Data Rearrangement

HI Dila,
You have to order it afterwards.
res <- transform(melt(dat,id.var=c("Year","Day")),Month=match(variable,month.abb),Amount=value)[,-c(3:4)]


res1 <- res[with(res,order(Year,Month,Day)),]
A.K.

Dear Arun,

Thank you so much for your concern.

It is slightly differ from what I want after I applied it on my data. My bad... It works by rearranging the data month by month.?

? ? Year Day Month Amount
1 ?2012 ? 1 ? ? 1 ? ?0.0
2 ?2012 ? 2 ? ? 1 ? ?0.0
3 ?2012 ? 3 ? ? 1 ? ?0.0
4 ?2012 ? 4 ? ? 1 ? ?0.0
5 ?2012 ? 5 ? ? 1 ? ?0.0
6 ?2012 ? 6 ? ? 1 ? ?0.0
7 ?2012 ? 7 ? ? 1 ? 11.0
8 ?2012 ? 8 ? ? 1 ? ?0.0
9 ?2013 ? 1 ? ? 1 ? ?0.0
10 2013 ? 2 ? ? 1 ? ?0.0
11 2013 ? 3 ? ? 1 ? ?0.0
12 2013 ? 4 ? ? 1 ? ?0.0
13 2013 ? 5 ? ? 1 ? ?0.0
14 2013 ? 6 ? ? 1 ? ?0.0
15 2013 ? 7 ? ? 1 ? 11.0
16 2013 ? 8 ? ? 1 ? ?0.0
17 2012 ? 1 ? ? 2 ? ?2.5
18 2012 ? 2 ? ? 2 ? ?6.5
19 2012 ? 3 ? ? 2 ? ?9.5
20 2012 ? 4 ? ? 2 ? ?0.0
21 2012 ? 5 ? ? 2 ? ?5.0
22 2012 ? 6 ? ? 2 ? ?4.0
23 2012 ? 7 ? ? 2 ? ?0.0
24 2012 ? 8 ? ? 2 ? ?5.0
25 2013 ? 1 ? ? 2 ? ?2.5
26 2013 ? 2 ? ? 2 ? ?6.5
27 2013 ? 3 ? ? 2 ? ?9.5
28 2013 ? 4 ? ? 2 ? ?0.0
29 2013 ? 5 ? ? 2 ? ?5.0
30 2013 ? 6 ? ? 2 ? ?4.0
31 2013 ? 7 ? ? 2 ? ?0.0
32 2013 ? 8 ? ? 2 ? ?5.0
33 2012 ? 1 ? ? 3 ? ?0.5
34 2012 ? 2 ? ? 3 ? ?0.0
35 2012 ? 3 ? ? 3 ? ?0.0
36 2012 ? 4 ? ? 3 ? ?8.0
37 2012 ? 5 ? ? 3 ? ?0.5
38 2012 ? 6 ? ? 3 ? ?3.5
39 2012 ? 7 ? ? 3 ? 12.5
40 2012 ? 8 ? ? 3 ? ?8.0
41 2013 ? 1 ? ? 3 ? ?0.5
42 2013 ? 2 ? ? 3 ? ?0.0
43 2013 ? 3 ? ? 3 ? ?0.0
44 2013 ? 4 ? ? 3 ? ?8.0
45 2013 ? 5 ? ? 3 ? ?0.5
46 2013 ? 6 ? ? 3 ? ?3.5
47 2013 ? 7 ? ? 3 ? 12.5
48 2013 ? 8 ? ? 3 ? ?8.0
49 2012 ? 1 ? ? 4 ? ?2.0
50 2012 ? 2 ? ? 4 ? 29.0
51 2012 ? 3 ? ? 4 ? ?0.0
52 2012 ? 4 ? ? 4 ? ?0.5
53 2012 ? 5 ? ? 4 ?110.5
54 2012 ? 6 ? ? 4 ? 22.0
55 2012 ? 7 ? ? 4 ? ?3.5
56 2012 ? 8 ? ? 4 ? 36.5
57 2013 ? 1 ? ? 4 ? ?2.0
58 2013 ? 2 ? ? 4 ? 29.0
59 2013 ? 3 ? ? 4 ? ?0.0
60 2013 ? 4 ? ? 4 ? ?0.5
61 2013 ? 5 ? ? 4 ?110.5
62 2013 ? 6 ? ? 4 ? 22.0
63 2013 ? 7 ? ? 4 ? ?3.5
64 2013 ? 8 ? ? 4 ? 36.5

