[R] a vectorized solution to some simple dataframe math?
David Winsemius
dwinsemius at comcast.net
Sat Mar 27 15:17:46 CET 2010
On Mar 27, 2010, at 6:53 AM, Dennis Murphy wrote:
> Hi:
>
> Does this do what you want?
>
> # Create some fake data...
>
> df <- data.frame(id = factor(rep(c('cell1', 'cell2'), each = 10)),
> cond = factor(rep(rep(c('A', 'B'), each = 5), 2)),
> time = round(rnorm(20, 350, 10), 2))
>
> # Create a function to subtract each element of a vector from its mean
> f <- function(x) x - mean(x)
A function, ave, already exists in base R for calculating means
within groups. Subtraction from the time variable is straightforward:
> df$dev <- df$time - ave(df$time, df$id, df$cond)
> df$dev
[1] -1.346 -8.586 -2.366 2.714 9.584 -12.108 13.052 0.742
-7.438
[10] 5.752 1.434 10.854 0.514 -21.166 8.364 4.128 -4.502
-5.602
[19] -4.322 10.298
Although the default for ave() is the mean function, other functions
can be used with the FUN= argument.
--
David.
> # Load the plyr package, which contains the function ddply():
> library(plyr)
> df2 <- ddply(df, .(id, cond), transform, dev = f(time))
>
> # output
>> df2
> id cond time dev
> 1 cell1 A 353.01 7.226
> 2 cell1 A 351.06 5.276
> 3 cell1 A 343.59 -2.194
> 4 cell1 A 341.50 -4.284
> 5 cell1 A 339.76 -6.024
> 6 cell1 B 351.18 0.644
> 7 cell1 B 340.53 -10.006
> 8 cell1 B 345.09 -5.446
> 9 cell1 B 347.44 -3.096
> 10 cell1 B 368.44 17.904
> 11 cell2 A 343.48 -3.776
> 12 cell2 A 352.35 5.094
> 13 cell2 A 350.78 3.524
> 14 cell2 A 340.38 -6.876
> 15 cell2 A 349.29 2.034
> 16 cell2 B 364.45 15.524
> 17 cell2 B 354.52 5.594
> 18 cell2 B 350.41 1.484
> 19 cell2 B 345.78 -3.146
> 20 cell2 B 329.47 -19.456
>
> # cell means
>> with(df, aggregate(time, list(id = id, cond = cond), mean))
> id cond x
> 1 cell1 A 345.784
> 2 cell2 A 347.256
> 3 cell1 B 350.536
> 4 cell2 B 348.926
>
> HTH,
> Dennis
>
> On Fri, Mar 26, 2010 at 1:31 PM, Dgnn <sharkbrainpdx at gmail.com> wrote:
>
>>
>> I have a data frame containing the results of time measurements
>> taken from
>> several cells. Each cell was measured in conditions A and B, and
>> there are
>> an arbitrary number of measurements in each condition. I am trying to
>> calculate the difference of each measurement from the mean of a
>> given cell
>> in a given condition without relying on loops.
>>
>>> my.df
>> id cond time
>> 1 cell1 A 343.5
>> 2 cell1 A 355.2
>> ...
>> 768 cell1 B 454.0
>> ...
>> 2106 cell2 A 433.9
>> ...
>>
>> as a first approach I tried:
>>
>>> mews<-aggregate(my.df$time, list(cond=data$id, id=data$cond), mean)
>> id cond time
>> cell1 A 352
>> cell1 B 446
>> cell2 A 244
>> cell2 B ...
>>
>> I then tried to use %in% to match id and cond of mews with my.df,
>> but I
>> haven't been able to get it to work.
>> Am I on the right track? What are some other solutions?
>>
>> Thanks for any help.
>>
>> jason
>>
>>
>> --
>> View this message in context:
>> http://n4.nabble.com/a-vectorized-solution-to-some-simple-dataframe-math-tp1692810p1692810.html
>> Sent from the R help mailing list archive at Nabble.com.
>>
>> ______________________________________________
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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.
>>
>
> [[alternative HTML version deleted]]
>
> ______________________________________________
> R-help at 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.
David Winsemius, MD
West Hartford, CT
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