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

Guys, sorry to bother you again:

I am running everything as before (see code below - before the line
with a lot of ######). But now I am getting an error:
Error in eval(expr, envir, enclos) : could not find function "na.locf"
I also noticed that after I run the 3rd line from the bottom: "wk <-
as.numeric(format(myframe$dates, "%Y.%W"))" - there are some weeks
that end with .00
And then, after I run the 2nd line from the bottom: "is.na(wk) <- wk
%% 1 == 0" those weeks turn into NAs.
Whether I run the second line or not - I get the same error about it
not finding the function "na.locf".
Do you know what might be going on?
Thanks a lot!
Dimitri

### 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)# write.csv(myframe,file="x.test.csv",row.names=F)

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 5:25 PM, Henrique Dallazuanna <wwwhsd at gmail.com> wrote:

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

--
Henrique Dallazuanna
Curitiba-Paran?-Brasil
25? 25' 40" S 49? 16' 22" O

--
Dimitri Liakhovitski
Ninah Consulting
www.ninah.com

--
Henrique Dallazuanna
Curitiba-Paran?-Brasil
25? 25' 40" S 49? 16' 22" O

--
Dimitri Liakhovitski
Ninah Consulting
www.ninah.com

--
Henrique Dallazuanna
Curitiba-Paran?-Brasil
25? 25' 40" S 49? 16' 22" O

--
Dimitri Liakhovitski
Ninah Consulting
www.ninah.com