similar to: assign index to colnames(matrix)

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

Just noticed that when we generate a custom lowering for some atomics that we forgot to transfer the MemOperand to the new instruction that touches memory. -- Mon Ping Index: lib/Target/X86/X86ISelLowering.cpp =================================================================== --- lib/Target/X86/X86ISelLowering.cpp (revision 53702) +++ lib/Target/X86/X86ISelLowering.cpp (working copy) @@

Dear Thomas, The head of my dataset > head(wsuv) parcel sp time censo treatment species 1 S8 Poecilanthe effusa ( Hub. ) Ducke. 1 1 1 1 2 S8 Poecilanthe effusa ( Hub. ) Ducke. 1 1 1 1 3 S8 Poecilanthe effusa ( Hub. ) Ducke. 1 1 1 1 4 S8 Poecilanthe effusa ( Hub. ) Ducke. 1 1 1

OK, Let's try this again! Here is the reproducible script; it is long because I had to copy the panel dataset here. My question is related to systemfit; I don't know how to get the result for the entire panel. #Reproducible script Empdata<- read.csv("/Users/ngwinuiazenui/Documents/UPLOADemp.csv") View(Empdata) install.packages("systemfit")

Sadly you failed to set your email program to send plain text and the data is corrupted at my end. I also think you need to reduce the size of the data set... the intent here is to increase your understanding, not debug your particular analysis. I will say that I am having a very challenging time understanding what you are trying to accomplish though. What are the equations that you think need

Dear all, The 'make check' step fails for the pacakge mva on IBM AIX. The tail of the Rout log file looks like: > for(factors in 2:4) print(update(Harman23.FA, factors = factors)) Call: factanal(factors = factors, covmat = Harman23.cor) Uniquenesses: height arm.span forearm lower.leg weight 0.170 0.107 0.166

Hello: In the example below (or for a censored data) using survfit.coxph, can anyone point me to a link or a pdf as to how the probabilities appearing in bold under "summary(pred$surv)" are calculated? Do these represent acumulative probability distribution in time (not including censored time)? Thanks very much, parmee *fit <- coxph(Surv(futime, fustat) ~ age, data = ovarian)*

... and the mailing list is picky about attachments... whatever you attached did not conform to the stringent requirements mentioned in the Posting Guide. Pasting the code right into the email is usually safest, though you DO have to post using plain text (as the Posting Guide indicates) or your code may get mangled by the automatic html format removal. On May 15, 2018 7:04:31 AM PDT, Bert Gunter

Dear R users, I have used the following function (in blue) aiming to find the linear regression between MOE and XLA and nesting my data by Species. I have obtained the following results (in green). model4<-lme(MOE~XLA, random = ~ XLA|Species, method="ML")summary(model4) Linear mixed-effects model fit by maximum likelihood Data: NULL         AIC     BIC   logLik  -1.040187 8.78533

Unless there is good reason not to, always cc the list -- there are lots of smarter folks than I on it who can help. I may or may not have time to look at this. Hopefully someone else will. -- 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

Hi, When I run the survfit function, I want to get the restricted mean value and the standard error also. I found out using the "print" function to do so, as shown below, print(km.fit,print.rmean=TRUE) Call: survfit(formula = Surv(diff, status) ~ 1, type = "kaplan-meier") records n.max n.start events *rmean *se(rmean) median 200.000

Hi All, simple question: do you know how to graph the following object/matrix in a 'surface manner': [,1] [,2] [,3] [,4] [,5] [,6] [1,] -0.154 -0.065 0.129 0.637 0.780 0.221 [2,] 0.236 0.580 0.448 0.729 0.859 0.475 [3,] 0.401 0.506 0.310 0.650 0.822 0.448 [4,] 0.548 0.625 0.883 0.825 0.945 0.637 [5,] 0.544 0.746 0.823 0.877 0.861 0.642 [6,] 0.262 0.399

Hi, I am trying to make a multi-class classification tree by using rpart. I used MASS package'd data: fgl to test and it works well. However, when I used my small-sampled data as below, the program seems to take forever. I am not sure if it is due to slowness or there is something wrong with my codes or data manipulation. Please be advised ! The data is described as the output from str()

Hi, I'm trying to use sapply to compute the min of several variables, each of them stored in data.frames, grouped as a list: Is it normal that mean() and min() produce different objects dimensions? > str(dats) List of 5 $ log20:'data.frame': 83 obs. of 5 variables: ..$ DATE : int [1:83] 2001081500 2001081512 2001081600 2001081612 2001081700 2001081712

Hi, im little bit confused about Cohen's Kappa and i should be look into the Kappa function code. Is the easy formula really wrong? kappa=agreement-chance/(1-chance) many thanks christian ############################################################################### true-negativ:7445 false-positive:3410 false-negativ:347 true-positiv:772 classification-aggrement:68,6%

Hello, I'm a grad student in the Intelligent Transportation Systems lab at Portland State Univ. in Portland, OR, USA. I'm trying to learn the basics of R to run a negative binomial in the near future, and so I ran a test regression on roadway crash data obtained from "Statistical and Econometric Methods for Transportation Data Analysis" by Washington et al (p. 250). I ran the

Hi list members, I saw a couple lavaan posts here so I think I?m sending this to the correct list. I am trying to run a multigroup analysis with lavaan in order to compare behavioural correlations across two populations. I?m following the method suggested in the paper by Dingemanse et al. (2010) in Behavioural Ecology. In one of the groups, lavaan returns negative variance for one path and I?m

Hi R User, Would you give me your hand to standardize some columns in a matrix? I have included the example table. > dput(test) structure(list(X = c(1, 2, 3, 4, 5, 6), Y = c(4, 5, 6, 7, 8, 9), temp = c(0.818, 0.113, 0.256, 0.587, 0.955, 0.207), oxy = c(0.797, 0.487, 0.727, 0.128, 0.514, 0.031)), .Names = c("X", "Y", "temp", "oxy"), row.names =

Hello all! I have a problem with R. I try to merge data like this: structure(c(2.1785, 1.868, 2.1855, 2.5175, 2.025, 2.435, 1.809, 1.628, 1.327, 1.3485, 1.4335, 2.052, 2.2465, 2.151, 1.7945, 1.79, 1.6055, 1.616, 1.633, 1.665, 2.002, 2.152, 1.736, 1.7985, 1.9155, 1.7135, 1.548, 1.568, 1.713, 2.079, 1.875, 2.12, 2.072, 1.906, 1.4645, 1.3025, 1.407, 1.5445, 1.437, 1.463, 1.5235, 1.609, 1.738, 1.478,

Hello all! I have a problem to extract values greater that for example 1820. I try this code: x[x[,1]>1820,]->x1 Please help me! Thank you! The data structure is: structure(c(2.576, 1.728, 3.434, 2.187, 1.928, 1.886, 1.2425, 1.23, 1.075, 1.1785, 1.186, 1.165, 1.732, 1.517, 1.4095, 1.074, 1.618, 1.677, 1.845, 1.594, 1.6655, 1.1605, 1.425, 1.099, 1.007, 1.1795, 1.3855, 1.4065, 1.138, 1.514,