please help generate a square correlation matrix
It sounds as though you have null hypothesis "x records independent
Bernoulli trials with the same (unknown) success probability p_x, y
records independent Bernoulli trials with the same (unknown) success
probability p_y, x and y are independent" and alternative hypothesis
"x and y succeed less often than they would under the null
hypothesis". The obvious way to do that is to fit \hat{p_x} =
sum(x)/length(x), \hat{p_y| = sum(y)/length(y), and then compute the
lower tail Pr(number of times x and y succeed <= sum(xy) | p_x =
\hat{p_x} \& p_y = \hat{p_y}, and unless I am completely off my head
with sleepiness, this is just
pbinom(sum(x*y), length(x), mean(x)*mean(y))
So I don't quite see why you wanted correlations.
Since you say that "WE measured ..." the various Signor-Lipps-like
scenarios I was thinking of probably don't apply. There are other
threats to validity:
- the presence of two (or more) mutations may be hard for your
equipment to detect
- patients with multiple mutations may die faster so may be less
likely to be captured for your study
- cell division rates decrease with age
https://www.ncbi.nlm.nih.gov/pmc/articles/PMC6789572/ so mutations
whose likelihood depends on *rate* might tend to occur earlier in
life while mutations that depend on *accumulated* error might tend to
occur later in life, so "x occurs SOME time in a patient's life" and
"y occurs SOME time in a patient's life" might be independent while "x
and y occur at the SAME time in a patient's life" might be unlikely.
It would be interesting to check whether the frequency of each
mutation is independent of patient age, because you might want to
stratify the pbinom test by age in that case. Exposure to
environmental mutagens is also likely to vary with age.
Looking at supermarket data in the past primed me to expect rates to
vary with age. Sunscreen and cough mixture are negatively associated
(:-).
On Sun, 28 Jul 2024 at 12:40, Yuan Chun Ding <ycding at coh.org> wrote:
HI Bert,
Thank you for extra help!!
Yes, exactly, your interpretation is perfectly correct and your R code is what I should look for.
after generated all those negative values of correlation,
I thought about the extremely small p values associated with those negative correlation, which is not meaningful as I truncated my data.
When examining the exclusiveness of mutation pairs, what I first thought about is correlation, so stepped into a more complicated correlation journey.
However, what Richard share is very helpful to explain why I got negative correlation values for all pairs.
In my case, we measured all mutations for all 1000 samples using an exactly same sequencing method, so no issue of never-reporting.
I am very grateful for help and comments from Rui, Richard and Bert!!
Ding
From: Bert Gunter <bgunter.4567 at gmail.com>
Sent: Saturday, July 27, 2024 4:50 PM
To: Yuan Chun Ding <ycding at coh.org>
Cc: Richard O'Keefe <raoknz at gmail.com>; r-help at r-project.org
Subject: Re: [R] please help generate a square correlation matrix
Your expanded explanation helps clarify your intent. Herewith some comments. Of course, feel free to ignore and not respond. And, as always, my apologies if I have failed to comprehend your intent. 1. I would avoid any notion of "statistical
Your expanded explanation helps clarify your intent. Herewith some
comments. Of course, feel free to ignore and not respond. And, as
always, my apologies if I have failed to comprehend your intent.
1. I would avoid any notion of "statistical significance" like the
plague. This is a purely exploratory exercise.
2. My understanding is that you want to know the proportion of rows in
a pair of columns/vectors in which only 1 values of the pair is 1 out
of the number of pairs where 1 or 2 values is 1. In R syntax, this is
simply:
sum(xor(x, y)) / sum(x | y) ,
where x and y are two columns of 1's and 0's
Better yet might be to report both this *and* sum(x|y) to help you
judge "meaningfulness".
Here is a simple function that does this
## first, define a function that does above calculation:
assoc <- \(z){
x <- z[,1]; y <- z[,2]
n <- sum(x|y)
c(prop = sum(xor(x, y))/n, N = n)
}
## Now a function that uses it for the various combinations:
somecor <- function(dat, func = assoc){
dat <- as.matrix(dat)
indx <- seq_len(ncol(dat))
rbind(w <- combn(indx,2),
combn(indx, 2, FUN = \(m)func(dat[,m]) )) |>
t() |> round(digits =2) |>
'dimnames<-'(list(rep.int('',ncol(w)), c("","", "prop","N")))
}
# Now apply it to your example data:
somecor(dat)
## which gives
prop N
1 2 0.67 6
1 3 0.60 5
1 4 0.57 7
2 3 0.60 5
2 4 0.33 6
3 4 0.71 7
This seems more interpretable and directly useful to me. Bigger values
of prop for bigger N are the more interesting, assuming I have
interpreted you correctly.
Cheers,
Bert
On Sat, Jul 27, 2024 at 12:54?PM Yuan Chun Ding <ycding at coh.org> wrote:
Hi Richard,
Nice to know you had similar experience.
Yes, your understanding is right. all correlations are negative after removing double-zero rows.
It is consistent with a heatmap we generated.
1 is for a cancer patient with a specific mutation. 0 is no mutation for the same mutation type in a patient.
a pair of mutation type (two different mutations) are exclusive for most of patients in heatmap or oncoplots.
If we include all 1000 patients, 900 of patients with no mutations in both mutation types, then the correlation is not significant at all.
But eyeball the heatmap (oncoplots) for mutation (row) by patient (column), mutations are exclusive for most of patients,
so I want to measure how strong the exclusiveness between two specific mutation types across those patients with at least one mutation type.
Then put the pair of mutations with strong negative mutations on the top rows by order of negative mutation values.
Regarding a final application, maybe there are some usage for my case.
If one develops two drugs specific to the two negative correlated mutations, the drug treatment for cancer patients is usually only for those patients carrying the specific mutation,
then it is informative to know how strong the negative correlation when considering different combination of treatment strategies.
Ding
From: R-help <r-help-bounces at r-project.org> On Behalf Of Richard O'Keefe
Sent: Saturday, July 27, 2024 4:47 AM
To: Bert Gunter <bgunter.4567 at gmail.com>
Cc: r-help at r-project.org
Subject: Re: [R] please help generate a square correlation matrix
Curses, my laptop is hallucinating again. Hope I can get through this. So we're talking about correlations between binary variables. Suppose we have two 0-1-valued variables, x and y. Let A <- sum(x*y) # number of cases where x and y are
Curses, my laptop is hallucinating again. Hope I can get through this.
So we're talking about correlations between binary variables.
Suppose we have two 0-1-valued variables, x and y.
Let A <- sum(x*y) # number of cases where x and y are both 1.
Let B <- sum(x)-A # number of cases where x is 1 and y is 0
Let C <- sum(y)-A # number of cases where y is 1 and x is 0
Let D <- sum(!x * !y) # number of cases where x and y are both 0.
(also D = length(x)-A-B-C)
All the information is summarised in the 2-by-2 contingency table.
Some years ago, Nathan Rountree and I supervised Yung-Sing Koh's
data-mining PhD.
She surveyed the data mining literature and found some 37 different
"interestingness measures" for two-variable associations -- if I
remember correctly; there were a lot of them. They fell into a much
smaller number of qualitatively similar groups.
At any rate, the Pearson correlation between x and y is
(A*D - B*C)/sqrt((A+B)*(C+D)*(A+C)*(B+D))
So what happens when we delete the rows where x = 0 and y = 0?
Right, it forces D to 0, leaving A B C unchanged.
And looking at the numerator,
If you delete rows with x = 0 y = 0 you MUST get a negative correlation.
Quite a modest "true" correlation (based on all the data) like -0.2
can masquerade as quite a strong "zero-suppressed" correlation like
-0.6. Even +0.2 can turn into -0.4. (These figures are from a
particular simulation run and may not apply in your case.)
Now one of the reasons why Yun-Sing Koh, Nathan Rountree, and I were
interested in interestingness measures is perhaps coincidentally
related to the file drawer/underreporting problem: it's quite common
for rows where x = 0 and y = 0 never to have been reported to you, so
we were hoping there were measures immune to that. I have argued for
years that "till record analysis" for supermarkets &c is badly flawed
by two facts: (a) it is hard to measure how much of a product people
WOULD have bought if only you had offered it for sale (although you
can make educated guesses) and (b) till records provide no evidence on
what the people who walked out without buying anything wanted (was the
price too high? could they not find it?). Problem (a) leads to a
commercial variant of the Signor-Lipps effect: "when x and/or y were
available for purchase" is not the same as "the period for which data
were recorded", thus inflating D, perhaps massively. Methods
developed for handling the Signor-Lipps effect in paleontology can be
used to estimate when x and y were available helping you to recover a
more realistic N=A+B+C+D. I really should have published that.
All of which is a long-winded way of saying that
- Pearson correlations on binary columns can be computed very efficiently
- the rows with x=0 and y=0 may be very informative, even essential for analysis
- delete them at your peril.
- really, delete them at your peril.
On Sat, 27 Jul 2024 at 23:07, Richard O'Keefe <raoknz at gmail.com> wrote:
Let's go back to the original posting.
in each column, less than 10% values are 1, most of them are 0;
so I want to remove a row with value of zero in both columns when calculate correlation between two columns.
So we're talking about correlations between binary variables.
Suppose we have two 0-1-valued variables, x and y.
Let A <- sum(x*y) # number of cases where x and y are both 1.
Let B <- sum(x)-a # number of cases where x is 1 and y is 0
Let C <- sum(y)-a # number of cases where y is 1 and x is 0
Let D <- sum(!x * !y) # number of cases where x and y are both 0.
N
On Fri, 26 Jul 2024 at 12:07, Bert Gunter <bgunter.4567 at gmail.com> wrote:
If I have understood the request, I'm not sure that omitting all 0
pairs for each pair of columns makes much sense, but be that as it
may, here's another way to do it by using the 'FUN' argument of combn
to encapsulate any calculations that you do. I just use cor() as the
calculation -- you can use anything you like that takes two vectors of
0's and 1's and produces fixed length numeric results (or fromm which
you can extract such).
I encapsulated it all in a little function. Note that I first
converted the data frame to a matrix. Because of their generality,
data frames carry a lot of extra baggage that can slow purely numeric
manipulations down.
Anyway, here's the function, 'somecors' (I'm a bad name picker :( ! )
somecors <- function(dat, func = cor){
dat <- as.matrix(dat)
indx <- seq_len(ncol(dat))
combn(indx, 2, FUN = \(z) {
i <- z[1]; j <- z[2]
k <- dat[, i ] | dat[, j ]
c(z,func(dat[k,i ], dat[k,j ]))
})
}
Results come out as a matrix with combn(ncol(dat),2) columns, the
first 2 rows giving the pair of column numbers for each column,and
then 1 or more rows (possibly extracted) from whatever func you use.
Here's the results for your data formatted to 2 decimal places:
round(somecors(dat),2)
[,1] [,2] [,3] [,4] [,5] [,6]
[1,] 1.0 1.00 1.00 2.00 2 3.00
[2,] 2.0 3.00 4.00 3.00 4 4.00
[3,] -0.5 -0.41 -0.35 -0.41 NA -0.47
Warning message:
In func(dat[k, i], dat[k, j]) : the standard deviation is zero
The NA and warning comes in the 2,4 pair of columns because after
removing all zero rows in the pair, dat[,4] is all 1's, giving a zero
in the denominator of the cor() calculation -- again, assuming I have
correctly understood your request. If so, this might be something you
need to worry about.
Again, feel free to ignore if I have misinterpreterd or this does not suit.
Cheers,
Bert
On Thu, Jul 25, 2024 at 2:01?PM Rui Barradas <ruipbarradas at sapo.pt> wrote:
?s 20:47 de 25/07/2024, Yuan Chun Ding escreveu:
Hi Rui,
You are always very helpful!! Thank you,
I just modified your R codes to remove a row with zero values in both column pair as below for my real data.
Ding
dat<-gene22mut.coded
r <- P <- matrix(NA, nrow = 22L, ncol = 22L,
dimnames = list(names(dat), names(dat)))
for(i in 1:22) {
#i=1
x <- dat[[i]]
for(j in (1:22)) {
#j=2
if(i == j) {
# there's nothing to test, assign correlation 1
r[i, j] <- 1
} else {
tmp <-cbind(x,dat[[j]])
row0 <-rowSums(tmp)
tem2 <-tmp[row0!=0,]
tmp3 <- cor.test(tem2[,1],tem2[,2])
r[i, j] <- tmp3$estimate
P[i, j] <- tmp3$p.value
}
}
}
r<-as.data.frame(r)
P<-as.data.frame(P)
From: R-help <r-help-bounces at r-project.org> On Behalf Of Yuan Chun Ding via R-help
Sent: Thursday, July 25, 2024 11:26 AM
To: Rui Barradas <ruipbarradas at sapo.pt>; r-help at r-project.org
Subject: Re: [R] please help generate a square correlation matrix
HI Rui, Thank you for the help! You did not remove a row if zero values exist in both column pair, right? Ding From: Rui Barradas <ruipbarradas@?sapo.?pt> Sent: Thursday, July 25, 2024 11:?15 AM To: Yuan Chun Ding <ycding@?coh.?org>;
HI Rui,
Thank you for the help!
You did not remove a row if zero values exist in both column pair, right?
Ding
From: Rui Barradas <ruipbarradas at sapo.pt<mailto:ruipbarradas at sapo.pt>>
Sent: Thursday, July 25, 2024 11:15 AM
To: Yuan Chun Ding <ycding at coh.org<mailto:ycding at coh.org>>; r-help at r-project.org<mailto:r-help at r-project.org>
Subject: Re: [R] please help generate a square correlation matrix
?s 17:?39 de 25/07/2024, Yuan Chun Ding via R-help escreveu: > Hi R users, > > I generated a square correlation matrix for the dat dataframe below; > dat<-data.?frame(g1=c(1,0,0,1,1,1,0,0,0), > g2=c(0,1,0,1,0,1,1,0,0), > g3=c(1,1,0,0,0,1,0,0,0),
?s 17:39 de 25/07/2024, Yuan Chun Ding via R-help escreveu:
Hi R users,
I generated a square correlation matrix for the dat dataframe below;
dat<-data.frame(g1=c(1,0,0,1,1,1,0,0,0),
g2=c(0,1,0,1,0,1,1,0,0),
g3=c(1,1,0,0,0,1,0,0,0),
g4=c(0,1,0,1,1,1,1,1,0))
library("Hmisc")
dat.rcorr = rcorr(as.matrix(dat))
dat.r <-round(dat.rcorr$r,2)
however, I want to modify this correlation calculation;
my dat has more than 1000 rows and 22 columns;
in each column, less than 10% values are 1, most of them are 0;
so I want to remove a row with value of zero in both columns when calculate correlation between two columns.
I just want to check whether those values of 1 are correlated between two columns.
Please look at my code in the following;
cor.4gene <-matrix(0,nrow=4*4, ncol=4)
for (i in 1:4){
#i=1
for (j in 1:4) {
#j=1
d <-dat[,c(i,j)]%>%
filter(eval(as.symbol(colnames(dat)[i]))!=0 |
eval(as.symbol(colnames(dat)[j]))!=0)
c <-cor.test(d[,1],d[,2])
cor.4gene[i*j,]<-c(colnames(dat)[i],colnames(dat)[j],
c$estimate,c$p.value)
}
}
cor.4gene<-as.data.frame(cor.4gene)%>%filter(V1 !=0)
colnames(cor.4gene)<-c("gene1","gene2","cor","P")
Can you tell me what mistakes I made?
first, why cor is NA when calculation of correlation for g1 and g1, I though it should be 1.
cor.4gene$cor[is.na(cor.4gene$cor)]<-1
cor.4gene$cor[is.na(cor.4gene$P)]<-0
cor.4gene.sq <-pivot_wider(cor.4gene, names_from = gene1, values_from = cor)
Then this line of code above did not generate a square matrix as what the HMisc library did.
