# lagging over consecutive pairs of rows in dataframe

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## lagging over consecutive pairs of rows in dataframe

 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]] ______________________________________________ [hidden email] mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-helpPLEASE do read the posting guide http://www.R-project.org/posting-guide.htmland provide commented, minimal, self-contained, reproducible code.
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## Re: lagging over consecutive pairs of rows in dataframe

 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 <[hidden email]> 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]] > > ______________________________________________ > [hidden email] mailing list -- To UNSUBSCRIBE and more, see > 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. >         [[alternative HTML version deleted]] ______________________________________________ [hidden email] mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-helpPLEASE do read the posting guide http://www.R-project.org/posting-guide.htmland provide commented, minimal, self-contained, reproducible code.
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## Re: lagging over consecutive pairs of rows in dataframe

 On 3/17/2017 12:58 PM, Ulrik Stervbo 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) > > Indeed -- thanks very much!         [[alternative HTML version deleted]] ______________________________________________ [hidden email] mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-helpPLEASE do read the posting guide http://www.R-project.org/posting-guide.htmland provide commented, minimal, self-contained, reproducible code.
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## Re: lagging over consecutive pairs of rows in dataframe

 In reply to this post by Ulrik Stervbo-2 Evan: You misunderstand the concept of a lagged variable. 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) 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 9:58 AM, Ulrik Stervbo <[hidden email]> 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 <[hidden email]> 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]] >> >> ______________________________________________ >> [hidden email] mailing list -- To UNSUBSCRIBE and more, see >> 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. >> > >         [[alternative HTML version deleted]] > > ______________________________________________ > [hidden email] mailing list -- To UNSUBSCRIBE and more, see > 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 -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-helpPLEASE do read the posting guide http://www.R-project.org/posting-guide.htmland provide commented, minimal, self-contained, reproducible code.
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## Re: lagging over consecutive pairs of rows in dataframe

 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 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 <[hidden email]> 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 <[hidden email]> 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]] >>> >>> ______________________________________________ >>> [hidden email] mailing list -- To UNSUBSCRIBE and more, see >>> 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. >>> >>          [[alternative HTML version deleted]] >> >> ______________________________________________ >> [hidden email] mailing list -- To UNSUBSCRIBE and more, see >> 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 -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-helpPLEASE do read the posting guide http://www.R-project.org/posting-guide.htmland provide commented, minimal, self-contained, reproducible code.
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## Re: lagging over consecutive pairs of rows in dataframe

 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 <[hidden email]> 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 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 <[hidden email]> >> 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 <[hidden email]> 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]] >>>> >>>> ______________________________________________ >>>> [hidden email] mailing list -- To UNSUBSCRIBE and more, see >>>> 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. >>>> >>>          [[alternative HTML version deleted]] >>> >>> ______________________________________________ >>> [hidden email] mailing list -- To UNSUBSCRIBE and more, see >>> 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 -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-helpPLEASE do read the posting guide http://www.R-project.org/posting-guide.htmland provide commented, minimal, self-contained, reproducible code.
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## Re: lagging over consecutive pairs of rows in dataframe

 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 <[hidden email]> 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 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 <[hidden email]> >>> 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 <[hidden email]> 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]] >>>>> >>>>> ______________________________________________ >>>>> [hidden email] mailing list -- To UNSUBSCRIBE and more, see >>>>> 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. >>>>> >>>>           [[alternative HTML version deleted]] >>>> >>>> ______________________________________________ >>>> [hidden email] mailing list -- To UNSUBSCRIBE and more, see >>>> 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 -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-helpPLEASE do read the posting guide http://www.R-project.org/posting-guide.htmland provide commented, minimal, self-contained, reproducible code.