similar to: matrix manipulation with a for loop

Hi, I have this dataset where columns z1.3 and z1.4 are full of NA's. I want to perform some calculations in the remaining columns, but after doing this, I want to recontruct the original matrix. I can with: out <- which( colMeans( is.na( z ) ) == 1 ) gd<-z[, - out] select the columns full of NA's and those without that pattern. Then after doing the calculus I need to

Hi, Having an zoo object I can subset it to obtain the days where I have the values within some range: is.zoo(z) TRUE subset(z[,1], z[,1]>=5 & z[,1]<= 10) #Yields: Year(day) 1988(13) 1988(14) 1988(16) 1988(20) 1988(21) 1988(22) 1988(25) 1988(26) 7.973946 9.933518 7.978227 7.512960 6.641862 5.667780 5.721358 6.863729 1988(27) 1988(28) 1988(29) 1988(30) 1988(32)

Hi, I have a object 'zoo': dim(zz) [1] 720 5551 where some columns only have NA's values (representing land data in a sea surface temperature dataset) I find straightforward the use of 'na.approx' for individual columns from the zz matrix, but when applied to the whole matrix: zz.approx<-na.approx(zz) Erro en approx(along[!na], y[!na], along[na], ...) : need

Hi, Having a matrix F(189,6575) I want to do this: z1<-subset(F[,1], F[,1] >= 5 & F[,1] <= 10) . . . z189<-subset(F[,189], F[,189] >= 5 & F[,189] <= 10) I would prefer to have an empty matrix, say 'z' in order to fill its columns with the output of subsetting F. But each of the subsets can differ in length. It's to say: length(z1) 1189 length(z2) 1238

Hi, I have a list of 20 elements, each of them of variable length and with a structure like this: lasker[[1]][1:10,] Var1 Freq 1 1988-02 3 2 1988-03 1 3 1988-04 1 4 1988-05 2 5 1988-06 3 6 1988-07 1 7 1988-08 1 8 1988-09 1 9 1989-03 1 10 1989-04 1 How do I can insert in this list: 1988-01 0 1988-10 0 1988-11 0 1988-12 0 1989-01 0

Hi, Having several daily wind speed time series I want to extract those consecutive days over and below certain values (i.e. 5 < x <8) Don't know which funtion to use (aggregate, lapply?) and how to do it. Thanks in advance Antonio

Hey, All In principal component analysis (PCA), we want to know how many percentage the first principal component explain the total variances among the data. Assume the data matrix X is zero-meaned, and I used the following procedures: C = covriance(X) %% calculate the covariance matrix; [EVector,EValues]=eig(C) %% L = diag(EValues) %%L is a column vector with eigenvalues as the elements percent

Hello, Does anyone know of a function that will determine whether or not a formula object has a left hand side? I.e., can differentiate between y ~ x + z and ~ x + z Perhaps I'm overlooking the obvious... Thanks!

Hi, I swear I have read almost all the posted messages about this issue, but it's evident I couldn't find an answer (surely esay) to my problem. What I want is the following: Make 8 days aggregates from a daily series like this (dput output): structure(c(6.91777181625366, 0.79051125049591, 9.00625133514404, 9.86966037750244, 14.4326181411743, 3.70155477523804, 9.67768573760986,

Creo que esto me da para DK, y luego veré como aplicar el barplot ess %>% filter(cntry %in% c("DK")) %>% count (stflife) %>% mutate (freq = (n /sum(n)*100))%>% print 2017-06-18 19:01 GMT-05:00 Antonio Rodriguez Andres < antoniorodriguezandres70 en gmail.com>: > He conseguido el total para un país, pero no me deja usar percent = > count() /sum(count),

Dear all, I often use dates and times in analyses. I just can't figure out how to format my date or time column in R. So, apparently R sees the date as something other than date (character). Let's say I am opening a CSV file, one of the columns of which is a date or time. How do I specify that when opening the file? Thanks for the help, Jim --

#Simple table con frecuencias absolutas y crear relativas count =table(ess$stflife) percent = 100* (count)/sum(count) Carlos he creado a esto a nivel general en vez de usar prop.table. Según lo que dices o entiendo, debo de usar la función ddply para hacerlo a nivel de todos los paises, y entiendo que ddply (, c(""), debo de indicar los paises? Saludos 2017-06-18 17:37 GMT-05:00

Running R.app on Mac OS X 10.4 > version _ platform powerpc-apple-darwin8.6.0 arch powerpc os darwin8.6.0 system powerpc, darwin8.6.0 status major 2 minor 3.1 year 2006

Running R.app on Mac OS X 10.4 > version _ platform powerpc-apple-darwin8.6.0 arch powerpc os darwin8.6.0 system powerpc, darwin8.6.0 status major 2 minor 3.1 year 2006

Hi all, Please, how could I calculate the time that two time segments has in common? Is there any function to perform this calculation? For instance, given four POSIXlt objects... endPeriod<-as.POSIXlt("2008-09-30") startPeriod<-as.POSIXlt("2007-10-01") endProject<-as.POSIXlt("2007-05-31") startProject<-as.POSIXlt("2006-12-01") that limit

Gracias. Alguna idea de que usar para calcular los porcentajes y almacenarlos. Se puede usar flat table? El 18/06/2017 4:50 p. m., "Carlos J. Gil Bellosta" <cgb en datanalytics.com> escribió: > Los porcentajes que obtienes con tu código son sobre todas las facetas, no > país a país. > > Calcula los porcentajes previamente a por país y representa esa columna en >

Hi All, How do I can delete from a matrix (or array) only those columns which have all their values set to NA? Cheers Antonio Rodriguez ---

In my Excel file, I have data in following format : Feb-07 38.49 Mar-07 39.95 Apr-07 37.47 May-07 35.77 Jun-07 32.96 Jul-07 33.27 I tried to copy this data as a time series using following code : library(zoo) dat <- read.zoo(file="clipboard", format="%m-%y") However getting following error : Error in read.zoo(file = "clipboard", format = "%m-%y") :

Newbie question - mechanical engineer trying to learn R I've had success with plotting time series data and even made a heat map using R Graphs Cookbook by Mittal. I have a new problem - I need to align a number of time series data "columns" to the desired regular exact time stamp vector. The target time vector could be finer or more course than the typical time stamp spacing found.

Muchas gracias. Me puede recomendar algún libro donde poder empezar. Gracias El 18/06/2017 9:46 p. m., <javier.ruben.marcuzzi en gmail.com> escribió: > > > Estimado Antonio Rodriguez Andres > > > > Usted dice que está en sus inicios con R, y desea realizar un gráfico. > Esto no es problema, el inconveniente es que está mezclando librerías, es > lógico, se puede