similar to: Count non-zero values in excluding NA Values

Hello, Your attachment didn't came through, R-Help strips off most types of files, including CSV. Anyway, the following will do what I understand of your question. Tested with a fake dataset. set.seed(3026) # make the results reproducible data <- matrix(1:100, ncol = 10) data[sample(100, 15)] <- 0 data[sample(100, 10)] <- NA data <- as.data.frame(data) zero <-

What was suggested by Eric and Rui works well, but here is a short and may be simpler answer provided your data is similar what Eric posted. It should work for your l data too. aa <- is.na(data)|data==0 nrow(data)-colSums(aa) EK On Sun, Oct 29, 2017 at 6:25 AM, Engin YILMAZ <ispanyolcom at gmail.com> wrote: > Dear R Staff > > You can see my data.csv file in the annex. >

Since i could not see your data, the easiest thing comes to mind is court values excluding NAs, is something like this sum(!is.na(x)) Best of luck--EK On Sun, Oct 29, 2017 at 6:25 AM, Engin YILMAZ <ispanyolcom at gmail.com> wrote: > Dear R Staff > > You can see my data.csv file in the annex. > > I try to count non-zero values in dataset but I need to exclude NA in this >

Dear R Staff This is my file (www.fiscalforecasting.com/data.csv) if you don't download this file, my dataset same as following Year Month A B C D E 2005 July 0 *4* NA NA *1* 2005 July 0 NA NA 0 *9* 2005 July NA *4* 0 *1* 0 2005 July *4* 0 *2* *9* NA I try to count non-zero values which are not NA values for every *column* *Sincerely* *Engin YILMAZ*

> On 27 Nov 2017, at 11:56, Engin YILMAZ <ispanyolcom at gmail.com> wrote: > > Dear > > I try to realize one scatter matrix which draws *one single variable to all > variables* with *regression line* . You can see my eviews version in the > annex . > > How can I draw this graph with R studio? A tiny note; You do calculations in R not RSudio. RStudio is a tool

Dear I try to realize one scatter matrix which draws *one single variable to all variables* with *regression line* . You can see my eviews version in the annex . How can I draw this graph with R studio? Sincerely Engin YILMAZ

Good morning. I've ran a test against MP3 format. Code: (first convert tested audio file to 16 bit 48khz with sox.exe if needed) lame.exe -b 320 48khzfilein.wav -o fileout.mp3 lame --decode fileout.mp3 -o fileout.mp3.wav opusenc.exe --bitrate 320 48khzfilein.wav fileout.opus opusdec.exe fileout.opus fileout.opus.wav wavdiff.exe 48khzfilein.wav fileout.mp3.wav -diff fileout.mp3.delta.wav

This is what I was looking for. Thank you everyone! Sincerely, Milu <https://www.avast.com/sig-email?utm_medium=email&utm_source=link&utm_campaign=sig-email&utm_content=webmail> Mail priva di virus. www.avast.com <https://www.avast.com/sig-email?utm_medium=email&utm_source=link&utm_campaign=sig-email&utm_content=webmail>

Oops, wrong language ;D Okay Rowland. Thank you very much for this help. To the next. <https://www.avast.com/sig-email?utm_medium=email&utm_source=link&utm_campaign=sig-email&utm_content=webmail> Livre de v?rus. www.avast.com <https://www.avast.com/sig-email?utm_medium=email&utm_source=link&utm_campaign=sig-email&utm_content=webmail>.

Hola buenos días hice un gráfico combinado de líneas, puntos y barras en ggplot2, pero no sé cómo puedo poner la leyenda de eso gráfico para que me represente para las líneas con puntos los valores estimados por un modelo y observados. este es mi código: ggplot(MLM,aes(x=Individuo)) + geom_bar(aes(y=Observada), stat = "identity", color= "gray") + geom_line(aes(y=Estimada),

On 22/06/2020 14:00, Fernando Gon?alves wrote: > Good morning Rowland. > > As you may have noticed, I am no expert in deploying SAMBA in an AD > domain. > Could you give me a link with a tutorial that explains in a simple way > the procedure for this? You could start here: https://wiki.samba.org/index.php/Setting_up_Samba_as_a_Domain_Member > > Just to not leave

Dear all, I have monthly data in wide format, I am only providing data (at the bottom of the email) for the first 24 columns but I have 2880 columns in total. I would like to take max of every 12 columns. I have taken the mean of every 12 columns with the following code: byapply <- function(x, by, fun, ...) { # Create index list if (length(by) == 1) { nc <- ncol(x)

You do not appear to have read the Posting Guide mentioned at the bottom if this and every posting on the mailing list. Only a very few attachment types are allowed through the mailing list... and due to the way many email programs fail to identify them properly, even those few types may not make it through. Also, this is a plain text email list... any time you send HTML-formatted email it gets

Ista, et. al: efficiency? (Note: I needed to correct my previous post: do.call() is required for pmax() over the data frame) > x <- data.frame(matrix(runif(12e6), ncol=12)) > system.time(r1 <- do.call(pmax,x)) user system elapsed 0.049 0.000 0.049 > identical(r1,r2) [1] FALSE > system.time(r2 <- apply(x,1,max)) user system elapsed 2.162 0.045 2.207 ##

Good day Sir/Ma'am! This is Alyssa Fatmah S. Mastura taking up Master of Science in Statistics at Mindanao State University-Iligan Institute Technology (MSU-IIT), Philippines. I am currently working on my master's thesis titled "Comparing the Three Estimation Methods for the Four Parametric Logistic (4PL) Item Response Model". While I am looking for a package about Markov chain

<div dir="auto">For what it's worth, it's working in NZ/vodafone/android firefox for me<br></div><div style="line-height:1.5"><br><br>-------- Original message --------<br>From: Gavin Stephens <gavin@stephens.net.nz><br>Date: Mon, 15 Jun 2020, 15:59<br>To: icecast@xiph.org<br>Subject: Re: [Icecast]

Don't do this (sorry Thierry)! max() already does this -- see ?max > x <- data.frame(a =rnorm(10), b = rnorm(10)) > max(x) [1] 1.799644 > max(sapply(x,max)) [1] 1.799644 Cheers, Bert Bert Gunter "The trouble with having an open mind is that people keep coming along and sticking things into it." -- Opus (aka Berkeley Breathed in his "Bloom County" comic

Jean-Mark sarkasm. Jean-Markasm. (Bonus points for providing an actual noisy WAV! ^_^) On 30/10/2017 20:28, Jean-Marc Valin wrote: Hi, Before I comment on the graphics you posted to visualize the difference between two audio signals, I'd like to ask for your help in evaluating my JPEG encoder. I've encoded an image with JPEG and then computed the difference with the original. I then

Hi Milu, byapply(df, 12, function(x) apply(x, 1, max)) You might also be interested in the matrixStats package. Best, Ista On Tue, Feb 20, 2018 at 9:55 AM, Miluji Sb <milujisb at gmail.com> wrote: > Dear all, > > I have monthly data in wide format, I am only providing data (at the bottom > of the email) for the first 24 columns but I have 2880 columns in total. > > I

On Tue, Feb 20, 2018 at 11:58 AM, Bert Gunter <bgunter.4567 at gmail.com> wrote: > Ista, et. al: efficiency? > (Note: I needed to correct my previous post: do.call() is required for > pmax() over the data frame) > > > x <- data.frame(matrix(runif(12e6), ncol=12)) > > > system.time(r1 <- do.call(pmax,x)) > user system elapsed > 0.049 0.000