Proof of last assertion - sorry, I just had to follow through...
library(data.table)
DT0 <- data.table(DF0, key = 'A, B')
DT0[, sum(C), by = 'A, B']
A B V1
[1,] 52 1 124
[2,] 52 59 38
[3,] 52 97 75
[4,] 57 1 64
[5,] 57 6 26
[6,] 57 114 12
[7,] 89 1 192
[8,] 89 6 20
[9,] 89 29 137
[10,] 89 52 13
[11,] 89 57 10
[12,] 89 97 23
> aggregate(C ~ A + B, data = DF0, FUN = sum)
A B C
1 52 1 124
2 57 1 64
3 89 1 192
4 57 6 26
5 89 6 20
6 89 29 137
7 89 52 13
8 89 57 10
9 52 59 38
10 52 97 75
11 89 97 23
12 57 114 12
> library(doBy)
> summaryBy(C ~ A + B, data = DF0, FUN = sum)
A B C.sum
1 52 1 124
2 52 59 38
3 52 97 75
4 57 1 64
5 57 6 26
6 57 114 12
7 89 1 192
8 89 6 20
9 89 29 137
10 89 52 13
11 89 57 10
12 89 97 23
That should give you enough options :)
Dennis
On Wed, Oct 20, 2010 at 3:24 AM, Dennis Murphy <djmu...@gmail.com> wrote:
> Or even better (doh!)...
>
> library(plyr)
> > ddply(DF0, .(A, B), summarise, C = sum(C))
>
> A B C
> 1 52 1 124
> 2 52 59 38
> 3 52 97 75
> 4 57 1 64
> 5 57 6 26
> 6 57 114 12
> 7 89 1 192
> 8 89 6 20
> 9 89 29 137
> 10 89 52 13
> 11 89 57 10
> 12 89 97 23
>
> which means that aggregate(), summaryBy() in the doBy package and several
> more functions/packages can do this quite easily.
>
> Dennis
>
>
> On Wed, Oct 20, 2010 at 3:21 AM, Dennis Murphy <djmu...@gmail.com> wrote:
>
>> Hi:
>>
>> Here's one way, although it can be improved a bit.
>>
>> d1 <- aggregate(C ~ A, data = subset(DF0, B == 1), FUN = sum)
>> d2 <- subset(DF0, B != 1)
>> # B not in d1, so need to replace it
>> > d1
>> A C
>> 1 52 124
>> 2 57 64
>> 3 89 192
>> d1$B <- rep(1, nrow(d1))
>> d1 <- d1[, c(1, 3, 2)] # reorder columns to permit cbinding
>> DF1 <- rbind(d1, d2)
>> > DF1[order(DF1$A, DF1$B), ]
>> A B C
>> 1 52 1 124
>> 19 52 59 38
>> 20 52 97 75
>> 2 57 1 64
>> 26 57 6 26
>> 24 57 114 12
>> 3 89 1 192
>> 21 89 6 20
>> 31 89 29 137
>> 4 89 52 13
>> 5 89 57 10
>> 6 89 97 23
>>
>> HTH,
>> Dennis
>>
>>
>> On Wed, Oct 20, 2010 at 2:42 AM, xtracto <b2017...@lhsdv.com> wrote:
>>
>>>
>>> Hello,
>>>
>>> I am trying to achieve something which I *think* is possible using
>>> rowsum,
>>> but a little help should be useful:
>>>
>>> Consider the following dataframe DF0:
>>> A B C
>>> 89 1 140
>>> 89 06 20
>>> 89 29 137
>>> 89 52 13
>>> 89 57 10
>>> 89 97 23
>>> 89 1 37
>>> 89 1 12
>>> 89 1 3
>>> 52 1 11
>>> 52 1 31
>>> 52 1 16
>>> 52 1 6
>>> 52 1 10
>>> 52 1 13
>>> 52 1 10
>>> 52 1 25
>>> 52 1 2
>>> 52 59 38
>>> 52 97 75
>>> 57 1 14
>>> 57 1 13
>>> 57 1 14
>>> 57 114 12
>>> 57 1 23
>>> 57 06 26
>>>
>>>
>>> I need create a new dataframe containing the sums of all the rows where B
>>> =
>>> 1 for the different values of A, keeping the rows with other B values the
>>> same. That is, for this data sample, the result I expect is something
>>> like
>>> this (the order of the rows does not matter):
>>>
>>> A B C
>>> 89 1 192 #From adding up: [140 + 37 + 12 + 3]
>>> 89 06 20
>>> 89 29 137
>>> 89 52 13
>>> 89 57 10
>>> 89 97 23
>>> 52 1 124 # From adding up: [11 + 31 + 16 + 6 + 10 + 13 + 10
>>> + 25 + 2]
>>> 52 59 38
>>> 52 97 75
>>> 57 1 64 #From adding up: [14 +13 +14 +23]
>>> 57 114 12
>>> 57 06 26
>>>
>>>
>>> Now, I now it should be possible to first separate the data in two sets,
>>> where
>>> DF1 <- DF0[DF0$B != 1,]
>>> DF2 <- DF0[DF0$B == 1,]
>>>
>>> Then I should apply sumrow to DF2 with some "group" vector, but I do not
>>> know where to go from here.
>>>
>>> Can anyone help?
>>> Thanks in advance!
>>>
>>>
>>> --
>>> View this message in context:
>>> http://r.789695.n4.nabble.com/rowsum-tp3003551p3003551.html
>>> Sent from the R help mailing list archive at Nabble.com.
>>>
>>> ______________________________________________
>>> R-help@r-project.org mailing list
>>> https://stat.ethz.ch/mailman/listinfo/r-help
>>> PLEASE do read the posting guide
>>> http://www.R-project.org/posting-guide.html
>>> and provide commented, minimal, self-contained, reproducible code.
>>>
>>
>>
>
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