similar to: converting DOB format to age

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  

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]]

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]]

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

Hi, ?In the example you provided, it looks like the dates in Date2 happens first.? So, I changed it a bit.? DataA<- read.table(text=" ID,Status,Date1,Date2 ??? ??? ?????? 1,A,3-Feb-01,15-May-01 ??? ??? 1,B,15-May-01,16-May-01 ??? ??? 1,A,16-May-01,3-Sep-01 ??? ??? ??? ??? ??? 1,B,3-Sep-01,13-Sep-01 ??? ??? ??? ??? ??? 1,C,13-Sep-01,26-Feb-04 ??? ??? ??? ??? ???

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")

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]]

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

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

Hi, I wrote a little helper that calulates someone''s age based on his/her date of birth. def age(dob) diff = Date.today - dob age = (diff / 365.25).floor age.to_s end It works fine, but it''s not completely accurate as it just takes the average days in a year. It should be able to calculate this more accurately, right? I can''t work it out

Hi,? I have a data set with various date formats in one column and not sure how to unify it.Here is a few formats: 02091702/22/170221201703/17/160015-08-239/2/1500170806May-2-201522-March-2014 I tried parse_date_time from lubridate library but it failed.Thanks so much.?Best,Farnoosh [[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 frame with 400 columns and wanted to filter character columns with "$" in it.For example:?>? x <- c("$5", "$89", "$10", "$34")??>? y <- c(1:4)>? My.Data <- data.frame (x,y)> My.Data? ? x y1? $5 12 $89 23 $10 34 $34 4 I want to detect the columns with $ and remove the $ from the selected columns.I have

Hello,? I have a data set similar to below and I wanted to keep the observations after the first occurrence of these department: "B", "D", "F".For example for ID=2, the observation with deps=B and anything after will be kept in the data. For ID=3, observations with deps=D and anything after will be included. Subject<- c("2", "2", "2",

Hi, Suppose you created a dataframe like this: set.seed(28) ?dat1<-as.data.frame(simplify2array(list(letters[1:5],sample(1:20,5,replace=TRUE),6:10)),stringsAsFactors=FALSE) ?str(dat1) #'data.frame':??? 5 obs. of? 3 variables: # $ V1: chr? "a" "b" "c" "d" ... # $ V2: chr? "1" "2" "10" "18" ... # $ V3: chr?

Hey, Are all the dates connected? So no comma or space btw? Regards, Christoph > On 29 Jun 2017, at 2:02 pm, Farnoosh Sheikhi via R-help <r-help at r-project.org> wrote: > > Hi, > I have a data set with various date formats in one column and not sure how to unify it.Here is a few formats: > 02091702/22/170221201703/17/160015-08-239/2/1500170806May-2-201522-March-2014

HI, You can do this in many ways: dat1<-read.table(text=" med1,med2,med3???? ?1,0,1?????? 0,1,1??? 2,0,0 ",sep=",",header=TRUE)?? #1st method library(reshape) dat2<-melt(dat1) dat3<-aggregate(dat2$value,by=list(dat2$variable),sum) ?colnames(dat3)<-c("name","sum(n11)") ?dat3 #? name sum(n11) #1 med1??????? 3 #2 med2??????? 1 #3 med3??????? 2

Hi Farnoosh, Perhaps this will help: drop_dollar<-function(x) return(as.numeric(as.character(gsub("\\$","",x)))) sapply(My.Data,drop_dollar) Jim On Thu, Apr 19, 2018 at 7:23 AM, Farnoosh Sheikhi via R-help <r-help at r-project.org> wrote: > Hello, > I have a data frame with 400 columns and wanted to filter character columns with "$" in it.For

i know it''s probably really simple, how do i work out someone''s age if i have their d.o.b. stored as a date in my db. cheers -- Posted via http://www.ruby-forum.com/.

Hi,? I have the below function which returns confidence intervals.?I wanted to pass different sample sizes through the function, but for some reason it's not working.?n ? <- seq(from=40, to=300, by=2o) I was also wondering how I can return a plot for different sample sizes.?plot(m~d, main="n(i)")? Any help or suggestion is appreciated.