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 <zjoanna2...@gmail.com>
To: arun <smartpink...@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 <smartpink...@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 <zjoanna2...@gmail.com>
>To: arun <smartpink...@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 <smartpink...@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: "zjoanna2...@gmail.com" <zjoanna2...@gmail.com>
>>To: smartpink...@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: "zjoanna2...@gmail.com" <zjoanna2...@gmail.com>
>>To: smartpink...@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 <zjoanna2...@gmail.com>
>>To: r-help@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+s789695n4657196...@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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>>
>>
>>--
>>View this message in context:
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>>Sent from the R help mailing list archive at Nabble.com.
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>>______________________________________________
>>R-help@r-project.org mailing list
>>https://stat.ethz.ch/mailman/listinfo/r-help
>>PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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>>______________________________________________
>>R-help@r-project.org mailing list
>>https://stat.ethz.ch/mailman/listinfo/r-help
>>PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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>></quote>
>>Quoted from:
>>http://r.789695.n4.nabble.com/cumulative-sum-by-group-and-under-some-criteria-tp4657074p4657360.html
>>
>>
>>______________________________________________
>>R-help@r-project.org mailing list
>>https://stat.ethz.ch/mailman/listinfo/r-help
>>PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
>>and provide commented, minimal, self-contained, reproducible code.
>>
>></quote>
>>Quoted from:
>>http://r.789695.n4.nabble.com/cumulative-sum-by-group-and-under-some-criteria-tp4657074p4657582.html
>>
>>
>
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
