Thank you John for your help and advice.
On Fri, Aug 26, 2022 at 11:04 AM John Fox <j...@mcmaster.ca> wrote:
>
> Dear Val,
>
> On 2022-08-26 10:41 a.m., Val wrote:
> > Hi John and Timothy
> >
> > Thank you for your suggestion and help. Using the sample data, I did
> > carry out a test run and found a difference in the correlation result.
> >
> > Option 1.
> > data_cor <- cor(dat[ , colnames(dat) != "x1"], # Calculate correlations
> > dat$x1, method = "pearson", use = "complete.obs")
> > resulted
> > [,1]
> > x2 -0.5845835
> > x3 -0.4664220
> > x4 0.7202837
> >
> > Option 2.
> > for(i in colnames(dat)){
> > print(cor.test(dat[,i], dat$x1, method = "pearson", use =
> > "complete.obs")$estimate)
> > }
> > [,1]
> > x2 -0.7362030
> > x3 -0.04935132
> > x4 0.85766290
> >
> > This was crosschecked using Excel and other softwares and all matches
> > with option 2.
> > One of the factors that contributed for this difference is loss of
> > information when we are using na.rm(). This is because that if x2 has
> > missing value but x3 and x4 don’t have then na.rm() removed entire
> > row information including x3 and x4.
>
> Yes, I already explained that in my previous message.
>
> As well, cor() is capable of computing pairwise-complete correlations --
> see ?cor.
>
> There's not an obvious right answer here, however. Using
> pairwise-complete correlations can produce inconsistent (i.e.,
> non-positive semi-definite) correlation matrices because correlations
> are computed on different subsets of the data.
>
> There are much better ways to deal with missing data.
>
> >
> > My question is there a way to extract the number of rows (N) used in
> > the correlation analysis?.
>
> I'm sure that there are many ways, but here is one that is very
> simple-minded and should be reasonably efficient for ~250 variables:
>
> > (nc <- ncol(dat))
> [1] 4
>
> > R <- N <- matrix(NA, nc, nc)
> > diag(R) <- 1
> > for (i in 1:(nc - 1)){
> + for (j in (i + 1):nc){
> + R[i, j] <- R[j, i] <-cor(dat[, i], dat[, j], use="complete.obs")
> + N[i, j] <- N[j, i] <- nrow(na.omit(dat[, c(i, j)]))
> + }
> + }
>
> > round(R, 3)
> [,1] [,2] [,3] [,4]
> [1,] 1.000 -0.736 -0.049 0.858
> [2,] -0.736 1.000 0.458 -0.428
> [3,] -0.049 0.458 1.000 0.092
> [4,] 0.858 -0.428 0.092 1.000
>
> > N
> [,1] [,2] [,3] [,4]
> [1,] NA 8 8 8
> [2,] 8 NA 8 8
> [3,] 8 8 NA 8
> [4,] 8 8 8 NA
>
> > round(cor(dat, use="pairwise.complete.obs"), 3) # check
> x1 x2 x3 x4
> x1 1.000 -0.736 -0.049 0.858
> x2 -0.736 1.000 0.458 -0.428
> x3 -0.049 0.458 1.000 0.092
> x4 0.858 -0.428 0.092 1.000
>
> More generally, I think that it's a good idea to learn a little bit
> about R programming if you intend to use R in your work. You'll then be
> able to solve problems like this yourself.
>
> I hope this helps,
> John
>
> > Thank you,
> >
> > On Mon, Aug 22, 2022 at 1:00 PM John Fox <j...@mcmaster.ca> wrote:
> >>
> >> Dear Val,
> >>
> >> On 2022-08-22 1:33 p.m., Val wrote:
> >>> For the time being I am assuming the relationship across variables
> >>> is linear. I want get the values first and detailed examining of
> >>> the relationship will follow later.
> >>
> >> This seems backwards to me, but I'll refrain from commenting further on
> >> whether what you want to do makes sense and instead address how to do it
> >> (not, BTW, because I disagree with Bert's and Tim's remarks).
> >>
> >> Please see below:
> >>
> >>>
> >>> On Mon, Aug 22, 2022 at 12:23 PM Ebert,Timothy Aaron <teb...@ufl.edu>
> >>> wrote:
> >>>>
> >>>> I (maybe) agree, but I would go further than that. There are assumptions
> >>>> associated with the test that are missing. It is not clear that the
> >>>> relationships are all linear. Regardless of a "significant outcome" all
> >>>> of the relationships need to be explored in more detail than what is
> >>>> provided in the correlation test.
> >>>>
> >>>> Multiplicity adjustment as in :
> >>>> https://www.sciencedirect.com/science/article/pii/S0197245600001069 is
> >>>> not an issue that I can see in these data from the information provided.
> >>>> At least not in the same sense as used in the link.
> >>>>
> >>>> My first guess at the meaning of "multiplicity adjustment" was closer to
> >>>> the experimentwise error rate in a multiple comparison procedure.
> >>>> https://dictionary.apa.org/experiment-wise-error-rateEssentially, the
> >>>> type 1 error rate is inflated the more test you do and if you perform
> >>>> enough tests you find significant outcomes by chance alone. There is
> >>>> great significance in the Redskins rule:
> >>>> https://en.wikipedia.org/wiki/Redskins_Rule.
