[R] lagging over consecutive pairs of rows in dataframe

Mathew Guilfoyle mrguilfoyle at gmail.com
Sat Mar 18 17:18:07 CET 2017


If you are strict about your data formatting then the following is a fast way of calculating the differences, based on reshaping the data column:

A = matrix(mydata$rslt, nrow=2)
data.frame(exp=1:ncol(A), diff=A[2,]-A[1,])

alternatively, if the 'exp' values are not guaranteed to be sequential you can reshape an index:

A = matrix(1:nrow(mydata), nrow=2)
data.frame(exp=mydata$exp[A[1,]], diff=mydata$rslt[A[2,]]-mydata$rslt[A[1,]])

However, I would suggest that you have a further variable to label 'control' and 'treatment' groups and explicitly use this for the calculation.  Otherwise, if at any time you sort or reorder the data you will run into problems or produce erroneous results (but more than likely won't generate any actual R errors to alert you):

data.frame(exp = unique(mydata$exp), diff = as.vector(by(mydata, mydata$exp, function(x) x$rslt[x$type=='treatment']-x$rslt[x$type=='control'])))

The efficiency of the various options that have been suggested in this thread piqued my interest so a quick benchmark seemed in order (see below, including a 'safer' method).  Of course, this is probably only relevant if you have huge datasets that you are repeatedly performing this calculation on.


library('microbenchmark')

#create some example data similar to the OP
ndata = 1000
mydata = data.frame(exp = cumsum(rep(c(1,0),ndata)),rslt=sample(1:50, size = ndata*2, replace = TRUE), type=rep(c('control','treatment'),ndata))

#various suggested options
diff.BG = function(mydata) {
  evens <-  (seq_len(nrow(mydata)) %% 2) == 0
  data.frame(exp = mydata[evens,1 ], diff = diff(mydata[,2])[evens[-1]])
}

diff.US = function(mydata) {
  aggregate(mydata$rslt, by = list(group = mydata$exp), FUN = diff)
}

diff.MG1 = function(mydata) {
  A = matrix(mydata$rslt, nrow=2)
  data.frame(exp=1:ncol(A), diff=A[2,]-A[1,])
}

diff.MG2 = function(mydata) {
  A = matrix(1:nrow(mydata), nrow=2)
  data.frame(exp=mydata$exp[A[1,]], diff=mydata$rslt[A[2,]]-mydata$rslt[A[1,]])
}

diff.safe = function(mydata) {
  data.frame(exp = unique(mydata$exp), diff = as.vector(by(mydata, mydata$exp, function(x) x$rslt[x$type=='treatment']-x$rslt[x$type=='control'])))
}

#benchmark
microbenchmark(BG=diff.BG(mydata), US=diff.US(mydata), MG1=diff.MG1(mydata), MG2=diff.MG2(mydata), safe=diff.safe(my data))

Unit: microseconds
 expr       min          lq        mean      median          uq        max neval
   BG   273.837    299.0015    351.0377    316.7400    350.5220   2385.289   100
   US  9872.457  10511.1065  11555.6048  11108.0790  12471.8060  17609.518   100
  MG1   168.783    196.8635    229.9329    210.9370    249.4895    471.381   100
  MG2   221.303    237.0480    265.5097    254.3895    280.7815    418.728   100
 safe 97869.540 104164.5130 109579.9834 107199.7715 110315.8590 170028.377   100



> On 17 Mar 2017, at 14:54, Evan Cooch <evan.cooch at 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))
> 
> 
> 
> 	[[alternative HTML version deleted]]
> 
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