d2<- data.frame()
for (m1 in 2:3) {
for (n1 in 2:2) {
for (x1 in 0:(m1-1)) {
for (y1 in 0:(n1-1)) {
for (m in (m1+2): (7-n1)){
for (n in (n1+2):(9-m)){
for (x in x1:(x1+m-m1)){
d2<- rbind(d2,c(m1,n1,x1,y1,m,n,x))
}}}}}}}
colnames(d2)<-c("m1","n1","x1","y1","m","n","x")
res<-do.call(rbind,lapply(2:3,function(m1)
do.call(rbind,lapply(2:2,function(n1)
do.call(rbind,lapply(0:(m1-1),function(x1)
do.call(rbind,lapply(0:(n1-1),function(y1)
do.call(rbind,lapply((m1+2):(7-n1),function(m)
do.call(rbind,lapply((n1+2):(9-m),function(n)
do.call(rbind,lapply(x1:(x1+m-m1), function(x) expand.grid(m1,n1,x1,y1,m,n,x))
)))))))))))))
names(res)<- c("m1","n1","x1","y1","m","n","x")
attr(res,"out.attrs")<-NULL
identical(d2,res)
#[1] TRUE
A.K.
________________________________
From: Joanna Zhang <zjoanna2...@gmail.com>
To: arun <smartpink...@yahoo.com>
Sent: Saturday, February 16, 2013 8:46 PM
Subject: Re: [R] cumulative sum by group and under some criteria
Hi,
What I need is to expand each row by adding several columns and . Let me
restate the question. I have a dataset d, I want to expand it to d2 showed
below.
d<-data.frame()
for (m1 in 2:3) {
for (n1 in 2:2) {
for (x1 in 0:(m1-1)) {
for (y1 in 0:(n1-1)) {
d<-rbind(d,c(m1,n1,x1,y1))
}
}
}}
colnames(d)<-c("m1","n1","x1","y1")
d
m1 n1 x1 y1
1 2 2 0 0
2 2 2 0 1
3 2 2 1 0
4 2 2 1 1
5 3 2 0 0
6 3 2 0 1
7 3 2 1 0
8 3 2 1 1
9 3 2 2 0
10 3 2 2 1
I want to expand it as follows:
for (m in (m1+2): (7-n1){
for (n in (n1+2):(9-m){
for (x in x1:(x1+m-m1){
}}}
so for the first row,
m1 n1 x1 y1
1 2 2 0 0
it should be expanded as
m1 n1 x1 y1 m n x
2 2 0 0 4 4 0
2 2 0 0 4 4 1
2 2 0 0 4 4 2
2 2 0 0 4 5 0
2 2 0 0 4 5 1
2 2 0 0 4 5 2
On Tue, Feb 12, 2013 at 8:19 PM, arun <smartpink...@yahoo.com> wrote:
Hi,
>
>Saw your reply again in Nabble. I thought I sent you the solution previously.
>
>
>res3new<- aggregate(.~m1+n1,data=res2[,c(1:2,12:13)],max)
>
> d2<-res3new[res3new[,3]<0.01 & res3new[,4]<0.01,]
>
>
>m1<- 3 #from d2
>maxN<- 9
>n1<- 2 #from d2
>
>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 <zjoanna2...@gmail.com>
>To: r-help@r-project.org
>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] <
>ml-node+s789695n4657773...@n4.nabble.com> 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
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>> >>>
>> >>>
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