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. > > ------------------------------ > If you reply to this email, your message will be added to the discussion > below: > > http://r.789695.n4.nabble.com/cumulative-sum-by-group-and-under-some-criteria-tp4657074p4657196.html > To unsubscribe from cumulative sum by group and under some criteria, click > here<http://r.789695.n4.nabble.com/template/NamlServlet.jtp?macro=unsubscribe_by_code&node=4657074&code=WmpvYW5uYTIwMTNAZ21haWwuY29tfDQ2NTcwNzR8LTE3NTE1MDA0MzY=> > . > NAML<http://r.789695.n4.nabble.com/template/NamlServlet.jtp?macro=macro_viewer&id=instant_html%21nabble%3Aemail.naml&base=nabble.naml.namespaces.BasicNamespace-nabble.view.web.template.NabbleNamespace-nabble.view.web.template.NodeNamespace&breadcrumbs=notify_subscribers%21nabble%3Aemail.naml-instant_emails%21nabble%3Aemail.naml-send_instant_email%21nabble%3Aemail.naml> > -- View this message in context: http://r.789695.n4.nabble.com/cumulative-sum-by-group-and-under-some-criteria-tp4657074p4657315.html Sent from the R help mailing list archive at Nabble.com. 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