Displaying 20 results from an estimated 90 matches similar to: "object gstat"
2012 Jun 08
2
help with rle function on paired data
Dear R Community - I hope you might be able to provide some guidance
regarding the use of the rle function. I have a set of time-series data
where a measured value is recorded every 30 seconds after the start of an
experiment. Many of the measured values repeat and I am interested only in
the values when there is a change. If I turn the measured values into a
vector, the rle function works
2011 Aug 18
2
Selecting subset of factor levels
Dear r-help,
I would like to select a subset of levels from a factor variable in a data frame and return a data frame.
The data set consists of 3 variables, 2 of which are factors (Site, Fish) and one numeric (Datavalue) as follows:
Site Fish Datavalue
AB 2-1 2.3
AB 2-1 2.4
AB 2-1 2.2
AB 2-2 2.6
AB 2-2 2.5
AB 2-2 2.7
AB 2-3
2014 Sep 01
1
Correlation Matrix with a Covariate
R Help -
I'm trying to run a correlation matrix with a covariate of "age" and will
at some point will also want to covary other variables concurrently.
I'm using the "psych" package and have tried other methods such as writing
a loop to extract semi-partial correlations, but it does not seem to be
working. How can I accomplish this?
library(psych)
> set.cor(y =
2005 Jun 14
2
ordinary polynomial coefficients from orthogonal polynomials?
How can ordinary polynomial coefficients be calculated
from an orthogonal polynomial fit?
I'm trying to do something like find a,b,c,d from
lm(billions ~ a+b*decade+c*decade^2+d*decade^3)
but that gives: "Error in eval(expr, envir, enclos) :
Object "a" not found"
> decade <- c(1950, 1960, 1970, 1980, 1990)
> billions <- c(3.5, 5, 7.5, 13, 40)
> #
2011 Oct 04
2
About stepwise regression problem
First of all, I have GAMs
noxd<-gam(newNOX~pressure+maxtemp+s(avetemp,bs="cr")+s(mintemp,bs="cr")+s(RH,bs="cr")+s(solar,bs="cr")+s(windspeed,bs="cr")+s(transport,bs="cr"),family=gaussian
(link=log),groupD,methods=REML)
Then I type " summary(noxd)". and show
Family: gaussian
Link function: log
Formula:
newNO2 ~ pressure
2006 May 24
2
data.frame
Dear all,
Does any one knows why should I get the following error message, when trying
to do a simple data.frame??
DataF<-data.frame(Subject,BiomR,Spp,Capas,Litter,Herbs,LitterD,MaxCanH,DDifS
p,DSSp,Slope, CanDens,NearestSp)
Erro em data.frame(Subject, BiomR, Spp, Capas, Litter, Herbs, LitterD, :
arguments imply differing number of rows: 202, 0
The data I am using
2013 Apr 11
2
Read the data from a text file and reshape the data
I have a data set for different time intervals. The data has three comment
lines before data for each time interval. For each time interval there are
500 data points. I want to change the dataset such that I have the following
format:
t1 t2 t3 ................
0.00208 0.00417 0.00625 .................
a1 a2 a3 ...................
2012 Sep 22
0
timeIsInterval function not found
hello i'm trying the script of gstat course of munich "Spatial and spacetime
classes in R
" in http://geostat-course.org/Topic_Bivand_2012
the code is follow ... but a in timeIsInterval(dts) the script failed....
the message is "timeIsInterval function not found"
i'm updated R and package spacetime but has not worked
any suggestions?
thanks
setwd("C:/Documents
2006 Feb 20
2
formatting results from a function argument
Hello all,
I have a simple function which calculates summary statistics of a dataset
in terms of a factor (say area).
> x = data.frame(Area = c(rep("cleanup", 5), rep("ref", 5)), TcCB =
c(rnorm(5)+2, rnorm(5)));x
Area TcCB
1 cleanup 2.5829747
2 cleanup 2.6796868
3 cleanup 2.5437094
4 cleanup 2.8453616
5 cleanup 1.1789683
6 ref 1.0140391
7
2011 Jun 13
1
glm with binomial errors - problem with overdispersion
Dear all,
I am new to R and my question may be trivial to you...
I am doing a GLM with binomial errors to compare proportions of species in
different categories of seed sizes (4 categories) between 2 sites.
In the model summary the residual deviance is much higher than the degree
of freedom (Residual deviance: 153.74 on 4 degrees of freedom) and even
after correcting for overdispersion by
2011 Apr 20
2
survexp with weights
Hello,
I probably have a syntax error in trying to generate an expected
survival curve from a weighted cox model, but I can't see it. I used
the help sample code to generate a weighted model, with the addition
of a "weights=albumin" argument (I only chose albumin because it had
no missing values, not because of any real relevance). Below are my
code with the resulting error
2010 Nov 17
3
stacking consecutive columns
I have a file, each column of which is a separate year, and each row of each column is mean precipitation for that month. Looks like this (except it goes back to 1964).
month X2000 X2001 X2002 X2003 X2004 X2005 X2006 X2007 X2008 X2009
1 1.600 1.010 4.320 2.110 0.925 3.275 3.460 0.675 1.315 2.920
2 2.960 3.905 3.230 2.380 2.720 1.880 2.430 1.380
2008 Aug 08
2
aggregate
Dear All-
I have a dataset that is comprised of the following:
doy yr mon day hr hgt1 hgt2 hgt3 co21 co22 co23 sig1 sig2 sig3 dif flag
244.02083 2005 09 01 00 2.6 9.5 17.8 375.665 373.737 373.227 3.698 1.107
0.963 -0.509 PRE
244.0625 2005 09 01 01 2.6 9.5 17.8 393.66 384.773 379.466 15.336 11.033
5.76 -5.307 PRE
244.10417 2005 09 01 02 2.6 9.5 17.8 411.162 397.866 387.755 6.835 5.61
6.728
2008 Jun 05
1
choosing an appropriate linear model
I am trying to model the observed leaching of wood preservative chemicals
from treated wood during an outdoor experiment where leaching is caused by
rainfall events. For each rainfall event, the amount of rainfall was
recorded as well as the amount of preservative chemical leached. A number
of climatic variables were measured, but the most important is the amount of
rainfall.
I have tried a
2009 Apr 01
1
Noobie ANOVA intercept question
Dear R list,
I've been attempting to interpret the results from a three-way ANOVA. I
think I understand contrasts and the R defaults for these (treatment
contrasts). My question is: what is the intercept in this test? As far as I
can tell, its NOT the expected value of a point that belongs to the first
level of all three explanatory factors (because there is only one point that
satisfies these
2023 Dec 08
1
Convert two-dimensional array into a three-dimensional array.
Colleagues
I want to convert a 10x2 array:
# create a 10x2 matrix.
datavals <- matrix(nrow=10,ncol=2)
datavals[,] <- rep(c(1,2),10)+c(rnorm(10),rnorm(10))
datavals
into a 10x3 array, ThreeDArray, dim(10,2,10).
The values storede in ThreeDArray's first dimensions will be the data stored in datavalues.
ThreeDArray[i,,] <- datavals[i,]
The values storede in ThreeDArray's second
2009 Feb 17
2
Chromatogram deconvolution and peak matching
Hi,
I'm trying to match peaks between chromatographic runs.
I'm able to match peaks when they are chromatographed with the same method,
but not when there are different methods are used and spectra comes in to
play.
While searching I found the ALS package which should be usefull for my
application, but I couldn't figure it out.
I made some dummy chroms with R, which mimic my actual
2009 Aug 26
0
Trying to make Nas 0
I have an lm object called mro
A summary gives
> summary(mro)
Call:
lm(formula = REGRESSIONSTRING, data = wData)
Residuals:
Min 1Q Median 3Q Max
-8.18077 -1.06867 -0.09387 1.03153 11.20201
Coefficients: (1 not defined because of singularities)
Estimate Std. Error t value Pr(>|t|)
(Intercept) 7.2096 1.0345 6.969 5.37e-11 ***
2013 May 03
1
Likelihood
Hi all,
I have run a regression and want to calculate the likelihood of obtaining
the sample.
Is there a way in which I can use R to get this likelihood value?
Appreciate your help on this.
The following are the details:
raw_ols1=lm(data$LOSS~data$GDP+data$HPI+data$UE)
summary(raw_ols1)
Call:
lm(formula = data$LOSS ~ data$GDP + data$HPI + data$UE)
Residuals:
Min 1Q
2011 Feb 03
3
interpret significance from the contr.poly() function
Hello R-help
I don’t know how to interpret significance from the contr.poly() function . From
the example below
: how can I tell if data has a significant Linear/quadratic/cubic trend?
> contr.poly(4, c(1,2,4,8))
.L .Q .C
[1,] -0.51287764 0.5296271 -0.45436947
[2,] -0.32637668 -0.1059254 0.79514657
[3,] 0.04662524 -0.7679594 -0.39757328
[4,] 0.79262909