I think NppToR may be a good choice.
http://sourceforge.net/projects/npptor/
On Tue, Aug 11, 2009 at 6:37 AM, Farley, Robert wrote:
>
>
> Does anyone have an R Syntax Highlighting file {userDefineLang.xml} for
> NotePad++?? I've started one, but I'm not so happy with it.
>
>
>
>
> Robert Farley
Try this:
> y[matrix(c(seq_along(x), x), ncol = 2)]
[1] 2 16 12
On Fri, Sep 11, 2009 at 4:17 PM, Luca Braglia wrote:
> Hello R-users
>
> I have a situation like this
>
> x=c(1,3,2)
>
> y=data.frame("a"=1:3, "b"=4:6, "c"=7:9)*2
>
> So we have
>
>> t(t(x))
> [,1]
> [1,] 1
> [2,] 3
> [3,]
To achieve this goal, it seems there are several ways, such as
WP-Syntax (http://wordpress.org/extend/plugins/wp-syntax/),
SyntaxHighlighter2
(http://mohanjith.com/2009/03/syntaxhighlighter2.html), etc. However,
these plugins seem not support R language by default, so you may have
to write some cod
I have saw it now. Thank you for your excellent works.
On Fri, Feb 12, 2010 at 9:15 PM, Tal Galili wrote:
> Hi Linlin,
> I am afraid I wasn't clear.
> In my post I fixed the current WP-Syntax plugin so it WILL support the R
> syntax :)
> The link to the article is:
> http://www.r-statistics.com/2
format(x, nsmall = 2)
On Mon, Feb 15, 2010 at 5:41 PM, wrote:
>
> Hi there,
>
> i'm not getting along with the following problem.
> I'd like to print a real number, e.g.
> x <- 12.3
> with exactly two digits after the decimal point, e.g.
> 12.30
> I've tried the whole format(), formatC() and pret
It's not an array, but a list:
lapply(2:n, function(x) matrix(1:x^2,x))
On Sat, Feb 20, 2010 at 3:17 PM, song song wrote:
> maybe its not an array
>
> like
>
> m[1]=matrix(1:4,2,2)
> m[2]=matrix(1:9,3,3)
> .
> .
> .
> m[k]=matrix(1:k^2,k,k)
>
> likes
>
> for (k in 1:n){
> s=matrix(1:k^2,k,k)
> }
You can just rep() it, and c() with extra data, and then matrix() it again:
> m <- matrix(c(1,4,3,6),2)
> matrix(c(rep(m, 3), c(2, 5)), nrow(m))
[,1] [,2] [,3] [,4] [,5] [,6] [,7]
[1,]1313132
[2,]4646465
On Sun, Feb 21, 2010 at 10:58 AM,
For general purpose of recursion formula, you could do it like this:
> make.vector <- function(w, n, a, b) c(w, sapply(1:(n-1), function(x) w <<- w
> * a / (b + x)))
> make.vector(w = 1, n = 4, a = 24, b = 1)
[1] 1 12 96 576
On Fri, Feb 26, 2010 at 11:23 PM, wrote:
> Hello all,
>
> I want t
Try this:
df[!duplicated(df[,'x']),]
On Sun, Feb 28, 2010 at 8:56 AM, Juliet Ndukum wrote:
> I wish to rearrange the matrix, df, such that all there are not repeated x
> values. Particularly, for each value of x that is reated, the corresponded y
> value should fall under the appropriate column
Maybe you can create a helper vector first:
> helper <- structure(names = unlist(j), rep(names(j), sapply(j, length)))
> helper
acbd
"j1" "j1" "j2" "j2"
> helper[i]
aabbbccd
"j1" "j1" "j2" "j2" "j2" "j1" "j1" "j2"
On Sat, Mar 6, 2010 at 1:42 AM, Carlos
Nice shot of cumsum(). Just improve it a little:
> x <- c(0,0,1,2,3,0,0,4,5,6)
> x.groups <- split(x, (x != 0) * cumsum(x == 0))[-1]
> x.groups
$`2`
[1] 1 2 3
$`4`
[1] 4 5 6
> lapply(x.groups, mean)
$`2`
[1] 2
$`4`
[1] 5
On Mon, Mar 8, 2010 at 11:02 AM, jim holtman wrote:
> Try this:
>
>> x <
> DF
V1 V2 V3
1 10:03:13 3.4 1002
2 10:03:14 5.6 1001
3 10:05:27 7.2 999
4 10:05:33 8.2 998
> DF2 <- t(sapply(split(DF[,-1], gsub('(.{5}).*', '\\1:00', DF$V1)), colSums))
> data.frame(V1 = rownames(DF2), DF2)
V1 V2 V3
10:03:00 10:03:00 9.0 2003
10:05:00 10:05:00 15.
It seems that the names of original data frames have not changed in
this way. I guess textConnection() could help, like this:
for (name in objects(pattern = "df[0-9]"))
eval(parse(textConnection(paste('names(', name, ') <-
column_names'
On Thu, Mar 11, 2010 at 9:25 PM, Henrique Dallazuanna w
Amazing! I haven't seen usage of calling `names<-` like this before.
Thanks so much!
On Thu, Mar 11, 2010 at 9:50 PM, Henrique Dallazuanna wrote:
> Yes, just in the list.
> If they want change the name in Environment GlobalEnv:
>
> for(i in ls(pattern = "DF[0-9]"))
> assign(i, `names<-`(ge
Did you mean this:
> n <- c('a', 'b')
> structure(list(1, 2), names = n)
$a
[1] 1
$b
[1] 2
On Fri, Mar 12, 2010 at 5:28 PM, Rune Schjellerup Philosof
wrote:
> I often find myself making lists similar to this
> list(var1=var1, var2=var2)
>
> It doesn't seem list has an option, to make it use th
How about like this:
for (i in seq_along(a)) {
result <- as.list(a[1:i])
cat("iterator", i, ":\n")
print(result)
}
On Sat, Jul 25, 2009 at 6:48 AM, Alberto Lora M wrote:
> Hi Everybody
>
> I have the following problem
>
> suppose that we
>
> a<-c("uno","dos","tres")
>
> I am working with a
rep(A, each=2)
On Thu, Jul 30, 2009 at 12:15 AM, Inchallah
Yarab wrote:
> Hi ,
>
> i have a vector A=(a1,a2,a3,a4) and i want to create another vector
> B=(a1,a1,a2,a2,a3,a3,a4,a4) !!!
> i know that it is simple but i begin with R so i nned your help!!
>
> thank you for your help !!!
>
>
>
>
Did you mean this:
> m <- matrix(1:12, 3, 4)
> m / max(m)
[,1] [,2] [,3] [,4]
[1,] 0.0833 0.333 0.583 0.833
[2,] 0.1667 0.417 0.667 0.917
[3,] 0.2500 0.500 0.750 1.000
On Thu, Jul 30, 2009 at 12:52 PM, Sam wrote:
> Hi,
> th
How about using operator ==
On Sat, Oct 31, 2009 at 5:00 AM, bamsel wrote:
>
> Dear R users:
> I need to compare character strings stored in 2 separate data frames. I need
> an exact match, so finding "a" in "animal" is no good.
> I've tried regexpr, match, and grepl, but to no avail.
> Anybody k
Try this:
> x[, colnames(x) != 'a']
[1] 3 4
On Tue, Nov 3, 2009 at 9:31 AM, Peng Yu wrote:
> I can exclude columns by column number using '-'. But I wondering if
> there is an easy way to exclude some columns by column names.
>
>> x=cbind(c(1,2),c(3,4))
>> x
> [,1] [,2]
> [1,] 1 3
> [2,
Try this:
> gsub("([a-z]*\\s[a-z]*).*", "\\1", nam)
[1] "Smith John" "Smith David" "Smith Ryan"
On Fri, Nov 6, 2009 at 4:11 PM, johannes rara wrote:
> How to split everything after second whitespace char using regular
> expression? I want to remove A, B, C and D from these names:
>
> nam <- c("S
Regular expression needs double the '\' again, so try this:
gsub('/','',string)
On Mon, Nov 16, 2009 at 7:35 AM, Peng Yu wrote:
> My question was from replacing a pattern by '\\'. How to replace '/'
> in string by '\'?
>
> string='abc/efg'
> gsub('/','\\',string)
> ___
Hope this help:
> m <- matrix(c(2,1,3,9,5,7,7,8,1,8,6,5,6,2,2,7),4,4)
> p <- c(2, 6)
> apply(m == p[1], 1, any) & apply(m == p[2], 1, any)
[1] TRUE FALSE TRUE FALSE
If you want the number of rows which contain the pair, sum() could be used:
> sum(apply(m == p[1], 1, any) & apply(m == p[2], 1,
I don't think this function is same as gcc's option -MM. Because gcc
checks pre-compile command #include, in which the filename can be
fetched definitely. But in your scenario, the filename may be from
some variables, which can not be determined by the R script only.
Maybe you can write a tool by y
Try this:
gsub("[?]", " ", x)
On Mon, Nov 23, 2009 at 7:01 AM, Steven Kang wrote:
> Hi all,
>
>
> I get an error message when trying to replace *+* or *?* signs (with empty
> space) from a string.
>
> x <- "asdf+,jkl?"
>
> gsub("?", " ", x)
>
>
> Error message:
>
> Error in
> gsub("?", " ", x) :
On Tue, Nov 24, 2009 at 1:01 AM, Henrique Dallazuanna wrote:
> Try this:
>
> test[grep("d2", names(test))]
>
Or it can be simply like this:
test[names(test) == "d2"]
> On Mon, Nov 23, 2009 at 2:53 PM, Rajasekaramya
> wrote:
>>
>> Hi There,
>>
>> I have a named List object.I want to access all
Try these:
sapply(lst, nrow) # get row numbers
which(sapply(lst, nrow) < 3) # get the index of rows which has less than 3 rows
lst <- lst[-which(sapply(lst, nrow) < 3)] # remove the rows from the list
On Sun, Nov 29, 2009 at 4:36 PM, Tim Clark wrote:
> Dear List,
>
> I have a list containing data
There is no year() function. Maybe you can try format() instead.
On Sun, Nov 29, 2009 at 8:44 PM, DispersionMap wrote:
>
> i have a column of dates in this format:
>
> data[,"Raised.Date"] <- as.Date(data[,"Raised.Date"], "%d/%m/%Y");
> data[1:10,"Raised.Date"]
> [1] "2006-07-07" "2006-07-07" "20
(sapply(lst, nrow) < 3)]
> list()
>>
>
> Notice the list is now empty. Instead use:
>
>> lst[sapply(lst, nrow) >=3]
> [[1]]
> letter number
> 1 A 1
> 2 B 2
> 3 C 3
> 4 D 4
> 5 E 5
>
> [[2]]
>
It means that R does have the lazy copy mechanism, which I didn't
know, and I think it can be very useful to make R running more
quickly.
On Tue, Dec 15, 2009 at 12:15 PM, Peng Yu wrote:
>> a=1:10
>> b=a
>> a=1:10
>> tracemem(a)# I assume the following is address 'a' points to
> [1] "<0x05cf2798>
Try this:
> f <- function(x) length(gregexpr("[[:digit:]]", as.character(x))[[1]])
> f(3.14)
[1] 3
> f(3.1415)
[1] 5
> f(3.14159265)
[1] 9
On Wed, Dec 16, 2009 at 1:39 PM, Xiang Wu wrote:
> Is there a function in R that could find the significant digit of a specific
> number? Such as for 3.1415,
I am afraid that although in same literally, they are indeed different
functions: as.Date.POSIXct and as.Date.POSIXlt. But I am not sure why
they are designed like this, which causes the confusion as you
mentioned.
On Thu, Dec 24, 2009 at 11:02 PM, MAL wrote:
> Mark, not sure that's the answer.
>
Does every 100 numbers in rnorm(100 * 1000, 0, 1) have the N(0,1) distribution?
On Wed, May 13, 2009 at 11:13 PM, Debbie Zhang wrote:
>
>
> Dear R users,
>
> Can anyone please tell me how to generate a large number of samples in R,
> given certain distribution and size.
