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I've created a short program to print a table of learning curve factors. However, I cannot figure out how to format the table to: 1) Get rid of the [1]s in the first column and replace it with the values of N. 2) Line up the first row with the factors (decimal fractions). Thanks for any help. The complete program and output is as follows: > Lc<-seq(0.70,0.95,0.05) #Specify learning

I am trying to use t.test on the following data: date type INTERVAL nCASES MTF SDF MTO SDO nFST MF nOBS MO MB BIASCV BIASEV ME MAE RMSE CRCF 2001-06-15 avn GE1.00 4385 0.246 0.300 1.502 0.556 1367 1.373 4385 1.502 1.471 0.285 0.164 -1.256 1.266 1.399 0.056 2001-06-15 avn

Dear R users I have a big matrix like 6021 1188 790 290 1174 1015 1990 6613 6288 100714 6021 1 0.658 0.688 0.474 0.262 0.163 0.137 0.32 0.252 0.206 1188 0.658 1 0.917 0.245 0.331 0.122 0.148 0.194 0.168 0.171 790 0.688 0.917 1 0.243 0.31 0.122 0.15 0.19 0.171 0.174 290 0.474

Hello, i'm a french student of medical oncology and i'm working on breast cancer. I have a variable with the histologic type of tumor wich is between 1 and 5. I use as.factor function to make some variable with level between 1 and 5. When i put it in the cox model i have only the level between 2 and 5. The level 1 doesn't appear. I think i have to change the number of level but i

I am wondering how to interpret the parameter estimates that lm() reports in this sort of situation: y = round(rnorm(n=24,mean=5,sd=2),2) A = gl(3,2,24,labels=c("one","two","three")) B = gl(4,6,24,labels=c("i","ii","iii","iv")) # Make both observations for A=1, B=4 missing y[19] = NA y[20] = NA data.frame(y,A,B) nonadd = lm(y ~

Hi - How can I 'manually' reproduce the results in 'pred1' below? My attempt is pred_manual, but is not correct. Any help is much appreciated. library(splines) set.seed(12345) y <- rgamma(1000, shape =0.5) age <- rnorm(1000, 45, 10) glm1 <- glm(y ~ ns(age, 4), family=Gamma(link=log)) dd <- data.frame(age = 16:80) mm <- model.matrix( ~ ns(dd$age, 4)) pred1 <-

I am hoping for some advice regarding limiting decimal points to 3. 'Round' produces the desired results except for the 97.5% confidence interval. Any advice as to how I modify the code to obtain output to 3 decimal points for all ouput is appreciated, regards Bob Green mod.multgran <-multinom(offence ~ grandiose * violent.convictions, data = kc, na.action = na.omit)

Hi Xavier, the reason you observe this feature is that in the 'constraint' argument you should specify the values under the additive parameterization, i.e., when in the second column of the matrix supplied in 'constraint' you specify 2, then you need to provide the easiness parameters (not the difficulty parameters) in the third column. Check the Details section of ?tpm() and

We are two french students and we have a problem concerning an exercize. We don't know how to resolve it. It would be fantastic if someone can help us. Thanks. Description: This study examined how the metabolic cost of locomotion varied with speed, stride frequency and body mass. Cost was determined by measuring oxygen consumption (?vo2?), analyzing the oxygen content in air inhaled and

Dear R list As suggested by Prof Brian Ripley, I have tried to read acf literature. The main problem is I am not the statistician and hence have some problem in understanding the concepts immediately. I came across one literature (http://www.stat.nus.edu.sg/~staxyc/REG32.pdf) on auto-correlation giving the methodology. As per that literature, the auto-correlation is arrived at as per following.

Sorry for the frantic nature of this message, but we need to allow pings on our firewall so our ISP can test things. I''ve done this, and it still doesn''t work: (I am now at v.2.0.10) rules: AllowPing net fw AllowPing sls fw show indicates some matches, so where are they? Chain AllowPing (4 references) pkts bytes target prot opt in out source

Dear R user, I want to do sparse principal component analysis (spca). I am using elastic net package for this and spca() and the code is following from the example. My question is How can I decide the *K =? *and *para=c(7,4,4,1,1,1)) . So, here k=6 i.e the no of Principal Components. and each pcs say , * ** pc1 number of non zero loading is 7 pc2 number of non zero loading

Hello, I am a very new R user and not a statistician so please excuse any over explanation, I'm just trying to be as clear as possible. I have performed a cross correlation of two time series (my columns) in a single data setusing: ccf(ts(A[rows,columnX]),(A[rows,columnY]), lag=NULL, type="correlation",plot=F) I?am able to get the results (for example): Autocorrelations of

Hi, I am getting a warning message when I am fitting a generalized mixed model (mod_2) and I don't understand why because when I add just an interaction factor the model works perfectly (mod_1). Does anyone know what it happpens ? Thanks, Aïda   > mod_1<-lmer(sur15~soeviv15_4plus+frviv15_4plus+frat_15death+dad_class_new+soeviv15_4plus:dad_class_new +frviv15_4plus:dad_class_new+

Thanks in advance. I want to derive correlations of variables in a dataset Specifically library(Ecdat) data(Housing) attach(Housing) cor(lotsize, bathrooms) this code results only the correlationship between two variables. But I want to examine all the combinations of variables in this dataset. And I will finally make a table in Latex. How can I test correlations for all combinations of

Full_Name: Adrian Custer Version: 0.90.0 OS: Linux on Thinkpad (pentium) and desktop (K6) Submission from: (NULL) (128.32.251.234) When I create a boxplot, and then try to overlay a lowess fit or just the points, the points do not appear in the highest level and the lowess curve does not reach the highest level. However, if I add one to each of the models, the problem is solved. I tried this

I have a data matrix with genes as columns and samples as rows. I want to create all possible gene ratios.Is there an elegant and fast way to do it in R and write it to a dataframe? Thanks for any help. Som. -- View this message in context: http://r.789695.n4.nabble.com/Calculating-all-possible-ratios-tp4627405.html Sent from the R help mailing list archive at Nabble.com. [[alternative HTML

Hi there, I am doing multiple comparisons for data that is not normally distributed. For this purpose I tried both functions kruskal{agricolae} and kruskalmc{pgirmess}. It confuses me that these functions do not yield the same results although they are doing the same thing, don't they? Can anyone tell my why this happens and which function I can trust? kruskalmc() tells me that there are no

Hi, I was trying to have R read files faster with readChar(). That was before I noticed that readChar() is not that bad! In any case, below I suggest a few simple changes that will make readChar slightly faster. I followed readChar(useBytes=T), and tried to identify all O(N) operations, where N is the size of the file. The assumption is that for LARGE files we want to avoid any O(N) operations,

Hi: I am using a subset of the below dataset to predict PRED_SUIT for the whole dataset but I am having trouble with 'newdata'. The model was created with 153 records and want to predict for 208 records. wolf2 <- structure(list(gridcell = c(367L, 444L, 533L, 587L, 598L, 609L, 620L, 629L, 641L, 651L, 662L, 674L, 684L, 695L, 738L, 748L, 804L, 805L, 872L, 919L, 929L, 938L, 950L, 958L,