Displaying 20 results from an estimated 1000 matches similar to: "cor() on sets of vectors"
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
2012 Feb 10
2
the value of the last expression
Is there an analogue of common lisp "*" variable which contains the
value of the last expression?
E.g., in lisp:
> (+ 1 2)
3
> *
3
I wish I could recover the value of the last expression without
re-evaluating it.
thanks
--
Sam Steingold (http://sds.podval.org/) on Ubuntu 11.10 (oneiric) X 11.0.11004000
http://www.childpsy.net/ http://camera.org http://ffii.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,]
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
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
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/
2012 Feb 24
1
count.fields inconsistent with read.table?
Hi,
batch is a vector of lines returned by readLines from a
NL-line-terminated file, here is the relevant section:
=========================================================
AA BB CC DD EE FF
GG H
H JJ KK LL MM
=========================================================
as you can see, a line is corrupt; two CRLF's are inserted.
This is okay, I drop the bad lines, at least I hope I do:
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
2012 Feb 10
2
naiveBayes: slow predict, weird results
I did this:
nb <- naiveBayes(users, platform)
pl <- predict(nb,users)
nrow(users) ==> 314781
ncol(users) ==> 109
1. naiveBayes() was quite fast (~20 seconds), while predict() was slow
(tens of minutes). why?
2. the predict results were completely off the mark (quite the opposite
of the expected overfitting). suffice it to show the tables:
pl:
android blackberry ipad
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
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 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 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))$x)
alas, aggregate has been running for ~30 minute, RSS is 14G, VIRT is
24.3G, and no end in sight.
both V1 and V2 are characters (not factors).
Is there anything I could do to speed this up?
Thanks.
--
Sam Steingold
2011 Feb 15
1
summary for factors is not very informative
summary() for a factor prints:
ColName
SNDK : 72
VXX : 36
MWW : 30
ACI : 28
FRO : 28
(Other):1801
it would have been much more useful if it additionally
printed frequency stats as if by
summary(aggregate(frame$ColName,by=list(frame$ColName),FUN=length)$x)
--
Sam Steingold (http://sds.podval.org/) on CentOS release 5.3 (Final)
http://jihadwatch.org
2012 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 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
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 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 =
2012 Mar 20
2
igraph: decompose.graph: Error: protect(): protection stack overflow
I just got this error:
> library(igraph)
> comp <- decompose.graph(gr)
Error: protect(): protection stack overflow
Error: protect(): protection stack overflow
>
what can I do?
the digraph is, indeed, large (300,000 vertexes), but there are very
many very small components (which I would rather not discard).
PS. the doc for decompose.graph does not say which mode is the default.
--
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")