Hi,
d3<-structure(list(m1 = c(2, 3, 2), n1 = c(2, 2, 3), cterm1_P0L = c(0.9025,
0.857375, 0.9025), cterm1_P1L = c(0.64, 0.512, 0.64), cterm1_P0H = c(0.9025,
0.9025, 0.857375), cterm1_P1H = c(0.64, 0.64, 0.512)), .Names = c("m1",
"n1", "cterm1_P0L", "cterm1_P1L", "cterm1_P0H", "cterm1_P1H"), row.names = c(NA,
3L), class = "data.frame")
d2<- data.frame()
for (m1 in 2:3) {
for (n1 in 2:3) {
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)){
for(y in y1:(y1+n-n1)){
d2<- rbind(d2,c(m1,n1,x1,y1,m,n,x,y))
}}}}}}}}
colnames(d2)<-c("m1","n1","x1","y1","m","n","x","y")
#or
res1<-do.call(rbind,lapply(unique(d3$m1),function(m1)
do.call(rbind,lapply(unique(d3$n1),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)
do.call(rbind,lapply(y1:(y1+n-n1), function(y)
expand.grid(m1,n1,x1,y1,m,n,x,y)) )))))))))))))))
names(res1)<- c("m1","n1","x1","y1","m","n","x","y")
attr(res1,"out.attrs")<-NULL
res1[]<- sapply(res1,as.integer)
library(plyr)
res2<- join(res1,d3,by=c("m1","n1"),type="inner")
#Assuming that these are the values you used:
p0L<-0.05
p0H<-0.05
p1L<-0.20
p1H<-0.20
res2<- within(res2,{p1<- x/m; p2<- y/n;term2_p0<-dbinom(x1,m1, p0L, log=FALSE)*
dbinom(y1,n1,p0H, log=FALSE)*dbinom(x-x1,m-m1, p0L, log=FALSE)*
dbinom(y-y1,n-n1,p0H, log=FALSE);term2_p1<- dbinom(x1,m1, p1L, log=FALSE)*
dbinom(y1,n1,p1H, log=FALSE)*dbinom(x-x1,m-m1, p1L, log=FALSE)*
dbinom(y-y1,n-n1,p1H, log=FALSE);Pm2<-rbeta(240, 0.2+x,
0.8+m-x);Pn2<-rbeta(240, 0.2+y, 0.8+n-y)})
Fm2<- ecdf(res2$Pm2)
Fn2<- ecdf(res2$Pn2)
res3<- within(res2,{Fmm2<-Fm2(p1);Fnn2<- Fn2(p2);R2<- (Fmm2+Fnn2)/2}) #not sure
about this step especially the Fm2() or Fn2()
res3$Fmm_f2<-apply(res3[,c("R2","Fmm2")],1,min)
res3$Fnn_f2<-apply(res3[,c("R2","Fnn2")],1,max)
res3<- within(res3,{Qm2<- 1-Fmm_f2;Qn2<- 1-Fnn_f2})
head(res3)
# m1 n1 x1 y1 m n x y cterm1_P0L cterm1_P1L cterm1_P0H cterm1_P1H Pn2
#1 2 2 0 0 4 4 0 0 0.9025 0.64 0.9025 0.64 0.001084648
#2 2 2 0 0 4 4 0 1 0.9025 0.64 0.9025 0.64 0.504593909
#3 2 2 0 0 4 4 0 2 0.9025 0.64 0.9025 0.64 0.541379357
#4 2 2 0 0 4 4 1 0 0.9025 0.64 0.9025 0.64 0.138947785
#5 2 2 0 0 4 4 1 1 0.9025 0.64 0.9025 0.64 0.272364957
#6 2 2 0 0 4 4 1 2 0.9025 0.64 0.9025 0.64 0.761635059
# Pm2 term2_p1 term2_p0 p2 p1 R2 Fnn2 Fmm2
#1 1.212348e-05 0.16777216 0.6634204313 0.00 0.00 0.0000000 0.0000000 0.0
#2 1.007697e-03 0.08388608 0.0698337296 0.25 0.00 0.1791667 0.3583333 0.0
#3 1.106946e-05 0.01048576 0.0018377297 0.50 0.00 0.3479167 0.6958333 0.0
# 2.086758e-01 0.08388608 0.0698337296 0.00 0.25 0.2000000 0.0000000 0.4
#5 2.382179e-01 0.04194304 0.0073509189 0.25 0.25 0.3791667 0.3583333 0.4
#6 4.494673e-01 0.00524288 0.0001934452 0.50 0.25 0.5479167 0.6958333 0.4
# Fmm_f2 Fnn_f2 Qn2 Qm2
#1 0.0000000 0.0000000 1.0000000 1.0000000
#2 0.0000000 0.3583333 0.6416667 1.0000000
#3 0.0000000 0.6958333 0.3041667 1.0000000
#4 0.2000000 0.2000000 0.8000000 0.8000000
#5 0.3791667 0.3791667 0.6208333 0.6208333
#6 0.4000000 0.6958333 0.3041667 0.6000000
A.K.
