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Hi. This is likely a trivial problem but have not found a solution. Imagine the following dataframe: Lat Lon x1 x2 x3 01 10 NA NA .1 01 11 NA .2 .3 01 12 .4 .5 .6 I want to generate another column that consist of the first value in each row from columns x1 to x3. That is NewColumn .1 .2 .4 Any input greatly appreciated, Thanks, Camilo Camilo Mora, Ph.D.

Hi everyone- I have a dataset with multiple "pre" and "post" variables I want to compare. The variables are named "apple_pre" or "pre_banana" with the corresponding post variables named "apple_post" or "post_banana". The variables are in no particular order. apple_pre orange_pre orange_post pre_banana apple_post post_banana person_1

...n of Fisher?s exact test in any package which can handle lower probabilities, or have other suggestions as to how I can compare the probabilities? I am for instance comparing the following two: dat2<-matrix(c(29,0,29,0,12,0,18,0,0,29,0,16,0,19), nrow=2) fisher.test(dat2, workspace=30000000) dat3<-matrix(c(29,0,0,29,0,0,12,0,0,17,0,1,0,29,0,0,15,1,0,0,19), nrow=3) fisher.test(dat3, workspace=30000000) Which both result in p-value < 2.2e-16 Kind regards, S?ren

...se questions, or perhaps suggest one of the other R-lists? I have two questions: 1. Why am I getting this warning? 2. Why is the second example "Point Forecast" the same value, I do not see that in previous attempts with similar but different data sets as in example 1? Example1: dat3 <- structure(c(3539122.86, 3081383.87, 4158672.31, 4137518.78, 4123682.08, 4819375.2, 4342687.77, 5028674.58, 4472145.07, 4967277.73, 4516240.31, 4876194.63, 4816446.59, 4887399.37, 5478504.85, 4871385.27, 5487543.68, 5464193.69, 5252591.03, 7071416.89, 5524350.89, 6107166.69...

Hi >i have this data > X [1] 5.79 1579.52 2323.70 68.85 426.07 110.29 108.29 1067.60 17.05 22.66 [11] 21.02 175.88 139.07 144.12 20.46 43.40 194.90 47.30 7.74 0.40 [21] 82.85 9.88 89.29 215.10 1.75 0.79 15.93 3.91 0.27 0.69 [31] 100.58 27.80 13.95 53.24 0.96 4.15 0.19 0.78 8.01 31.75 [41] 7.35 6.50

Hi List, I have one column of beginning dates and one column of ending dates, I want to find their difference. And I want to ignore the trailing zeros, basically everything after the first colon mark. Begin_date End_date 01JAN2000:00:00:00:000 02FEB2002:00:00:00:000 24MAR2012:00:00:00:000 18MAY2012:00:00:00:000 01OCT2003:00:00:00:000

...oss classified model using GLMM in lme4. I have, for potential use, created a second coding of transect with levels 1-5 for site 1 and 6-10 for site2. Likewise, if a groupedData object is necessary, there are als ts1 and ts2 dummy variables, as was necessary in the old lme..... > str(dat3) `data.frame': 100 obs. of 7 variables: $ site : Factor w/ 2 levels "Here","There": 1 1 1 1 1 1 1 1 1 1 ... $ day : Factor w/ 10 levels "1","2","3","4",..: 1 1 1 1 1 2 2 2 2 2 ... $ trans : Factor w/ 5 levels "1",...

[My previous message rejected, therefore I am sending same one with some modification] I have 3 vectors with object name : dat1, dat2, dat3 Now I want to create a loop, like : for (i in 1:3) { cat(sd(dati)) } How I can do this in R? Regards,

My guess would be that if you inspect the output from ma(dat3[1:28], order=3) you will find some NAs in it. And then forecast() doesn't like NAs. But I can't check, because I can't find the ma() and forecast() functions. I assume they come from some package you installed; it would be helpful to say which package. -Don -- Don MacQueen Lawrence L...

HI Elisa, You could use ?cut() 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) 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) paste(i[1],"<x<=",i[2],sep="")))

Hello I have dataframe How to remove dates in hourly time seriesThe example time series likedate value 2000-01-05 00:00:00 1.0 2000-01-05 01:00:00 1.0 2000-01-05 05:00:00 3.6 2000-01-05 06:00:00 3.6 2000-01-05 07:00:00 2.2 2000-01-05 08:00:00 2.2 2000-01-05 09:00:00 2.2 2000-01-05 10:00:00

Hi all Say I have a function: myname=function(dat,x=5,y=6){ res<<-x+y-dat } for various input such as myname(dat1) myname(dat2) myname(dat3) myname(dat4) myname(dat5) how should I modify the 'res' line, to have new informative variable name correspondingly, such as dat1.res dat2.res dat3.res dat4.res dat5.res stored in the workspace. This is only an example of a complex function I have written. Thanks in advance! Casper...

