similar to: How to have columns lined up?

Displaying 20 results from an estimated 200 matches similar to: "How to have columns lined up?"

2012 Sep 19
2
Help reproducing a contour plot
Hi All, I am trying to reproduce this using R instead. [image: Full-size image (38 K)] I tried using the following code *SChla <- read.csv("SM_Chla_data.csv")* *Atlantis <- SChla[16:66,]* *head(Atlantis)* * * Seamount Station Depth Pico Nano Micro Total_Ch dbar Latitude Longitud 16 Atlantis 1217 Surface 0.0639 0.1560 0.0398 0.2597 2.082 -32.71450 57.29733
2012 Feb 13
2
Error in apply(x2, 1, diff) : dim(X) must have a positive length
Anyone knows hat might be the cause of this error? Thanks for any help! >library(MASS) > dif.mns = function(x2,tr1=.2,tr2=.3){ + #generates four different 'means' using + #difference scores from x2, an n x 2 matrix + #for use w/ bootstrap comparisons + diffs = apply(x2,1,diff) + mn1=mean(diffs) + mn2=mean(diffs,tr=.2) + mn3=mean(diffs,tr=.3) +
2001 Feb 08
2
Test for multiple contrasts?
Hello, I've fitted a parametric survival model by > survreg(Surv(Week, Cens) ~ C(Treatment, srmod.contr), > data = poll.surv.wo3) where srmod.contr is the following matrix of contrasts: prep auto poll self home [1,] 1 1 1.0000000 0.0 0 [2,] -1 0 0.0000000 0.0 0 [3,] 0 -1 0.0000000 0.0 0 [4,] 0 0 -0.3333333 1.0 0 [5,] 0 0
2011 Sep 09
2
Different results with arima in R 2.12.2 and R 2.11.1
Hello , I have estimated the following model, a sarima: p=9 d=1 q=2 P=0 D=1 Q=1 S=12 In R 2.12.2 Call: arima(x = xdata, order = c(p, d, q), seasonal = list(order = c(P, D, Q), period = S), optim.control = list(reltol = tol)) Coefficients: ar1 ar2 ar3 ar4 ar5 ar6 ar7 ar8 ar9 0.3152 0.8762 -0.4413 0.0152 0.1500 0.0001 -0.0413 -0.1811
2007 Nov 24
5
how to calculate the return?
Hi, R-users, data is a matrix like this AMR BS GE HR MO UK SP500 1974 -0.3505 -0.1154 -0.4246 -0.2107 -0.0758 0.2331 -0.2647 1975 0.7083 0.2472 0.3719 0.2227 0.0213 0.3569 0.3720 1976 0.7329 0.3665 0.2550 0.5815 0.1276 0.0781 0.2384 1977 -0.2034 -0.4271 -0.0490 -0.0938 0.0712 -0.2721 -0.0718 1978 0.1663 -0.0452 -0.0573 0.2751 0.1372 -0.1346
2004 Jul 11
2
Interpreting Results of Bootstrapping
I tried to bootstrap the correlation between two variables x1 and x2. The resulting distribution has two distinct peaks, how should I interprete it? The original code is attached. Y. C. Tao ---------------- library(boot); my.correl<-function(d, i) cor(d[i,1], d[i,2]) x1<-c(-2.612,-0.7859,-0.5229,-1.246,1.647,1.647,0.1811,-0.07097,0.8711,0.4323,0.1721,2.143,
2007 Jul 25
2
using contrasts on matrix regressions (using gmodels, perhaps)
Hi, I want to test for a contrast from a regression where I am regressing the columns of a matrix. In short, the following. X <- matrix(rnorm(50),10,5) Y <- matrix(rnorm(50),10,5) lm(Y~X) Call: lm(formula = Y ~ X) Coefficients: [,1] [,2] [,3] [,4] [,5] (Intercept) 0.3350 -0.1989 -0.1932 0.7528 0.0727 X1 0.2007 -0.8505 0.0520
2006 Jun 01
1
why does arima returns "NAN" standard error?
