similar to: Help with lattice, regressions and respective lines

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) { #

Dear R user I have data created using code below b<-matrix(2:21,nrow=4) b[,1:3]=NA b[4,2]=5 b[3,1]=6 Now the data is > b [,1] [,2] [,3] [,4] [,5] [1,] NA NA NA 14 18 [2,] NA NA NA 15 19 [3,] 6 NA NA 16 20 [4,] NA 5 NA 17 21 I want to keep data in column 4 greater than 15 and the value in column 1 & 2 either greater than 4

?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),

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

David Bear <David.Bear@asu.edu> wrote : > I have read through some of the info on using dfs roots and I am > needing some advice. Since a unc is still \\servername\ based it > occurs to me that the only way to do this properly is to create a > smb.conf file that publishes a netbios name like \\dfsroot -- Then, to > create a failover system, I would take that config file and

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

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

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

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

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> Sent: Thursday, July 25, 2024 11:15 AM To: Yuan Chun Ding <ycding at coh.org>; 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

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

Dear list, here are my two data frames: av <- structure(list(DESCRIPTION = c("COFFEE C Sep/10", "COPPER Sep/10", "CORN Dec/10", "CRUDE OIL miNY Sep/10", "GOLD Aug/10", "HENRY HUB NATURAL GAS Sep/10", "PALLADIUM Sep/10", "SILVER Sep/10", "SOYBEANS Nov/10", "SPCL HIGH GRADE ZINC USD", "SUGAR

Greetings all. I am banging my head about this one, I will try to be as specific as possible, bear with me please. I have a W2KDC ADC, and trying to join a Samba 3 linux workstation to it. What works: net join:? succeeded wbinfo -t:? checking the trust secret via RPC calls succeeded wbinfo -m: return to prompt, no output wbinfo -u: correct list of local + AD members wbinfo -g: correct list

Hi, I think that it would be a good thing to know more about those 2 projects (and also the future patent free format). I think that many people as me know about Lame, but not about Vorbis, and vice-versa. It would be fine that someone (perhaps the maintainer) of every project would introduce to both group of people those projects. 2 things would be interesting (to my mind): - to know about the

Hello, Just wondering why I am unable to run this in parallel. A dput of my dataset is attached at the end. Please use to create my data object. I want to run this function in parallel (not sure if this is an efficient implementation): #Function to calculate the time to maturity for the option require(fCalendar,quietly=TRUE) #Trying to calculate the trading days

Hi, I need some help with running a recursive function. I like to run funlp2 recursively. When I try to run recursive function in another function named "calclp" I get this "Error: any(!dat2$norm_sd) >= 1 is not TRUE". I have never built a recursive function before so having trouble executing it in this case. I would appreciate any help or guidance to resolve this issue.

Hello everybody, I have a problem and would like to start with an example: library(snow) library(tseries) fn <- function(x) adf.test(x) clusterApply(cl=cl, x=x , fun=fn) R cannot find the function adf.test() because it is inside the function fn(). This problem does not occur when, for example, fn <- function(x) mean(x) holds. Therefore, I think the package tseries has to be put somehow

I know folks have different opinions about Network Manager, but while researching how to change the name of a NIC without rebooting, I also came across the following page from RedHat that I?ll share as I find it?s very clear and concise explanations on how these services interact on CentOS/RHEL 7:

Hi all, I find myself sometimes in the situation where I lapply over a list and in the particular function I'd like to use the name and or position of the respective list item. What I usually do is to use mapply on the list and the names of the list / a position list: o <- list(A=1:3, B=1:2, C=1) mapply(function (item, name) paste(name, sum(item), sep="="), o, names(o))

My own typo ... whoops ... !( any(dat2$norm_sd >= 1 )) On Thu, Dec 14, 2017 at 3:43 PM, Eric Berger <ericjberger at gmail.com> wrote: > You seem to have a typo at this expression (and some others like it) > > Namely, you write > > any(!dat2$norm_sd) >= 1 > > when you possibly meant to write > > !( any(dat2$norm_sd) >= 1 ) > > i.e. I think your !