this is the data I expected
suppose that I have a dataset 'd'
m1 n1 A B C D
1 2 2 0.902500 0.640 0.9025 0.64
2 3 2 0.857375 0.512 0.9025 0.64
I want to add x1 (from 0 to m1), y1(from 0 to n1), m (range from m1+2 to
7-n1), n(from n1+2 to 9-m), x (x1 to x1+m-m1), y(y1 to y1+n-n1), expanding
to another dataset 'd2' based on each row (combination of m1 and n1)
so for the first row,
m1 n1 A B C D
1 2 2 0.902500 0.640 0.9025 0.64
it should be expanded as
m1 n1 x1 y1 m n x y A B C D
2 2 0 0 4 4 0 0 0.902500 0.640 0.9025 0.64
2 2 0 0 4 4 0 1 0.902500 0.640 0.9025 0.64
2 2 0 0 4 4 0 2 0.902500 0.640 0.9025 0.64
2 2 0 0 4 4 1 0 0.902500 0.640 0.9025 0.64
2 2 0 0 4 4 1 1 0.902500 0.640 0.9025 0.64
2 2 0 0 4 4 1 2 0.902500 0.640 0.9025 0.64
2 2 0 0 4 4 2 0 0.902500 0.640 0.9025 0.64
2 2 0 0 4 4 2 1 0.902500 0.640 0.9025 0.64
2 2 0 0 4 4 2 2 0.902500 0.640 0.9025 0.64
2 2 0 0 4 5 0 0 0.902500 0.640 0.9025 0.64
2 2 0 0 4 5 0 1 0.902500 0.640 0.9025 0.64
2 2 0 0 4 5 0 2 0.902500 0.640 0.9025 0.64
2 2 0 0 4 5 1 0 0.902500 0.640 0.9025 0.64
2 2 0 0 4 5 1 1 0.902500 0.640 0.9025 0.64
2 2 0 0 4 5 1 2 0.902500 0.640 0.9025 0.64
2 2 0 0 4 5 2 0 0.902500 0.640 0.9025 0.64
2 2 0 0 4 5 2 1 0.902500 0.640 0.9025 0.64
2 2 0 0 4 5 2 2 0.902500 0.640 0.9025 0.64
2 2 0 1 4 4 0 0 0.902500 0.640 0.9025 0.64
2 2 0 1 4 4 0 1 0.902500 0.640 0.9025 0.64
2 2 0 1 4 4 0 2 0.902500 0.640 0.9025 0.64
2 2 0 1 4 4 1 0 0.902500 0.640 0.9025 0.64
2 2 0 1 4 4 1 1 0.902500 0.640 0.9025 0.64
2 2 0 1 4 4 1 2 0.902500 0.640 0.9025 0.64
2 2 0 1 4 4 2 0 0.902500 0.640 0.9025 0.64
2 2 0 1 4 4 2 1 0.902500 0.640 0.9025 0.64
2 2 0 1 4 4 2 2 0.902500 0.640 0.9025 0.64
2 2 0 1 4 5 0 0 0.902500 0.640 0.9025 0.64
2 2 0 1 4 5 0 1 0.902500 0.640 0.9025 0.64
2 2 0 1 4 5 0 2 0.902500 0.640 0.9025 0.64
2 2 0 1 4 5 1 0 0.902500 0.640 0.9025 0.64
2 2 0 1 4 5 1 1 0.902500 0.640 0.9025 0.64
2 2 0 1 4 5 1 2 0.902500 0.640 0.9025 0.64
2 2 0 1 4 5 2 0 0.902500 0.640 0.9025 0.64
2 2 0 1 4 5 2 1 0.902500 0.640 0.9025 0.64
2 2 0 1 4 5 2 2 0.902500 0.640 0.9025 0.64
2 2 0 2 4 4 0 0 0.902500 0.640 0.9025 0.64
2 2 0 2 4 4 0 1 0.902500 0.640 0.9025 0.64
2 2 0 2 4 4 0 2 0.902500 0.640 0.9025 0.64
.
.
.
.
2 2 1 0 4 4 0 0 0.902500 0.640 0.9025 0.64
2 2 1 1 4 4 0 1 0.902500 0.640 0.9025 0.64
2 2 1 2 4 4 0 2 0.902500 0.640 0.9025 0.64
.
.
.
On Tue, Feb 19, 2013 at 11:59 AM, arun kirshna [via R] <
ml-node+s789695n4659078...@n4.nabble.com> wrote:
>
>
> Hi,
>
> Try this:
> res1<- do.call(rbind,lapply(paste(d3$m1,d3$n1),function(m1)
> do.call(rbind,lapply(0:(as.numeric(substr(m1,1,1))-1),function(x1)
> do.call(rbind,lapply(0:(as.numeric(substr(m1,3,3))-1),function(y1)
> do.call(rbind,lapply((as.numeric(substr(m1,1,1))+2):(7-as.numeric(substr(m1,3,3))),function(m)
> do.call(rbind,lapply((as.numeric(substr(m1,3,3))+2):(9-m),function(n)
> do.call(rbind,lapply(x1:(x1+m-as.numeric(substr(m1,1,1))), function(x)
> do.call(rbind,lapply(y1:(y1+n-as.numeric(substr(m1,3,3))), function(y)
> expand.grid(m1,x1,y1,m,n,x,y)) )))))))))))))
> names(res1)<- c("m1n1","x1","y1","m","n","x","y")
> res1$m1<- NA; res1$n1<- NA
> res1[,8:9]<-do.call(rbind,lapply(strsplit(as.character(res1$m1n1),"
> "),as.numeric))
> res2<- res1[,c(8:9,3:7)]
> library(plyr)
> res2<-join(res1,d3,by=c("m1","n1"),type="full") #Instead of this step, you
> can paste() the whole row of d3 and make suitable changes to the code above
>
> tail(res2)
> # m1n1 x1 y1 m n x y m1 n1 cterm1_P0L cterm1_P1L cterm1_P0H cterm1_P1H
> #235 2 3 1 2 4 5 2 2 2 3 0.9025 0.64 0.857375 0.512
> #236 2 3 1 2 4 5 2 3 2 3 0.9025 0.64 0.857375 0.512
> #237 2 3 1 2 4 5 2 4 2 3 0.9025 0.64 0.857375 0.512
> #238 2 3 1 2 4 5 3 2 2 3 0.9025 0.64 0.857375 0.512
> #239 2 3 1 2 4 5 3 3 2 3 0.9025 0.64 0.857375 0.512
> #240 2 3 1 2 4 5 3 4 2 3 0.9025 0.64 0.857375 0.512
>
> A.K.
