[R] Using cumsum with 'group by' ?
peter dalgaard
pdalgd at gmail.com
Tue Nov 27 18:16:58 CET 2012
On Nov 27, 2012, at 14:30 , TheRealJimShady wrote:
> Thanks everyone for your help. All good now. Once I'd got the data I
> ended up doing this (replaced with real column names rather than the
> fakes before).
>
> microaeth_data$date<-as.Date(microaeth_data$date_time,format="%Y-%m-%d
> %H:%M:%S")
>
> dat2<-microaeth_data[order(microaeth_data[,13],microaeth_data[,14]),]
>
> dat2$cumsum<-ave(dat2$bc,list(dat2$person_id,dat2$date),FUN=cumsum)
>
> If someone can be bothered I'd appreciate a breakdown of the final
> line of code so that I understood what I did, but no problem if not.
>
OK. ave() is originally designed to replace observations by their per-group average. It works by splitting the input vector into groups according to one or more factors, performing FUN on each group, and putting the result back in the original positions. The default for FUN is mean, but it doesn't need to be a scalar function; it also works with a FUN that returns a vector the same length as the input, such as FUN=cumsum.
The specification does not seem to allow a list() of grouping factors. It does work, but as it is unauthorised, it might not keep working.
Elaborating on the help page example:
> attach(warpbreaks)
> ave(breaks,wool,tension)
[1] 44.55556 44.55556 44.55556 44.55556 44.55556 44.55556 44.55556 44.55556
[9] 44.55556 24.00000 24.00000 24.00000 24.00000 24.00000 24.00000 24.00000
[17] 24.00000 24.00000 24.55556 24.55556 24.55556 24.55556 24.55556 24.55556
[25] 24.55556 24.55556 24.55556 28.22222 28.22222 28.22222 28.22222 28.22222
[33] 28.22222 28.22222 28.22222 28.22222 28.77778 28.77778 28.77778 28.77778
[41] 28.77778 28.77778 28.77778 28.77778 28.77778 18.77778 18.77778 18.77778
[49] 18.77778 18.77778 18.77778 18.77778 18.77778 18.77778
> ave(breaks,wool,tension, FUN=sum)
[1] 401 401 401 401 401 401 401 401 401 216 216 216 216 216 216 216 216 216 221
[20] 221 221 221 221 221 221 221 221 254 254 254 254 254 254 254 254 254 259 259
[39] 259 259 259 259 259 259 259 169 169 169 169 169 169 169 169 169
> ave(breaks,wool,tension, FUN=cumsum)
[1] 26 56 110 135 205 257 308 334 401 18 39 68 85 97 115 150 180 216 36
[20] 57 81 99 109 152 180 195 221 27 41 70 89 118 149 190 210 254 42 68
[39] 87 103 142 170 191 230 259 20 41 65 82 95 110 125 141 169
> ave(breaks,list(wool,tension), FUN=cumsum) # not actually supposed to work
[1] 26 56 110 135 205 257 308 334 401 18 39 68 85 97 115 150 180 216 36
[20] 57 81 99 109 152 180 195 221 27 41 70 89 118 149 190 210 254 42 68
[39] 87 103 142 170 191 230 259 20 41 65 82 95 110 125 141 169
> Thanks
>
> James
>
> On 23 November 2012 20:06, arun kirshna [via R]
> <ml-node+s789695n4650584h30 at n4.nabble.com> wrote:
>> HI,
>>
>> If that is the case, this should work:
>> dat1<-read.table(text="
>> id, x, date
>> 1, 5, 2012-06-05 12:01
>> 1, 10, 2012-06-05 12:02
>> 1, 45, 2012-06-05 12:03
>> 2, 5, 2012-06-05 12:01
>> 2, 3, 2012-06-05 12:03
>> 2, 2, 2012-06-05 12:05
>> 3, 5, 2012-06-05 12:03
>> 3, 5, 2012-06-05 12:04
>> 3, 8, 2012-06-05 12:05
>> 1, 5, 2012-06-08 13:01
>> 1, 9, 2012-06-08 13:02
>> 1, 3, 2012-06-08 13:03
>> 2, 0, 2012-06-08 13:15
>> 2, 1, 2012-06-08 13:18
>> 2, 8, 2012-06-08 13:20
>> 2, 4, 2012-06-08 13:21
>> 3, 6, 2012-06-08 13:15
>> 3, 2, 2012-06-08 13:16
>> 3, 7, 2012-06-08 13:17
>> 3, 2, 2012-06-08 13:18
>> ",sep=",",header=TRUE,stringsAsFactors=FALSE)
>> dat1$date<-as.Date(dat1$date,format="%Y-%m-%d %H:%M")
>> dat2<-dat1[order(dat1[,1],dat1[,3]),]
>> dat2$Cumsum<-ave(dat2$x,list(dat2$id,dat2$date),FUN=cumsum)
>>
>> head(dat2)
>> # id x date Cumsum
>> #1 1 5 2012-06-05 5
>> #2 1 10 2012-06-05 15
>> #3 1 45 2012-06-05 60
>> #10 1 5 2012-06-08 5
>> #11 1 9 2012-06-08 14
>> #12 1 3 2012-06-08 17
>> #or
>> with(dat2,aggregate(x,by=list(id=id,date=date),cumsum))
>> # id date x
>> #1 1 2012-06-05 5, 15, 60
>> #2 2 2012-06-05 5, 8, 10
>> #3 3 2012-06-05 5, 10, 18
>> #4 1 2012-06-08 5, 14, 17
>> #5 2 2012-06-08 0, 1, 9, 13
>> #6 3 2012-06-08 6, 8, 15, 17
>> A.K.
>>
>>
>>
>> ----- Original Message -----
>> From: TheRealJimShady <[hidden email]>
>> To: [hidden email]
>> Cc:
>> Sent: Friday, November 23, 2012 6:04 AM
>> Subject: Re: [R] Using cumsum with 'group by' ?
>>
>> Hi Arun & everyone,
>>
>> Thank you very much for your helpful suggestions. I've been working
>> through them, but have realised that my data is a little more
>> complicated than I said and that the solutions you've kindly provided
>> don't work. The problem is that there is more than one day of data for
>> each person. It looks like this:
>>
>> id x date
>> 1 5 2012-06-05 12:01
>> 1 10 2012-06-05 12:02
>> 1 45 2012-06-05 12:03
>> 2 5 2012-06-05 12:01
>> 2 3 2012-06-05 12:03
>> 2 2 2012-06-05 12:05
>> 3 5 2012-06-05 12:03
>> 3 5 2012-06-05 12:04
>> 3 8 2012-06-05 12:05
>> 1 5 2012-06-08 13:01
>> 1 9 2012-06-08 13:02
>> 1 3 2012-06-08 13:03
>> 2 0 2012-06-08 13:15
>> 2 1 2012-06-08 13:18
>> 2 8 2012-06-08 13:20
>> 2 4 2012-06-08 13:21
>> 3 6 2012-06-08 13:15
>> 3 2 2012-06-08 13:16
>> 3 7 2012-06-08 13:17
>> 3 2 2012-06-08 13:18
>>
>> So what I need to do is something like this (in pseudo code anyway):
>>
>> - Order the data by the id field and then the date field
>> - add a new variable called cumsum
>> - calculate this variable as the cumulative value of X, but grouping
>> by the id and date (not date, not date and time).
>>
>> Thank you
>>
>> James
>>
>>
>>
>>
>>
>> On 23 November 2012 03:54, arun kirshna [via R]
>> <[hidden email]> wrote:
>>
>>> Hi,
>>> No problem.