But, I want it to be year by year. ?I meant 2012 (Jan-April) followed by 2013 (Jan - April)

Your help is really helpful, but I have to do some alteration which is I stuck in rearranging it year by year.

Regards,?
Dila

HI DIla,

Isn't this the same question you asked earlier?


transform(melt(dat,id.var=c("Year","Day")),Month=match(variable,month.abb),Amount=value)[,-c(3:4)]
A.K.





On Thursday, February 27, 2014 10:24 PM, dila radi <dilaradi21 at gmail.com> wrote:
Hi all,

I know this is easy, but I really do not have any idea to solve it.

I have this kind of data set:

dat <- read.table(text="Day Year Jan Feb Mar Apr
? 1? 2012 0 2.5 0.5 2
? 2? 2012 0 6.5 0 29
? 3? 2012 0 9.5 0 0
? 4? 2012 0 0 8 0.5
? 5? 2012 0 5 0.5 110.5
? 6? 2012 0 4 3.5 22
? 7? 2012 11 0 12.5 3.5
? 8? 2012 0 5 8 36.5
? 1? 2013 0 2.5 0.5 2
? 2? 2013 0 6.5 0 29
? 3? 2013 0 9.5 0 0
? 4? 2013 0 0 8 0.5
? 5? 2013 0 5 0.5 110.5
? 6? 2013 0 4 3.5 22
? 7? 2013 11 0 12.5 3.5
? 8? 2013 0 5 8 36.5",sep="",header=TRUE)

and I want it to be in this form:

? ? Year? ? ? ?Day Month Amount? 2012 1 1 0? 2012 2 1 0? 2012 3 1 0? 2012 4
1 0? 2012 5 1 0? 2012 6 1 0? 2012 7 1 11? 2012 8 1 0? 2012 1 2 2.5? 2012 2 2
6.5? 2012 3 2 9.5? 2012 4 2 0? 2012 5 2 5? 2012 6 2 4? 2012 7 2 0? 2012 8 2
5? 2012 1 3 0.5? 2012 2 3 0? 2012 3 3 0? 2012 4 3 8? 2012 5 3 0.5? 2012 6 3
3.5? 2012 7 3 12.5? 2012 8 3 8? 2012 1 4 2? 2012 2 4 29? 2012 3 4 0? 2012 4
4 0.5? 2012 5 4 110.5? 2012 6 4 22? 2012 7 4 3.5? 2012 8 4 36.5? ? ?2013 1 1
0? 2013 2 1 0? 2013 3 1 0? 2013 4 1 0? 2013 5 1 0? 2013 6 1 0? 2013 7 1 11
2013 8 1 0? 2013 1 2 2.5? 2013 2 2 6.5? 2013 3 2 9.5? 2013 4 2 0? 2013 5 2 5
2013 6 2 4? 2013 7 2 0? 2013 8 2 5? 2013 1 3 0.5? 2013 2 3 0? 2013 3 3 0
2013 4 3 8? 2013 5 3 0.5? 2013 6 3 3.5? 2013 7 3 12.5? 2013 8 3 8? 2012 1 4
2? 2012 2 4 29? 2012 3 4 0? 2012 4 4 0.5? 2012 5 4 110.5? 2012 6 4 22? 2012
7 4 3.5? 2012 8 4 36.5? 2013 1 4 2? 2013 2 4 29? 2013 3 4 0? 2013 4 4 0.5
2013 5 4 110.5? 2013 6 4 22? 2013 7 4 3.5? 2013 8 4 36.5
I want to rearrange the data according to the YEAR (year by year)

Thank you.

Regards,
Dila


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