> ________________________________
> From: Joanna Zhang <[hidden
> email]<http://user/SendEmail.jtp?type=node&node=4659078&i=0>>
>
> To: arun <[hidden
> email]<http://user/SendEmail.jtp?type=node&node=4659078&i=1>>
>
> Sent: Tuesday, February 19, 2013 11:43 AM
> Subject: Re: [R] cumulative sum by group and under some criteria
>
>
> Thanks. I can get the data I expected (get rid of the m1=3, n1=3) using
> the join and 'inner' code, but just curious about the way to expand the
> data. There should be a way to expand the data based on each row
> (combination of the variables), unique(d3$m1 & d3$n1) ?.
>
> or is there a way to use 'data.frame' and 'for' loop to expand directly
> from the data? like res1<-data.frame (d3) for () {....
>
>
> On Tue, Feb 19, 2013 at 9:55 AM, arun <[hidden
> email]<http://user/SendEmail.jtp?type=node&node=4659078&i=2>>
> wrote:
>
> If you can provide me the output that you expect with all the rows of the
> combination in the res2, I can take a look.
>
> >
> >
> >
> >
> >
> >
> >________________________________
> >
> >From: Joanna Zhang <[hidden
> >email]<http://user/SendEmail.jtp?type=node&node=4659078&i=3>>
>
> >To: arun <[hidden
> >email]<http://user/SendEmail.jtp?type=node&node=4659078&i=4>>
>
> >
> >Sent: Tuesday, February 19, 2013 10:42 AM
> >
> >Subject: Re: [R] cumulative sum by group and under some criteria
> >
> >
> >Thanks. But I thougth the expanded dataset 'res1' should not have
> combination of m1=3 and n1=3 because it is based on dataset 'd3' which
> doesn't have m1=3 and n1=3, right?>
> >>In the example that you provided:
> >> (m1+2):(maxN-(n1+2))
> >>#[1] 5
> >> (n1+2):(maxN-5)
> >>#[1] 4
> >>#Suppose
> >> x1<- 4
> >> y1<- 2
> >> x1:(x1+5-m1)
> >>#[1] 4 5 6
> >> y1:(y1+4-n1)
> >>#[1] 2 3 4
> >>
> >> datnew<-expand.grid(5,4,4:6,2:4)
> >> colnames(datnew)<- c("m","n","x","y")
> >>datnew<-within(datnew,{p1<- x/m;p2<-y/n})
> >>res<-cbind(datnew,d2[rep(1:nrow(d2),nrow(datnew)),])
> >> row.names(res)<- 1:nrow(res)
> >> res
> >># m n x y p2 p1 m1 n1 cterm1_P1L cterm1_P0H
> >>#1 5 4 4 2 0.50 0.8 3 2 0.00032 0.0025
> >>#2 5 4 5 2 0.50 1.0 3 2 0.00032 0.0025
> >>#3 5 4 6 2 0.50 1.2 3 2 0.00032 0.0025
> >>#4 5 4 4 3 0.75 0.8 3 2 0.00032 0.0025
> >>#5 5 4 5 3 0.75 1.0 3 2 0.00032 0.0025
> >>#6 5 4 6 3 0.75 1.2 3 2 0.00032 0.0025
> >>#7 5 4 4 4 1.00 0.8 3 2 0.00032 0.0025
> >>#8 5 4 5 4 1.00 1.0 3 2 0.00032 0.0025
> >>#9 5 4 6 4 1.00 1.2 3 2 0.00032 0.0025
> >>
> >>A.K.
> >>
> >>
> >>
> >>
> >>
> >>----- Original Message -----
> >>From: Zjoanna <[hidden
> >>email]<http://user/SendEmail.jtp?type=node&node=4659078&i=5>>
>
> >>To: [hidden email]<http://user/SendEmail.jtp?type=node&node=4659078&i=6>
> >>Cc:
> >>
> >>Sent: Sunday, February 10, 2013 6:04 PM
> >>Subject: Re: [R] cumulative sum by group and under some criteria
> >>
> >>
> >>Hi,
> >>How to expand or loop for one variable n based on another variable? for
> >>example, I want to add m (from m1 to maxN- n1-2) and for each m, I want
> to
> >>add n (n1+2 to maxN-m), and similarly add x and y, then I need to do
> some
> >>calculations.
> >>
> >>d3<-data.frame(d2)
> >> for (m in (m1+2):(maxN-(n1+2)){
> >> for (n in (n1+2):(maxN-m)){
> >> for (x in x1:(x1+m-m1)){
> >> for (y in y1:(y1+n-n1)){
> >> p1<- x/m
> >> p2<- y/n
> >>}}}}
> >>
> >>On Thu, Feb 7, 2013 at 12:16 AM, arun kirshna [via R] <
> >>[hidden email] <http://user/SendEmail.jtp?type=node&node=4659078&i=7>>
> wrote:
> >>
> >>> Hi,
> >>>
> >>> Anyway, just using some random combinations:
> >>> dnew<- expand.grid(4:10,5:10,6:10,3:7,4:5,6:8)
> >>> names(dnew)<-c("m","n","x1","y1","x","y")
> >>> resF<- cbind(dnew,d2[rep(1:nrow(d2),nrow(dnew)),])
> >>>
> >>> row.names(resF)<- 1:nrow(resF)
> >>> head(resF)
> >>> # m n x1 y1 x y m1 n1 cterm1_P1L cterm1_P0H
> >>> #1 4 5 6 3 4 6 3 2 0.00032 0.0025
> >>> #2 5 5 6 3 4 6 3 2 0.00032 0.0025
> >>> #3 6 5 6 3 4 6 3 2 0.00032 0.0025
> >>> #4 7 5 6 3 4 6 3 2 0.00032 0.0025
> >>> #5 8 5 6 3 4 6 3 2 0.00032 0.0025
> >>> #6 9 5 6 3 4 6 3 2 0.00032 0.0025
> >>>
> >>> nrow(resF)
> >>> #[1] 6300
> >>> I am not sure what you want to do with this.
