similar to: removing NA from a data frame

Displaying 20 results from an estimated 1000 matches similar to: "removing NA from a data frame"

2011 Feb 15
1
all.equal: subscript out of bounds
When I do > all(all$X.Time == all$Y.Time); [1] TRUE as expected, but > all.equal(all$X.Time,all$Y.Time); Error in target[[i]] : subscript out of bounds why? thanks! -- Sam Steingold (http://sds.podval.org/) on CentOS release 5.3 (Final) http://mideasttruth.com http://honestreporting.com http://dhimmi.com http://jihadwatch.org http://pmw.org.il http://ffii.org The dark past once was the
2006 May 11
3
cannot turn some columns in a data frame into factors
Hi, I have a data frame df and a list of names of columns that I want to turn into factors: df.names <- attr(df,"names") sapply(factors, function (name) { pos <- match(name,df.names) if (is.na(pos)) stop(paste(name,": no such column\n")) df[[pos]] <- factor(df[[pos]]) cat(name,"(",pos,"):",is.factor(df[[pos]]),"\n")
2011 Jul 11
1
plot means ?
Hi, I need this plot: given: x,y - numerical vectors of length N plot xi vs mean(yj such that |xj - xi|<epsilon) (running mean?) alternatively, discretize X as if for histogram plotting and plot mean y over the center of the histogram group. is there a simple way? thanks! -- Sam Steingold (http://sds.podval.org/) on CentOS release 5.6 (Final) X 11.0.60900031 http://thereligionofpeace.com
2013 Jan 04
4
non-consing count
Hi, to count vector elements with some property, the standard idiom seems to be length(which): --8<---------------cut here---------------start------------->8--- x <- c(1,1,0,0,0) count.0 <- length(which(x == 0)) --8<---------------cut here---------------end--------------->8--- however, this approach allocates and discards 2 vectors: a logical vector of length=length(x) and an
2011 Jul 12
3
when to use `which'?
when do I need to use which()? > a <- c(1,2,3,4,5,6) > a [1] 1 2 3 4 5 6 > a[a==4] [1] 4 > a[which(a==4)] [1] 4 > which(a==4) [1] 4 > a[which(a>2)] [1] 3 4 5 6 > a[a>2] [1] 3 4 5 6 > seems unnecessary... -- Sam Steingold (http://sds.podval.org/) on CentOS release 5.6 (Final) X 11.0.60900031 http://jihadwatch.org http://palestinefacts.org http://mideasttruth.com
2012 Dec 04
3
list to matrix?
How do I convert a list to a matrix? --8<---------------cut here---------------start------------->8--- list(c(50000, 101), c(1e+05, 46), c(150000, 31), c(2e+05, 17), c(250000, 19), c(3e+05, 11), c(350000, 12), c(4e+05, 25), c(450000, 19), c(5e+05, 16)) as.matrix(a) [,1] [1,] Numeric,2 [2,] Numeric,2 [3,] Numeric,2 [4,] Numeric,2 [5,] Numeric,2 [6,] Numeric,2 [7,]
2012 Feb 23
5
cor() on sets of vectors
suppose I have two sets of vectors: x1,x2,...,xN and y1,y2,...,yN. I want N correlations: cor(x1,y1), cor(x2,y2), ..., cor(xN,yN). my sets of vectors are arranged as data frames x & y (vector=column): x <- data.frame(a=rnorm(10),b=rnorm(10),c=rnorm(10)) y <- data.frame(d=rnorm(10),e=rnorm(10),f=rnorm(10)) cor(x,y) returns a _matrix_ of all pairwise correlations: cor(x,y)
2017 Nov 09
2
[R-pkgs] Release of ess 0.0.1
> * Jorge Cimentada <pvzragnqnw at tznvy.pbz> [2017-11-09 00:02:53 +0100]: > > I'm happy to announce the release of ess 0.0.1 a package designed to > download data from the European Social Survey Given the existence of ESS (Emacs Speaks Statistics - https://ess.r-project.org/) the package name "ess" seems unfortunate. -- Sam Steingold (http://sds.podval.org/) on
2013 Jan 18
5
select rows with identical columns from a data frame
I have a data frame with several columns. I want to select the rows with no NAs (as with complete.cases) and all columns identical. E.g., for --8<---------------cut here---------------start------------->8--- > f <- data.frame(a=c(1,NA,NA,4),b=c(1,NA,3,40),c=c(1,NA,5,40)) > f a b c 1 1 1 1 2 NA NA NA 3 NA 3 5 4 4 40 40 --8<---------------cut
2012 Feb 08
4
