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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