OK. Thanks to all. Suppose I have two vectors, x and y. Is there a way to do the covariance matrix with ?apply?. The matrix I need really contains the deviation products divided by the degrees of freedom (n-1). That is, the elements (1,1), (1,2),...,(1,n) (2,1), (2,2),...., (2,n) .... (n,1),(n,2),...,(n,n).> Hello, > > This doesn't make sense, if you have only one vector you can estimate > its variance with > > var(x) > > > but there is no covariance, the joint variance of two rv's. "co" or > joint with what if you have only x? > Note that the variance of x[1] or any other vector element is zero, > it's only one value therefore it does not vary. A similar reasonong > can be applied to cov(x[1], x[2]), etc. > > Hope this helps, > > Rui Barradas > > ?s 12:14 de 04/10/2024, Steven Yen escreveu: >> Hello >> >> I have a vector: >> >> set.seed(123) > n<-3 > x<-rnorm(n); x [1] -0.56047565 -0.23017749 >> 1.55870831 I like to create a matrix with elements containing >> variances and covariances of x. That is var(x[1]) cov(x[1],x[2]) >> cov(x[1],x[3]) cov(x[2],x[1]) var(x[2]) cov(x[2],x[3]) cov(x[3],x[1]) >> cov(x[3],x[2]) var(x[3]) And I like to do it with "apply". Thanks. >> >> On 10/4/2024 6:35 PM, Rui Barradas wrote: >>> Hello, >>> >>> If you have a numeric matrix or data.frame, try something like >>> >>> cov(mtcars) >>> >>> Hope this helps, >>> >>> Rui Barradas >>> >>> >>> ?s 10:15 de 04/10/2024, Steven Yen escreveu: >>>> On 10/4/2024 5:13 PM, Steven Yen wrote: >>>> >>>>> Pardon me!!! >>>>> >>>>> What makes you think this is a homework question? You are not >>>>> obligated to respond if the question is not intelligent enough for >>>>> you. >>>>> >>>>> I did the following: two ways to calculate a covariance matrix but >>>>> wonder how I might replicate the results with "apply". I am not too >>>>> comfortable with the online documentation of "apply". >>>>> >>>>>> set.seed(122345671) > n<-3 > x<-rnorm(n); x [1] 0.92098449 >>>>>> 0.80940115 >>>>> 0.60374785 > cov1<-outer(x-mean(x),x-mean(x))/(n-1); cov1 [,1] [,2] >>>>> [,3] [1,] 0.0102159207 0.00224105983 -0.0124569805 [2,] 0.0022410598 >>>>> 0.00049161983 -0.0027326797 [3,] -0.0124569805 -0.00273267965 >>>>> 0.0151896601 > cov2<-(x-mean(x))%*%t((x-mean(x)))/(n-1); cov2 [,1] >>>>> [,2] [,3] [1,] 0.0102159207 0.00224105983 -0.0124569805 [2,] >>>>> 0.0022410598 0.00049161983 -0.0027326797 [3,] -0.0124569805 >>>>> -0.00273267965 0.0151896601 > >>>>> On 10/4/2024 4:57 PM, Uwe Ligges wrote: >>>>>> Homework questions are not answered on this list. >>>>>> >>>>>> Best, >>>>>> Uwe Ligges >>>>>> >>>>>> >>>>>> >>>>>> On 04.10.2024 10:32, Steven Yen wrote: >>>>>>> The following line calculates standard deviations of a column >>>>>>> vector: >>>>>>> >>>>>>> se<-apply(dd,1,sd) >>>>>>> >>>>>>> How can I calculate the covariance matrix using apply? Thanks. >>>>>>> >>>>>>> ______________________________________________ >>>>>>> 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 >>>>>>> https://www.R-project.org/posting-guide.html >>>>>>> and provide commented, minimal, self-contained, reproducible code. >>>> ????[[alternative HTML version deleted]] >>>> >>>> ______________________________________________ >>>> 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 https://www.R-project.org/posting- >>>> guide.html >>>> and provide commented, minimal, self-contained, reproducible code. >>> >>> > >
? Fri, 4 Oct 2024 20:28:01 +0800 Steven Yen <styen at ntu.edu.tw> ?????:> Suppose I have two vectors, x and y. Is there a way > to do the covariance matrix with ?apply?.There is no covariance matrix for just two samples (vectors) 'x' and 'y'. You can only get one covariance value for these. If you had a pair of vectors of _random variates_, the situation would be different, but those are more abstract mathematical concepts. You would need to sample every random variate, producing two matrices 'x' and 'y' in order to calculate a covariance matrix for them. -- Best regards, Ivan
