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] <
[email protected]> 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.
> >>>>
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> >>>
> >>>
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