>>> One more method if you wanted to try:
>>> library(data.table)
>>> dat2<-data.table(dat1)
>>> dat2[,list(x,time,Cumsum=cumsum(x)),list(id)]
>>> # id x time Cumsum
>>> #1: 1 5 12:01 5
>>> #2: 1 14 12:02 19
>>> #3: 1 6 12:03 25
>>> #4: 1 3 12:04 28
>>> #5: 2 98 12:01 98
>>> #6: 2 23 12:02 121
>>> #7: 2 1 12:03 122
>>> #8: 2 4 12:04 126
>>> #9: 3 5 12:01 5
>>> #10: 3 65 12:02 70
>>> #11: 3 23 12:03 93
>>> #12: 3 23 12:04 116
>>>
>>>
>>> A.K.
>>>
>>>
>>>
>>> ----- Original Message -----
>>> From: TheRealJimShady <[hidden email]>
>>> To: [hidden email]
>>> Cc:
>>> Sent: Thursday, November 22, 2012 12:27 PM
>>> Subject: Re: [R] Using cumsum with 'group by' ?
>>>
>>> Thank you very much, I will try these tomorrow morning.
>>>
>>> On 22 November 2012 17:25, arun kirshna [via R]
>>> <[hidden email]> wrote:
>>>
>>>> HI,
>>>> You can do this in many ways:
>>>> dat1<-read.table(text="
>>>> id time x
>>>> 1 12:01 5
>>>> 1 12:02 14
>>>> 1 12:03 6
>>>> 1 12:04 3
>>>> 2 12:01 98
>>>> 2 12:02 23
>>>> 2 12:03 1
>>>> 2 12:04 4
>>>> 3 12:01 5
>>>> 3 12:02 65
>>>> 3 12:03 23
>>>> 3 12:04 23
>>>> ",sep="",header=TRUE,stringsAsFactors=FALSE)
>>>> dat1$Cumsum<-ave(dat1$x,dat1$id,FUN=cumsum)
>>>> #or
>>>> unlist(tapply(dat1$x,dat1$id,FUN=cumsum),use.names=FALSE)
>>>> # [1] 5 19 25 28 98 121 122 126 5 70 93 116
>>>> #or
>>>> library(plyr)
>>>> ddply(dat1,.(id),function(x) cumsum(x[3]))[,2]
>>>> # [1] 5 19 25 28 98 121 122 126 5 70 93 116
>>>> head(dat1)
>>>> # id time x Cumsum
>>>> #1 1 12:01 5 5
>>>> #2 1 12:02 14 19
>>>> #3 1 12:03 6 25
>>>> #4 1 12:04 3 28
>>>> #5 2 12:01 98 98
>>>> #6 2 12:02 23 121
>>>> A.K.
>>>>
>>>>
>>>>
>>>>
>>>> ________________________________
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>>>>
>>>>
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>>>> To unsubscribe from Using cumsum with 'group by' ?, click here.
>>>> NAML
>>>
>>>
>>>
>>>
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>>
>>
>>
>>
>> --
>> View this message in context:
>> http://r.789695.n4.nabble.com/Using-cumsum-with-group-by-tp4650457p4650538.html
>>
>> Sent from the R help mailing list archive at Nabble.com.
>> [[alternative HTML version deleted]]
>>
>> ______________________________________________
>> [hidden email] 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.
>>
>>
>> ______________________________________________
>> [hidden email] mailing list
>> https://stat.ethz.ch/mailman/listinfo/r-help
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>> and provide commented, minimal, self-contained, reproducible code.
>>
>>
>> ________________________________
>> If you reply to this email, your message will be added to the discussion
>> below:
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>> NAML
>
>
>
>
> --
> View this message in context: http://r.789695.n4.nabble.com/Using-cumsum-with-group-by-tp4650457p4650959.html
> Sent from the R help mailing list archive at Nabble.com.
> [[alternative HTML version deleted]]
>
> ______________________________________________
> R-help at r-project.org mailing list
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> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
--
Peter Dalgaard, Professor,
Center for Statistics, Copenhagen Business School
Solbjerg Plads 3, 2000 Frederiksberg, Denmark
Phone: (+45)38153501
Email: pd.mes at cbs.dk Priv: PDalgd at gmail.com
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