Hi everyone, ----------------------------- Coefficients: ar1 ar2 ma1 ma2 sar1 intercept drift 1.5283 -0.7189 -1.9971 0.9999 0.3982 0.0288 -9e-04 s.e. 0.0869 0.0835 0.0627 0.0627 0.1305 NaN NaN sigma^2 estimated as 0.04383: log likelihood = 4.34, aic = 7.32 Warning message: NaNs produced in: sqrt(diag(object$var.coef))
2008 Dec 11
1
candisc plotting
Hello, I have a file with two dependent variables (three and five) and one independent variable. I do i.mod <- lm(cbind(three, five) ~ species, data=i.txt) and get the following output: Coefficients: three five (Intercept) 9.949 9.586 species -1.166 -1.156 I do a" i.can<-candisc(i.mod,data=i): and get the following output: Canonical Discriminant Analysis
2011 Nov 14
1
2^k*r (with replications) experimental design question
Hello, I have one replication (r=1 of the 2^k*r) of a 2^k experimental design in the context of performance analysis i.e. my response variables are Throughput and Response Time. I use the "aov" function and the results look ok: > str(throughput) 'data.frame': 286 obs. of 7 variables: $ Time : int 6 7 8 9 10 11 12 13 14 15 ... $ Throughput : int 42 44 33 41
2012 Jun 30
2
Significance of interaction depends on factor reference level - lmer/AIC model averaging
Dear R users, I am using lmer combined with AIC model selection and averaging (in the MuMIn package) to try and assess how isotope values (which indicate diet) vary within a population of animals. I have multiple measures from individuals (variable 'Tattoo') and multiple individuals within social groups within 4 locations (A, B, C ,D) crucially I am interested if there are
2018 Mar 08
0
Names of variables needed in newdata for predict.glm
Hi, Some try: > names(mi$xlevels) [1] "f" > all.vars(mi$formula) [1] "D" "x" "f" "Y" > names(mx$xlevels) [1] "f" > all.vars(mx$formula) [1] "D" "x" "f" When offset is indicated out of the formula, it does not work... Marc Le 07/03/2018 ? 06:20, Bendix Carstensen a ?crit?: > I would like
2008 Mar 24
1
Great difference for piecewise linear function between R and SAS
Dear Rusers, I am now using R and SAS to fit the piecewise linear functions, and what surprised me is that they have a great differrent result. See below. #R code--Knots for distance are 16.13 and 24, respectively, and Knots for y are -0.4357 and -0.3202 m.glm<-glm(mark~x+poly(elevation,2)+bs(distance,degree=1,knots=c(16.13,24)) +bs(y,degree=1,knots=c(-0.4357,-0.3202
2011 Jan 07
1
Currency return calculations
Dear sir, I am extremely sorry for messing up the logic asking for help w.r.t. my earlier mails   I have tried to explain below what I am looking for.     I have a database (say, currency_rates) storing datewise currency exchange rates with some base currency XYZ.   currency_rates <- data.frame(date = c("12/31/2010", "12/30/2010", "12/29/2010",
2009 Sep 25
1
error while plotting
I am getting the following errors when I am trying to plot the data below. I cannot figure out the error. Error in plot.window(...) : need finite 'xlim' values In addition: Warning messages: 1: In min(x) : no non-missing arguments to min; returning Inf 2: In max(x) : no non-missing arguments to max; returning -Inf 3: In min(x) : no non-missing arguments to min; returning Inf 4: In max(x) :
2012 Aug 09
1
Factor moderators in metafor
I'm puzzled by the behaviour of factors in rma models, see example and comments below. I'm sure there's a simple explanation but can't see it... Thanks for any input John Hodgson ------------------------------------- code/selected output ----------------- library(metafor) ## Set up data (from Lenters et al A Meta-analysis of Asbestos and Lung Cancer... ##
2018 Mar 07
3
Names of variables needed in newdata for predict.glm
I would like to extract the names, modes [numeric/factor] and levels of variables needed in a data frame supplied as newdata= argument to predict.glm() Here is a small example illustrating my troubles; what I want from (both of) the glm objects is the vector c("x","f","Y") and an indication that f is a factor: library( splines ) dd <- data.frame( D =
2012 May 15
1
Regression Analysis or Anova?
Dear all, I hope to be the clearest I can. Let's say I have a dataset with 10 variables, where 4 of them represent for me a certain phenomenon that I call Y. The other 6 represent for me another phenomenon that I call X. Each one of those variables (10) contains 37 units. Those units are just the respondents of my analysis (a survey). Since all the questions are based on a Likert scale, they
2005 Jun 22
1
chisq test and fisher exact test
Hi, I have a text mining project and currently I am working on feature generation/selection part. My plan is selecting a set of words or word combinations which have better discriminant capability than other words in telling the group id's (2 classes in this case) for a dataset which has 2,000,000 documents. One approach is using "contrast-set association rule mining" while the
2010 Jan 19
1
splitting a factor in an analysis of deviance table (negative binomial model)
Dears useRs, I have 2 factors, (for the sake of explanation - A and B), with 4 levels each. I've already fitted a negative binomial generalized linear model to my data, and now I need to split the factors in two distinct analysis of deviance table:  - A within B1, A within B2, A within B3 and A within B4  - B within A1, B within A2, B within A3 and B within A4 Here is a code that illustrates