> >>> A.K.
> >>> ________________________________
> >>> From: Joanna Zhang <[hidden email]<
> http://user/SendEmail.jtp?type=node&node=4657773&i=0>>
> >>>
> >>> To: arun <[hidden email]<
> http://user/SendEmail.jtp?type=node&node=4657773&i=1>>
> >>
> >>>
> >>> Sent: Wednesday, February 6, 2013 10:29 AM
> >>> Subject: Re: cumulative sum by group and under some criteria
> >>>
> >>>
> >>> Hi,
> >>>
> >>> Thanks! I need to do some calculations in the expended data, the
> expended
> >>> data would be very large, what is an efficient way, doing calculations
> >>> while expending the data, something similiar with the following, or
> >>> expending data using the code in your message and then add
> calculations in
> >>> the expended data?
> >>>
> >>> d3<-data.frame(d2)
> >>> for .......{
> >>> for {
> >>> for .... {
> >>> for .....{
> >>> p1<- x/m
> >>> p2<- y/n
> >>> ..........
> >>> }}
> >>> }}
> >>>
> >>> I also modified your code for expending data:
> >>> dnew<-expand.grid((m1+2):(maxN-(n1+2)),(n1+2):(maxN-m),0:m1,0:n1,
> >>> x1:(x1+m-m1),y1:(y1+n-n1))
> >>> names(dnew)<-c("m","n","x1","y1","x","y")
> >>> dnew
> >>> resF<-cbind(dnew[,c(2,1)],d2[rep(1:nrow(d2),nrow(dnew)),]) # this
> is
> >>> not correct, how to modify it.
> >>> resF
> >>> row.names(resF)<-1:nrow(resF)
> >>> resF
> >>>
> >>>
> >>>
> >>>
> >>> On Tue, Feb 5, 2013 at 2:46 PM, arun <[hidden email]<
> http://user/SendEmail.jtp?type=node&node=4657773&i=2>>
> >>
> >>> wrote:
> >>>
> >>> Hi,
> >>>
> >>> >
> >>> >You can reduce the steps to reach d2:
> >>> >res3<-
> >>> with(res2,aggregate(cbind(cterm1_P1L,cterm1_P0H),by=list(m1,n1),max))
> >>> >
> >>> >#Change it to:
> >>> >res3new<- aggregate(.~m1+n1,data=res2[,c(1:2,12:13)],max)
> >>> >res3new
> >>> > m1 n1 cterm1_P1L cterm1_P0H
> >>> >1 2 2 0.01440 0.00273750
> >>> >2 3 2 0.00032 0.00250000
> >>> >3 2 3 0.01952 0.00048125
> >>> >d2<-res3new[res3new[,3]<0.01 & res3new[,4]<0.01,]
> >>> >
> >>> > dnew<-expand.grid(4:10,5:10)
> >>> > names(dnew)<-c("n","m")
> >>> >resF<-cbind(dnew[,c(2,1)],d2[rep(1:nrow(d2),nrow(dnew)),])
> >>> >
> >>> >row.names(resF)<-1:nrow(resF)
> >>> > head(resF)
> >>> ># m n m1 n1 cterm1_P1L cterm1_P0H
> >>> >#1 5 4 3 2 0.00032 0.0025
> >>> >#2 5 5 3 2 0.00032 0.0025
> >>> >#3 5 6 3 2 0.00032 0.0025
> >>> >#4 5 7 3 2 0.00032 0.0025
> >>> >#5 5 8 3 2 0.00032 0.0025
> >>> >#6 5 9 3 2 0.00032 0.0025
> >>> >
> >>> >A.K.
> >>> >
> >>> >________________________________
> >>> >From: Joanna Zhang <[hidden email]<
> http://user/SendEmail.jtp?type=node&node=4657773&i=3>>
> >>>
> >>> >To: arun <[hidden email]<
> http://user/SendEmail.jtp?type=node&node=4657773&i=4>>
> >>
> >>>
> >>> >Sent: Tuesday, February 5, 2013 2:48 PM
> >>> >
> >>> >Subject: Re: cumulative sum by group and under some criteria
> >>> >
> >>> >
> >>> > Hi ,
> >>> >what I want is :
> >>> >m n m1 n1 cterm1_P1L cterm1_P0H
> >>> > 5 4 3 2 0.00032 0.00250000
> >>> > 5 5 3 2 0.00032 0.00250000
> >>> > 5 6 3 2 0.00032 0.00250000
> >>> > 5 7 3 2 0.00032 0.00250000
> >>> > 5 8 3 2 0.00032 0.00250000
> >>> > 5 9 3 2 0.00032 0.00250000
> >>> >5 10 3 2 0.00032 0.00250000
> >>> >6 4 3 2 0.00032 0.00250000
> >>> >6 5 3 2 0.00032 0.00250000
> >>> >6 6 3 2 0.00032 0.00250000
> >>> >6 7 3 2 0.00032 0.00250000
> >>> >.....