> >>>>
> >>>> A simple solution is to apply a Bonferroni correction where alpha is
> >>>> divided by the number of comparisons. If there are 250, then 0.05/250 =
> >>>> 0.0002. Another approach is to try to discuss the outcomes in a way that
> >>>> makes sense. What is the connection between a football team's last home
> >>>> game an the election result that would enable me to take another team
> >>>> and apply their last home game result to the outcome of a different
> >>>> election?
> >>>>
> >>>> Another complication is if variables x2 through x250 are themselves
> >>>> correlated. Not enough information was provided in the problem to know
> >>>> if this is an issue, but 250 orthogonal variables in a real dataset
> >>>> would be a bit unusual considering the experimentwise error rate
> >>>> previously mentioned.
> >>>>
> >>>> Large datasets can be very messy.
> >>>>
> >>>>
> >>>> Tim
> >>>>
> >>>> -----Original Message-----
> >>>> From: Bert Gunter <bgunter.4...@gmail.com>
> >>>> Sent: Monday, August 22, 2022 12:07 PM
> >>>> To: Ebert,Timothy Aaron <teb...@ufl.edu>
> >>>> Cc: Val <valkr...@gmail.com>; r-help@R-project.org
> >>>> (r-help@r-project.org) <r-help@r-project.org>
> >>>> Subject: Re: [R] Correlate
> >>>>
> >>>> [External Email]
> >>>>
> >>>> ... But of course the p-values are essentially meaningless without some
> >>>> sort of multiplicity adjustment.
> >>>> (search on "multiplicity adjustment" for details). :-(
> >>>>
> >>>> -- Bert
> >>>>
> >>>>
> >>>> On Mon, Aug 22, 2022 at 8:59 AM Ebert,Timothy Aaron <teb...@ufl.edu>
> >>>> wrote:
> >>>>>
> >>>>> A somewhat clunky solution:
> >>>>> for(i in colnames(dat)){
> >>>>> print(cor.test(dat[,i], dat$x1, method = "pearson", use =
> >>>>> "complete.obs")$estimate)
> >>>>> print(cor.test(dat[,i], dat$x1, method = "pearson", use =
> >>>>> "complete.obs")$p.value) }
> >>
> >> Because of missing data, this computes the correlations on different
> >> subsets of the data. A simple solution is to filter the data for NAs:
> >>
> >> D <- na.omit(dat)
> >>
> >> More comments below:
> >>
> >>>>>
> >>>>> Rather than printing you could set up an array or list to save the
> >>>>> results.
> >>>>>
> >>>>>
> >>>>> Tim
> >>>>>
> >>>>> -----Original Message-----
> >>>>> From: R-help <r-help-boun...@r-project.org> On Behalf Of Val
> >>>>> Sent: Monday, August 22, 2022 11:09 AM
> >>>>> To: r-help@R-project.org (r-help@r-project.org) <r-help@r-project.org>
> >>>>> Subject: [R] Correlate
> >>>>>
> >>>>> [External Email]
> >>>>>
> >>>>> Hi all,
> >>>>>
> >>>>> I have a data set with ~250 variables(columns). I want to calculate
> >>>>> the correlation of one variable with the rest of the other variables
> >>>>> and also want the p-values for each correlation. Please see the
> >>>>> sample data and my attempt. I have got the correlation but unable to
> >>>>> get the p-values
> >>>>>
> >>>>> dat <- read.table(text="x1 x2 x3 x4
> >>>>> 1.68 -0.96 -1.25 0.61
> >>>>> -0.06 0.41 0.06 -0.96
> >>>>> . 0.08 1.14 1.42
> >>>>> 0.80 -0.67 0.53 -0.68
> >>>>> 0.23 -0.97 -1.18 -0.78
> >>>>> -1.03 1.11 -0.61 .
> >>>>> 2.15 . 0.02 0.66
> >>>>> 0.35 -0.37 -0.26 0.39
> >>>>> -0.66 0.89 . -1.49
> >>>>> 0.11 1.52 0.73 -1.03",header=TRUE)
> >>>>>
> >>>>> #change all to numeric
> >>>>> dat[] <- lapply(dat, function(x) as.numeric(as.character(x)))
> >>
> >> This data manipulation is unnecessary. Just specify the argument
> >> na.strings="." to read.table().
> >>
> >>>>>
> >>>>> data_cor <- cor(dat[ , colnames(dat) != "x1"], dat$x1, method =
> >>>>> "pearson", use = "complete.obs")
> >>>>>
> >>>>> Result
> >>>>> [,1]
> >>>>> x2 -0.5845835
> >>>>> x3 -0.4664220
> >>>>> x4 0.7202837
> >>>>>
> >>>>> How do I get the p-values ?
> >>
> >> Taking a somewhat different approach from cor.test(), you can apply
> >> Fisher's z-transformation (recall that D is the data filtered for NAs):
> >>
> >> > 2*pnorm(abs(atanh(data_cor)), sd=1/sqrt(nrow(D) - 3), lower.tail=FALSE)
> >> [,1]
> >> x2 0.2462807
> >> x3 0.3812854
> >> x4 0.1156939
> >>
> >> I hope this helps,
> >> John
> >>
> >>>>>
> >>>>> Thank you,
> >>>>>
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> >>> ______________________________________________
> >>> 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.
> >> --
> >> John Fox, Professor Emeritus
> >> McMaster University
> >> Hamilton, Ontario, Canada
> >> web: https://socialsciences.mcmaster.ca/jfox/
> >>
> --
> John Fox, Professor Emeritus
> McMaster University
> Hamilton, Ontario, Canada
> web: https://socialsciences.mcmaster.ca/jfox/
>
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