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