Thanks very much. I suspect 50% of my time in R is spent translating from what I know how to do in SAS (25+ years of heavy use), to what is equivalent in SAS. So far, I haven't found anything I can do in SAS that I can't do in R, with some help. ;-)

Cheers...

On 3/17/2017 1:51 PM, Bert Gunter wrote:
Evan:

Yes, I stand partially corrected. You have the concept correct, but R
implements it differently than SAS.

I think what you want for your approach is diff():

evens <-  (seq_len(nrow(mydata)) %% 2) == 0
newdat <-data.frame(exp=mydata[evens,1 ],reslt= diff(mydata[,2])[evens[-1]])

... which seems neater to me than what I offered previously.

Cheers,
Bert

Bert Gunter

"The trouble with having an open mind is that people keep coming along
and sticking things into it."
-- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )


On Fri, Mar 17, 2017 at 10:25 AM, Evan Cooch <evan.co...@gmail.com> wrote:

On 3/17/2017 1:19 PM, Bert Gunter wrote:
Evan:

You misunderstand the concept of a lagged variable.

Well, lag in R, perhaps (and by my own admission). In SAS, thats exactly how
it works.:

data test;
input exp rslt;
cards;
<data in the data frame in OP>
     *;


     data test2; set test; by exp;
     diff=rslt-lag(rslt);
       if last.exp;

Ulrik:

Well, yes, that is certainly a general solution that works. However,
given the *specific* structure described by the OP, an even more
direct (maybe more efficient?) way to do it just uses (logical)
subscripting:

odds <-  (seq_len(nrow(mydata)) %% 2) == 1
newdat <-data.frame(mydata[odds,1 ],mydata[!odds,2] - mydata[odds,2])
names(newdat) <- names(mydata)

Interesting - thanks!


Bert Gunter

"The trouble with having an open mind is that people keep coming along
and sticking things into it."
-- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )


On Fri, Mar 17, 2017 at 9:58 AM, Ulrik Stervbo <ulrik.ster...@gmail.com>
wrote:
Hi Evan

you can easily do this by applying diff() to each exp group.

Either using dplyr:
library(dplyr)
mydata %>%
    group_by(exp) %>%
    summarise(difference = diff(rslt))

Or with base R
aggregate(mydata, by = list(group = mydata$exp), FUN = diff)

HTH
Ulrik


On Fri, 17 Mar 2017 at 17:34 Evan Cooch <evan.co...@gmail.com> wrote:

Suppose I have a dataframe that looks like the following:

n=2
mydata <- data.frame(exp = rep(1:5,each=n), rslt =
c(12,15,7,8,24,28,33,15,22,11))
mydata
      exp rslt
1    1   12
2    1   15
3    2    7
4    2    8
5    3   24
6    3   28
7    4   33
8    4   15
9    5   22
10   5   11

The variable 'exp' (for experiment') occurs in pairs over consecutive
rows -- 1,1, then 2,2, then 3,3, and so on. The first row in a pair is
the 'control', and the second is a 'treatment'. The rslt column is the
result.

What I'm trying to do is create a subset of this dataframe that consists
of the exp number, and the lagged difference between the 'control' and
'treatment' result.  So, for exp=1, the difference is (15-12)=3. For
exp=2,  the difference is (8-7)=1, and so on. What I'm hoping to do is
take mydata (above), and turn it into

        exp  diff
1   1      3
2   2      1
3   3      4
4   4      -18
5   5      -11

The basic 'trick' I can't figure out is how to create a lagged variable
between the second row (record) for a given level of exp, and the first
row for that exp.  This is easy to do in SAS (which I'm more familiar
with), but I'm struggling with the equivalent in R. The brute force
approach  I thought of is to simply split the dataframe into to (one
even rows, one odd rows), merge by exp, and then calculate a difference.
But this seems to require renaming the rslt column in the two new
dataframes so they are different in the merge (say, rslt_cont n the odd
dataframe, and rslt_trt in the even dataframe), allowing me to calculate
a difference between the two.

While I suppose this would work, I'm wondering if I'm missing a more
elegant 'in place' approach that doesn't require me to split the data
frame and do every via a merge.

Suggestions/pointers to the obvious welcome. I've tried playing with
lag, and some approaches using lag in the zoo package,  but haven't
found the magic trick. The problem (meaning, what I can't figure out)
seems to be conditioning the lag on the level of exp.

Many thanks...


mydata <-*data.frame*(x = c(20,35,45,55,70), n = rep(50,5), y =
c(6,17,26,37,44))



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