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arrange data

4 messages · Roslina Zakaria, Rui Barradas, arun

arun
# 
Hi,

I hope this helps you.
?I created a small dataset: 3 replications per month for 1998:2000. 

set.seed(1)
dat1<-data.frame(Tahun=rep(c(98:99,00),each=36),Bahun=rep(rep(1:12,times=3),each=3), x=sample(1:500,108,replace=TRUE))
dat2<-within(dat1,{Tahun<-factor(Tahun,levels=c(98,99,0))}) 


agg_dt1<-aggregate(x=dat2[,3],by=dat2[,c(1,2)],FUN=sum)
?head(agg_dt1)
#? Tahun Bahun??? x
#1??? 98???? 1 1252
#2??? 99???? 1? 680
#3???? 0???? 1? 687
#4??? 98???? 2? 761
#5??? 99???? 2? 860
#6???? 0???? 2? 786
I guess this is what you wanted.


In addition, you can also use ddply() with a different way of grouping: but with the same result. 
library(plyr)
?dd_dt1<-ddply(dat2,.(Tahun,Bahun),summarize, sum(x))
?head(dd_dt1)
#? Tahun Bahun? ..1
#1??? 98???? 1 1252
#2??? 98???? 2? 761
#3??? 98???? 3? 440
#4??? 98???? 4? 597
#5??? 98???? 5? 987
#6??? 98???? 6? 692
?tail(dd_dt1)
#?? Tahun Bahun? ..1
#31???? 0???? 7? 685
#32???? 0???? 8? 504
#33???? 0???? 9? 633
#34???? 0??? 10? 553
#35???? 0??? 11? 914
#36???? 0??? 12 1039

A.K.






----- Original Message -----
From: Roslina Zakaria <zroslina at yahoo.com>
To: "r-help at r-project.org" <r-help at r-project.org>
Cc: 
Sent: Friday, October 5, 2012 8:09 PM
Subject: [R] arrange data

Dear r-users,

I have dailly rainfall data from year 1971 to 2000. I use aggregate to form monthly rainfall data. ?What I don't understand is that the data for the year 2000 become on the top, instead of year 1971. ?Here are some codes and output:


agg_dt1 ? ? <- aggregate(x=dt1[,4],by=dt1[,c(1,2)],FUN=sum)
? ?Tahun Bulan ? ? x
1 ? ? ?0 ? ? 1 398.6
2 ? ? 71 ? ? 1 934.9
3 ? ? 72 ? ? 1 107.2
4 ? ? 73 ? ? 1 236.4
5 ? ? 74 ? ? 1 ?10.5
6 ? ? 75 ? ? 1 744.6
7 ? ? 76 ? ? 1 ? 9.2
8 ? ? 77 ? ? 1 108.7
9 ? ? 78 ? ? 1 251.5
10 ? ?79 ? ? 1 197.3
11 ? ?80 ? ? 1 144.1
12 ? ?81 ? ? 1 104.5
13 ? ?82 ? ? 1 ?17.7
14 ? ?83 ? ? 1 151.8
...

Thank you so much for your help.

Roslina
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Rui Barradas
# 
Hello,

Using Arun's data example, instead of creating a factor "convert" to 4 
digits years.

set.seed(1)
dat1 <- data.frame(Tahun=rep(c(98:99,00),each=36),
             Bahun=rep(rep(1:12,times=3),each=3),
             x=sample(1:500,108,replace=TRUE))

dat2 <- dat1  # operate on a copy
dat2$Tahun <- with(dat2, ifelse(Tahun < 71, 2000 + Tahun, 1900 + Tahun))

agg_dt1 <- aggregate(x=dat2[,3],by=dat2[,c(1,2)],FUN=sum)
head(agg_dt1)

Hope this helps,

Rui Barradas
Em 06-10-2012 03:38, arun escreveu:
arun
# 
Hi Roslina,

Extending Rui's solution if you want only the last two digits for Year.
?

agg_dt1$Tahun<-as.numeric(gsub("\\d{2}(\\d+)","\\1",agg_dt1$Tahun))
?head(agg_dt1)
#? Tahun Bahun??? x
#1??? 98???? 1? 607
#2??? 99???? 1? 814
#3???? 0???? 1? 580
#4??? 98???? 2 1006
#5??? 99???? 2? 941
#6???? 0???? 2 1075A.K.





----- Original Message -----
From: Rui Barradas <ruipbarradas at sapo.pt>
To: arun <smartpink111 at yahoo.com>
Cc: Roslina Zakaria <zroslina at yahoo.com>; R help <r-help at r-project.org>
Sent: Saturday, October 6, 2012 7:22 AM
Subject: Re: [R] arrange data

Hello,

Using Arun's data example, instead of creating a factor "convert" to 4 
digits years.

set.seed(1)
dat1 <- data.frame(Tahun=rep(c(98:99,00),each=36),
? ? ? ? ? ?  Bahun=rep(rep(1:12,times=3),each=3),
? ? ? ? ? ?  x=sample(1:500,108,replace=TRUE))

dat2 <- dat1? # operate on a copy
dat2$Tahun <- with(dat2, ifelse(Tahun < 71, 2000 + Tahun, 1900 + Tahun))

agg_dt1 <- aggregate(x=dat2[,3],by=dat2[,c(1,2)],FUN=sum)
head(agg_dt1)

Hope this helps,

Rui Barradas
Em 06-10-2012 03:38, arun escreveu: