similar to: Calculate area under a curve

I have a model Q=K*A*(R^r)*(S^s) A, R, and S are data I have and K is a curve fitting parameter. I have linearized as log(Q)=log(K)+log(A)+r*log(R)+s*log(S) I have taken the log of the data that I have and this is the model formula without the K part lm(Q~offset(A)+R+S, data=x) What is the formula that I should use? Thanks for all of your help. I can provide a subset of data if necessary.

I would like to be able to grab x and y columns out of a dataframe and then grab all of the columns that are not equal to x or y. I am sure that I am missing something easy. ftbr_UTM_downstream <- (structure(list(site = c("Jennie_Creek_Main_Stem", "Wolf_Pit_Creek_Main_Stem", "Little_Rockfish_Main_Stem_North", "Big_Muddy_Creek_Main_Stem",

I have just upgraded to R 2.13 and have library(ggplot2) in my .Rprofile (among other things). when i start R I get an error message. Has something in the start up scripts changed? Is there a better way to specify the library calls in .Rprofile? Thanks for all of the help in advance. Error: Loading required package: grid Loading required package: proto Error in rename(x, .base_to_ggplot) :

Hi All, I am trying to lable the y-axis on my scatterplot with the following: "Soil moisture content (m3m-3)" I am using the following coding for plotting the graph: plot(soilmoisture~gradientlevel, xlab="Levels of droughting gradient", ylab="Soil moisture content (m3m-3)", bty="l", font.main="2", pch=16, las=1, cex.lab="1.13") I have

s <- 1.00 max(s) returns 1 is there anyway that I can get it to return 1.00. I am using the results of this max statement in a grep statement and it returns the wrong numbers, I will provide more information and code if it would make more sense in context. -- Stephen Sefick ____________________________________ | Auburn University? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?? | | Department of

Hello all R listers, I'm struggling to select an appropriate statistical method for my data set. I have collected soil moisture measurements every hour for 2 years. There are 75 sensors taking these automated measurements, spread evenly across 4 treatments and a control. I'm not interested in being able to predict soil future soil moisture trends, but rather in knowing whether the

I am trying to it a particular nonlinear model common in Soil Science to moisture release data from soil. I have written the function as shown below according to the logist example in Ch8 of Pinheiro & Bates. I am getting the following error (R version 2.1.1) *Error in qr(attr(rhs, "gradient")) : NA/NaN/Inf in foreign function call (arg 1)* Below is the function and data. /#

R 2.12.2 on Scientific Linux 6.4 #works chron(times.="15:00:00", format=c(times="h:m:s")) #doesn't work chron(times.="15:00", format=c(times="h:m")) From chron Manual: The times format can be any permutation of "h", "m", and "s" separated by any one non-special character. The default is "h:m:s". what am I

These data represent stream channel cross-sectional surveys. I would like to be able to find the measurement on the tape (measure) where the Bank Full Depth (bkf_depths) is 0. This will happen twice because the channel has two sides. I thought fitting a loess line to these data and then predicting the measurment number would do it. I was wrong. Below is my failed attempt. My naive thought is

Hi everyone, I have a question about calculating r-squared in R. I have tried searching the archives and couldn't find what I was looking for - but apologies if there is somewhere I can find this... I carried out a droughting experiment to test plant competition under limited water. I had: - 7 different levels of watering treatment (1 -7 - from most watered to least watered/) - 15

When I use this I am getting following warning at the time of opening R from that shortcut : "-max-mem-size=2048MB:too large and taken as 2047M" Why I am getting this? I have 3GB ram installed and using within Vista. Thanks -- View this message in context: http://r.789695.n4.nabble.com/How-to-allocate-more-memories-to-R-tp2271714p2272436.html Sent from the R help mailing list archive

I have a DF ID VALUE 100 120 101 100 102 100 103 104 105 .... when i calculate the sum of the values, it returned NA. should I populate the blank value as 0? Thanks, -- View this message in context: http://r.789695.n4.nabble.com/MISSING-VALUE-IN-R-tp2299586p2299586.html Sent from the R help mailing list archive at Nabble.com.

POS=sum(x[-1][x[-1]>0],na.rm=TRUE) is this the correct syntax? -- View this message in context: http://r.789695.n4.nabble.com/na-rm-TRUE-tp2299596p2299596.html Sent from the R help mailing list archive at Nabble.com.

Hi user's Does anybody work with wavelets on R? Please I need some help.....!!!! Atte Marize Sim?es -- View this message in context: http://r.789695.n4.nabble.com/Package-wavelets-tp2526023p2526023.html Sent from the R help mailing list archive at Nabble.com.

By any chance is there an R package that may contain functionality similar to the frestimate in Matlab/Simulink. Here is the URL for a description of the frestimate functionality: http://www.mathworks.com/help/toolbox/slcontrol/ug/frestimate.html Thank you again for any feedback.

Dear all, I'm just wondering how to submit a package to cran? I followed the instruction, using anonymous as username and my email address as password, but it didn't connect. Any hints? Yan [[alternative HTML version deleted]]

The dataframe is id salary 100 500 101 600 102 700 103 800 how can i generate a subsets if salary>600? -- View this message in context: http://r.789695.n4.nabble.com/Filtering-in-R-tp2299603p2299603.html Sent from the R help mailing list archive at Nabble.com.

Hi, I have a data frame with columns as follows: tree species (independent variable) and several dependent variables (e.g. carbon, nitrogen, phosphorus). Each row represents one tree sample. Some tree samples are unique species in the data frame while other species were replicated (i.e. rows 1,2,3 may be identical for the "tree species" column but have different values for the

the date were created with chron with this argument format=c(dates="Y/m/d", times="H:M:S")) so I have the dates being displayed as (10/06/22 12:00:00) I would like to have them displayed as "2010-06-22 12:00:00" or "%Y-%m-%d %H:%M:%S" and then I can convert these for mergeing with another data frame x <- (structure(c(14464, 14464.0104166667,

Hi, I have been unable to find how to go about changing the significance level in R for hypothesis testing and regression analysis. R has a default setting to alpha=95% (a significance level of 95%), but I want to be able to decrease or increase this level when necessary. If you could give me instructions on how to change the significance level in R for hypothesis testing and regression analysis