search for: trending

Displaying 20 results from an estimated 1376 matches for "trending".

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2009 Nov 08
2
linear trend line and a quadratic trend line.
Dear list users How is it possible to visualise both a linear trend line and a quadratic trend line on a plot of two variables? Here my almost working exsample. data(Duncan) attach(Duncan) plot(prestige ~ income) abline(lm(prestige ~ income), col=2, lwd=2) Now I would like to add yet another trend line, but this time a quadratic one. So I have two trend lines. One linear trend line
2005 Jun 14
1
using forecast() in dse2 with an ARMA model having a trend component
(My apologies if this is a repeated posting. I couldn't find any trace of my previous attempt in the archive.) I'm having trouble with forecast() in the dse2 package. It works fine for me on a model without a trend, but gives me NaN output for the forecast values when using a model with a trend. An example: # Set inputs and outputs for the ARMA model fit and test periods
2011 Dec 01
3
Assign name to object for each iteration in a loop.
Hi R-users, I'm trying to produce decompositions of a multiple time-series, grouped by a factor (called "area"). I'm modifying the code in the STLperArea function of package ndvits, as this function only plots produces stl plots, it does not return the underlying data. I want to extract the trend component of each decomposition ("x$time.series[,trend]), assign a name
2011 Jul 21
1
for/if loop in R
hi, Can someone please help me figure out where I am making a mistake in my for/if loop: I have a data frame (112 rows) called pricedata with 3 columns: date, prices, return. Now, I want to add a 4th column, trend, which can have 2 values 0 or 1. if return>1%, trend=1 else trend=0. so, this is what I did: >trend<-numeric(nrow(pricedata)) >cbind(pricedata,trend) >for(i in
2009 Jul 15
0
problems in resampling time series, block length, trend.test
Hi, I have a time series (say "x") of 13 years showing an evident increase. I want to exclude two observations (the fourth and 10th), so I do: > trend.test(x[-c(4,10)]) where: > x[-c(4,10)] [1] 7 37 79 72 197 385 636 705 700 1500 1900 and I get: Spearman's rank correlation rho data: x[-c(4, 10)] and time(x[-c(4, 10)]) S = 4, p-value < 2.2e-16
2009 Jul 15
0
FW: problems in resampling time series, block length, trend.test
Hi, I have a time series (say "x") of 13 years showing an evident increase. I want to exclude two observations (the fourth and 10th), so I do: > trend.test(x[-c(4,10)]) where: > x[-c(4,10)] [1] 7 37 79 72 197 385 636 705 700 1500 1900 and I get: Spearman's rank correlation rho data: x[-c(4, 10)] and time(x[-c(4, 10)]) S = 4, p-value < 2.2e-16
2003 Sep 30
2
non-linear trends in kriging model
Hi I am struggling to fit a non-linear trend using the likfit function in geoR. Specifically I want a sigmoidal function, something like SSfpl in the nls package to fit the trend. But it seems trend.spatial in geoR only works with lm or glm type models. Any ideas how I can specify the model to calculate the kriging parameters using REML, including the parameters of a sigmoidal trend function
2017 Jan 21
2
mail-trends+Dovecot
Hi, Out of curiosity, has anyone managed to use mail-trends[0] to analyse their e-mails. I thought about trying it out, but got stuck midstream. The mail-trends scripts work very well with gmail, but because they say it is supposed to work with _any_ IMAP server, I thought I could get it running with Dovecot too. I know this is NOT a mail-trends support group though, but I believe there is
2006 Jul 06
2
KPSS test
Hi, Am I interpreting the results properly? Are my conclusions correct? > KPSS.test(df) ---- ---- KPSS test ---- ---- Null hypotheses: Level stationarity and stationarity around a linear trend. Alternative hypothesis: Unit root. ---- Statistic for the null hypothesis of level stationarity: 1.089 Critical values: 0.10 0.05 0.025 0.01 0.347 0.463
2004 Apr 14
7
trend turning points
Hi, does anybody know of a nice test to detect trend turning points in time series? Possibly with reference? Thanks, joerg
2011 Oct 28
0
problem with glsm.krige: trendd and trend l must have similar specifications error
Hello, I used glsm.mcmc and likfit.glsm to create model. Now I want to predict at different locations, but I can't get glsm.krige to work. I keep getting the error that trend.d and trend.l must have similar specifications. I have tried numerous ways to include the covariates in the glsm.krige model, and I keep getting the same error message. The bolded part is the part that doesn't work.
2008 Apr 30
1
apply and monthly time series (PR#11352)
Full_Name: Stephen McIntyre Version: 2.7 OS: Windows XP Submission from: (NULL) (99.231.2.44) When I use the apply function to calculate a trend for a matrix of monthly time series, it yields a different answer than when the trend is calculated one at a time (by a factor of 12) rather than the identical answer as it should. Here's an example:
2012 Feb 19
1
time-series trend classification
Hello everyone, I was looking for a way to classify time-series based on the curve-fit. I try to campute several trends so i was thinking to link each trend with a function. increase with exponential for example, increase and decrease with a gaussian etc. The possiblities are endless though and it seems that is not always working well, especially if you work on small time-series (of 5-10 points
2007 Jul 13
1
spatstat - Fitting a Strauss model with trend determined by kernel density smoother
Dear r-help, I would like to use the 'ppm' function of the 'spatstat' package to fit a Strauss inhibition model. I understand that I can specify a parametric model for the "background" trend, but how would I specify a trend which is estimated using a Kernel density smoother? In particular, I would like to use the 'kde' function of the 'ks' package to
2009 May 15
1
Dickey-Fuller Tests with no constant and no trend
R has a Dickey-Fuller Test implementation (adf.test) that tests for unit roots in an autoregressive process with a constant and linear trend. Is there a DF implementation that doesn't use the constant or trend? Thanks, Jake. -- View this message in context: http://www.nabble.com/Dickey-Fuller-Tests-with-no-constant-and-no-trend-tp23565210p23565210.html Sent from the R help mailing list
2010 Aug 19
1
How to include trend (drift term) in arima.sim
I have been trying to simulate from a time series with trend but I don't see how to include the trend in the arima.sim() call. The following code illustrates the problem: # Begin demonstration program x <- c(0.168766559, 0.186874000, 0.156710548, 0.151809531, 0.144638812, 0.142106888, 0.140961714, 0.134054659, 0.138722419, 0.134037018, 0.122829846, 0.120188714,
2008 Apr 21
2
Trend test for survival data
Hello, is there a R package that provides a log rank trend test for survival data in >=3 treatment groups? Or are there any comparable trend tests for survival data in R? Thanks a lot Markus -- Dipl. Inf. Markus Kreuz Universitaet Leipzig Institut fuer medizinische Informatik, Statistik und Epidemiologie (IMISE) Haertelstr. 16-18 D-04107 Leipzig Tel. +49 341 97 16 276 Fax. +49 341 97 16
2007 Feb 27
2
ts; decompose; plot and title
Is there any way to give a "decent" title after I plot something generated by decompose? For example: # generate something with period 12 x <- rnorm(600) + sin(2 * pi * (1:600) / 12) # transform to a monthy time series y <- ts(x, frequency=12, start=c(1950,1)) # decompose z <- decompose(y) # plot plot(z) Now, the title is the ugly "Decomposition of additive time
2003 May 28
1
Test for trend?
Hello R community, I would like to test for learning effects by subjects in my experiment. Each subject participates in six consecutive auction rounds of the same treatment. The response variable is the efficiency of an auction outcome measured by a real number. Since the efficiency increases over the six rounds, I suppose that subjects learn about the rules of the auction institution, but I
2004 Mar 04
2
adding trend to an arima model
Hi, Does anyone know a method for adding a linear/polynominal trend to a simulated arima model using the arima.sim function? Any help will be greatly appreciated. Cheers, Sam.