similar to: convert date to a factor

I was wondering why I get <NA> instead of the timestamp in the following. Thanks. > dataDir <- file.path(wd) > localRaw <- read.csv(file.path(dataDir,"LOCAL.csv"), as.is=T,stringsAsFactors = FALSE) > localRaw[1:2,] Year Month Day hour minute second Temp1mab Temp7mab Temp14mab Salinity1mab 1 2009 10 5 0 0 0 11.288 13.675 13.743 33.513

Hello, This should be pretty simple but I cannot get it right. Please point to the right code. Thanks. > last <- read.csv(file.path(dataDir,"plot1.csv"), as.is=T,stringsAsFactors = FALSE) > last date r_wvht 1 8/6/2008 0.9766667 2 8/8/2008 0.7733333 3 8/11/2008 1.4833333 4 8/13/2008 1.5766667 5 8/14/2008 1.3900000 6 8/18/2008 0.7800000 7 8/20/2008

Hi I got the following error using as.xts Error in xts(x, order.by = order.by, frequency = frequency, ...) : NROW(x) must match length(order.by) Here is how the data looks like > d1 <- read.csv(file.path(dataDir,"AppendixA-FishCountsTable-2009.csv"), as.is=T) > d1[1:3,] dive_id date time species count size site depth level TRANSECT VIS_M 1 62 10/12/2009

Hello, I looked in the R-help but could not find an archive addressing the following. I would like to convert a character to numeric after reading a file with RDS extension. After using as.numeric, I checked if it is numeric. It was not converted. Please help. Here is my code >Report <- readRDS(file="RDS/Report.RDS") > Report[1:2,] dive_id date

Thanks Marc. It never occurred to me that I would need a ""stringsAsFactors" expression in a data.frame.? I could have sworn I never did before when mocking up some data but clearly I was wrong or there has been a change in R v. 3.4.1 which seems unlikely. On Friday, July 7, 2017, 10:37:29 AM EDT, Marc Schwartz <marc_schwartz at me.com> wrote: > On Jul 7, 2017, at 6:03

> On Jul 7, 2017, at 7:03 PM, John Kane <jrkrideau at yahoo.ca> wrote: > > Thanks Marc. > It never occurred to me that I would need a ""stringsAsFactors" expression in a data.frame. I could have sworn I never did before when mocking up some data but clearly I was wrong or there has been a change in R v. 3.4.1 which seems unlikely. Welcome John. Going back to

> On Jul 7, 2017, at 6:03 AM, John Kane via R-help <r-help at r-project.org> wrote: > > This is not serious problem but I just wonder if someone can explain what is happening. > The same command within a dataframe is giving me a factor and as a plain vector is giving me a character. It's probably something simple that I have read and forgotten but I thought I'd ask.

All, I am having trouble with a "read.table()" function that is inside of another function. But if I call the function by itself, it works fine. Moreover, if I run the script on a Mac OS X (with the default Mac OS X version of R installed, rev 2.8), it works fine. But it does not work if I run it on windows vista (also default Windows version of R, rev. 2.8). Again, both

Hi, Here is my data, the first 10 rows > u=regCond_all[1:10,] > dput(u) structure(c(999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 99, 99, 99, 99, 99, 99, 99, 99, 99, 99, 99, 99, 99, 99, 99, 99, 99, 99, 99, 99, 1.9, 2, 1.97, 1.99, 1.83, 1.78, 1.6, 1.52, 1.52, 1.36, 10.53, 9.88, 9.88, 10.53, 10.53, 10.53, 5.26, 9.88, 10.53, 10.53, 5.4, 5.57, 5.46, 5.34, 5.5, 5.59, 5.62, 5.76, 6.23, 6.19,

This is not? serious problem but I just wonder if someone can explain what is happening. The same command within a dataframe is giving me a factor and as a plain vector is giving me a character.? It's probably something simple that I have read and forgotten but I thought I'd ask. Thanks #================================================ dat1 <- data.frame(aa = letters[1:10]) str(dat1)

