similar to: NA-values and logical operation

In a number of places internal to R, we need to know which files have changed (e.g. after building a vignette). I've just written a general purpose function "changedFiles" that I'll probably commit to R-devel. Comments on the design (or bug reports) would be appreciated. The source for the function and the Rd page for it are inline below. ----- changedFiles.R: changedFiles

Hello, I have a csv-file which looks like: #### pwt6_r.csv #### code;year;rgdpch AGO;1998;1234 ALB;1998;3576 ARG;1998;#NA SVN;1996;13439 SWE;1996;21492 AGO;1960;#NA ALB;1960;2345 ARG;1960;4634 #### pwt6_r.csv #### To import this file i call: pwt6<-read.csv("d:/pwt6_r.csv",header=T,na.strings="#NA",sep=";") Now I want to generate a new data.frame which include

Dear list, I'm trying to replace NA-values with the preceding values in that column. This code works, but I am sure there is a more elegant way... df <- data.frame(id = c("A1", NA, NA, NA, "B1", NA, NA, "C1", NA, NA, NA, NA), value = c(1:12)) rn <- c(rownames(df[!is.na(df$id),]), nrow(df)+1) rn <-

Hi... I have a dataframe with n columns and n rows. I need to find how many rows contains zero raw read count across all column. Thanks -- *Yogesh Gupta* *Postdoctoral Researcher* *Department of Biological Science* *Seoul National University* *Seoul, South Korea* [[alternative HTML version deleted]]

Dear list users, let me ask you this trivial question. I worked on that for a long time, by now. Suppose to have a data frame with NAs and to sum some columns with rowSums: df <- data.frame(A = runif(10), B = runif(10), C = rnorm(10)) df[1, ] <- NA rowSums(df[ , which(names(df) %in% c("A","B"))], na.rm=T) If all the elements of the selected columns are NA, rowSums

In R 3.3.3, I observe the following on Ubuntu 16.04 (when building from source as well as for the sudo apt r-base build): > x <- c(NA, NaN) > mean(x) [1] NA > mean(rev(x)) [1] NaN > rowMeans(matrix(x, nrow = 1, ncol = 2)) [1] NA > rowMeans(matrix(rev(x), nrow = 1, ncol = 2)) [1] NaN > .rowMeans(x, m = 1, n = 2) [1] NA > .rowMeans(rev(x), m = 1, n = 2) [1] NaN >

What do you mean by "should not"? NULL means "missing object" in R. The result of the sum function is always expected to be numeric... so NA_real or NA_integer could make sense as possible return values. But you cannot compute on NULL so no, that doesn't work. See the note under the "Value" section of ?sum as to why zero is returned when all inputs are removed.

Should not the result be NULL if you have removed the NA with na.rm=TRUE ? B. > On Mar 21, 2018, at 11:44 AM, Stefano Sofia <stefano.sofia at regione.marche.it> wrote: > > Dear list users, > let me ask you this trivial question. I worked on that for a long time, by now. > Suppose to have a data frame with NAs and to sum some columns with rowSums: > > df <-

Hi Martin, Henrik, Thanks for the follow up. @Martin: I vote for 2) without *any* hesitation :-) (and uniformity could be restored at some point in the future by having prod(), rowSums(), colSums(), and others align with the behavior of length() and sum()) Cheers, H. On 01/27/2018 03:06 AM, Martin Maechler wrote: >>>>>> Henrik Bengtsson <henrik.bengtsson at gmail.com>

Dear All, I am trying to fill in a blank vector ("a") with one value at a time, with the value of the number of rows in a randomized dataset with rowSums=0. Below is the code I've got so far, but what I want to be as the last line is " a[1st NA,]=nz" such that this will run until all of my NAs are replaced with an integer value based on randomizations.

Do you heve any idea why I get after this instruction everywhere false? > seq (0, 1, by=0.1) == 0.3 [1] FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE But after different step it's ok: > seq(0, 1, by=0.1) == 0.4 [1] FALSE FALSE FALSE FALSE TRUE FALSE FALSE FALSE FALSE FALSE FALSE -- View this message in context:

On 21/03/2018 11:44 AM, Stefano Sofia wrote: > Dear list users, > let me ask you this trivial question. I worked on that for a long time, by now. > Suppose to have a data frame with NAs and to sum some columns with rowSums: > > df <- data.frame(A = runif(10), B = runif(10), C = rnorm(10)) > df[1, ] <- NA > rowSums(df[ , which(names(df) %in%

No. The empty sum is zero. Adding it to another sum should not change it. Nothing audacious about that. This is consistent; other definitions just cause trouble. -pd > On 21 Mar 2018, at 18:05 , Boris Steipe <boris.steipe at utoronto.ca> wrote: > > Surely the result of summation of non-existent values is not defined, is it not? And since the NA values have been _removed_,

Surely the result of summation of non-existent values is not defined, is it not? And since the NA values have been _removed_, there's nothing left to sum over. In fact, pretending the the result in that case is zero would appear audacious, no? Cheers, Boris > On Mar 21, 2018, at 12:58 PM, Jeff Newmiller <jdnewmil at dcn.davis.ca.us> wrote: > > What do you mean by

Dear everyone, I have looked all over the internet but I cannot find a way to solve my problem. In my data I want to sum a couple of variables. Some of these variables have NA values, and when I add them together, the result is NA dat <- data.frame( id = gl(5,1), var1 = rnorm(5, 10), var2 = rnorm(5, 7), var3 = rnorm(5, 6), var4 = rnorm(5, 3), var5 = rnorm(5, 8) ) dat[3,3] <- NA dat[4,5]

I see: consistency with additive identity. That makes sense. Thanks. B. > On Mar 21, 2018, at 1:22 PM, peter dalgaard <pdalgd at gmail.com> wrote: > > No. The empty sum is zero. Adding it to another sum should not change it. Nothing audacious about that. This is consistent; other definitions just cause trouble. > > -pd > >> On 21 Mar 2018, at 18:05 , Boris

On Fri, Mar 31, 2017 at 10:14 PM, Prof Brian Ripley <ripley at stats.ox.ac.uk> wrote: > From ?NA > > Numerical computations using ?NA? will normally result in ?NA?: a > possible exception is where ?NaN? is also involved, in which case > either might result. > > and ?NaN > > Computations involving ?NaN? will return ?NaN? or perhaps ?NA?: >

Hi I have a data frame df with 3 columns. Some rows are NA across all 3 columns. How can I remove rows with NA across all columns? df=data.frame(col1=c(1:3,NA,NA,4),col2=c(7:9,NA,NA,NA),col3=c(2:4,NA,NA,4)) Thanks Joseph ____________________________________________________________________________________ Be a better friend, newshound, and [[alternative HTML version deleted]]

"I see: consistency with additive identity. " Ummm, well: > 1+NULL numeric(0) > sum(1,NULL) [1] 1 Of course, there could well be something here I don't get, but that doesn't look very consistent to me. However, as I said privately, so long as the corner case behavior is documented, which it is, I don't care. Cheers, Bert Bert Gunter "The trouble with

Hi, A noobie question: I'm simply trying to run a conditional statement that evaluates if a substring is found within a larger string. I find that if it IS found, my function returns TRUE (great!), but if not, the condition does not evaluate to FALSE. ex): if( grep("hi", "hop", fixed = TRUE) ) print('yes, your substring is in your string') else