[R] cumulative sum by group and under some criteria
arun
smartpink111 at yahoo.com
Wed Feb 6 20:15:27 CET 2013
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 <zjoanna2013 at gmail.com>
To: arun <smartpink111 at yahoo.com>
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 <smartpink111 at yahoo.com> 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 <zjoanna2013 at gmail.com>
>To: arun <smartpink111 at yahoo.com>
>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 <smartpink111 at yahoo.com> 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 <zjoanna2013 at gmail.com>
>>To: arun <smartpink111 at yahoo.com>
>>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 <smartpink111 at yahoo.com> 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: "Zjoanna2013 at gmail.com" <Zjoanna2013 at gmail.com>
>>>To: smartpink111 at yahoo.com
>>>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: "Zjoanna2013 at gmail.com" <Zjoanna2013 at gmail.com>
>>>To: smartpink111 at yahoo.com
>>>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 <Zjoanna2013 at gmail.com>
>>>To: r-help at r-project.org
>>>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] <
>>>ml-node+s789695n4657196h87 at n4.nabble.com> 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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