similar to: as.data.frame(do.call(rbind,lapply)) produces something weird

Displaying 20 results from an estimated 4000 matches similar to: "as.data.frame(do.call(rbind,lapply)) produces something weird"

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 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
2011 Feb 15
1
summary for factors is not very informative
summary() for a factor prints: ColName SNDK : 72 VXX : 36 MWW : 30 ACI : 28 FRO : 28 (Other):1801 it would have been much more useful if it additionally printed frequency stats as if by summary(aggregate(frame$ColName,by=list(frame$ColName),FUN=length)$x) -- Sam Steingold (http://sds.podval.org/) on CentOS release 5.3 (Final) http://jihadwatch.org
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,]
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 Dec 27
4
vectorization & modifying globals in functions
I have the following code: --8<---------------cut here---------------start------------->8--- d <- rep(10,10) for (i in 1:100) { a <- sample.int(length(d), size = 2) if (d[a[1]] >= 1) { d[a[1]] <- d[a[1]] - 1 d[a[2]] <- d[a[2]] + 1 } } --8<---------------cut here---------------end--------------->8--- it does what I want, i.e., modified vector d 100 times.
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)
2012 Jul 13
1
LiblineaR: read/write model files?
How do I read/write liblinear models to files? E.g., if I train a model using the command line interface, I might want to load it into R to look the histogram of the weights. Or I might want to train a model in R and then apply it using a command line interface. -- Sam Steingold (http://sds.podval.org/) on Ubuntu 12.04 (precise) X 11.0.11103000 http://www.childpsy.net/
2013 Apr 21
1
cedta decided 'igraph' wasn't data.table aware
Hi, what does this mean? --8<---------------cut here---------------start------------->8--- > graph <- graph.data.frame(merged[!v,], vertices=ve, directed=FALSE) cedta decided 'igraph' wasn't data.table aware cedta decided 'igraph' wasn't data.table aware cedta decided 'igraph' wasn't data.table aware cedta decided 'igraph' wasn't
2012 Oct 18
3
how to concatenate factor vectors?
How do I concatenate two vectors of factors? --8<---------------cut here---------------start------------->8--- > a <- factor(5:1,levels=1:9) > b <- factor(9:1,levels=1:9) > str(c(a,b)) int [1:14] 5 4 3 2 1 9 8 7 6 5 ... > str(unlist(list(a,b),use.names=FALSE)) Factor w/ 9 levels "1","2","3","4",..: 5 4 3 2 1 9 8 7 6 5 ...
2012 Sep 19
2
drop zero slots from table?
I find myself doing --8<---------------cut here---------------start------------->8--- tab <- table(...) tab <- tab[tab > 0] tab <- sort(tab,decreasing=TRUE) --8<---------------cut here---------------end--------------->8--- all the time. I am wondering if the "drop 0" (and maybe even sort?) can be effected by some magic argument to table() which I fail to discover
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 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
2006 Mar 17
6
removing NA from a data frame
Hi, It appears that deal does not support missing values (NA), so I need to remove them (NAs) from my data frame. how do I do this? (I am very new to R, so a detailed step-by-step explanation with code samples would be nice). Some columns (variables) have quite a few NAs, so I would rather drop the whole column than sacrifice all the rows (observations) which have NA in that column. How do I
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 Sep 16
2
multi-column factor
I have a data frame with columns which draw on the same underlying universe, so I want them to be factors with the same level set: --8<---------------cut here---------------start------------->8--- > z <- data.frame(a=c("a","b","c"),b=c("b","c","d"),stringsAsFactors=FALSE) > str(z) 'data.frame': 3 obs. of 2
2012 Aug 28
5
variable scope
At the end of a for loop its variables are still present: for (i in 1:10) { x <- vector(length=100000000) } ls() will print "i" and "x". this means that at the end of the for loop body I have to write rm(x) gc() is there a more elegant way to handle this? Thanks. -- Sam Steingold (http://sds.podval.org/) on Ubuntu 12.04 (precise) X 11.0.11103000
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$Y.Time); Error in target[[i]] : subscript out of bounds why? thanks! -- Sam Steingold (http://sds.podval.org/) on CentOS release 5.3 (Final) http://mideasttruth.com http://honestreporting.com http://dhimmi.com http://jihadwatch.org http://pmw.org.il http://ffii.org The dark past once was the
2012 Sep 14
3
aggregate() runs out of memory
I have a large data.frame Z (2,424,185,944 bytes, 10,256,441 rows, 17 columns). I want to get the result of table(aggregate(Z$V1, FUN = length, by = list(id=Z$V2))$x) alas, aggregate has been running for ~30 minute, RSS is 14G, VIRT is 24.3G, and no end in sight. both V1 and V2 are characters (not factors). Is there anything I could do to speed this up? Thanks. -- Sam Steingold
2012 Apr 04
2
plot with a regression line(s)
I am sure a common need is to plot a scatterplot with some fitted line(s) and maybe save to a file. I have this: plot.glm <- function (x, y, file = NULL, xlab = deparse(substitute(x)), ylab = deparse(substitute(y)), main = NULL) { m <- glm(y ~ x) if (!is.null(file)) pdf(file = file) plot(x, y, xlab = xlab, ylab = ylab, main = main) lines(x, y =