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

arun smartpink111 at yahoo.com
Tue Feb 5 15:53:22 CET 2013

```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 <zjoanna2013 at gmail.com>
To: arun <smartpink111 at yahoo.com>
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 <smartpink111 at yahoo.com> wrote:

>
>Hi,
>
>
>A.K.
>
>
>----- Original Message -----
>
>From: arun <smartpink111 at yahoo.com>
>To: "Zjoanna2013 at gmail.com" <Zjoanna2013 at gmail.com>
>Cc: R help <r-help at r-project.org>
>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: "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.
>>
>> ------------------------------
>> below:
>>
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