search for: farnoosh_81

Hello, I have a data as follow: ID    Visit xa1 xa2 yb1 yc23 yb33   I want to look at frequency of visit for ID and create a new column as response .  For example my response would be 2 for x and 3 for y. I think I need to write a loop, but I don't know how. I really appreciate your help. Thanks a lot. Best,Farnoosh Sheikhi [[alternative HTML version deleted]]

...ataNew1,DataB,by=c("Date.Accident","ID"),type="right") ?res1 #? Date.Accident ID Status #1????? 3-Sep-01? 1????? A #2???? 20-Jan-05? 1?? <NA> #3???? 26-Feb-04? 1????? C #4????? 6-Mar-02? 2????? A A.K. ________________________________ From: farnoosh sheikhi <farnoosh_81 at yahoo.com> To: "smartpink111 at yahoo.com" <smartpink111 at yahoo.com> Sent: Friday, April 12, 2013 5:40 PM Subject: Comparison of Date format ?Hi there, Hope all is well. I have a complicated data and I need to create a new variable based on the date and description...

Hi, I have a column of DOB in mm/dd/yyyy. I want to convert this format to age. Thanks a lot.   Best,Farnoosh Sheikhi [[alternative HTML version deleted]]

Hello, I have a data with following format:      Predictors          n11     n12     n21     n22     Odds.Ratio    log.ratio    se.log.odds. 1 ProcOR respirato     2 ProcVaric vein      3 DiagCardiac anom   4 DiagAllergy        5  DiagOth skin dx    6    DiagGastritis          I want to plot odds ratio by command: forestplot in rmeta library, but I get the following error constantly.  Error in

Hello, I have a data which looks like below: Some of the patients have multiple diagnosis. ID(200 patients)   Diag (100 unique Diag-200 in general)   Proc (50 uniqe Proc)  DOB (200)   Gender (200)    a                           daig1 b                           diag2 c                            diag1 I want to reformat this data to : ID   diag1 diag 2 diag 3..  diagx   proc1   proc2  

Hi, I have a R output that looks as follow: Rad:0 Rad1:2 Rad3:3 I want to make a new matrix that looks like : sample size is 2400 Variable    n11  n12 Rad            0     2400-0=2400 Rad1          2       2400-2 Rad3  3      2400-3   Thanks a lot for your time and help:) Best,Farnoosh Sheikhi [[alternative HTML version deleted]]

Hi , I want to apply the following code fo my data with 400 predictors. I was wondering if there ia an alternative way instead of typing 400 predictors for the following code. I really appreciate your help. fit0<-lm(Y~1, data= mydata) fit.final<- lm(Y~X1+X2+X3+.....+X400, data=mydata) ??? step(fit0, scope=list(lower=fit0, upper=fit.final), data=mydata, direction="forward")

...??????? -1.9002023? 0.10667909 -17.81232 1.023637e-05 #X4??????????? 1.4421062? 0.09790384? 14.72982 2.606788e-05 #X7??????????? 0.5726324? 0.05396720? 10.61075 1.285606e-04 #X8?????????? -1.9560052? 0.14204298 -13.77052 3.625118e-05 A.K. ________________________________ From: farnoosh sheikhi <farnoosh_81 at yahoo.com> To: arun <smartpink111 at yahoo.com> Sent: Wednesday, November 21, 2012 2:47 PM Subject: update fit (removing insignificant variables) Hi there, I have a question for removing insignificant predictors from fit. The output looks like below but with 170 predictors. Lab_1 ?...

Hi Arun, I have a question about choosing a subset of data. I have a matrix of 5000 in 3500. I want to choose specific variables such as proc1 to proc1000 and Lab1 to Lab1600 and put it into a new matrix to run some correlation analysis. Since I have 3500 variables, I don't know from what column my lab or procedures start. I really appreciate your help. Thanks.   Best,Farnoosh Sheikhi

...ction(x) c(diff(x),NA))),use.names=FALSE) #[1] 126 140 445 224? NA 715? NA #or unlist(lapply(split(dat1,dat1$ID),function(x) c(diff(as.Date(x$Date,format="%d-%b-%y")),NA)),use.names=FALSE) #[1] 126 140 445 224? NA 715? NA A.K. ________________________________ From: farnoosh sheikhi <farnoosh_81 at yahoo.com> To: "smartpink111 at yahoo.com" <smartpink111 at yahoo.com> Sent: Tuesday, July 23, 2013 3:11 PM Subject: subtracting rows for unique Hi there, Hope you are doing well. I have a data set which looks like below, I want to create a new variable (x3-here) which s...

...... dat1[,2:3]<- lapply(dat1[,2:3],as.numeric) ?str(dat1) #'data.frame':??? 5 obs. of? 3 variables: # $ V1: chr? "a" "b" "c" "d" ... # $ V2: num? 1 2 10 18 2 # $ V3: num? 6 7 8 9 10 A.K. ________________________________ From: farnoosh sheikhi <farnoosh_81 at yahoo.com> To: "smartpink111 at yahoo.com" <smartpink111 at yahoo.com> Sent: Monday, August 26, 2013 1:48 PM Subject: Loop for converting character columns to Numeric Hi there, I want to change the columns of my data from Character to numeric. Since the number of predic...

