# [R] Correlate

Val v@|kremk @end|ng |rom gm@||@com
Fri Aug 26 22:06:19 CEST 2022

```Thank you John for your help and advice.

On Fri, Aug 26, 2022 at 11:04 AM John Fox <jfox using 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 <jfox using 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).
> >>
> >>
> >>>
> >>> On Mon, Aug 22, 2022 at 12:23 PM Ebert,Timothy Aaron <tebert using 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.4567 using gmail.com>
> >>>> Sent: Monday, August 22, 2022 12:07 PM
> >>>> To: Ebert,Timothy Aaron <tebert using ufl.edu>
> >>>> Cc: Val <valkremk using gmail.com>; r-help using R-project.org (r-help using r-project.org) <r-help using 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 <tebert using 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)
> >>
> >>
> >>>>>
> >>>>> Rather than printing you could set up an array or list to save the results.
> >>>>>
> >>>>>
> >>>>> Tim
> >>>>>
> >>>>> -----Original Message-----
> >>>>> From: R-help <r-help-bounces using r-project.org> On Behalf Of Val
> >>>>> Sent: Monday, August 22, 2022 11:09 AM
> >>>>> To: r-help using R-project.org (r-help using r-project.org) <r-help using 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
> >>
> >>>>>
> >>>>>       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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> >>>>>
> >>>>> ______________________________________________
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> >>>>> and provide commented, minimal, self-contained, reproducible code.
> >>>
> >>> ______________________________________________
> >>> R-help using r-project.org mailing list -- To UNSUBSCRIBE and more, see
> >>> https://stat.ethz.ch/mailman/listinfo/r-help
> >>> and provide commented, minimal, self-contained, reproducible code.
> >> --
> >> John Fox, Professor Emeritus
> >> McMaster University