> >>> >6 10 3 2 0.00032 0.00250000
> >>> >
> >>> >
> >>> >
> >>> >On Tue, Feb 5, 2013 at 1:12 PM, arun <[hidden email]<
> http://user/SendEmail.jtp?type=node&node=4657773&i=5>>
> >>
> >>> wrote:
> >>> >
> >>> >Hi,
> >>> >>
> >>> >>Saw your message on Nabble.
> >>> >>
> >>> >>
> >>> >>"I want to add some more columns based on the results. Is the
> following
> >>> code good way to create such a data frame and How to see the column m
> and n
> >>> in the updated data?
> >>> >>
> >>> >>d2<- reres3[res3[,3]<0.01 & res3[,4]<0.01,]
> >>> >># should be a typo
> >>> >>
> >>> >>colnames(d2)[1:2]<- c("m1","n1");
> >>> >>d2 #already a data.frame
> >>> >>
> >>> >>d3<-data.frame(d2)
> >>> >> for (m in (m1+2):10){
> >>> >> for (n in (n1+2):10){
> >>> >> d3<-rbind(d3, c(d2))}}" #this is not making much sense to me.
> >>> Especially, you mentioned you wanted add more columns.
> >>> >>#Running this step gave error
> >>> >>#Error: object 'm1' not found
> >>> >>
> >>> >>Not sure what you want as output.
> >>> >>Could you show the ouput that is expected:
> >>> >>
> >>> >>A.K.
> >>> >>
> >>> >>
> >>> >>
> >>> >>
> >>> >>
> >>> >>
> >>> >>
> >>> >>
> >>> >>________________________________
> >>> >>From: Joanna Zhang <[hidden email]<
> http://user/SendEmail.jtp?type=node&node=4657773&i=6>>
> >>>
> >>> >>To: arun <[hidden email]<
> http://user/SendEmail.jtp?type=node&node=4657773&i=7>>
> >>
> >>>
> >>> >>Sent: Tuesday, February 5, 2013 10:23 AM
> >>> >>
> >>> >>Subject: Re: cumulative sum by group and under some criteria
> >>> >>
> >>> >>
> >>> >>Hi,
> >>> >>
> >>> >>Yes, I changed code. You answered the questions. But how can I put
> two
> >>> criteria in the code, if both the maximum value of cterm1_p1L <= 0.01
> and
> >>> cterm1_p1H <=0.01, the output the m1,n1.
> >>> >>
> >>> >>
> >>> >>
> >>> >>
> >>> >>On Tue, Feb 5, 2013 at 8:47 AM, arun <[hidden email]<
> http://user/SendEmail.jtp?type=node&node=4657773&i=8>>
> >>
> >>> wrote:
> >>> >>
> >>> >>
> >>> >>>
> >>> >>> HI,
> >>> >>>
> >>> >>>
> >>> >>>I am not getting the same results as yours: You must have changed
> the
> >>> dataset.
> >>> >>> res2[,1:2][res2$cterm1_P1L<0.6 & res2$cterm1_P0H<0.95,]
> >>> >>> m1 n1
> >>> >>>1 2 2
> >>> >>>2 2 2
> >>> >>>3 2 2
> >>> >>>4 2 2
> >>> >>>5 2 2
> >>> >>>6 2 2
> >>> >>>7 2 2
> >>> >>>8 2 2
> >>> >>>9 2 2
> >>> >>>10 3 2
> >>> >>>11 3 2
> >>> >>>12 3 2
> >>> >>>13 3 2
> >>> >>>14 3 2
> >>> >>>15 3 2
> >>> >>>16 3 2
> >>> >>>17 3 2
> >>> >>>18 3 2
> >>> >>>19 3 2
> >>> >>>20 3 2
> >>> >>>21 3 2
> >>> >>>22 2 3
> >>> >>>23 2 3
> >>> >>>24 2 3
> >>> >>>25 2 3
> >>> >>>26 2 3
> >>> >>>27 2 3
> >>> >>>28 2 3
> >>> >>>29 2 3
> >>> >>>30 2 3
> >>> >>>31 2 3
> >>> >>>32 2 3
> >>> >>>33 2 3
> >>> >>>
> >>> >>>
> >>> >>>Regarding the maximum value within each block, haven't I answered
> in
> >>> the earlier post.
> >>> >>>
> >>> >>>aggregate(cterm1_P1L~m1+n1,data=res2,max)
> >>> >>># m1 n1 cterm1_P1L
> >>> >>>#1 2 2 0.01440
> >>> >>>#2 3 2 0.00032
> >>> >>>#3 2 3 0.01952
> >>> >>>
> >>> >>>
> >>> >>>
> with(res2,aggregate(cbind(cterm1_P1L,cterm1_P0H),by=list(m1,n1),max))
> >>> >>># Group.1 Group.2 cterm1_P1L cterm1_P0H
> >>> >>>#1 2 2 0.01440 0.00273750
> >>> >>>#2 3 2 0.00032 0.00250000
> >>> >>>#3 2 3 0.01952 0.00048125
> >>> >>>
> >>> >>>
> >>> >>>A.K.
> >>> >>>
> >>> >>>
> >>> >>>----- Original Message -----
> >>
> >>> >>>From: "[hidden email]<
> http://user/SendEmail.jtp?type=node&node=4657773&i=9>";;;
> >>> <[hidden email] <http://user/SendEmail.jtp?type=node&node=4657773&i=10>>
>
> >>> >>>To: [hidden email]<
> http://user/SendEmail.jtp?type=node&node=4657773&i=11>
> >>> >>>Cc:
> >>> >>>
> >>> >>>Sent: Tuesday, February 5, 2013 9:33 AM
> >>> >>>Subject: Re: cumulative sum by group and under some criteria
> >>> >>>
> >>> >>>Hi,
> >>> >>>If use this
> >>> >>>
> >>> >>>res2[,1:2][res2$cterm1_P1L<0.6 & res2$cterm1_P0H<0.95,]
> >>> >>>
> >>> >>>the results are the following, but actually only m1=3, n1=2 sastify
> the
> >>> criteria, as I need to look at the row with maximum value within each
> >>> block,not every row.
