# [R] cumulative sum by group and under some criteria

arun smartpink111 at yahoo.com
Tue Feb 5 21:46:44 CET 2013

```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)
#  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,
>
>
>
>"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");
>
>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
>
>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
>>
>>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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