similar to: Strange dataframe behavior

Dear all, A question related to the following has been asked on R-help before, but I could not find any answer to it. Input will be much appreciated. I got an unexpected sign of the "slope" parameter associated with a covariate (diam) using zeroinfl(). It led me to compare the estimates given by zeroinfl() and hurdle(): The (significant) negative estimate here is surprising, given

Dear Sir or Madam? I am a doctor of urology,and I am engaged in developing a nomogram of bladder cancer. May I ask for your help on below issue? I set up a dataset which include 317 cases. I got the Binary Logistic Regression model by SPSS.And then I try to reconstruct the model ?lrm(RECU~Complication+T.Num+T.Grade+Year+TS)? by R-Project,and try to internal validate the model through

Target: FreeBSD 6.2-STABLE on i386 autoconf says: configure:2122: checking build system type configure:2140: result: i386-unknown-freebsd6.2 [...] configure:2721: gcc -v >&5 Using built-in specs. Configured with: FreeBSD/i386 system compiler Thread model: posix gcc version 3.4.6 [FreeBSD] 20060305 [...] objdir != srcdir, for both llvm and gcc. Release build. llvm-gcc 4.2 from source.

Greetings everyone, I'm trying to add a specific table or a specific number of rows (e.g.44) to a table with no success. This is my basic table > head(dataA) year plot spp prop.B DCA1 DCA2 DCA3 DCA4 1 2000 1 a1 0.031079 -0.0776 -0.0009 0.0259 -0.0457 2 2000 1 a2 0.968921 -0.0448 0.1479 -0.1343 0.1670 3 2000 2 a1 0.029218

Hi, I am trying to construct a logistic regression model from my data (104 patients and 25 events). I build a full model consisting of five predictors with the use of penalization by rms package (lrm, pentrace etc) because of events per variable issue. Then, I tried to approximate the full model by step-down technique predicting L from all of the componet variables using ordinary least squares

If I have a table (we'll call it, "test") containing two columns (as below): i x1 x2 x3 x4 x5 y 0 1.125 0.232 7.160 0.0859 8.905 1.5563 7 0.920 0.268 8.804 0.0865 7.388 0.8976 15 0.835 0.271 8.108 0.0852 5.348 0.7482 22 1.000 0.237 6.370 0.0838 8.056 0.7160 29 1.150 0.192 6.441 0.0821 6.960 0.3130 37 0.990 0.202 5.154 0.0792 5.690 0.3617 44 0.840 0.184 5.896 0.0812 6.932 0.1139 58

Hello, I would be very grateful if someone could give me a hand with my split plot design problems. So here is my design : I am studying the crossed-effects of water (wet/dry) and mowing (mowed/not-mowed = nm) on plant height (PH) within 2 types of plant communities (Xerobromion and Mesobromion) : - Within each type of communities, I have localised 4 blocks - In each block, I have defined

Hello R-community, I am trying to populate a column (doy) in a large dataset with the first column number that exceeds the value in another column (thold) using the 'apply' function. Sample data: pt D1 D17 D33 D49 D65 D81 D97 D113 D129 D145 D161 D177 D193 D209 D225 D241 D257 1 39177 0 0 0 0 0 0 0 0 0.4336 0.4754 0.5340667 0.5927334 0.6514 0.6966

Hello, assume I have an "unstructured" text line from a connection. Unfortunately, it is in string format: R> x [1] "\talpha0\t-0.638\t0.4043\t0.4043\t-2.215\t-0.5765\t-0.137\t501\t2000" How can I extract the data included in this string object "x" in order to get the elements for the parameter vector called "alpha0", i.e. -0.638 0.4043 0.0467

On 23.11.2012, at 15:12, john skaller <skaller at users.sourceforge.net> wrote: > > On 23/11/2012, at 5:46 PM, Sean Silva wrote: > >> Adding LLVMdev, since this is intimately related to the optimization passes. >> >>> I think this is roughly because some function level optimisations are >>> worse than O(N) in the number of instructions. >>

Hi All, Please pardon me if I am missing something obvious here. How do I get the Kaplan-Meier estimate function that is created by survfit and plotted by the code. fit <- survfit(Surv(time, status) , data=aml) plot(fit) That is, I need a function that will give me the survival estimate at a given time: \hat{S}(t). Thanks in advance. Ritwik Sinha ritwik.sinha at gmail.com | +12033042111 |

Hi All, I've been searching the help archives but haven't found a workable solution to this problem. I'm running an lme model with the following call: >lme.fnl <- lme(Max ~ S + Tr + Yr + Tr:Yr, random = ~1 |TID) > anova(lme.fnl) numDF denDF F-value p-value (Intercept) 1 168 19255.389 <.0001 S 1 168 5.912 0.0161 Tr

2017-02-05 23:12 GMT+01:00 Michael Stone <mstone at mathom.us>: > > It was probably because of this commit: > > http://cvsweb.openbsd.org/cgi-bin/cvsweb/src/usr.bin/ssh/sshd.c.diff?r1=1.472&r2=1.473 > Yes here the combination cr and lf is removed. > Which removed support for protocols older than 2 but perhaps failed to > account for the fact that newline had been

I want to use caret to build a model with an algorithm that actually has no parameters to find. How do I stop it from repeatedly building the same model 25 times? library(caret) data(mdrr) LOGISTIC_model <- train(mdrrDescr,mdrrClass ,method='glm' ,family=binomial(link="logit") ) LOGISTIC_model 528

Dear all,   I did ANOVA for many response variables (Var1, Var2, ....Var75000), and i got the result of p-value like below. Now, I want to select those predictors, which have pvalue less than or equal to 0.05 for each response variable. For example, X1, X2, X3, X4, X5 and X6 in case of Var1, and similarly, X1, X2.......X5 in case of Var2, only X1 in case of Var3 and none of the predictors in case

Hello, I am attempting to train a random forest model using the randomForest package on 500,000 rows and 8 columns (7 predictors, 1 response). The data set is the first block of data from the UCI Machine Learning Repo dataset "Record Linkage Comparison Patterns" with the slight modification that I dropped two columns with lots of NA's and I used knn imputation to fill in other gaps.

Hi, I'm sure this is simple, but I haven't been able to find this in TFM, say I have some data in R like this (pasted here: http://pastebin.com/raw.php?i=sjS9Zkup): > head(df) gender age smokes disease Y 1 female 65 ever control 0.18 2 female 77 never control 0.12 3 male 40 state1 0.11 4 female 67 ever control 0.20 5 male 63 ever state1 0.16

I am having a hard time understanding how to perform a "repeated measures" type of ANOVA with R. When reading the document found here: http://cran.r-project.org/doc/contrib/Lemon-kickstart/kr_repms.html I find that there is a reference to a function make.rm () that is supposed to rearrange a "one row per person" type of frame to a "one row per observation" type

Hi, In the following results I interpret exp(coef) as the factor that multiplies the base hazard rate if the corresponding variable is TRUE. For example, when the bucket is ks008 and fidelity <= 3, then the rate, compared to the base rate h_0(t), is h(t) = 0.200 h_0(t). My question is then, to what case does the base hazard rate correspond to? I would expect the reference to be the first

Dear R-helpers, I'm stuck with a little problem that surely has an easy solution but I can't think of a way to solve it. I'd really appreciate any help you can offer me! I'll provide a small example. Given a dataframe data.txt that looks like this: ID freq Var Var_mean Ratio_mean Var_median Ratio_median Var_sum Ratio_min Var_max Ratio_max Var_min