[R] aggregate
David L Carlson
dcarlson at tamu.edu
Wed Aug 24 22:24:29 CEST 2016
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:gangchen6 at 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 <dcarlson at 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:gangchen6 at 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 <dcarlson at 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:gangchen6 at 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$Z), function(x) crossprod(x[, 1], x[, 2]))))
>>
>>
>> On Wed, Aug 24, 2016 at 10:37 AM, David L Carlson <dcarlson at tamu.edu> wrote:
>>> Thank you for the reproducible example, but it is not clear what cross product you want. Jim's solution gives you the cross product of the 2-column matrix with itself. If you want the cross product between the columns you need something else. The aggregate function will not work since it will treat the columns separately:
>>>
>>>> A <- as.matrix(myData[myData$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
>>>
>>> $B
>>> [,1]
>>> [1,] 10
>>>
>>> -------------------------------------
>>> David L Carlson
>>> Department of Anthropology
>>> Texas A&M University
>>> College Station, TX 77840-4352
>>>
>>>
>>> -----Original Message-----
>>> From: R-help [mailto:r-help-bounces at r-project.org] On Behalf Of Jim Lemon
>>> Sent: Tuesday, August 23, 2016 6:02 PM
>>> To: Gang Chen; r-help mailing list
>>> Subject: Re: [R] aggregate
>>>
>>> Hi Gang Chen,
>>> If I have the right idea:
>>>
>>> for(zval in levels(myData$Z))
>>> crossprod(as.matrix(myData[myData$Z==zval,c("X","Y")]))
>>>
>>> Jim
>>>
>>> On Wed, Aug 24, 2016 at 8:03 AM, Gang Chen <gangchen6 at 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 at 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 at 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.
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