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.
>>>
>>>
>
>