It's still hard to figure out what you want. If you have two vectors you can compute their (2x2) covariance matrix using cov(cbind(x,y)). If you want to compute all pairwise squared differences between elements of x and y you could use outer(x, y, "-")^2. Can you explain a little bit more about (1) the context for your question and (2) why you want/need to use apply() ? On 2024-10-04 8:28 a.m., Steven Yen wrote:> OK. Thanks to all. Suppose I have two vectors, x and y. Is there a way > to do the covariance matrix with ?apply?. The matrix I need really > contains the deviation products divided by the degrees of freedom (n-1). > That is, the elements > > (1,1), (1,2),...,(1,n) > > (2,1), (2,2),...., (2,n) > > .... > > (n,1),(n,2),...,(n,n). > >> Hello, >> >> This doesn't make sense, if you have only one vector you can estimate >> its variance with >> >> var(x) >> >> >> but there is no covariance, the joint variance of two rv's. "co" or >> joint with what if you have only x? >> Note that the variance of x[1] or any other vector element is zero, >> it's only one value therefore it does not vary. A similar reasonong >> can be applied to cov(x[1], x[2]), etc. >> >> Hope this helps, >> >> Rui Barradas >> >> ?s 12:14 de 04/10/2024, Steven Yen escreveu: >>> Hello >>> >>> I have a vector: >>> >>> set.seed(123) > n<-3 > x<-rnorm(n); x [1] -0.56047565 -0.23017749 >>> 1.55870831 I like to create a matrix with elements containing >>> variances and covariances of x. That is var(x[1]) cov(x[1],x[2]) >>> cov(x[1],x[3]) cov(x[2],x[1]) var(x[2]) cov(x[2],x[3]) cov(x[3],x[1]) >>> cov(x[3],x[2]) var(x[3]) And I like to do it with "apply". Thanks. >>> >>> On 10/4/2024 6:35 PM, Rui Barradas wrote: >>>> Hello, >>>> >>>> If you have a numeric matrix or data.frame, try something like >>>> >>>> cov(mtcars) >>>> >>>> Hope this helps, >>>> >>>> Rui Barradas >>>> >>>> >>>> ?s 10:15 de 04/10/2024, Steven Yen escreveu: >>>>> On 10/4/2024 5:13 PM, Steven Yen wrote: >>>>> >>>>>> Pardon me!!! >>>>>> >>>>>> What makes you think this is a homework question? You are not >>>>>> obligated to respond if the question is not intelligent enough for >>>>>> you. >>>>>> >>>>>> I did the following: two ways to calculate a covariance matrix but >>>>>> wonder how I might replicate the results with "apply". I am not too >>>>>> comfortable with the online documentation of "apply". >>>>>> >>>>>>> set.seed(122345671) > n<-3 > x<-rnorm(n); x [1] 0.92098449 >>>>>>> 0.80940115 >>>>>> 0.60374785 > cov1<-outer(x-mean(x),x-mean(x))/(n-1); cov1 [,1] [,2] >>>>>> [,3] [1,] 0.0102159207 0.00224105983 -0.0124569805 [2,] 0.0022410598 >>>>>> 0.00049161983 -0.0027326797 [3,] -0.0124569805 -0.00273267965 >>>>>> 0.0151896601 > cov2<-(x-mean(x))%*%t((x-mean(x)))/(n-1); cov2 [,1] >>>>>> [,2] [,3] [1,] 0.0102159207 0.00224105983 -0.0124569805 [2,] >>>>>> 0.0022410598 0.00049161983 -0.0027326797 [3,] -0.0124569805 >>>>>> -0.00273267965 0.0151896601 > >>>>>> On 10/4/2024 4:57 PM, Uwe Ligges wrote: >>>>>>> Homework questions are not answered on this list. >>>>>>> >>>>>>> Best, >>>>>>> Uwe Ligges >>>>>>> >>>>>>> >>>>>>> >>>>>>> On 04.10.2024 10:32, Steven Yen wrote: >>>>>>>> The following line calculates standard deviations of a column >>>>>>>> vector: >>>>>>>> >>>>>>>> se<-apply(dd,1,sd) >>>>>>>> >>>>>>>> How can I calculate the covariance matrix using apply? Thanks. >>>>>>>> >>>>>>>> ______________________________________________ >>>>>>>> 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 >>>>>>>> https://www.R-project.org/posting-guide.html >>>>>>>> and provide commented, minimal, self-contained, reproducible code. >>>>> ????[[alternative HTML version deleted]] >>>>> >>>>> ______________________________________________ >>>>> 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 https://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 https://www.R-project.org/posting- > guide.html > and provide commented, minimal, self-contained, reproducible code.-- Dr. Benjamin Bolker Professor, Mathematics & Statistics and Biology, McMaster University Director, School of Computational Science and Engineering > E-mail is sent at my convenience; I don't expect replies outside of working hours.
