similar to: help splitting a data frame

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 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.

Hi , Does any one know how to split a character vector , I have a vector X that looks like this and each row has 3 characters X ASK DGH ASG AUJ FRT I would like to split the vector into 3 vectors that look like this X1 X2 X3 A S K D G H A S G A U J U R T thanks -- View this message in context: http://www.nabble.com/How-to-split-a-character-vector-into-3-vectors-tp21939492p21939492.html

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) :

I would like to merge ?data1 ?that contains 100 unique ID?s with another data set ?data 2? with 150 ID?s and the age of those individuals ( the ID in data1 is a subset of the ID in data 2) I would like to merge these data1 with data2 and have the result of the merge to have the ID ordered as in data1. Can this be done in R? -- View this message in context:

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

I have a problem with ifelse(), I do not understand how it works. > X<-c(2,2,1,1,0,0) > str(X) num [1:6] 2 2 1 1 0 0 > Y<-ifelse(X>0,1,0) > Y [1] 1 1 1 1 0 0 > Can some one explain what is going on, I do not understand what ifelse is doing in this case. Can someone explain the output Y. Thanks -- View this message in context:

Hi All, I always have a problem with write.csv when I want the column names to be ignored, when I specify col.names=F, I get a header of V1 V2 V3 V4 etc. for example I tried write.csv(mydata, file="data.csv", quote=FALSE, row.names=F, col.names=F) Warning message: In write.csv(mydata, file = "data.csv", quote = FALSE, : attempt to set 'col.names' ignored >

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

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