similar to: Can package plyr also calculate the mode?

1)I am wandering how the following SQL statement can be written in R language w/o using sqldf: create table detail2 as select a.* from detail a, pdetail b where a.TDATE=b.TDATE and (a.STIM >= b.STIM and a.STIM <=b.MAXTIM) 2) when try if then in R it only applies to the 1st row & not to whole dataset like in SAS. How do you get round that? in SAS: data summary; set all1;

Hello, I was wandering if it is possible to perform on a dataframe called 'all' a shapiro wilk normality test for COUNTS by variable Group ACTIVITY? Could it be done using plyer? I saw an eg that applies to an array but not to a dataframe: lapply(split(dataset1$Height,dataset1$Group),shapiro.test) Any thoughts would be much appreciated. My dataframe is in shape: dat ACTIVIT

Hello, I have a column called max_date in my data frame and I only want to keep the bigger values for the same activity. How can I do that? data frame: activity max_dt A 2013-03-05 B 2013-03-28 A 2013-03-28 C 2013-03-28 B 2013-03-01 Thank you for your help -- View this message in context:

I am using the describe function in prettyR. I would like to add the 25% 75% quartiles to the summary table how do I do this I have tried describe(x.f, num.desc=c("mean", "median", "sd", "min", "max", "skewness", "quantile(x.f, na.rm=T, probs=seq(0.25, 0.75))", "valid.n")) help -- Let's not spend our time

Hello, I have a variable in a data frame that contains NA values. I just want to subset so that I get the obs where that variable is missing. In SAS I would do: data missing; set test; if myvalue=' '; run; How can I perform this simple task in R? Thanks in advance for your help. -- View this message in context:

Hello, I am trying to re-code all my programs from SAS into R. In SAS I use the following code: proc sort data=upper; by tdate stock_symbol expire strike; run; data upper1; set upper; by tdate stock_symbol expire strike; if first.expire then output; rename strike=astrike; run; on the following data set: tdate stock_symbol expiration strike 9/11/2012 C 9/16/2012

Hello, I am a newbie to R coming from SAS background. I am trying to program the following: I have a monthly data frame with 2 variables: client pct_total A 15% B 10% C 10% D 9% E 8% F 6% G 4% I need to come up w/ a monthly list of clients that make 50% or just above it every month so I can pass them to the rest of the program.

Hi R users, The following code worked well to summarize four data groups in a dataframe for three variables (t_depth, t_alt_count, t_alt_ratio), 12 columns of summary, see attached. However, after running another 2000 lines of R codes using functions from more than 10 other R libraries, then it only generated one column of summary. Do you know why? Thank you, Yuan Chun Ding summary_anno1148ft

I am sorry that I know I should provide a dataset that allows to replicate my problem. It is a research dataset and quite large, so I can not share. Both Bert and Tim guessed my problem correctly. I also thought about the conflicting issue between different packages and function masking. I just hope to that someone has similar experience, so providing me suggestion. For conflicting issue,

#is there a way to get NA in the table of descriptive statistics instead of the function stopping Thank you in advance #data x.f <- structure(list(Site = structure(c(9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L), .Label = c("BC", "HC", "RM119", "RM148", "RM179", "RM185",

hi dovecot: when i login 100 users with imap protocol ,Someusers appeared a BUG in maillog like : Sep 10 09:05:22 auth(default): Error: BUG: Worker sent reply with id 393, expected 394 Sep 10 09:05:22 auth(default): Error: worker-server(q82 at t.com,10.12.80.3): Aborted: Worker is buggy Sep 10 09:05:22 auth(default): Error: BUG: Worker sent reply with id 31, expected 32 Sep 10 09:05:22

Hello, Inline. ?s 19:03 de 12/06/2024, Yuan Chun Ding via R-help escreveu: > I am sorry that I know I should provide a dataset that allows to replicate my problem. > > It is a research dataset and quite large, so I can not share. > > Both Bert and Tim guessed my problem correctly. I also thought about the conflicting issue between different packages and function masking. > I

Hello, I have a dataframe w/ 3 variables of interest: transaction,date(tdate) & time(event_tim). How could I create a 4th variable (last_trans) that would flag the last transaction of the day for each day? In SAS I use: proc sort data=all6; by tdate event_tim; run; /*Create last transaction flag per day*/ data all6; set all6; by tdate event_tim; last_trans=last.tdate; Thanks

Hi Rui, Thank you very much! Yes, I verified using real data, it worked correctly as expected after adding tidyr:: to the pivot_longer function and dplyr:: to the group_by and summarize Function. I did not know how to assign the tidyr and dplyr to the three functions because I do not really understand well the three functions and just got the code from a google search. I also tried your

Hello, I know how to increment a date by calendar date: ticker$ldate <- ticker$tdate + days(5) How do I increment it by business days only so that week-ends are not counted? So for example friday november 2 + 5days becomes friday november 9 & not wednesday nov 7. Thanks for your help. -- View this message in context:

I am new to R and I have the following SAS statements: if otype='M' and ocond='1' and entry='a.Prop' then MOC=1; else MOC=0; How would I translate that into R code? Thanks in advance -- View this message in context: http://r.789695.n4.nabble.com/if-then-in-R-versus-SAS-tp4641225.html Sent from the R help mailing list archive at Nabble.com.

Hello, We lost our SAS licence & I am busy transfering my old SAS programs to R environment. I am very new to R. In 1 program I was creating SAS macro vars & passing them into a SQL query to run against the server. There are 3 variables firm, begindt, enddt. # of values for each varies month to month. Is there anyway I could do the same thing in R & pass the afore mentioned values

Hello all, I have a question: I am using the interquantile method to spot outliers & it gives me values of say 234 & -120 or for the higher & lower benchmarks. I don't have any issues w/ the higher end. However I don't have any negative values. My lowest possible value is 0. Should I consider 0 as an outlier? Thanks ahead for your thoughts -- View this message in

I get following error message: trying URL 'http://cran.stat.ucla.edu/bin/windows/contrib/2.15/xlsx_0.4.2.zip' Content type 'application/zip' length 365611 bytes (357 Kb) opened URL downloaded 357 Kb Error in read.dcf(file.path(pkgname, "DESCRIPTION"), c("Package", "Type")) : cannot open the connection In addition: Warning messages: 1: In

I sometimes think people on this list are quite rude to posters. I'm afraid I'm likely to join in with some rudeness? 1. "Here is some code that works but also doesn't" is probably not going to get you an answer 2. I provide no information about the data it works on or doesn't 3. I tell you I'm using a load of dependencies, but don't tell you what 4. I refer to