similar to: Convert chr pieces to numbers that have specific values defined by 2 vectors

Displaying 20 results from an estimated 3000 matches similar to: "Convert chr pieces to numbers that have specific values defined by 2 vectors"

2013 Sep 26
2
Sums based on values of other matrix
Dear all, I have a big problem: - I got two matrices, A and B - A shows identifies the value of B, however the values of B must be summed - For instance, 1 1 2 2 2 2 1 1 gives matrix a 3 4 2 1 1 1 2 2 gives matrix b Now the result for the value 1 would be 7 4 which are the rowsums of the values of matrix B given that matrix A has the value 1. How can I do this automatically? I
2013 Sep 02
1
R dataframe and looping help
HI, You may try this: dat1<- read.table(text=" CustID TripDate Store Bread Butter Milk Eggs 1 2-Jan-12 a 2 0 2 1 1 6-Jan-12 c 0 3 3 0 1 9-Jan-12 a 3 3 0 0 1 31-Mar-13 a 3 0 0 0 2 31-Aug-12 a 0 3 3 0 2 24-Sep-12 a 3 3 0 0 2 25-Sep-12 b 3 0 0 0 ",sep="",header=TRUE,stringsAsFactors=FALSE) dat2<- dat1[,-c(1:3)] res<- lapply(seq_len(ncol(dat2)),function(i)
2013 Feb 17
6
histogram
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="")))
2013 Apr 29
4
expanding a presence only dataset into presence/absence
Hello, I'm working with a very large dataset (250,000+ lines in its' current form) that includes presence only data on various species (which is nested within different sites and sampling dates). I need to convert this into a dataset with presence/absence for each species. For example, I would like to expand "My current data" to "Desired data": My current data
2012 Nov 30
2
missed values
Hello I have dataframe 101 2008-07 0.2898966 102 2008-08 0.3101667 103 2008-09 0.3730476 104 2008-10 0.2717037 105 2008-11 0.1344286 106 2008-12 0.1375000 107 2009-01 0.1781000 108 2009-02 0.2146667 109 2009-03 0.2808235 110 2009-04 0.4326250 111 2009-05 0.3420741 112 2009-06 0.2675238 113 2009-07 0.2478667 114 2009-08 0.3147000 115 2009-09 0.3437826 116 2009-10 0.2057391 117 2009-11 0.1824737 118
2012 Jul 24
3
Linear Model Prediction
I have data X and Y, and I want to predict what the very next point would be based off the model. This is what I have: >model=lm(x~y) I think I want to use the predict function, but I'm not exactly sure what to do. Thank you! -- View this message in context: http://r.789695.n4.nabble.com/Linear-Model-Prediction-tp4637644.html Sent from the R help mailing list archive at Nabble.com.
2013 May 07
4
create unique ID for each group
Hey All, I have a dataset(dat1) like this: ObsNumber ID Weight 1 0001 12 2 0001 13 3 0001 14 4 0002 16 5 0002 17 And another dataset(dat2) like this: ID Height 0001 3.2 0001 2.6 0001
2008 Mar 25
2
Compare two data sets
I would like to compare two data sets saved as text files (example below) to determine if both sets are identical(or if dat2 is missing information that is included in dat1) and if they are not identical list what information is different between the two sets(ie output "a1", "a3" as the differing information). The overall purpose would be to remove "a1" and
2012 Jun 07
3
conditional statement to replace values in dataframe with NA
Hello and thanks for helping. #some data L3 <- LETTERS[1:3] dat1 <- data.frame(cbind(x=1, y=rep(1:3,2), fac=sample(L3, 6, replace=TRUE))) #When x==1 and y==1 I want to replace the 1 values with NA #I can select the rows I want: dat2<-subset(dat1,x==1 & y==1) #replace the 1 with NA dat2$x<-rep(NA,nrow(dat2) dat2$y<-rep(NA,nrow(dat2) #select the other rows and rbind
2013 Mar 27
9
conditional Dataframe filling
Hi everyone: This may be trivial but I just have not been able to figure it out. Imagine the following dataframe: a b c d TRUE TRUE TRUE TRUE FALSE FALSE FALSE TRUE FALSE TRUE FALSE FALSE I would like to create a new dataframe, in which TRUE gets 0 but if false then add 1 to the cell to the left. So the results for the example above should be something like: a b c
2010 Mar 22
1
Replacing elements of list
Dear all, I have following two list object, both are basically collection of matrices : dat1 <- matrix(rnorm(25*6), ncol=6) dat1 <- split(dat1, seq(5,25,by=5)) dat1 <- lapply(dat1, matrix, ncol=6) dat2 <- matrix(rnorm(25*4), ncol=4) dat2 <- split(dat2, seq(5,25,by=5)) dat2 <- lapply(dat2, matrix, ncol=4) Now I want to replace last 4 columns of each matrix at "dat1"
2012 Jul 01
2
list to dataframe conversion-testing for identical
HI R help, I was trying to get identical data frame from a list using two methods. #Suppose my list is: listdat1<-list(rnorm(10,20),rep(LETTERS[1:2],5),rep(1:5,2)) #Creating dataframe using cbind dat1<-data.frame(do.call("cbind",listdat1)) colnames(dat1)<-c("Var1","Var2","Var3") #Second dataframe conversion
2012 Aug 13
4
if else elseif for data frames
Hi all, It seems like I cannot use normal 'if' for data frames. What would be the best way to do the following. if data$col1='high' data$col2='H' else if data$col1='Neutral' data$col2='N' else if data$col='low' data$col2='L' else #chuch a warning? Note that col2 was not an existing column and was newly assigned for this
2013 Apr 12
2
split date and time
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
2010 Mar 30
2
Need help to split a given matrix is a "sequential" way
I need to split a given matrix in a sequential order. Let my matrix is : > dat <- cbind(sample(c(100,200), 10, T), sample(c(50,100, 150, 180), 10, > T), sample(seq(20, 200, by=20), 10, T)); dat [,1] [,2] [,3] [1,] 200 100 80 [2,] 100 180 80 [3,] 200 150 180 [4,] 200 50 140 [5,] 100 150 60 [6,] 100 50 60 [7,] 100 100 100 [8,] 200 150 100 [9,]
2007 Jun 23
2
Names of objects passed as ... to a function?
Dear list, I have a function whose first argument is '...'. Each element of '...' is a data frame, and there will be at least 2 data frames in '...'. The function processes each of the data frames in '...' and returns a list, whose components are the processed data frames. I would like to name the components of this returned list with the names of the original data
2004 Jul 16
3
sas to r
I would be incredibly grateful to anyone who'll help me translate some SAS code into R code. Say for example that I have a dataset named "dat1" that includes five variables: wshed, site, species, bda, and sla. I can calculate with the following SAS code the mean, CV, se, and number of observations of "bda" and "sla" for each combination of
2012 Jul 24
5
First value in a row
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.
2012 Oct 04
3
R combining vectors into a data frame but without a continuous common variable
Hello, I have two different files which I'd like to combine to make one data frame but I've no idea how to do it! The first file has two columns; one is the date, the following is a binary code for debris flow events. Then my other file has also two columns; the date and then precipitation data. The thing is, is that the two date columns don't all contain the same dates. The binary
2005 Dec 29
1
S4 classes: referencing slots with other slots
For those who suggest other ways to do this, I ALREADY HAVE ANOTHER DESIGN SOLUTION, DESCRIBED AT THE END. That being said, I want to know if it's possible to reference a slot in an S4 class from another slot, i.e. I'd like to have the "self.*" semantics of Python so that I can reuse a slot. That is, for various reasons it would be nice to be able to do something like: