[R] searching for data.frame rows / processing of rows

[chao gai]

I tought that aggregate was the way to go, but only for large dataframes it is 
faster.

>   df <- read.table(stdin(),header=TRUE)
0: Location   Time    X    Y
1:   1          0      1.6  9.3
2:   1          3      4.2  10.4
3:   1          6      2.7  16.3
4:   2          0      0.5  2.1
5:   2          3      NA   3.6
6:   2          3      5.0  0.06
7:   2          6      3.4  14.0
8:
> aggregate(df[,3:4],df[,1:2],FUN=mean,na.rm=TRUE)
  Location Time   X     Y
1        1    0 1.6  9.30
2        2    0 0.5  2.10
3        1    3 4.2 10.40
4        2    3 5.0  1.83
5        1    6 2.7 16.30
6        2    6 3.4 14.00
>  system.time(  aggregate(df[,3:4],df[,1:2],FUN=mean,na.rm=TRUE)  )
[1] 0.008 0.000 0.008 0.000 0.000
>
>  system.time(  do.call(rbind, by(df, df[2:1], colMeans, na.rm = TRUE)))
[1] 0.005 0.000 0.005 0.000 0.000
>
>
> df <- data.frame(Location=rep(1:50,50),
+                  Time=sample(rep(1:50,each=10),2500,replace=TRUE),
+                  X=runif(2500),Y=runif(2500))
>
>  system.time(  aggregate(df[,3:4],df[,1:2],FUN=mean,na.rm=TRUE)  )
[1] 0.162 0.000 0.163 0.000 0.000
>
>  system.time(  do.call(rbind, by(df, df[2:1], colMeans, na.rm = TRUE)))
[1] 2.179 0.006 2.216 0.000 0.000


Kees

Gabor Grothendieck wrote:
> Grouping the data frame by the first two columns, apply colMeans
> and then rbind the resulting by-structure together:
> 
> do.call(rbind, by(DF, DF[2:1], colMeans, na.rm = TRUE))
> 
> 
> On 10/5/06, Greg Tarpinian <sasprog474 at yahoo.com> wrote:
>> R 2.3.1, WinXP:
>>
>> I have a puzzling problem that I suspect may be solved using
>> grep or a regular expression but am at a loss how to actually do it...
>> My data.frame looks like
>>
>>  Location   Time    X    Y
>>  --------   ----   ---  ---
>>  1          0      1.6  9.3
>>  1          3      4.2  10.4
>>  1          6      2.7  16.3
>>  2          0      0.5  2.1
>>  2          3      NA   3.6
>>  2          3      5.0  0.06
>>  2          6      3.4  14.0
>>
>> and so forth.  I would like to search for duplicate Time values
>> within a Location and take the numerical average (where possible)
>> of the elements in X and Y.  These numerical averages should
>> then be used to create a single row where multiple rows once
>> existed.  So, I would like to obtain
>>
>>  2          3      5.0  1.83
>>
>> for the two Time = 3 rows for Location = 2 and use it to replace
>> these two rows.  Ideally, avoiding for(i in 1:blah) loops would be
>> nice because the data.frame has about 10,000 rows that need to
>> be searched and processed.  My intent is to do some comparing of
>> SAS to R -- the DATA step processing in SAS is quite fast and
>> using the RETAIN statement along with the LAG( ) function allows
>> this sort of thing to be done rapidly.
>>
>>
>> Thanks in advance,
>>
>>       Greg
>>
>> ______________________________________________
>> R-help at stat.math.ethz.ch 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.
>>
> 
> ______________________________________________
> R-help at stat.math.ethz.ch 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.
>