similar to: Having trouble using data returned by Ajax.request

Displaying 20 results from an estimated 100 matches similar to: "Having trouble using data returned by Ajax.request"

2013 Jul 25
1
GGplot 2 – cannot get histogram and box plot axis to match.
Problem: I am trying to get the histogram and box plot x axis to match. I?ve tried using the expand_limits function to make the axis match but that didn?t make the axis match. The histogram?s axis are still consistently larger than the ones for the box plot (though the function did help). Does anyone have a suggestion as to what I should do instead? Background: I am building a Shiny app that
2010 Nov 04
1
Best Fit line trouble with rsruby
Hello, I am using R, through rsruby, to create a graph and best fit line for a set of data points, regarding data collected in a Chemistry class. The problem is that although the graph functions perfectly properly, the best fit line will not work. I initially used code I pretty much copied from a website with a tutorial on this, which was: graphData.png("/code/Beer's-Law
2018 Apr 05
2
Obtain gradient at multiple values for exponetial decay model
Readers, Data set: t,c 0,100 40,78 80,59 120,38 160,25 200,21 240,16 280,12 320,10 360,9 400,7 graphdata<-read.csv('~/tmp/data.csv') graphmodeld<-lm(log(graphdata[,2])~graphdata[,1]) graphmodelp<-exp(predict(graphmodeld)) plot(graphdata[,2]~graphdata[,1]) lines(graphdata[,1],graphmodelp) Please what is the function and syntax to obtain gradient values for the model curve at
2008 Aug 14
1
Graphing: plot 3rd variable based on color gradient
Hello, I am searching for the best method to plot two variables with points whose output color depends on the size of a third variable. For example, the darkness of the x-y point would increase incrementally based on the size of the z value, similar to the colramp parameter in geneplotter. This would be analagous to symbols(), except changing the selection from the color gradient rather than the
2018 Apr 06
3
Obtain gradient at multiple values for exponential decay model
> On Apr 6, 2018, at 8:03 AM, David Winsemius <dwinsemius at comcast.net> wrote: > > >> On Apr 6, 2018, at 3:43 AM, g l <gnulinux at gmx.com> wrote: >> >>> Sent: Friday, April 06, 2018 at 5:55 AM >>> From: "David Winsemius" <dwinsemius at comcast.net> >>> >>> >>> Not correct. You already have
2012 Feb 17
1
basic help: graph multivariate analysis.
Hey guys, I'd really appreciate any help. I have a multivariate analysis done, the output of which is: > GraphData <-read.table("eigen.coa") > GraphData V1 V2 V3 V4 1 1 0.371970 0.8552 0.8552 2 2 0.061785 0.1420 0.9972 3 3 0.001211 0.0028 1.0000 4 4 0.000000 0.0000 1.0000 > summary(GraphData) V1 V2 V3
2018 Apr 06
0
Obtain gradient at multiple values for exponential decay model
> On Apr 6, 2018, at 3:43 AM, g l <gnulinux at gmx.com> wrote: > >> Sent: Friday, April 06, 2018 at 5:55 AM >> From: "David Winsemius" <dwinsemius at comcast.net> >> >> >> Not correct. You already have `predict`. It is capale of using the `newdata` values to do interpolation with the values of the coefficients in the model. See:
2018 Apr 06
2
Obtain gradient at multiple values for exponential decay model
> Sent: Friday, April 06, 2018 at 5:55 AM > From: "David Winsemius" <dwinsemius at comcast.net> > > > Not correct. You already have `predict`. It is capale of using the `newdata` values to do interpolation with the values of the coefficients in the model. See: > > ?predict > The ? details did not mention interpolation explicity; thanks. > The
2018 Apr 07
0
Obtain gradient at multiple values for exponential decay model
I have never found the R symbolic differentiation helpful because my functions are typically quite complicated, but was prompted by Steve Ellison's suggestion to try it out in this case: ################# reprex (see reprex package) graphdta <- read.csv( text = "t,c 0,100 40,78 80,59 120,38 160,25 200,21 240,16 280,12 320,10 360,9 400,7 ", header = TRUE ) nd <- c( 100, 250,
2010 Apr 24
1
help please: predict error code
Hello,   I am trying to calculate predicted values derived from one dataset into a hypothetical dataset. I tried this line of code:   graphdata$fmgpredvalues <- predict(Acs250.3.4, graphdata)   and received the following error message:   ERROR: ZXend[1], drop = FALSE] %*%lmeFit$beta   I have made sure all variable names are the same between the two datasets and all factors are appropriately
2009 Apr 07
4
Re ading Excel 5.0 files with RODBC?
Hi, i'm trying to read some data from excel files but it seems that neither xlsReadWrite nor sqlFetch (RODBC) doesn't like the format (Excel 5.0). When i open the file in Excel and save it in a new format Excel 97 -2003 everything works fine. Is it possible to use ODBC connection to open old format files, or i guess i will have to open and save every file in Excel in new format, which
2018 Apr 06
1
Obtain gradient at multiple values for exponential decay model
> Sent: Friday, April 06, 2018 at 1:44 PM > From: "Jeff Newmiller" <jdnewmil at dcn.davis.ca.us> > > You did not try my suggestion. You tried David's, which has a leftover mistake from your guesses about what the argument to coef should be. Yes, sorry for the mistake. coef(graphmodeld) (Intercept) graphdata[, 1] 4.513544204 -0.006820623 This corresponds
2018 Apr 06
2
Obtain gradient at multiple values for exponential decay model
> Sent: Friday, April 06, 2018 at 4:53 AM > From: "Jeff Newmiller" <jdnewmil at dcn.davis.ca.us> > To: "g l" <gnulinux at gmx.com> > coef( graphmodeld ) > coef(graphmodelp) Error: $ operator is invalid for atomic vectors A quick search engine query revealed primarily references to the dollar sign ($) operator which does not seem relevant to this
2018 Apr 06
0
Obtain gradient at multiple values for exponential decay model
You did not try my suggestion. You tried David's, which has a leftover mistake from your guesses about what the argument to coef should be. -- Sent from my phone. Please excuse my brevity. On April 6, 2018 3:30:10 AM PDT, g l <gnulinux at gmx.com> wrote: >> Sent: Friday, April 06, 2018 at 4:53 AM >> From: "Jeff Newmiller" <jdnewmil at dcn.davis.ca.us>
2018 Apr 05
0
Obtain gradient at multiple values for exponetial decay model
This smells like homework, which the Posting Guide indicates is off topic. I am not aware of "the function" that will solve this, but if you know what a gradient is analytically then you should be able to put together a solution very similar to the code you already have with the addition of using the coef function. -- Sent from my phone. Please excuse my brevity. On April 5, 2018
2011 Apr 17
7
possibly an inter-process comm problem
I have been running a program with Wine for over a year. With its version 3 there has been basically no problem. Lately, the program is upgraded to version 4 and it crashes with Wine. I tried to print out debug messages but can't seem to get any clue about what is wrong. While I was testing with the debug messages, an interesting thing I found is that if I run it with the following debug
2007 May 10
13
REST own action
I can sent my own action to a users_controller stating in routes.rb map.resources :users, :member => { :network => get } I defined in my users_controller the network action.... so I can have an url like "http://localhost:3000/users/168.xml;network" now I would like to have another action network_plus which is a little bit different fron the network.. sure, I can state another
2007 Dec 20
1
Available environment variables
Hello, I am trying to set my environment to streamline the downloading / updating of packages. I have been through R-intro.html (10.8 Customizing-the-environment), R-FAQ (5.2 How can add-on packages be installed?), rw-FAQ, and help pages for Sys.setenv {base}, download.packages {utils}, etc.,. I am looking for something similar to the download.packages destdir argument on a global environment
2004 Mar 26
0
Package update: 'urca' version 0.3-3
Dear R-list member, an update of package 'urca' has been uploaded to CRAN (Mirror: Austria). In the updated release unit root and cointegration tests encountered in applied econometric analysis are implemented. The package is written in 'pure' R and utilises S4 classes. In particular, the Johansen procedure with likelihood ratio tests for the inclusion of a linear trend,
2010 Oct 15
0
nomianl response model
Is there a way to estimate a nominal response model? To be more specific let's say I want to calibrate: \pi_{v}(\theta_j)=\frac{e^{\xi_{v}+\lambda_{v}\theta_j}}{\sum_{h=1}^m e^{\xi_{h}+\lambda_{h}\theta_j}} Where $\theta_j$ is a the dependent variable and I need to estimate $\xi_{h}$ and $\lambda_{h}$ for $h \in {1...,m}$. Thank you, Mauricio Romero Quantil S.A.S. Cel: 3112231150