similar to: colmeans not working

Displaying 10 results from an estimated 10 matches similar to: "colmeans not working"

2012 Dec 24
2
whats wrong in my codes???
Dear R family,i am trying to plot and save, simultaneously, about 1000. the name of each plot is contained in "names" file. when i run this loop, i get an error. "Error in plot.new() : Unable to open file 'C:/R/SAVEHERE/myplot_Tak.jpg' for writing". could you please correct the mistake in the loop? >names<-(names(sp)) >for(a in seq_along(names)){ >mypath
2013 Mar 14
0
reshape
Hi Elisa, You need to check your data.? For some 'st', the data is repeated/duplicated (I am assuming, didn't check it) especially for a particular year. dat1<-read.csv("elisa.csv",sep="\t") dat1$st<- as.character(dat1$st) ?str(dat1) #'data.frame':??? 506953 obs. of? 5 variables: # $ st?????? : chr? "AGOMO" "AGOMO"
2011 Mar 28
0
Bootstrapping the colMeans statistic
I have a fair bit of experience with S-Plus and have been asked to port some of my S-Plus bootstrapping functions to R, to which I am relatively new, Needless to say, I've run into some problems. In particular, I need to perform bootstrap resampling of the colMeans function using a moving blocks bootstrap, where the blocks are row indices. It's easy to build a function that allows me
2008 Mar 27
2
colMeans in a data.frame with numeric and character data
Hi all, I would like to know if it is posible by, someway, to get colMeans from a data.frame with numeric as well as character data, dispersed all over the object. Note that I would like to get colMeans neglecting character data. I am really in need of some function proceeding in that way… All the best Diogo André Alagador [[alternative HTML version deleted]]
2009 Sep 28
1
colMeans()
Hello, I use the function rowMeans(x,na.rm=T). The result is the mean of valid values in each row, with NA removed. A for me very important information is, from how many valid "n" this mean has computed. The thing is, that I apply this function on many millions of rows, so the time it takes is multiplied by this factor! (so getting this with "rowSums(is.na(x))" it takes
2005 Apr 01
2
Error in colMeans ... what's wrong with my data?
Having searched and searched I still haven't found what's the problem with my data (I've attached the relevant file). Every time I tried to use the CANCOR-Function I got error messages. So I turned to check my two sets of variables separately by using the "colMeans"-Function. With one set no problem .. but with the other. My input : > struktur <-
2007 Sep 04
2
Howto sort dataframe columns by colMeans
I read from external data source containing several columns. Each column represents value of a metric. The columns are time series data. I want to sort the resulting dataframe such that the column with the largest mean is the leftmost column, descending in colMean values to the right. I see many solutions for sorting rows based on some column characteristic, but haven't found any
2005 Jan 11
1
Please use colMeans()! was: Re: Calculate Mean of Column Vectors?
There are indeed speed advantages in using colSums etc. However the disadvantage is that the newbie doesn't always find the power inherent in the apply, sapply, tapply and mapply. For many things that I do, the speed is the least of my worries; although I take the point that using apply for means or sums in packages that are distibuted to others is not the way to go. As many of us have found
2012 Dec 17
3
mean of each month in data
Dear R users, [in case the format of email is changed or you dont finf it easy to understand, i have attached a text file of my question] i have the data in the following format and i want to convert it in the format given at the end. Ta ans Sa are the names of certain cities. there are 69 cities in my data. column 1 is representing station name (i am writing the data of only two cities for
2004 Apr 07
4
Problems with rlm
Dear all, When calling rlm with the following data, I get an error. (R v.1.8.1, WinXP Pro 2002 with service pack 1.) > d <- na.omit(data.frame(CPRATIO, HEIGHTZ, FAMILYID)) > c <- tapply(d$CPRATIO, d$FAMILYID, mean) > h <- tapply(d$HEIGHTZ, d$FAMILYID, mean) > c 1 2 3 6 7 9 10 11 6.000000 2.500000 3.250000