Displaying 20 results from an estimated 1000 matches similar to: "write.matrix.csr data conversion"
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 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 Mar 14
2
sum(hist$density) == 2 ?!
> x <- rnorm(1000)
> h <- hist(x,plot=FALSE)
> sum(h$density)
[1] 2 ----------------------------- shouldn't it be 1?!
> h <- hist(x,plot=FALSE, breaks=(-4:4))
> sum(h$density)
[1] 1 ----------------------------- now it's 1. why?!
--
Sam Steingold (http://sds.podval.org/) on Ubuntu 11.10 (oneiric) X 11.0.11004000
http://www.childpsy.net/ http://www.memritv.org
2012 Aug 27
1
matrix.csr %*% matrix --> matrix
When a sparse matrix is multiplied by a regular one, the result is
usually not sparse. However, when matrix.csr is multiplied by a regular
matrix in R, a matrix.csr is produced.
Is there a way to avoid this?
Thanks!
--
Sam Steingold (http://sds.podval.org/) on Ubuntu 12.04 (precise) X 11.0.11103000
http://www.childpsy.net/ http://palestinefacts.org http://truepeace.org
2012 Sep 19
4
where are these NAs coming from?
I see this:
--8<---------------cut here---------------start------------->8---
> length(which(is.na(z$language)))
[1] 0
> locals <- z[z$country == mycountry,]
> length(which(is.na(locals$language)))
[1] 229
--8<---------------cut here---------------end--------------->8---
where are those locals without the language coming from?!
--
Sam Steingold (http://sds.podval.org/) on
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 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/)
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")
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 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 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 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 Aug 24
2
SparseM buglet
read.matrix.csr does not close the connection:
> library('SparseM')
Package SparseM (0.96) loaded.
> read.matrix.csr(foo)
...
Warning message:
closing unused connection 3 (foo)
>
--
Sam Steingold (http://sds.podval.org/) on Ubuntu 12.04 (precise) X 11.0.11103000
http://www.childpsy.net/ http://truepeace.org http://camera.org
http://pmw.org.il http://think-israel.org
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 Sep 06
2
merge a list of data frames
I have a list of data frames:
> str(data)
List of 4
$ :'data.frame': 700773 obs. of 3 variables:
..$ V1: chr [1:700773] "200130446465779" "200070050127778" "200030633708779" "200010587002779" ...
..$ V2: int [1:700773] 0 0 0 0 0 0 0 0 0 0 ...
..$ V3: num [1:700773] 1 1 1 1 1 ...
$ :'data.frame': 700773 obs. of 3 variables:
..$
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,]
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$FOO),
FUN = mean);
I expect a data frame with 4
2013 Sep 18
2
strsplit with a vector split argument
Hi,
I find this behavior unexpected:
--8<---------------cut here---------------start------------->8---
> strsplit(c("a,b;c","d;e,f"),c(",",";"))
[[1]]
[1] "a" "b;c"
[[2]]
[1] "d" "e,f"
--8<---------------cut here---------------end--------------->8---
I thought that it should be identical to this:
2011 Feb 15
1
all.equal: subscript out of bounds
When I do
> all(all$X.Time == all$Y.Time);
[1] TRUE
as expected, but
> all.equal(all$X.Time,all$Y.Time);
Error in target[[i]] : subscript out of bounds
why?
thanks!
--
Sam Steingold (http://sds.podval.org/) on CentOS release 5.3 (Final)
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The dark past once was the
2011 Feb 15
2
strptime format = "%H:%M:%OS6"
I read a dataset with times in them, e.g., "09:31:29.18761".
I then parse them:
> all$X.Time <- strptime(all$X.Time, format = "%H:%M:%OS6");
and get a vector of NAs (how do I check that except for a visual inspection?)
then I do
> options("digits.secs"=6);
> all$X.Time <- strptime(all$X.Time, format = "%H:%M:%OS");
and it, apparently, works: