search for: col8

...sing read.table. The file consists of 100 tables, each of which is headed by two lines of characters. The first of these lines is: TABLE NO. 1 The second is a list of column headers. For example: TABLE NO. 1 COL1 COL2 COL3 COL4 COL5 COL6 COL7 COL8 COL9 COL10 COL11 COL12 1.0010E+05 0.0000E+00 1.0000E+00 1.0000E+03 -1.0000E+00 1.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 1.0010E+05 1.0001E+01 1.0000E+00 1.0000E+03 -1.0000E+00 1.0000E+00 2.2737E-14 -2.2737E-14 0.0000E...

Dear useRs, i want to split a matrix having 1116rows and 12 columns. i want to split that matrix into 36 small matrices each having 12 columns and 31 rows. The big matrix should be splitted row wise. which means that the first small matrix should copy values which are in first 31 rows and 12 columns of the big matrix. similarly 2nd small matrix should contain values from 32nd to 63rd row of the

...1) the number of values in each row of columns 6-10 that were NA??s (2) the sum of all values on columns 1-5 for which there were no missing values in the corresponding cells of columns 6-10. Example: (let??s call the data frame "data") Col1 Col2 Col3 Col4 Col5 Col6 Col7 Col8 Col9 Col10 1 2 5 2 3 NA 5 NA 1 4 3 1 4 5 2 6 NA 4 NA 1 The result would then be (for the first row) (1) "There were 2 NA??s in columns 6-10." (2) The mean of Columns 1-5 was 2+2+3=7" (because t...

dear useRs, i have a list containing 16 matrices. i want to calculate the column mean of each of them. i tried >sr <- lapply(s,function(x) colMeans(x, na.rm=TRUE)) but i am getting the following error >Error in colMeans(x, na.rm = TRUE) : 'x' must be numeric can it be done in any other way? and why i am getting this error?? thanks in advance.. elisa [[alternative

...gc()[,2])/1024, 3), "\n") #GB = 0.397 colnames(DF) = NULL system.time(nm1<-colnames(DF, FALSE)) # user system elapsed # 22.158 0.299 22.498 print(nm1) #[1] "col1" "col2" "col3" "col4" "col5" "col6" "col7" "col8" "col9" ### restart R colnames <- function (x, do.NULL = TRUE, prefix = "col") { if (is.data.frame(x)) { nm <- names(x) if (do.NULL || !is.null(nm)) return(nm) else return(paste0(prefix, seq_along(x))) } dn...

Hi, I have two loosely related questions which could make my live again a bit easier: 1) Is there a simple way to select a range of columns in a data frame using column names? I am thinking of something like mydf[1,"col4":"col8"] 2) I have a data frame with many columns and they all have short variable names which is good in most cases but sometimes it would be nice to have also a longer descriptive name / label attached to the variable which could then be used for printing and latex output. Has anybody come up with...

Hi, is there a way I can calculate a summary statistics for a columns matrix let say we have this matrix x <- matrix(sample(1:8000),nrow=100) colnames(x)<- paste("Col",1:ncol(x),sep="") if I used summary summary(x) i get the output for each column but I need the output to be in matrix with rownames and all the columns beside it this how I want it

...ad.table('in.dat') colnames(All) <- paste('col',1:ncol(All),sep='') # Val = target values, matrix of n rows X 3 columns Val <- read.table('target.dat') # create dataframe dt <- data.frame(Val,All) nn <- neuralnet(Val ~ col1+col2+col3+col4+col5+col6+col7+col8,data=dt) Is there someone kind enough to correct my code ? Many many thanks in advance !! Luc ---------------------------------------------------------------- Dr. Luc Moulinier, Laboratoire de Biologie et G?nomique Int?gratives IGBMC 1, rue Laurent Fries 67 404 ILLKIRCH Cedex Phone : +33 (0...