________________________________
From: Joanna Zhang <zjoanna2...@gmail.com>
To: arun <smartpink...@yahoo.com>
Sent: Friday, February 22, 2013 11:02 AM
Subject: Re: [R] cumulative sum by group and under some criteria
Thanks! Then I need to create new variables based on the res2. I can't find
Fmm_f1, Fnn_f2, R2, Qm2, Qn2 until running the code several times and the
values of Fnn_f2, Fmm_f2 are correct.
attach(res2)
res2$p1<-x/m
res2$p2<-y/n
res2$term2_p0 <- dbinom(x1,m1, p0L, log=FALSE)* dbinom(y1,n1,p0H,
log=FALSE)*dbinom(x-x1,m-m1, p0L, log=FALSE)* dbinom(y-y1,n-n1,p0H, log=FALSE)
res2$term2_p1 <- dbinom(x1,m1, p1L, log=FALSE)* dbinom(y1,n1,p1H,
log=FALSE)*dbinom(x-x1,m-m1, p1L, log=FALSE)* dbinom(y-y1,n-n1,p1H, log=FALSE)
Pm2<-rbeta(1000, 0.2+x, 0.8+m-x)
Fm2<-ecdf(Pm2)
res2$Fmm2<-Fm2(x/m) #not correct, it comes out after running code two times
Pn2<-rbeta(1000, 0.2+y, 0.8+n-y)
Fn2<-ecdf(Pn2)
res2$Fnn2<-Fn2(y/n)
res2$R2<-(Fmm2+Fnn2)/2
res2$Fmm_f2<-min(R2,Fmm2) # not correct
res2$Fnn_f2<-max(R2,Fnn2)
res2$Qm2<-(1-Fmm_f2)
res2$Qn2<-(1-Fnn_f2)
detach(res2)
res2
head(res2)
On Tue, Feb 19, 2013 at 4:09 PM, arun <smartpink...@yahoo.com> wrote:
Hi,
>
>""suppose that I have a dataset 'd'
> m1 n1 A B C D
>1 2 2 0.902500 0.640 0.9025 0.64
>2 3 2 0.857375 0.512 0.9025 0.64
>I want to add x1 (from 0 to m1), y1(from 0 to n1), m (range from
>m1+2 to 7-n1), n(from n1+2 to 9-m), x (x1 to x1+m-m1), y(y1 to y1+n-n1),
>expanding to another dataset 'd2' based on each row (combination of m1
>and n1)""
>
>
>Try:
>
>
> d<-read.table(text="
>
>m1 n1 A B C D
>1 2 2 0.902500 0.640 0.9025 0.64
>2 3 2 0.857375 0.512 0.9025 0.64
>",sep="",header=TRUE)
>
>vec1<- paste(d[,1],d[,2],d[,3],d[,4],d[,5],d[,6])
>res1<- do.call(rbind,lapply(vec1,function(m1)
>do.call(rbind,lapply(0:(as.numeric(substr(m1,1,1))),function(x1)
>do.call(rbind,lapply(0:(as.numeric(substr(m1,3,3))),function(y1)
>do.call(rbind,lapply((as.numeric(substr(m1,1,1))+2):(7-as.numeric(substr(m1,3,3))),function(m)
> do.call(rbind,lapply((as.numeric(substr(m1,3,3))+2):(9-m),function(n)
>
> do.call(rbind,lapply(x1:(x1+m-as.numeric(substr(m1,1,1))), function(x)
> do.call(rbind,lapply(y1:(y1+n-as.numeric(substr(m1,3,3))), function(y)
> expand.grid(m1,x1,y1,m,n,x,y)) )))))))))))))
>
names(res1)<- c("group","x1","y1","m","n","x","y")
> res1$m1<- NA; res1$n1<- NA; res1$A<- NA; res1$B<- NA; res1$C<- NA;res1$D <- NA
>res1[,8:13]<-do.call(rbind,lapply(strsplit(as.character(res1$group),"
>"),as.numeric))
>res2<- res1[,c(8:9,2:7,10:13)]
>
>
> head(res2)
># m1 n1 x1 y1 m n x y A B C D
>#1 2 2 0 0 4 4 0 0 0.9025 0.64 0.9025 0.64
>#2 2 2 0 0 4 4 0 1 0.9025 0.64 0.9025 0.64
>#3 2 2 0 0 4 4 0 2 0.9025 0.64 0.9025 0.64
>#4 2 2 0 0 4 4 1 0 0.9025 0.64 0.9025 0.64
>#5 2 2 0 0 4 4 1 1 0.9025 0.64 0.9025 0.64
>#6 2 2 0 0 4 4 1 2 0.9025 0.64 0.9025 0.64
>
>
>
>
>
>
>________________________________
>From: Joanna Zhang <zjoanna2...@gmail.com>
>To: arun <smartpink...@yahoo.com>
>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 <smartpink...@yahoo.com> 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 <zjoanna2...@gmail.com>
>>To: arun <smartpink...@yahoo.com>
>>
>>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 <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
>>>> discussion
>>>> >>>> below:
>>>> >>>>
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