similar to: Plot unusual subset of data

Displaying 15 results from an estimated 15 matches similar to: "Plot unusual subset of data"

2013 Mar 12
5
extract values
Hello all! I have a problem to extract values greater that for example 1820. I try this code: x[x[,1]>1820,]->x1 Please help me! Thank you! The data structure is: structure(c(2.576, 1.728, 3.434, 2.187, 1.928, 1.886, 1.2425, 1.23, 1.075, 1.1785, 1.186, 1.165, 1.732, 1.517, 1.4095, 1.074, 1.618, 1.677, 1.845, 1.594, 1.6655, 1.1605, 1.425, 1.099, 1.007, 1.1795, 1.3855, 1.4065, 1.138, 1.514,
2013 Mar 13
2
merge datas
Hello all! I have a problem with R. I try to merge data like this: structure(c(2.1785, 1.868, 2.1855, 2.5175, 2.025, 2.435, 1.809, 1.628, 1.327, 1.3485, 1.4335, 2.052, 2.2465, 2.151, 1.7945, 1.79, 1.6055, 1.616, 1.633, 1.665, 2.002, 2.152, 1.736, 1.7985, 1.9155, 1.7135, 1.548, 1.568, 1.713, 2.079, 1.875, 2.12, 2.072, 1.906, 1.4645, 1.3025, 1.407, 1.5445, 1.437, 1.463, 1.5235, 1.609, 1.738, 1.478,
2004 Jan 13
3
How can I test if a not independently and not identically distributed time series residuals' are uncorrelated ?
I'm analizing the Argentina stock market (merv) I download the data from yahoo library(tseries) Argentina <- get.hist.quote(instrument="^MERV","1996-10-08","2003-11-03", quote="Close") merv <- na.remove(log(Argentina)) I made the Augmented Dickey-Fuller test to analyse if merv have unit root: adf.test(merv,k=13) Dickey-Fuller = -1.4645,
2009 Feb 16
1
Overdispersion with binomial distribution
I am attempting to run a glm with a binomial model to analyze proportion data. I have been following Crawley's book closely and am wondering if there is an accepted standard for how much is too much overdispersion? (e.g. change in AIC has an accepted standard of 2). In the example, he fits several models, binomial and quasibinomial and then accepts the quasibinomial. The output for residual
2010 Mar 29
2
Need help on matrix manipulation
Dear all, Ket say I have 3 matrices : mat1 <- matrix(rnorm(16), 4) mat2 <- matrix(rnorm(16), 4) mat3 <- matrix(rnorm(16), 4) Now I want to merge those three matrices to a single one with dimension 4*3=12 and 4 wherein on resulting matrix, row 1,4,7,10 will be row-1,2,3,4 of "mat1", row 2,5,8,11 will be row-1,2,3,4 of "mat2" and row 3,6,8,12 will be row-1,2,3,4 of
2013 Mar 21
1
multiple peak fit
Hi I went through some extensive search to find suitable method (package, function) to fit multiple peaks. The best I found is ALS package but it requires rather complicated input structure probably resulting from GC-MS experimental data and seems to be an overkill to my problem. I have basically simple two column data frame with columns time and sig dput(temp) structure(list(time = c(33, 34,
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
2010 Feb 07
0
[LLVMdev] [PATCH] FoldingSetNodeID: use MurmurHash2 instead of SuperFastHash
While I've not reviewed the patch in too much detail, it looks promising. Can you run some end-to-end benchmarks to make sure that cache pressure in the full program or other variables not accounted for in a micro-benchmark don't dominate performance? Specifically the nightly tester includes a number of real programs and machinery to measure total compile time. On Sat, Feb 6, 2010 at 7:09
2004 Jan 14
0
How can I test if a not independently and not identicallydistributed time series residuals' are uncorrelated ?
I'm analizing the Argentina stock market (merv) I download the data from yahoo library(tseries) Argentina <- get.hist.quote(instrument="^MERV","1996-10-08","2003-11-03", quote="Close") merv <- na.remove(log(Argentina)) I made the Augmented Dickey-Fuller test to analyse if merv have unit root: adf.test(merv,k=13) Dickey-Fuller = -1.4645,
2005 Jul 15
2
glm(family=binomial(link=logit))
Hi I am trying to make glm() work to analyze a toy logit system. I have a dataframe with x and y independent variables. I have L=1+x-y (ie coefficients 1,1,-1) then if I have a logit relation with L=log(p/(1-p)), p=1/(1+exp(L)). If I interpret "p" as the probability of success in a Bernouilli trial, and I can observe the result (0 for "no", 1 for
2010 Feb 07
3
[LLVMdev] [PATCH] FoldingSetNodeID: use MurmurHash2 instead of SuperFastHash
On Sat, Feb 06, 2010 at 04:51:15PM -0800, Chandler Carruth wrote: > While I've not reviewed the patch in too much detail, it looks > promising. Can you run some end-to-end benchmarks to make sure that > cache pressure in the full program or other variables not accounted > for in a micro-benchmark don't dominate performance? Specifically the > nightly tester includes a number
2010 May 18
2
[LLVMdev] Fast register allocation
As you may have noticed, I have been working on a fast register allocator in the last week. The new allocator is intended to replace the local allocator for debug builds. Both allocators use a local allocation strategy. Each basic block is scanned from top to bottom, and virtual registers are assigned to physical registers as they appear. There are no live registers between blocks. Everything is
2010 Feb 06
4
[LLVMdev] [PATCH] FoldingSetNodeID: use MurmurHash2 instead of SuperFastHash
Some additional info can be found at: http://murmurhash.googlepages.com/ http://en.wikipedia.org/wiki/MurmurHash http://www.codeproject.com/KB/recipes/hash_functions.aspx as well as in the patch description itself. Patch and benchmark attached. Gregory -------------- next part -------------- A non-text attachment was scrubbed... Name:
2012 Apr 10
3
nls function
Hi, I've got the following data: x<-c(1,3,5,7) y<-c(37.98,11.68,3.65,3.93) penetrationks28<-dataframe(x=x,y=y) now I need to fit a non linear function so I did: fit <- nls(y ~ I(a+b*exp(1)^(-c * x)), data = penetrationks28, start = list(a=0,b = 1,c=1), trace = T) The error message I get is: Error in nls(y ~ I(a + b * exp(1)^(-c * x)), data = penetrationks28, start = list(a =
2012 Jul 06
4
differences between survival models between STATA and R
Dear Community, I have been using two types of survival programs to analyse a data set. The first one is an R function called aftreg. The second one an STATA function called streg. Both of them include the same analyisis with a weibull distribution. Yet, results are very different. Shouldn't the results be the same? Kind regards, J -- View this message in context: