Correlate
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.
My question is there a way to extract the number of rows (N) used in
the correlation analysis?.
Thank you,
On Mon, Aug 22, 2022 at 1:00 PM John Fox <jfox at 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 <tebert at 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 at gmail.com> Sent: Monday, August 22, 2022 12:07 PM To: Ebert,Timothy Aaron <tebert at ufl.edu> Cc: Val <valkremk at gmail.com>; r-help at R-project.org (r-help at r-project.org) <r-help at 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 at 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")$p.value) }
Because of missing data, this computes the correlations on different subsets of the data. A simple solution is to filter the data for NAs: D <- na.omit(dat) More comments below:
Rather than printing you could set up an array or list to save the results.
Tim
-----Original Message-----
From: R-help <r-help-bounces at r-project.org> On Behalf Of Val
Sent: Monday, August 22, 2022 11:09 AM
To: r-help at R-project.org (r-help at r-project.org) <r-help at r-project.org>
Subject: [R] Correlate
[External Email]
Hi all,
I have a data set with ~250 variables(columns). I want to calculate
the correlation of one variable with the rest of the other variables
and also want the p-values for each correlation. Please see the
sample data and my attempt. I have got the correlation but unable to
get the p-values
dat <- read.table(text="x1 x2 x3 x4
1.68 -0.96 -1.25 0.61
-0.06 0.41 0.06 -0.96
. 0.08 1.14 1.42
0.80 -0.67 0.53 -0.68
0.23 -0.97 -1.18 -0.78
-1.03 1.11 -0.61 .
2.15 . 0.02 0.66
0.35 -0.37 -0.26 0.39
-0.66 0.89 . -1.49
0.11 1.52 0.73 -1.03",header=TRUE)
#change all to numeric
dat[] <- lapply(dat, function(x) as.numeric(as.character(x)))
This data manipulation is unnecessary. Just specify the argument na.strings="." to read.table().
data_cor <- cor(dat[ , colnames(dat) != "x1"], dat$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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-- John Fox, Professor Emeritus McMaster University Hamilton, Ontario, Canada web: https://socialsciences.mcmaster.ca/jfox/