Hi, Could you please advice some easy way to do the following for a dataframe (header=F) having unequal column- & row- length. 1. Combine/stack/join contents from - a) multiple rows into one column. b) multiple columns into one row. 2. Stack contents from multiple columns (or, rows) into one column (or, row). Thank you. Cheers, Santana [[alternative HTML

Hello, I have a data set that needs to be combined so that rows are summed by a group based on a certain variable. I'm pretty sure rowsum() or rowsums() can do this but it's difficult for me to figure out how it will work for my data based on the examples I've read. My data are structured like this: Plot SizeClass Stems 12 Class3 1 12 Class4

Hi, I have a table in which one column has the name of the objects as shown below. Name Budlamp-Woodcutter Complex - 15 to 60% slope (60/25/15) Budlamp-Woodcutter Complex - 15 to 60% slope (60/25/15) Terrarossa-Blacktail-Pyeatt Complex - 1 to 40% slope (40/35/15/10) Terrarossa-Blacktail-Pyeatt Complex - 1 to 40% slope (40/35/15/10) How can I split the single column into three columns

Hi, I have encountered an issue about finding a date closest to another date So this is how the data frame looks like: PT_ID IDX_DT OBS_DATE DAYS_DIFF OBS_VALUE CATEGORY 13 4549 2002-08-21 2002-08-20 -1 183 2 14 4549 2002-08-21 2002-11-14 85 91 1 15 4549 2002-08-21 2003-02-18 181 89 1 16 4549 2002-08-21 2003-05-15

...scription of the data: > dat1 PointEst TT1 1 3.6 TT2 2 5.0 TT3 3 5.8 TT4 4 11.5 TT5 5 7.5 TT5 6 8.7 TT7 7 17.4 > dat2 Lower TT1 1 1.0 TT2 2 2.2 TT3 3 2.7 TT4 4 6.7 TT5 5 3.9 TT5 6 4.6 TT7 7 11.5 > dat3 Upper TT1 1 12.3 TT2 2 11.2 TT3 3 12.1 TT4 4 19.1 TT5 5 14.2 TT5 6 15.6 TT7 7 25.6 > ooplot(dat1,type="barplot",col=rich.colors(7,"temperature"),names.arg=c("X","Y","Z","A","B","C","...

Hi guys, It would be great if you could help me with this one... I'm looking to create a script to convert a matrix of species abundance e.g: <http://r.789695.n4.nabble.com/file/n4643978/species_matrix.jpg> into two vectors e.g: <http://r.789695.n4.nabble.com/file/n4643978/communitylist.jpg> ---------------- If you feel there is no easy answer to this and that it would be

...rently has the value '&nbsp' as a space holder. My code is currently: # Loop through dataset, summarizing data based on Animal name wild_ID <-unique(wild$ID) Animal <-unique(wild$Name) filepath <-"c:/Jared/Data/Kenya/Wildebeest/Summary/" # Create an empty dataframe dat3 <- data.frame(Animal) for (i in c(1:length(wild_ID))) { # Create Subset dat <-subset(wild, ID == wild_ID[i]) # Calculate Sum of Movement, put in km dat2 <-(sum(dat$STEPLENGTH))/1000 dat3$Distance[i] <-dat2 # Write Value to CSV file outfile <-paste(filepath,"Animal_Dista...

...11 b???? a???????? 10 e????? c???????? 12 c????? e???????? 12 ",sep="",header=TRUE,stringsAsFactors=FALSE) ?dat2[apply(dat2[,-3],1,function(x) {x1<- order(x); x1[1]<x1[2]}),] ?# id1 id2 value #1?? a?? b??? 10 #2?? c?? d??? 11 #5?? c?? e??? 12 #or you have cases like these: dat3<- read.table(text=" id1?? id2?? value a????? b?????? 10 c????? d??????? 11 b???? a???????? 10 a????? b??????? 10 e????? c???????? 12 c????? e???????? 12 c????? d???????? 11 ",sep="",header=TRUE,stringsAsFactors=FALSE) ?dat3New<-dat3[apply(dat3[,-3],1,function(x) {x1<...