Hi everybody! I am sorry to bother you with a question so simple but I think there might be a better solution: I have a matrix of size 360x501 where I want to check the value of each 5th column of each row and replace it (and the 6th, 7th, 8th column) by zero if the value is less than 1000. I have written a double loop to do that but that requires a lot of time. Is there a faster way to

...> #GB = 0.397 > colnames(DF) = NULL > system.time(nm1<-colnames(DF, FALSE)) > # user system elapsed > # 22.158 0.299 22.498 > print(nm1) > #[1] "col1" "col2" "col3" "col4" "col5" "col6" "col7" "col8" "col9" > > ### restart R > > colnames <- function (x, do.NULL = TRUE, prefix = "col") > { > if (is.data.frame(x)) { > nm <- names(x) > if (do.NULL || !is.null(nm)) > return(nm) > else >...

...e(hcl,8)) # Change this, but how??? > > col1=rgb(0.164,0.043,0.850) # Set the colors > col2=rgb(0.150,0.306,1.000) > col3=rgb(0.250,0.630,1.000) > col4=rgb(0.450,0.853,1.000) > col5=rgb(0.670,0.973,1.000) > col6=rgb(0.880,1.000,1.000) > col7=rgb(1.000,1.000,0.750) > col8=rgb(1.000,0.880,0.600) > col9=rgb(1.000,0.679,0.450) > col10=rgb(0.970,0.430,0.370) > col11=rgb(0.850,0.150,0.196) > col12=rgb(0.650,0.000,0.130) > colTemperature=c(col1,col2,col3,col4,col5,col6,col7,col8,col9,col10,col11,col12) # Put all colors in a vector > > tempLT50_ACC=...

...NA Col2 0.69283 NA NA NA Col3 0.62603 NA NA NA Col4 0.27695 NA NA NA Col5 0.06056 NA NA NA Col6 -0.19582 NA NA NA Col7 -0.83044 NA NA NA Col8 -1.27914 NA NA NA Col9 -1.93235 NA NA NA Col10 -2.49709 NA NA NA When I omited 'weights=w' above table was filled in with numbers, but the results were wrong (because of taking zeros in regression). Could you tel...

Hi, In a function, I compute 10 (un-named) vectors of reasonable length (4471 in the particular example I have to hand) that I want to combine into a data frame object, that the function will return. This is very slow, so *I'm* doing something wrong if I want it to be quick and efficient, though I'm not sure what the best way to do this would be. I know it is the combining into data

...t read the last row of the table excell 2) Don' t take the hours My excell file call prueba4.xls and have the following rows: where prueba4.xls was make in excell (office xp) and have one spreadsheet call "Hoja1", you see each rows of she: D??a Hora col1 col2 col3 col4 col5 col6 col7 col8 15/12/2003 12:14:59 217 2760 8,2 35 79,6 86,4 15/12/2003 12:15:00 217 2764 8,2 35 79,6 86,4 15/12/2003 12:15:01 217 2758 8,3 35 79,6 86,4 15/12/2003 12:15:02 217 2760 8,3 35 79,6 86,4 15/12/2003 12:15:03 217 2755 8,3 35 79,6 86,4 15/12/2003 12:15:04 217 2766 8,3 35 79,6 86,4 15/12/2003 12:15:05 217...

...ake the hours See below >My excell file call prueba4.xls and have the following rows: >where prueba4.xls was make in excell (office xp) and have one spreadsheet >call "Hoja1", you see each rows of she: >D??a Hora col1 col2 col3 col4 col5 col6 col7 col8 >15/12/2003 12:14:59 217 2760 8,2 35 79,6 86,4 >15/12/2003 12:15:00 217 2764 8,2 35 79,6 86,4 >15/12/2003 12:15:01 217 2758 8,3 35 79,6 86,4 >15/12/2003 12:15:02 217 2760 8,3...

Hello, my question is about the data handling. I have a data set that is lined as: 4 1 17 1 1 -5.1536 -0.1668 -2.3412 -0.5062 0.9621 0.3640 0.3678 -0.5081 -0.2227 0.8142 -0.0389 -0.0445 -0.0578 -0.1175 -0.1232 0.8673 -0.1033 -0.0796 -0.0341 -0.1716 -0.1801 -0.7014 0.6578 0.5611 4 1 17 2 1 -5.1536 -0.1668 -2.3412 -0.5062 0.9621 0.3640 0.3678 -0.5081 -0.2227 0.8142 -0.0389 -0.0445