"unsparse" a vector
Suppose I have a vector of strings: c("A1B2","A3C4","B5","C6A7B8") [1] "A1B2" "A3C4" "B5" "C6A7B8" where each string is a sequence of <column><value> pairs (fixed width, in this example both value and name are 1 character, in reality the column name is 6 chars and value is 2 digits). I need to
2012 Sep 20
1
aggregate help
I want to count attributes of IDs: --8<---------------cut here---------------start------------->8--- z <- data.frame(id=c(10,20,10,30,10,20), a1=c("a","b","a","c","b","b"), a2=c("x","y","x","z","z","y"),
2012 Aug 27
1
write.matrix.csr data conversion
> write.matrix.csr(mx, y = y, file = file) > table(y) 0 1 5194394 23487 $ cut -d' ' -f1 f | sort | uniq -c 23487 2 5194394 1 i.e., 0 is written as 1 and 1 is written as 2. why? is there a way to disable this? -- Sam Steingold (http://sds.podval.org/) on Ubuntu 12.04 (precise) X 11.0.11103000 http://www.childpsy.net/ http://palestinefacts.org
2012 Apr 04
2
recover lost global function
Since R has the same namespace for functions and variables, > c <- 1 kills the global function, which can be restored by > c <- get("c",mode="function") Is there a way to prevent R from overriding globals or at least warning when I do that or at least warning when I replace a functional value with non-functional? thanks. -- Sam Steingold (http://sds.podval.org/)
2011 Aug 16
2
merge(join) problem
I have two datasets: A with columns Open and Name (and many others, irrelevant to the merge) B with columns Time and Name (and many others, irrelevant to the merge) I want the dataset AB with all these columns Open from A - a difftime (time of day) Time from B - a difftime (time of day) Name (same in A & B) - a factor, does NOT index rows, i.e., there are _many_ rows in both A & B with
2012 Nov 09
4
as.data.frame(do.call(rbind,lapply)) produces something weird
The following code: --8<---------------cut here---------------start------------->8--- > myfun <- function (x) list(x=x,y=x*x) > z <- as.data.frame(do.call(rbind,lapply(1:3,function(x) c(a=paste("a",x,sep=""),as.list(unlist(list(b=myfun(x),c=myfun(x*x*x)))))))) > z a b.x b.y c.x c.y 1 a1 1 1 1 1 2 a2 2 4 8 64 3 a3 3 9 27 729
2012 Aug 15
3
per-vertex statistics of edge weights
I have a graph with edge and vertex weights, stored in two data frames: --8<---------------cut here---------------start------------->8--- vertices <- data.frame(vertex=c("a","b","c","d"),weight=c(1,2,1,3)) edges <-
2012 Aug 30
3
apply --> data.frame
Is there a way for an apply-type function to return a data frame? the closest thing I think of is foo <- as.data.frame(sapply(...)) names(foo) <- c(....) is there a more "elegant" way? Thanks! -- Sam Steingold (http://sds.podval.org/) on Ubuntu 12.04 (precise) X 11.0.11103000 http://www.childpsy.net/ http://palestinefacts.org http://dhimmi.com http://honestreporting.com
2012 Feb 13
1
entropy package: how to compute mutual information?
suppose I have two factor vectors: x <- as.factor(c("a","b","a","c","b","c")) y <- as.factor(c("b","a","a","c","c","b")) I can compute their entropies: entropy(table(x)) [1] 1.098612 using library(entropy) but it is not clear how to compute their mutual information
2011 Feb 15
4
string parsing
I am trying to get stock metadata from Yahoo finance (or maybe there is a better source?) here is what I did so far: yahoo.url <- "http://finance.yahoo.com/d/quotes.csv?f=j1jka2&s="; stocks <- c("IBM","NOIZ","MSFT","LNN","C","BODY","F"); # just some samples socket <-
2011 Feb 14
3
help with aggregate()
Hi, I am trying to aggregate some data and I am confused by the results. I load a data frame "all" from a csv file, and then I do: (FOO,BAR,X,Y come from the header line in the csv file, BTW, how do I rename a column?) byFOO <- aggregate(list(all$BAR,all$QUUX,all$X/all$Y), by = list(FOO=all$FOO), FUN = mean); I expect a data frame with 4