Hello, You don't need apply, covariance calculations are so frequent that R or any other statistics package already has pre-programmed functions. This time with two vectors x and y. set.seed(123) n <- 3 x <- rnorm(n) y <- rnorm(n) # the two main diagonal values var(x) #> [1] 1.300025 var(y) #> [1] 0.8704518 # the secondary diagonal values cov(x, y) #> [1] 1.056885 # cov(x, y) == cov(y, x) cov(y, x) #> [1] 1.056885 # and the result you are after # (the covariance matrix is symmetric) cov(cbind(x, y)) #> x y #> x 1.300025 1.0568845 #> y 1.056885 0.8704518 Hope this helps, Rui Barradas ?s 13:28 de 04/10/2024, Steven Yen escreveu:> OK. Thanks to all. Suppose I have two vectors, x and y. Is there a way > to do the covariance matrix with ?apply?. The matrix I need really > contains the deviation products divided by the degrees of freedom (n-1). > That is, the elements > > (1,1), (1,2),...,(1,n) > > (2,1), (2,2),...., (2,n) > > .... > > (n,1),(n,2),...,(n,n). > >> Hello, >> >> This doesn't make sense, if you have only one vector you can estimate >> its variance with >> >> var(x) >> >> >> but there is no covariance, the joint variance of two rv's. "co" or >> joint with what if you have only x? >> Note that the variance of x[1] or any other vector element is zero, >> it's only one value therefore it does not vary. A similar reasonong >> can be applied to cov(x[1], x[2]), etc. >> >> Hope this helps, >> >> Rui Barradas >> >> ?s 12:14 de 04/10/2024, Steven Yen escreveu: >>> Hello >>> >>> I have a vector: >>> >>> set.seed(123) > n<-3 > x<-rnorm(n); x [1] -0.56047565 -0.23017749 >>> 1.55870831 I like to create a matrix with elements containing >>> variances and covariances of x. That is var(x[1]) cov(x[1],x[2]) >>> cov(x[1],x[3]) cov(x[2],x[1]) var(x[2]) cov(x[2],x[3]) cov(x[3],x[1]) >>> cov(x[3],x[2]) var(x[3]) And I like to do it with "apply". Thanks. >>> >>> On 10/4/2024 6:35 PM, Rui Barradas wrote: >>>> Hello, >>>> >>>> If you have a numeric matrix or data.frame, try something like >>>> >>>> cov(mtcars) >>>> >>>> Hope this helps, >>>> >>>> Rui Barradas >>>> >>>> >>>> ?s 10:15 de 04/10/2024, Steven Yen escreveu: >>>>> On 10/4/2024 5:13 PM, Steven Yen wrote: >>>>> >>>>>> Pardon me!!! >>>>>> >>>>>> What makes you think this is a homework question? You are not >>>>>> obligated to respond if the question is not intelligent enough for >>>>>> you. >>>>>> >>>>>> I did the following: two ways to calculate a covariance matrix but >>>>>> wonder how I might replicate the results with "apply". I am not too >>>>>> comfortable with the online documentation of "apply". >>>>>> >>>>>>> set.seed(122345671) > n<-3 > x<-rnorm(n); x [1] 0.92098449 >>>>>>> 0.80940115 >>>>>> 0.60374785 > cov1<-outer(x-mean(x),x-mean(x))/(n-1); cov1 [,1] [,2] >>>>>> [,3] [1,] 0.0102159207 0.00224105983 -0.0124569805 [2,] 0.0022410598 >>>>>> 0.00049161983 -0.0027326797 [3,] -0.0124569805 -0.00273267965 >>>>>> 0.0151896601 > cov2<-(x-mean(x))%*%t((x-mean(x)))/(n-1); cov2 [,1] >>>>>> [,2] [,3] [1,] 0.0102159207 0.00224105983 -0.0124569805 [2,] >>>>>> 0.0022410598 0.00049161983 -0.0027326797 [3,] -0.0124569805 >>>>>> -0.00273267965 0.0151896601 > >>>>>> On 10/4/2024 4:57 PM, Uwe Ligges wrote: >>>>>>> Homework questions are not answered on this list. >>>>>>> >>>>>>> Best, >>>>>>> Uwe Ligges >>>>>>> >>>>>>> >>>>>>> >>>>>>> On 04.10.2024 10:32, Steven Yen wrote: >>>>>>>> The following line calculates standard deviations of a column >>>>>>>> vector: >>>>>>>> >>>>>>>> se<-apply(dd,1,sd) >>>>>>>> >>>>>>>> How can I calculate the covariance matrix using apply? Thanks. >>>>>>>> >>>>>>>> ______________________________________________ >>>>>>>> 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 >>>>>>>> https://www.R-project.org/posting-guide.html >>>>>>>> and provide commented, minimal, self-contained, reproducible code. >>>>> ????[[alternative HTML version deleted]] >>>>> >>>>> ______________________________________________ >>>>> 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 https://www.R-project.org/posting- >>>>> guide.html >>>>> and provide commented, minimal, self-contained, reproducible code. >>>> >>>> >> >>-- Este e-mail foi analisado pelo software antiv?rus AVG para verificar a presen?a de v?rus. www.avg.com