similar to: add one variable to a data frame

Hi Sarah, Thank you so much!! I got your good ideas. Ding -----Original Message----- From: Sarah Goslee [mailto:sarah.goslee at gmail.com] Sent: Friday, May 11, 2018 11:40 AM To: Ding, Yuan Chun Cc: r-help mailing list Subject: Re: [R] add one variable to a data frame [Attention: This email came from an external source. Do not open attachments or click on links from unknown senders or

Hi, Here's one way to approach it, using the coercion of factor to numeric. Note that I changed your data.frame() statement to avoid coercing strings to factors, just to make it simpler to set the levels. dat1 <-data.frame(N=seq(1, 12,1), B=c("29_log","29_log", "29_log", "27_cat", "27_cat", "1_log", "1_log",

Sarah's solutions are good, and here's another, even more basic: tmp1 <- unique(dat1$B) tmp2 <- seq_along(tmp1) dat1$C <- tmp2[ match( dat1$B, tmp1) ] > dat1 N B C 1 1 29_log 1 2 2 29_log 1 3 3 29_log 1 4 4 27_cat 2 5 5 27_cat 2 6 6 1_log 3 7 7 1_log 3 8 8 1_log 3 9 9 1_log 3 10 10 1_log 3 11 11 3_cat 4 12 12 3_cat 4 As a single line

Sarah et. al.: As a matter of aesthetics (i.e. my personal ocd-ness) I prefer using the public API of an object, i.e. *not* to makes use of the representation of a factor as essentially an integer vector with labels, but rather to use its documented behavior. (Feel free to ignore this remark!) Anyway, >cumsum(!duplicated(dat1$B)) [1] 1 1 1 2 2 3 3 3 3 3 4 4 will do it. This is very

David: I believe your response on SO is incorrect. This is a standard OFAT (one factor at a time) design, so that assuming additivity (no interactions), the effects of drugA and drugB can be determined via the model you rejected: For example, if baseline control (no drugs) has a response of 0, drugA has an effect of 1, drugB has an effect of 2, and the effects are additive, with no noise we

Hi Richard, Thank you so much!! I understand the problem now, I assign a name to the "ggsurvplot" object and then add print(fig) at bottom of function definition, now figure gets printed on screen. Ding # function to generate RFS curves RFS <- function( inputfile, N ) { cluster<- survfit(Surv(RFS_days2, OV_Had_a_Recurrence_CODE) ~ clusters, data =

Dear R users, I need to analyze data generated from a partial two-by-two factorial design: two levels for drug A (yes, no), two levels for drug B (yes, no); however, data points are available only for three groups, no drugA/no drugB, yes drugA/no drugB, yes drugA/yes drug B, omitting the fourth group of no drugA/yes drugB. I think we can not investigate interaction between drug A and drug B,

Hi Bert and David, Thank you so much for willingness to spend some time on my problem!!! I have some statistical knowledge (going to get a master in applied statisitics), but do not have a chance to purse a phD for statistics, so I am always be careful before starting to do analysis and hope to gather supportive information from real statisticians. Sorry that I did not tell more info about

Hi Bert, I am very sorry to bother you again. For the following question, as you suggested, I posted it in both Biostars website and stackexchange website, so far no reply. I really hope that you can do me a great favor to share your points about how to explain the coefficients for drug A and drug B if run anova model (response variable = drug A + drug B). is it different from running three

That is certainly OK, but you can also just use print(ggsurvplot(...)) as your final statement. out <- RFS( ...) would then return the ggsurvplot object *and* graph it. Any good R tutorial or a web search will provide more details on function returns, which you might find useful. Cheers, Bert Bert Gunter "The trouble with having an open mind is that people keep coming along and

This list provides help on R programming (see the posting guide linked below for details on what is/is not considered on topic), and generally avoids discussion of purely statistical issues, which is what your query appears to be. The simple answer is yes, you can fit the model as described, but you clearly need the off topic discussion as to what it does or does not mean. For that, you might try

Thank you, your suggestion is simpler and logically better. I had impression that the last object in a function gets returned, so I did not add the print function at the bottom line of the function definition. Returning an object and graph the object are different process, I am a beginner for writing R function and need to find a good guide source about writing R functions. If you know a good

Hi Bert, I am sorry to bother you on weekend. I am still struggling on defining a correct function. I first defined the function RFS (see below), then run it by provide the two argument. m52.2cluster <-RFS(inputfile =allinfo_m52, N=2 ) I do not get error message, but no figure displays on screen. I do not know what is going on. Can you help me a little more on this issue? Thank you,

But of course the whole point of additivity is to decompose the combined effect as the sum of individual effects. "Mislead" is a subjective judgment, so no comment. The explanation I provided is standard. I used it for decades when I taught in industry. Cheers, Bert Bert Gunter "The trouble with having an open mind is that people keep coming along and sticking things into

> On Mar 5, 2018, at 8:52 AM, Ding, Yuan Chun <ycding at coh.org> wrote: > > Hi Bert, > > I am very sorry to bother you again. > > For the following question, as you suggested, I posted it in both Biostars website and stackexchange website, so far no reply. > > I really hope that you can do me a great favor to share your points about how to explain the

FAQ 7.22 You must print a ggplot object, for example with print(m52.2cluster) For the FAQ, run the line system.file("../../doc/FAQ") in R on your computer. Open up the resulting filepath in your favorite editor and scroll down to 7.22 On Sun, Jan 14, 2018 at 4:21 PM, Ding, Yuan Chun <ycding at coh.org> wrote: > Hi Bert, > > I am sorry to bother you on weekend. >

Hi All, I want to create a categorical variable, cat.pfoa, in the file of pfas.pheno (a data frame) based on log2pfoa values. I can do it using the following code. pfas.pheno <-within(pfas.pheno, {cat.pfoa<-NA cat.pfoa[pfas.pheno$log2pfoa <=quantile(pfas.pheno$log2pfoa,0.25, na.rm =T)]<-0 cat.pfoa[pfas.pheno$log2pfoa >=quantile(pfas.pheno$log2pfoa,0.75, na.rm =T)]<-2

> On Mar 5, 2018, at 3:04 PM, Bert Gunter <bgunter.4567 at gmail.com> wrote: > > But of course the whole point of additivity is to decompose the combined effect as the sum of individual effects. Agreed. Furthermore your encoding of the treatment assignments has the advantage that the default treatment contrast for A+B will have a statistical estimate associated with it. That was a

> On Mar 5, 2018, at 2:27 PM, Bert Gunter <bgunter.4567 at gmail.com> wrote: > > David: > > I believe your response on SO is incorrect. This is a standard OFAT (one factor at a time) design, so that assuming additivity (no interactions), the effects of drugA and drugB can be determined via the model you rejected: >> three groups, no drugA/no drugB, yes drugA/no drugB,

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