Dear list, I am using R.2.11.1 version on a PC. I downloaded successfully gplots_2.8.0 but I could not find where to download caTools > library("gplots") Loading required package: caTools Error: package 'caTools' could not be loaded In addition: Warning message: In library(pkg, character.only = TRUE, logical.return = TRUE, lib.loc = lib.loc) : there is no package called

Dear all, I think this is an easy task, but I don't know how to do it. Specifically, I have 69 columns with 300.000 rows. In each cell there is a code like "2E3", "4RR", etc. I now have a list that replaces this with values, e.g., old new 2E3 5 4RR 3 etc. The list ist about 1600 rows long, so also to extensive for normal solutions Do you how to do that? It would

Hello, I have the following design, counts were collected at different transects, different depths and different sites at different times. Time is continuous and assumed to be random, all the others are categorical fixed where transect is nested within depth which is nested within site. I would like an expert opinion about the following code where intercept is modeled as random (I am not sure

Hi R experts, For example I have a dataset looks like this: Number TimeStamp Value 1 1/1/2013 0:00 1 2 1/1/2013 0:01 2 3 1/1/2013 0:03 3 How can I split the "TimeStamp" Column into two and return a new table like this: Number Date Time Value 1 1/1/2013 0:00 1 2 1/1/2013 0:01 2 3 1/1/2013 0:03 3 Thank! [[alternative HTML version

Hi, I am not sure I understand your question correctly. dat1<- read.table(text=" id??????????? responsed_at???????????????? number_of_connection????????????????? scores 1????????????????? 12-01-2010?????????????????????????????????? 1????????????????????????????????????????????? 2 1????????????????? 15-02-2010??????????????????????????????????

Hi, Suppose you created a dataframe like this: set.seed(28) ?dat1<-as.data.frame(simplify2array(list(letters[1:5],sample(1:20,5,replace=TRUE),6:10)),stringsAsFactors=FALSE) ?str(dat1) #'data.frame':??? 5 obs. of? 3 variables: # $ V1: chr? "a" "b" "c" "d" ... # $ V2: chr? "1" "2" "10" "18" ... # $ V3: chr?

Hi, I have the following function: getDataFromDVFileCustom <- function (file, hasHeader = TRUE, separator = "\t") { DVdatatmp <- as.matrix(read.table(file, sep = "\t", fill = TRUE, comment.char = "#", as.is = TRUE, stringsAsFactors = FALSE, na.strings = "NA")) DVdatatmper <- as.matrix(DVdatatmp[ , c("datetime",

Dear Elisa, Try this: vec1<-c(33,18,13,47,30,10,6,21,39,25,40,29,14,16,44,1,41,4,15,20,46,32,38,5,31,12,48,27,36,24,34,2,35,11,42,9,8,7,26,22,43,17,19,28,23,3,49,37,50,45) vec2<-vec1[1:26] names(vec2)<-LETTERS[1:26] label1<-unlist(lapply(mapply(c,lapply(seq(0,45,5),function(x) x),lapply(seq(5,50,5),function(x) x),SIMPLIFY=FALSE),function(i)

Hi, May be this helps: set.seed(25) dat1<- data.frame(ID=c("1001#01","1001#02","1001#03","1002#01","1002#02"),val=rnorm(5),stringsAsFactors=FALSE) ?dat1$ID<-as.numeric(gsub("#.*","",dat1$ID)) ?dat1 #??? ID??????? val #1 1001 -0.2118336 #2 1001 -1.0415911 #3 1001 -1.1533076 #4 1002? 0.3215315 #5 1002 -1.5001299 A.K.

Hi, From your example data, dat1<- read.table(text=" id1?? id2?? value a????? b?????? 10 c????? d??????? 11 b???? a???????? 10 c????? e???????? 12 ",sep="",header=TRUE,stringsAsFactors=FALSE) #it is easier to get the output you wanted dat1[!duplicated(dat1$value),] #? id1 id2 value #1?? a?? b??? 10 #2?? c?? d??? 11 #4?? c?? e??? 12 But, if you have cases like the one