> >>> >>>
> >>> >>>
> >>> >>> m1 n1
> >>> >>>1 2 2
> >>> >>>10 3 2
> >>> >>>11 3 2
> >>> >>>12 3 2
> >>> >>>13 3 2
> >>> >>>14 3 2
> >>> >>>15 3 2
> >>> >>>16 3 2
> >>> >>>17 3 2
> >>> >>>18 3 2
> >>> >>>19 3 2
> >>> >>>20 3 2
> >>> >>>21 3 2
> >>> >>>22 2 3
> >>> >>>23 2 3
> >>> >>>
> >>> >>>
> >>> >>><quote author='arun kirshna'>
> >>> >>>
> >>> >>>
> >>> >>>
> >>> >>>Hi,
> >>> >>>Thanks. This extract every row that satisfy the condition, but I
> need
> >>> look
> >>> >>>at the last row (the maximum of cumulative sum) for each block
> (m1,n1).
> >>> for
> >>> >>>example, if I set the criteria
> >>> >>>
> >>> >>>res2$cterm1_P1L<0.6 & res2$cterm1_P0H<0.95, this should extract m1=
> 3,
> >>> n1 =
> >>> >>>2.
> >>> >>>
> >>> >>>
> >>> >>>Hi,
> >>> >>>I am not sure I understand your question.
> >>> >>>res2$cterm1_P1L<0.6 & res2$cterm1_P0H<0.95
> >>> >>> #[1] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE
> TRUE
> >>> TRUE
> >>> >>>TRUE
> >>> >>>#[16] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE
> TRUE
> >>> TRUE
> >>> >>>TRUE
> >>> >>>#[31] TRUE TRUE TRUE
> >>> >>>
> >>> >>>This will extract all the rows.
> >>> >>>
> >>> >>>
> >>> >>>res2[,1:2][res2$cterm1_P1L<0.01 & res2$cterm1_P1L!=0,]
> >>> >>># m1 n1
> >>> >>>#21 3 2
> >>> >>>This extract only the row you wanted.
> >>> >>>
> >>> >>>For the different groups:
> >>> >>>
> >>> >>>aggregate(cterm1_P1L~m1+n1,data=res2,max)
> >>> >>># m1 n1 cterm1_P1L
> >>> >>>#1 2 2 0.01440
> >>> >>>#2 3 2 0.00032
> >>> >>>#3 2 3 0.01952
> >>> >>>
> >>> >>> aggregate(cterm1_P1L~m1+n1,data=res2,function(x) max(x)<0.01)
> >>> >>> # m1 n1 cterm1_P1L
> >>> >>>#1 2 2 FALSE
> >>> >>>#2 3 2 TRUE
> >>> >>>#3 2 3 FALSE
> >>> >>>
> >>> >>>res4<-aggregate(cterm1_P1L~m1+n1,data=res2,function(x) max(x)<0.01)
> >>> >>>res4[,1:2][res4[,3],]
> >>> >>># m1 n1
> >>> >>>#2 3 2
> >>> >>>
> >>> >>>A.K.
> >>> >>>
> >>> >>>
> >>> >>>
> >>> >>>
> >>> >>>----- Original Message -----
> >>
> >>> >>>From: "[hidden email]<
> http://user/SendEmail.jtp?type=node&node=4657773&i=12>";;;
> >>> <[hidden email] <http://user/SendEmail.jtp?type=node&node=4657773&i=13>>
>
> >>> >>>To: [hidden email]<
> http://user/SendEmail.jtp?type=node&node=4657773&i=14>
> >>> >>>Cc:
> >>> >>>Sent: Sunday, February 3, 2013 3:58 PM
> >>> >>>Subject: Re: cumulative sum by group and under some criteria
> >>> >>>
> >>> >>>Hi,
> >>> >>>Let me restate my questions. I need to get the m1 and n1 that
> satisfy
> >>> some
> >>> >>>criteria, for example in this case, within each group, the maximum
> >>> >>>cterm1_p1L ( the last row in this group) <0.01. I need to extract
> m1=3,
> >>> >>>n1=2, I only need m1, n1 in the row.
> >>> >>>
> >>> >>>Also, how to create the structure from the data.frame, I am new to
> R, I
> >>> need
> >>> >>>to change the maxN and run the loop to different data.
> >>> >>>Thanks very much for your help!
> >>> >>>
> >>> >>><quote author='arun kirshna'>
> >>> >>>HI,
> >>> >>>
> >>> >>>I think this should be more correct:
> >>> >>>maxN<-9
> >>> >>>c11<-0.2
> >>> >>>c12<-0.2
> >>> >>>p0L<-0.05
> >>> >>>p0H<-0.05
> >>> >>>p1L<-0.20
> >>> >>>p1H<-0.20
> >>> >>>
> >>> >>>d <- structure(list(m1 = c(2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2,
> >>> >>>2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3),
> >>> >>> n1 = c(2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3,
> >>> >>> 3, 3, 3, 3, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2), x1 = c(0,
> >>> >>> 0, 0, 1, 1, 1, 2, 2, 2, 0, 0, 0, 0, 1, 1, 1, 1, 2, 2, 2,
> >>> >>> 2, 0, 0, 0, 1, 1, 1, 2, 2, 2, 3, 3, 3), y1 = c(0, 1, 2, 0,
> >>> >>> 1, 2, 0, 1, 2, 0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3, 0, 1,
> >>> >>> 2, 0, 1, 2, 0, 1, 2, 0, 1, 2), Fmm = c(0, 0, 0, 0.7, 0.59,
> >>> >>> 0.64, 1, 1, 1, 0, 0, 0, 0, 0.63, 0.7, 0.74, 0.68, 1, 1, 1,
> >>> >>> 1, 0, 0, 0, 0.62, 0.63, 0.6, 0.63, 0.6, 0.68, 1, 1, 1), Fnn =
> c(0,
> >>> >>> 0.64, 1, 0, 0.51, 1, 0, 0.67, 1, 0, 0.62, 0.69, 1, 0, 0.54,
> >>> >>> 0.62, 1, 0, 0.63, 0.73, 1, 0, 0.63, 1, 0, 0.7, 1, 0, 0.7,
> >>> >>> 1, 0, 0.58, 1), Qm = c(1, 1, 1, 0.65, 0.45, 0.36, 0.5, 0.165,
> >>> >>> 0, 1, 1, 1, 1, 0.685, 0.38, 0.32, 0.32, 0.5, 0.185, 0.135,
> >>> >>> 0, 1, 1, 1, 0.69, 0.37, 0.4, 0.685, 0.4, 0.32, 0.5, 0.21,
> >>> >>> 0), Qn = c(1, 0.36, 0, 0.65, 0.45, 0, 0.5, 0.165, 0, 1, 0.38,
> >>> >>> 0.31, 0, 0.685, 0.38, 0.32, 0, 0.5, 0.185, 0.135, 0, 1, 0.37,
> >>> >>> 0, 0.69, 0.3, 0, 0.685, 0.3, 0, 0.5, 0.21, 0), term1_p0 =
> >>> c(0.81450625,
> >>> >>> 0.0857375, 0.00225625, 0.0857375, 0.009025, 0.0002375,
> 0.00225625,
> >>> >>> 0.0002375, 6.25e-06, 0.7737809375, 0.1221759375,
> >>> 0.00643031249999999,
> >>> >>> 0.0001128125, 0.081450625, 0.012860625, 0.000676875,
> 1.1875e-05,
> >>> >>> 0.0021434375, 0.0003384375, 1.78125e-05, 3.125e-07,
> 0.7737809375,
> >>> >>> 0.081450625, 0.0021434375, 0.1221759375, 0.012860625,
> >>> 0.0003384375,
> >>> >>> 0.00643031249999999, 0.000676875, 1.78125e-05, 0.0001128125,
> >>> >>> 1.1875e-05, 3.125e-07), term1_p1 = c(0.4096, 0.2048, 0.0256,
> >>> >>> 0.2048, 0.1024, 0.0128, 0.0256, 0.0128, 0.0016, 0.32768,
> >>> >>> 0.24576, 0.06144, 0.00512, 0.16384, 0.12288, 0.03072, 0.00256,
> >>> >>> 0.02048, 0.01536, 0.00384, 0.00032, 0.32768, 0.16384, 0.02048,
> >>> >>> 0.24576, 0.12288, 0.01536, 0.06144, 0.03072, 0.00384, 0.00512,
> >>> >>> 0.00256, 0.00032)), .Names = c("m1", "n1", "x1", "y1", "Fmm",
> >>> >>>"Fnn", "Qm", "Qn", "term1_p0", "term1_p1"), row.names = c(NA,
> >>> >>>33L), class = "data.frame")
> >>> >>>
> >>> >>>library(zoo)
> >>> >>>lst1<- split(d,list(d$m1,d$n1))
> >>> >>>res2<-do.call(rbind,lapply(lst1[lapply(lst1,nrow)!=0],function(x){
> >>> >>>x[,11:14]<-NA;
> >>> >>>x[,11:12][x$Qm<=c11,]<-cumsum(x[,9:10][x$Qm<=c11,]);
> >>> >>>x[,13:14][x$Qn<=c12,]<-cumsum(x[,9:10][x$Qn<=c12,]);
> >>> >>>colnames(x)[11:14]<-
> >>> c("cterm1_P0L","cterm1_P1L","cterm1_P0H","cterm1_P1H");
> >>> >>>x1<-na.locf(x);
> >>> >>>x1[,11:14][is.na(x1[,11:14])]<-0;
> >>> >>>x1}))
> >>> >>>row.names(res2)<- 1:nrow(res2)
> >>> >>>
> >>> >>> res2
> >>> >>> # m1 n1 x1 y1 Fmm Fnn Qm Qn term1_p0 term1_p1
> >>> cterm1_P0L
> >>> >>>cterm1_P1L cterm1_P0H cterm1_P1H
> >>> >>>
> >>> >>>#1 2 2 0 0 0.00 0.00 1.000 1.000 0.8145062500 0.40960
> >>> 0.0000000000
> >>> >>> 0.00000 0.0000000000 0.00000
> >>> >>>#2 2 2 0 1 0.00 0.64 1.000 0.360 0.0857375000 0.20480
> >>> 0.0000000000
> >>> >>> 0.00000 0.0000000000 0.00000
> >>> >>>#3 2 2 0 2 0.00 1.00 1.000 0.000 0.0022562500 0.02560
> >>> 0.0000000000
> >>> >>> 0.00000 0.0022562500 0.02560
> >>> >>>#4 2 2 1 0 0.70 0.00 0.650 0.650 0.0857375000 0.20480
> >>> 0.0000000000
> >>> >>> 0.00000 0.0022562500 0.02560
> >>> >>>#5 2 2 1 1 0.59 0.51 0.450 0.450 0.0090250000 0.10240
> >>> 0.0000000000
> >>> >>> 0.00000 0.0022562500 0.02560
> >>> >>>#6 2 2 1 2 0.64 1.00 0.360 0.000 0.0002375000 0.01280
> >>> 0.0000000000