How to fix my code?
Thank you,
Ding
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Hello,
You are complicating the code, there's no need for as.symbol/eval, the
column numbers do exactly the same.
# create the two results matrices beforehand
r <- P <- matrix(NA, nrow = 4L, ncol = 4L, dimnames = list(names(dat),
names(dat)))
for(i in 1:4) {
x <- dat[[i]]
for(j in (1:4)) {
if(i == j) {
# there's nothing to test, assign correlation 1
r[i, j] <- 1
} else {
tmp <- cor.test(x, dat[[j]])
r[i, j] <- tmp$estimate
P[i, j] <- tmp$p.value
}
}
}
# these two results are equal up to floating-point precision
dat.rcorr$r
#> g1 g2 g3 g4
#> g1 1.0000000 0.1000000 0.3162278 0.1581139
#> g2 0.1000000 1.0000000 0.3162278 0.6324555
#> g3 0.3162278 0.3162278 1.0000000 0.0000000
#> g4 0.1581139 0.6324555 0.0000000 1.0000000
r
#> g1 g2 g3 g4
#> g1 1.0000000 0.1000000 3.162278e-01 1.581139e-01
#> g2 0.1000000 1.0000000 3.162278e-01 6.324555e-01
#> g3 0.3162278 0.3162278 1.000000e+00 1.355253e-20
#> g4 0.1581139 0.6324555 1.355253e-20 1.000000e+00
# these two results are equal up to floating-point precision
dat.rcorr$P
#> g1 g2 g3 g4
#> g1 NA 0.79797170 0.4070838 0.68452834
#> g2 0.7979717 NA 0.4070838 0.06758329
#> g3 0.4070838 0.40708382 NA 1.00000000
#> g4 0.6845283 0.06758329 1.0000000 NA
P
#> g1 g2 g3 g4
#> g1 NA 0.79797170 0.4070838 0.68452834
#> g2 0.7979717 NA 0.4070838 0.06758329
#> g3 0.4070838 0.40708382 NA 1.00000000
#> g4 0.6845283 0.06758329 1.0000000 NA
You can put these two results in a list, like Hmisc::rcorr does.
lst_rcorr <- list(r = r, P = P)
Hope this helps,
Rui Barradas
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Hello,
Here are two other ways.
The first is equivalent to your long format attempt.
library(tidyverse)
dat %>%
names() %>%
expand.grid(., .) %>%
apply(1L, \(x) {
tmp <- dat[rowSums(dat[x]) > 0, ]
tmp2 <- cor.test(tmp[[ x[1L] ]], tmp[[ x[2L] ]])
c(tmp2$estimate, P = tmp2$p.value)
}) %>%
t() %>%
as.data.frame() %>%
cbind(tmp_df, .) %>%
na.omit()
The second is, in my opinion the one that makes more sense. If you see
the results, cor is symmetric (as it should) so the calculations are
repeated. If you only run the cor.tests on the combinations of
names(dat) by groups of 2, it will save a lot of work. But the output is
a much smaller a data.frame.
cbind(
combn(names(dat), 2L) %>%
t() %>%
as.data.frame(),
combn(dat, 2L, FUN = \(d) {
d2 <- d[rowSums(d) > 0, ]
tmp2 <- cor.test(d2[[1L]], d2[[2L]])
c(tmp2$estimate, P = tmp2$p.value)
}) %>% t()
) %>% na.omit()
Hope this helps,
Rui Barradas
______________________________________________
R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see
and provide commented, minimal, self-contained, reproducible code.
______________________________________________
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and provide commented, minimal, self-contained, reproducible code.
______________________________________________
R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see
and provide commented, minimal, self-contained, reproducible code.