Yes, this works out perfectly! Thanks a lot, David. Have a wonderful day...
On Wed, Aug 24, 2016 at 4:24 PM, David L Carlson <dcarl...@tamu.edu> wrote: > This will work, but you should double-check to be certain that CP and > unique(myData[, 3:5]) are in the same order. It will fail if N is not > identical for all rows of the same S-Z combination. > >> CP <- sapply(split(myData, paste0(myData$S, myData$Z)), function(x) > + crossprod(x[, 1], x[, 2])) >> data.frame(CP, unique(myData[, 3:5])) > CP N S Z > S1A 22 2.1 S1 A > S1B 38 2.1 S1 B > S2A 38 3.2 S2 A > S2B 22 3.2 S2 B > > David C > > -----Original Message----- > From: Gang Chen [mailto:gangch...@gmail.com] > Sent: Wednesday, August 24, 2016 2:51 PM > To: David L Carlson > Cc: r-help mailing list > Subject: Re: [R] aggregate > > Thanks again for patiently offering great help, David! I just learned > dput() and paste0() now. Hopefully this is my last question. > > Suppose a new dataframe is as below (one more numeric column): > > myData <- structure(list(X = c(1, 2, 3, 4, 5, 6, 7, 8), Y = c(8, 7, 6, > 5, 4, 3, 2, 1), N =c(rep(2.1, 4), rep(3.2, 4)), S = structure(c(1L, > 1L, 1L, 1L, 2L, 2L, 2L, 2L > ), .Label = c("S1", "S2"), class = "factor"), Z = structure(c(1L, > 1L, 2L, 2L, 1L, 1L, 2L, 2L), .Label = c("A", "B"), class = "factor")), > .Names = c("X", > "Y", "N", "S", "Z"), row.names = c(NA, -8L), class = "data.frame") > >> myData > > X Y N S Z > 1 1 8 2.1 S1 A > 2 2 7 2.1 S1 A > 3 3 6 2.1 S1 B > 4 4 5 2.1 S1 B > 5 5 4 3.2 S2 A > 6 6 3 3.2 S2 A > 7 7 2 3.2 S2 B > 8 8 1 3.2 S2 B > > Once I obtain the cross product, > >> sapply(split(myData, paste0(myData$S, myData$Z)), function(x) crossprod(x[, >> 1], x[, 2])) > S1A S1B S2A S2B > 22 38 38 22 > > how can I easily add the other 3 columns (N, S, and Z) in a new > dataframe? For S and Z, I can play with the names from the cross > product output, but I have trouble dealing with the numeric column N. > > > > > On Wed, Aug 24, 2016 at 1:07 PM, David L Carlson <dcarl...@tamu.edu> wrote: >> You need to spend some time with a basic R tutorial. Your data is messed up >> because you did not use a simple text editor somewhere along the way. R >> understands ', but not ‘ or ’. The best way to send data to the list is to >> use dput: >> >>> dput(myData) >> structure(list(X = c(1, 2, 3, 4, 5, 6, 7, 8), Y = c(8, 7, 6, >> 5, 4, 3, 2, 1), S = structure(c(1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L >> ), .Label = c("S1", "S2"), class = "factor"), Z = structure(c(1L, >> 1L, 2L, 2L, 1L, 1L, 2L, 2L), .Label = c("A", "B"), class = "factor")), >> .Names = c("X", >> "Y", "S", "Z"), row.names = c(NA, -8L), class = "data.frame") >> >> Combining two labels just requires the paste0() function: >> >>> sapply(split(myData, paste0(myData$S, myData$Z)), function(x) crossprod(x[, >>> 1], x[, 2])) >> S1A S1B S2A S2B >> 22 38 38 22 >> >> David C >> >> -----Original Message----- >> From: Gang Chen [mailto:gangch...@gmail.com] >> Sent: Wednesday, August 24, 2016 11:56 AM >> To: David L Carlson >> Cc: Jim Lemon; r-help mailing list >> Subject: Re: [R] aggregate >> >> Thanks a lot, David! I want to further expand the operation a little >> bit. With a new dataframe: >> >> myData <- data.frame(X=c(1, 2, 3, 4, 5, 6, 7, 8), Y=c(8, 7, 6, 5, 4, >> 3, 2, 1), S=c(‘S1’, ‘S1’, ‘S1’, ‘S1’, ‘S2’, ‘S2’, ‘S2’, ‘S2’), >> Z=c(‘A’, ‘A’, ‘B’, ‘B’, ‘A’, ‘A’, ‘B’, ‘B’)) >> >>> myData >> >> X Y S Z >> 1 1 8 S1 A >> 2 2 7 S1 A >> 3 3 6 S1 B >> 4 4 5 S1 B >> 5 5 4 S2 A >> 6 6 3 S2 A >> 7 7 2 S2 B >> 8 8 1 S2 B >> >> I would like to obtain the same cross product between columns X and Y, >> but at each combination level of factors S and Z. In other words, the >> cross product would be still performed each two rows in the new >> dataframe myData. How can I achieve that? >> >> On Wed, Aug 24, 2016 at 11:54 AM, David L Carlson <dcarl...