> >>> >>> 0.00000 0.0024937500 0.03840
> >>> >>>#7 2 2 2 0 1.00 0.00 0.500 0.500 0.0022562500 0.02560
> >>> 0.0000000000
> >>> >>> 0.00000 0.0024937500 0.03840
> >>> >>>#8 2 2 2 1 1.00 0.67 0.165 0.165 0.0002375000 0.01280
> >>> 0.0002375000
> >>> >>> 0.01280 0.0027312500 0.05120
> >>> >>>#9 2 2 2 2 1.00 1.00 0.000 0.000 0.0000062500 0.00160
> >>> 0.0002437500
> >>> >>> 0.01440 0.0027375000 0.05280
> >>> >>>#10 3 2 0 0 0.00 0.00 1.000 1.000 0.7737809375 0.32768
> >>> 0.0000000000
> >>> >>> 0.00000 0.0000000000 0.00000
> >>> >>>#11 3 2 0 1 0.00 0.63 1.000 0.370 0.0814506250 0.16384
> >>> 0.0000000000
> >>> >>> 0.00000 0.0000000000 0.00000
> >>> >>>#12 3 2 0 2 0.00 1.00 1.000 0.000 0.0021434375 0.02048
> >>> 0.0000000000
> >>> >>> 0.00000 0.0021434375 0.02048
> >>> >>>#13 3 2 1 0 0.62 0.00 0.690 0.690 0.1221759375 0.24576
> >>> 0.0000000000
> >>> >>> 0.00000 0.0021434375 0.02048
> >>> >>>#14 3 2 1 1 0.63 0.70 0.370 0.300 0.0128606250 0.12288
> >>> 0.0000000000
> >>> >>> 0.00000 0.0021434375 0.02048
> >>> >>>#15 3 2 1 2 0.60 1.00 0.400 0.000 0.0003384375 0.01536
> >>> 0.0000000000
> >>> >>> 0.00000 0.0024818750 0.03584
> >>> >>>#16 3 2 2 0 0.63 0.00 0.685 0.685 0.0064303125 0.06144
> >>> 0.0000000000
> >>> >>> 0.00000 0.0024818750 0.03584
> >>> >>>#17 3 2 2 1 0.60 0.70 0.400 0.300 0.0006768750 0.03072
> >>> 0.0000000000
> >>> >>> 0.00000 0.0024818750 0.03584
> >>> >>>#18 3 2 2 2 0.68 1.00 0.320 0.000 0.0000178125 0.00384
> >>> 0.0000000000
> >>> >>> 0.00000 0.0024996875 0.03968
> >>> >>>#19 3 2 3 0 1.00 0.00 0.500 0.500 0.0001128125 0.00512
> >>> 0.0000000000
> >>> >>> 0.00000 0.0024996875 0.03968
> >>> >>>#20 3 2 3 1 1.00 0.58 0.210 0.210 0.0000118750 0.00256
> >>> 0.0000000000
> >>> >>> 0.00000 0.0024996875 0.03968
> >>> >>>#21 3 2 3 2 1.00 1.00 0.000 0.000 0.0000003125 0.00032
> >>> 0.0000003125
> >>> >>> 0.00032 0.0025000000 0.04000
> >>> >>>#22 2 3 0 0 0.00 0.00 1.000 1.000 0.7737809375 0.32768
> >>> 0.0000000000
> >>> >>> 0.00000 0.0000000000 0.00000
> >>> >>>#23 2 3 0 1 0.00 0.62 1.000 0.380 0.1221759375 0.24576
> >>> 0.0000000000
> >>> >>> 0.00000 0.0000000000 0.00000
> >>> >>>#24 2 3 0 2 0.00 0.69 1.000 0.310 0.0064303125 0.06144
> >>> 0.0000000000
> >>> >>> 0.00000 0.0000000000 0.00000
> >>> >>>#25 2 3 0 3 0.00 1.00 1.000 0.000 0.0001128125 0.00512
> >>> 0.0000000000
> >>> >>> 0.00000 0.0001128125 0.00512
> >>> >>>#26 2 3 1 0 0.63 0.00 0.685 0.685 0.0814506250 0.16384
> >>> 0.0000000000
> >>> >>> 0.00000 0.0001128125 0.00512
> >>> >>>#27 2 3 1 1 0.70 0.54 0.380 0.380 0.0128606250 0.12288
> >>> 0.0000000000
> >>> >>> 0.00000 0.0001128125 0.00512
> >>> >>>#28 2 3 1 2 0.74 0.62 0.320 0.320 0.0006768750 0.03072
> >>> 0.0000000000
> >>> >>> 0.00000 0.0001128125 0.00512
> >>> >>>#29 2 3 1 3 0.68 1.00 0.320 0.000 0.0000118750 0.00256
> >>> 0.0000000000
> >>> >>> 0.00000 0.0001246875 0.00768
> >>> >>>#30 2 3 2 0 1.00 0.00 0.500 0.500 0.0021434375 0.02048
> >>> 0.0000000000
> >>> >>> 0.00000 0.0001246875 0.00768
> >>> >>>#31 2 3 2 1 1.00 0.63 0.185 0.185 0.0003384375 0.01536
> >>> 0.0003384375
> >>> >>> 0.01536 0.0004631250 0.02304
> >>> >>>#32 2 3 2 2 1.00 0.73 0.135 0.135 0.0000178125 0.00384
> >>> 0.0003562500
> >>> >>> 0.01920 0.0004809375 0.02688
> >>> >>>#33 2 3 2 3 1.00 1.00 0.000 0.000 0.0000003125 0.00032
> >>> 0.0003565625
> >>> >>> 0.01952 0.0004812500 0.02720
> >>> >>>
> >>> >>>#Sorry, some values in my previous solution didn't look right. I
> >>> didn't
> >>> >>>A.K.
> >>> >>>
> >>> >>>
> >>> >>>
> >>> >>>
> >>> >>>
> >>> >>>----- Original Message -----
> >>> >>>From: Zjoanna <[hidden email]<
> http://user/SendEmail.jtp?type=node&node=4657773&i=15>>
> >>>
> >>> >>>To: [hidden email]<
> http://user/SendEmail.jtp?type=node&node=4657773&i=16>
> >>
> >>> >>>Cc:
> >>> >>>Sent: Friday, February 1, 2013 12:19 PM
> >>> >>>Subject: Re: [R] cumulative sum by group and under some criteria
> >>> >>>
> >>> >>>Thank you very much for your reply. Your code work well with this
> >>> example.
