Thanks very much for your help!
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,]
colnames(d2)[1:2]<- c("m1","n1");
d2
d3<-data.frame(d2)
for (m in (m1+2):10){
for (n in (n1+2):10){
d3<-rbind(d3, c(d2))}}
On Tue, Feb 5, 2013 at 9:32 AM, arun kirshna [via R] <
ml-node+s789695n4657614...@n4.nabble.com> wrote:
>
> Hi,
> res3<-with(res2,aggregate(cbind(cterm1_P1L,cterm1_P0H),by=list(m1,n1),max))
>
>
> res3[res3[,3]<0.6 & res3[,4]<0.95,] #this doesn't change the result as
> the conditions are not met
> # 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
>
> res3[res3[,3]<0.01 & res3[,4]<0.01,]
> # Group.1 Group.2 cterm1_P1L cterm1_P0H
> #2 3 2 0.00032 0.0025
> Hope it helps.
> A.K.
> ________________________________
> From: Joanna Zhang <[hidden
> email]<http://user/SendEmail.jtp?type=node&node=4657614&i=0>>
>
> To: arun <[hidden
> email]<http://user/SendEmail.jtp?type=node&node=4657614&i=1>>
>
> Sent: Tuesday, February 5, 2013 9:43 AM
> Subject: Re: cumulative sum by group and under some criteria
>
>
> Yes, it did answer my question, max(x) look at the maximum value. However,
> how to put both criteria in the code, like both the maximum value of
> cterm1_p1L <0.6 and the maximum value of cterm1_p0L<0.95. Thanks a lot!
>
>
> On Mon, Feb 4, 2013 at 4:26 PM, arun <[hidden
> email]<http://user/SendEmail.jtp?type=node&node=4657614&i=2>>
> wrote:
>
>
> >
> >Hi,
> >
> >Did this answered your question?
> >
> >A.K.
> >
> >
> >----- Original Message -----
> >
> >From: arun <[hidden
> >email]<http://user/SendEmail.jtp?type=node&node=4657614&i=3>>
>
> >To: "[hidden email]<http://user/SendEmail.jtp?type=node&node=4657614&i=4>"
> <[hidden email] <http://user/SendEmail.jtp?type=node&node=4657614&i=5>>
> >Cc: R help <[hidden
> >email]<http://user/SendEmail.jtp?type=node&node=4657614&i=6>>
>
> >Sent: Monday, February 4, 2013 4:44 PM
> >Subject: Re: cumulative sum by group and under some criteria
> >
> >
> >
> >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: "[hidden email]<http://user/SendEmail.jtp?type=node&node=4657614&i=7>"
> <[hidden email] <http://user/SendEmail.jtp?type=node&node=4657614&i=8>>
> >To: [hidden email] <http://user/SendEmail.jtp?type=node&node=4657614&i=9>
> >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 <[hidden
> >email]<http://user/SendEmail.jtp?type=node&node=4657614&i=10>>
>
> >To: [hidden email]<http://user/SendEmail.jtp?type=node&node=4657614&i=11>
> >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] <
> >[hidden email] <http://user/SendEmail.jtp?type=node&node=4657614&i=12>>
> 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
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> >>
> >
> >
> >
> >
> >--
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