@tamu.edu> wrote: >>> Your is fine, but it will be a little simpler if you use sapply() instead: >>> >>>> data.frame(Z=levels(myData$Z), CP=sapply(split(myData, myData$Z), >>> + function(x) crossprod(x[, 1], x[, 2]))) >>> Z CP >>> A A 10 >>> B B 10 >>> >>> David C >>> >>> >>> -----Original Message----- >>> From: Gang Chen [mailto:gangch...@gmail.com] >>> Sent: Wednesday, August 24, 2016 10:17 AM >>> To: David L Carlson >>> Cc: Jim Lemon; r-help mailing list >>> Subject: Re: [R] aggregate >>> >>> Thank you all for the suggestions! Yes, I'm looking for the cross >>> product between the two columns of X and Y. >>> >>> A follow-up question: what is a nice way to merge the output of >>> >>> lapply(split(myData, myData$Z), function(x) crossprod(x[, 1], x[, 2])) >>> >>> with the column Z in myData so that I would get a new dataframe as the >>> following (the 2nd column is the cross product between X and Y)? >>> >>> Z CP >>> A 10 >>> B 10 >>> >>> Is the following legitimate? >>> >>> data.frame(Z=levels(myData$Z), CP= unlist(lapply(split(myData, >>> myData$Z), function(x) crossprod(x[, 1], x[, 2])))) >>> >>> >>> On Wed, Aug 24, 2016 at 10:37 AM, David L Carlson <dcarl...@tamu.edu> wrote: >>>> Thank you for the reproducible example, but it is not clear what cross >>>> product you want. Jim's solution gives you the cross product of the >>>> 2-column matrix with itself. If you want the cross product between the >>>> columns you need something else. The aggregate function will not work >>>> since it will treat the columns separately: >>>> >>>>> A <- as.matrix(myData[myData$Z=="A", 1:2]) >>>>> A >>>> X Y >>>> 1 1 4 >>>> 2 2 3 >>>>> crossprod(A) # Same as t(A) %*% A >>>> X Y >>>> X 5 10 >>>> Y 10 25 >>>>> crossprod(A[, 1], A[, 2]) # Same as t(A[, 1] %*% A[, 2] >>>> [,1] >>>> [1,] 10 >>>>> >>>>> # For all the groups >>>>> lapply(split(myData, myData$Z), function(x) crossprod(as.matrix(x[, >>>>> 1:2]))) >>>> $A >>>> X Y >>>> X 5 10 >>>> Y 10 25 >>>> >>>> $B >>>> X Y >>>> X 25 10 >>>> Y 10 5 >>>> >>>>> lapply(split(myData, myData$Z), function(x) crossprod(x[, 1], x[, 2])) >>>> $A >>>> [,1] >>>> [1,] 10 >>>> >>>> $B >>>> [,1] >>>> [1,] 10 >>>> >>>> ------------------------------------- >>>> David L Carlson >>>> Department of Anthropology >>>> Texas A&M University >>>> College Station, TX 77840-4352 >>>> >>>> >>>> -----Original Message----- >>>> From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Jim Lemon >>>> Sent: Tuesday, August 23, 2016 6:02 PM >>>> To: Gang Chen; r-help mailing list >>>> Subject: Re: [R] aggregate >>>> >>>> Hi Gang Chen, >>>> If I have the right idea: >>>> >>>> for(zval in levels(myData$Z)) >>>> crossprod(as.matrix(myData[myData$Z==zval,c("X","Y")])) >>>> >>>> Jim >>>> >>>> On Wed, Aug 24, 2016 at 8:03 AM, Gang Chen <gangch...@gmail.com> wrote: >>>>> This is a simple question: With a dataframe like the following >>>>> >>>>> myData <- data.frame(X=c(1, 2, 3, 4), Y=c(4, 3, 2, 1), Z=c('A', 'A', 'B', >>>>> 'B')) >>>>> >>>>> how can I get the cross product between X and Y for each level of >>>>> factor Z? My difficulty is that I don't know how to deal with the fact >>>>> that crossprod() acts on two variables in this case. >>>>> >>>>> ______________________________________________ >>>>> R-help@r-project.org 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. >>>> >>>> ______________________________________________ >>>> R-help@r-project.org 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. ______________________________________________ R-help@r-project.org 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.