...Use ?as.Date() dat1<- read.table(text="V1 2012-01-12 08:23:00 2012-01-19 15:29:00",sep=",",header=TRUE,stringsAsFactors=FALSE) dat1$V1<- as.Date(dat1$V1) ?dat1 #????????? V1 #1 2012-01-12 #2 2012-01-19 A.K. ________________________________ From: farnoosh sheikhi <farnoosh_81 at yahoo.com> To: "smartpink111 at yahoo.com" <smartpink111 at yahoo.com> Sent: Wednesday, October 2, 2013 1:38 AM Subject: Removing time and min from Date Hi Arun, I have a variable including time and date together like below. I want to just keep the date part (2012-01-12 )...

Hi, I have a data set as follow: X         Z x1        102 x2        102 x2        102 x2        77 x3        23   I need to pivot this data as follows and assign the values based on frequency of column Z: X       Z.102   Z.77 Z.23 x1          1        0        0 x2          21  0 x3         00  1 Thanks. Best,Farnoosh Sheikhi [[alternative HTML version deleted]]

...ID,] ?DataAllExclude #?? ID?????? value #2?? G? 0.23508558 #4?? I -0.38736217 #6?? J -0.24331815 #7?? G -0.32137638 #13? J? 0.57385268 #14? F? 0.59750664 #15? G -0.01775669 #17? F? 0.11683731 #18? H -1.11311883 #20? H? 0.46184261 A.K. ________________________________ From: farnoosh sheikhi <farnoosh_81 at yahoo.com> To: "smartpink111 at yahoo.com" <smartpink111 at yahoo.com> Sent: Sunday, April 21, 2013 2:45 AM Subject: Excluding observations Hi Arun, I have a bit problem excluding some observations from the ID variable in the data. I really appreciate if you can help me....

Left as an exercise for the student. -- Sent from my phone. Please excuse my brevity. On June 29, 2017 7:25:36 PM EDT, Farnoosh Sheikhi <farnoosh_81 at yahoo.com> wrote: >Thanks Jeff. This is a nice way of solving this problem. What about the >cases with 0015-02-21?Many thanks.?Best,Farnoosh > > > >On Wednesday, June 28, 2017 10:49 PM, Jeff Newmiller ><jdnewmil at dcn.davis.ca.us> wrote: > > > I doubt yo...

...such homework on your own. Cheers, Bert Bert Gunter "The trouble with having an open mind is that people keep coming along and sticking things into it." -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip ) On Mon, Sep 25, 2017 at 3:28 PM, Farnoosh Sheikhi <farnoosh_81 at yahoo.com> wrote: > Hi Bert, > > Here is the code: > f<-function(n){ > m=runif(n, 50, 100) > d=rnorm(n, 0, 1) > se=sd(d)/sqrt (n) > > ci.u<-mean(d)+1.96*sd(d) > ci.l<-mean(d)-1.96*sd(d) > ci.w<-ci.u-ci.l > > w.ci.uu<-ci.u+(qt (0.975, df=n...

...d res<-data.frame(colSums(dat1)) ?names(res)<-"sum(n11)" ?res #???? sum(n11) #med1??????? 3 #med2??????? 1 #med3??????? 2 #3rd method ?do.call(rbind,lapply(dat1,sum)) #???? [,1] #med1??? 3 #med2??? 1 #med3??? 2 A.K. ________________________________ From: farnoosh sheikhi <farnoosh_81 at yahoo.com> To: arun <smartpink111 at yahoo.com> Sent: Monday, November 12, 2012 7:24 PM Subject: for loop Hi there, I want to calculate the odds ratio for a data like below. ?I want to compute the sum of each column as a new column and variable names as a new column. I have about 10...

Thanks Jeff. This is a nice way of solving this problem. What about the cases with 0015-02-21?Many thanks.?Best,Farnoosh On Wednesday, June 28, 2017 10:49 PM, Jeff Newmiller <jdnewmil at dcn.davis.ca.us> wrote: I doubt your actual file looks like the mess that made it to my email software (below) because you posted HTML-format email. Read the Posting Guide, and in particular

I doubt your actual file looks like the mess that made it to my email software (below) because you posted HTML-format email. Read the Posting Guide, and in particular figure out how to send plain text email. You might try the "anytime" contributed package, though I suspect it too will choke on your mess. Otherwise, that will pretty much leave only a brute-force series of regular

Hi, I have a 3d array as below, I want to make this array to a matrix of p=50(rows) and n=20(columns) with the coverage values . The code before the array is: library(binom) Loading required package: lattice pi.seq<-seq(from = 0.01, to = 0.5, by = 0.01) no.seq<-seq(from = 5, to = 100, by = 5) cp.all = binom.coverage( p = pi.seq, n = no.seq , conf.level = 0.95, method = "exact")