> >>> >>>I modified a little to fit my real data, I got an error massage.
> >>> >>>
> >>> >>>Error in split.default(x = seq_len(nrow(x)), f = f, drop = drop,
> ...) :
> >>> >>> Group length is 0 but data length > 0
> >>> >>>
> >>> >>>
> >>> >>>On Thu, Jan 31, 2013 at 12:21 PM, arun kirshna [via R] <
> >>> >>>[hidden email] <
> http://user/SendEmail.jtp?type=node&node=4657773&i=17>>
> >>
> >>> wrote:
> >>> >>>
> >>> >>>> Hi,
> >>> >>>> Try this:
> >>> >>>> colnames(d)<-c("m1","n1","x1","y1","p11","p12")
> >>> >>>> library(zoo)
> >>> >>>> res1<-
> >>> do.call(rbind,lapply(lapply(split(d,list(d$m1,d$n1)),function(x)
> >>> >>>> {x$cp11[x$x1>1]<- cumsum(x$p11[x$x1>1]);x$cp12[x$y1>1]<-
> >>> >>>> cumsum(x$p12[x$y1>1]);x}),function(x)
> >>> >>>> {x$cp11<-na.locf(x$cp11,na.rm=F);x$cp12<-
> >>> na.locf(x$cp12,na.rm=F);x}))
> >>> >>>> #there would be a warning here as one of the list element is
> NULL.
> >>> The,
> >>> >>>> warning is okay
> >>> >>>> row.names(res1)<- 1:nrow(res1)
> >>> >>>> res1[,7:8][is.na(res1[,7:8])]<- 0
> >>> >>>> res1
> >>> >>>> # m1 n1 x1 y1 p11 p12 cp11 cp12
> >>> >>>> #1 2 2 0 0 0.00 0.00 0.00 0.00
> >>> >>>> #2 2 2 0 1 0.00 0.50 0.00 0.00
> >>> >>>> #3 2 2 0 2 0.00 1.00 0.00 1.00
> >>> >>>> #4 2 2 1 0 0.50 0.00 0.00 1.00
> >>> >>>> #5 2 2 1 1 0.50 0.50 0.00 1.00
> >>> >>>> #6 2 2 1 2 0.50 1.00 0.00 2.00
> >>> >>>> #7 2 2 2 0 1.00 0.00 1.00 2.00
> >>> >>>> #8 2 2 2 1 1.00 0.50 2.00 2.00
> >>> >>>> #9 2 2 2 2 1.00 1.00 3.00 3.00
> >>> >>>> #10 3 2 0 0 0.00 0.00 0.00 0.00
> >>> >>>> #11 3 2 0 1 0.00 0.50 0.00 0.00
> >>> >>>> #12 3 2 0 2 0.00 1.00 0.00 1.00
> >>> >>>> #13 3 2 1 0 0.33 0.00 0.00 1.00
> >>> >>>> #14 3 2 1 1 0.33 0.50 0.00 1.00
> >>> >>>> #15 3 2 1 2 0.33 1.00 0.00 2.00
> >>> >>>> #16 3 2 2 0 0.67 0.00 0.67 2.00
> >>> >>>> #17 3 2 2 1 0.67 0.50 1.34 2.00
> >>> >>>> #18 3 2 2 2 0.67 1.00 2.01 3.00
> >>> >>>> #19 3 2 3 0 1.00 0.00 3.01 3.00
> >>> >>>> #20 3 2 3 1 1.00 0.50 4.01 3.00
> >>> >>>> #21 3 2 3 2 1.00 1.00 5.01 4.00
> >>> >>>> #22 2 3 0 0 0.00 0.00 0.00 0.00
> >>> >>>> #23 2 3 0 1 0.00 0.33 0.00 0.00
> >>> >>>> #24 2 3 0 2 0.00 0.67 0.00 0.67
> >>> >>>> #25 2 3 0 3 0.00 1.00 0.00 1.67
> >>> >>>> #26 2 3 1 0 0.50 0.00 0.00 1.67
> >>> >>>> #27 2 3 1 1 0.50 0.33 0.00 1.67
> >>> >>>> #28 2 3 1 2 0.50 0.67 0.00 2.34
> >>> >>>> #29 2 3 1 3 0.50 1.00 0.00 3.34
> >>> >>>> #30 2 3 2 0 1.00 0.00 1.00 3.34
> >>> >>>> #31 2 3 2 1 1.00 0.33 2.00 3.34
> >>> >>>> #32 2 3 2 2 1.00 0.67 3.00 4.01
> >>> >>>> #33 2 3 2 3 1.00 1.00 4.00 5.01
> >>> >>>> A.K.
> >>> >>>>
> >>> >>>> ------------------------------
> >>> >>>> If you reply to this email, your message will be added to the
> >>> discussion
> >>> >>>> below:
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> >>>
> >>> >>>>
> >>> >>>
> >>> >>>
> >>> >>>
> >>> >>>
> >>> >>>--
> >>> >>>View this message in context:
> >>> >>>
> >>>
> http://r.789695.n4.nabble.com/cumulative-sum-by-group-and-under-some-criteria-tp4657074p4657315.html
> >>> >>>Sent from the R help mailing list archive at Nabble.com.
> >>> >>> [[alternative HTML version deleted]]
> >>> >>>
> >>> >>>______________________________________________
> >>> >>>[hidden email] <
> http://user/SendEmail.jtp?type=node&node=4657773&i=18>mailing list
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> >>> >>>and provide commented, minimal, self-contained, reproducible code.
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> >>> >>>
> >>> >>>______________________________________________
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> >>> >>>Quoted from:
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> >>> >>>______________________________________________
> >>> >>>[hidden email] <
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> >>> >>>https://stat.ethz.ch/mailman/listinfo/r-help
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> >>> ______________________________________________
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