Creating a longer example data set:
> mydates<-rep(seq(as.Date("2008-12-29"), length = 500, by = "day"),2)
> myfactor<-c(rep("group.1",500),rep("group.2",500))
> set.seed(123)
> myvalues<-runif(1000,0,1)
> myframe<-data.frame(dates=mydates,group=myfactor,value=myvalues)
> (myframe)
> dim(myframe)
>
> ## Removing same rows (dates) unsystematically:
> set.seed(123)
> removed.group1<-sample(1:500,size=150,replace=F)
> set.seed(456)
> removed.group2<-sample(501:1000,size=150,replace=F)
> to.remove<-c(removed.group1,removed.group2);length(to.remove)
> to.remove<-to.remove[order(to.remove)]
> myframe<-myframe[-to.remove,]
> (myframe)
> dim(myframe)
> names(myframe)
>
> library(zoo)
> wk <- as.numeric(format(myframe$dates, '%W'))
> is.na(wk) <- wk == 0
> solution<-aggregate(value ~ group + na.locf(wk), myframe, FUN = sum)
> solution<-solution[order(solution$group),]
> write.csv(solution,file="test.csv",row.names=F)
>
>
>
> On Wed, Mar 30, 2011 at 4:45 PM, Henrique Dallazuanna <wwwhsd at gmail.com> wrote:
>> Try this:
>>
>> library(zoo)
>> wk <- as.numeric(format(myframe$dates, '%W'))
>> is.na(wk) <- wk == 0
>> aggregate(value ~ group + na.locf(wk), myframe, FUN = sum)
>>
>>
>>
>> On Wed, Mar 30, 2011 at 4:35 PM, Dimitri Liakhovitski
>> <dimitri.liakhovitski at gmail.com> wrote:
>>> Henrique, this is great, thank you!
>>>
>>> It's almost what I was looking for! Only one small thing - it doesn't
>>> "merge" the results for weeks that "straddle" 2 years. In my example -
>>> last week of year 2008 and the very first week of 2009 are one week.
>>> Any way to "join them"?
>>> Asking because in reality I'll have many years and hundreds of groups
>>> - hence, it'll be hard to do it manually.
>>>
>>>
>>> BTW - does format(dates,"%Y.%W") always consider weeks as starting with Mondays?
>>>
>>> Thank you very much!
>>> Dimitri
>>>
>>>
>>> On Wed, Mar 30, 2011 at 2:55 PM, Henrique Dallazuanna <wwwhsd at gmail.com> wrote:
>>>> Try this:
>>>>
>>>> aggregate(value ~ group + format(dates, "%Y.%W"), myframe, FUN = sum)
>>>>
>>>>
>>>> On Wed, Mar 30, 2011 at 11:23 AM, Dimitri Liakhovitski
>>>> <dimitri.liakhovitski at gmail.com> wrote:
>>>>> Dear everybody,
>>>>>
>>>>> I have the following challenge. I have a data set with 2 subgroups,
>>>>> dates (days), and corresponding values (see example code below).
>>>>> Within each subgroup: I need to aggregate (sum) the values by week -
>>>>> for weeks that start on a Monday (for example, 2008-12-29 was a
>>>>> Monday).
>>>>> I find it difficult because I have missing dates in my data - so that
>>>>> sometimes I don't even have the date for some Mondays. So, I can't
>>>>> write a proper loop.
>>>>> I want my output to look something like this:
>>>>> group   dates   value
>>>>> group.1 2008-12-29  3.0937
>>>>> group.1 2009-01-05  3.8833
>>>>> group.1 2009-01-12  1.362
>>>>> ...
>>>>> group.2 2008-12-29  2.250
>>>>> group.2 2009-01-05  1.4057
>>>>> group.2 2009-01-12  3.4411
>>>>> ...
>>>>>
>>>>> Thanks a lot for your suggestions! The code is below:
>>>>> Dimitri
>>>>>
>>>>> ### Creating example data set:
>>>>> mydates<-rep(seq(as.Date("2008-12-29"), length = 43, by = "day"),2)
>>>>> myfactor<-c(rep("group.1",43),rep("group.2",43))
>>>>> set.seed(123)
>>>>> myvalues<-runif(86,0,1)
>>>>> myframe<-data.frame(dates=mydates,group=myfactor,value=myvalues)
>>>>> (myframe)
>>>>> dim(myframe)
>>>>>
>>>>> ## Removing same rows (dates) unsystematically:
>>>>> set.seed(123)
>>>>> removed.group1<-sample(1:43,size=11,replace=F)
>>>>> set.seed(456)
>>>>> removed.group2<-sample(44:86,size=11,replace=F)
>>>>> to.remove<-c(removed.group1,removed.group2);length(to.remove)
>>>>> to.remove<-to.remove[order(to.remove)]
>>>>> myframe<-myframe[-to.remove,]
>>>>> (myframe)
>>>>>
>>>>>
>>>>>
>>>>> --
>>>>> Dimitri Liakhovitski
>>>>> Ninah Consulting
>>>>> www.ninah.com
>>>>>
>>>>> ______________________________________________
>>>>> R-help at r-project.org mailing list
>>>>> https://stat.ethz.ch/mailman/listinfo/r-help
>>>>> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
>>>>> and provide commented, minimal, self-contained, reproducible code.
>>>>>
>>>>
>>>>
>>>>
>>>> --
>>>> Henrique Dallazuanna
>>>> Curitiba-Paraná-Brasil
>>>> 25° 25' 40" S 49° 16' 22" O
>>>>
>>>
>>>
>>>
>>> --
>>> Dimitri Liakhovitski
>>> Ninah Consulting
>>> www.ninah.com
>>>
>>
>>
>>
>> --
>> Henrique Dallazuanna
>> Curitiba-Paraná-Brasil
>> 25° 25' 40" S 49° 16' 22" O
>>
>
>
>
> --
> Dimitri Liakhovitski
> Ninah Consulting
> www.ninah.com
>



-- 
Henrique Dallazuanna
Curitiba-Paraná-Brasil
25° 25' 40" S 49° 16' 22" O



More information about the R-help mailing list