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 =