>
> For example, if I wan
On Thu, May 14, 2009 at 2:16 PM, christiaan pauw wrote:
> Hi everybody.
> I want to identify not only duplicate number but also the original number
> that has been duplicated.
> Example:
> x=c(1,2,3,4,4,5,6,7,8,9)
> y=duplicated(x)
> rbind(x,y)
>
> gives:
> [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,
The operator %in% is very good! And that can be simpler like this:
x %in% x[duplicated(x)]
[1] FALSE FALSE FALSE TRUE TRUE FALSE FALSE FALSE FALSE FALSE
On Thu, May 14, 2009 at 4:43 PM, Andrej Blejec wrote:
> Try this
>
> x%in%x[which(y)]
>
> >From your example
>
>> x=c(1,2,3,4,4,5,6,7,8,9)
>>
Since you got the most suitable way to get x, why can't you get the
variances in the same way? Just like:
v = vector()
for (i in 1:length(x)) v[i] = var(x[[i]])
BTW, it is much better to use lapply, like this:
lapply(x, var)
On Thu, May 14, 2009 at 8:26 PM, Debbie Zhang wrote:
>
> Thanks f
On Sat, May 16, 2009 at 12:05 PM, Debbie Zhang wrote:
>
> Dear R users,
>
> Does anyone know how to write a function involving derivative?
>
> i.e. I want to implementing Newton's method in R, so my function is something
> like
>
> x<- x-y/y'
>
> I am not sure how to write y' in my function. Can
How about ceiling(x), which return the smallest integer not less than x?
On Sun, May 17, 2009 at 2:49 AM, Thomas Mang wrote:
> Hello,
>
> Suppose I have x, which is a variable of class numeric. The calculations
> performed to yield x imply that mathematically it should be an integer , but
> due t
aillist addres
r-h...@stat.math.ethz.ch.
>
> cheers and thanks,
> Thomas
>
>
> Linlin Yan wrote:
>>
>> I see. What you want is the integer with same sign as the original
>> numeral, and whose absolute value is the least integer which is not
>> less than a
> z <- paste(x, y, sep = '')
> z
[1] "A1" "B2" "C3" "D4" "E5" "F6"
On Mon, May 18, 2009 at 7:09 PM, Henning Wildhagen wrote:
> Dear users,
>
> a very simple question:
>
> Given two vectors x and y
>
> x<-as.character(c("A","B","C","D","E","F"))
> y<-as.factor(c("1","2","3","4","5","6"))
>
> i wan
It seems that "c(x,y)" is not correct:
> z<-c(x,y)
> z
[1] "A" "B" "C" "D" "E" "F" "1" "2" "3" "4" "5" "6"
On Mon, May 18, 2009 at 7:17 PM, Simon Pickett wrote:
> z<-c(x,y)
>
> cheers, Simon.
>
>
> - Original Message - From: "Henning Wildhagen"
> To:
> Sent: Monday, May 18, 2009 12:09
SmoothData$span is not an object which can be checked by exists(), but
part of an object which can be checked by is.null().
On Wed, May 20, 2009 at 12:07 AM, Žroutík wrote:
> Dear R-users,
>
> in a minimal example exists() gives FALSE on an object which obviously does
> exist. How can I check on
> t <- c(
+ "",
+ "01001001011011101100",
+ "1001001011010101",
+ "1101110100000011",
+ "000100100101001001011001",
+ "000101101101101001101001")
> {
+ cat ('rom_array := (\n');
+ for (i in 1:length(t)) {
+
Did you mean this:
> write.table(t, eol=",\n", row.names=FALSE, col.names=FALSE)
"",
"01001001011011101100",
"1001001011010101",
"1101110100000011",
"000100100101001001011001",
"000101101101101001101001",
Why did you use different variable names rather than index of list/data.frame?
On Wed, May 27, 2009 at 6:34 PM, Maithili Shiva
wrote:
> Dear R helpers,
>
> Following is a R script I am using to run the Fast Fourier Transform. The csv
> files has 10 columns with titles m1, m2, m3 .m10.
>
> Wh
Hope this helps:
> df <- data.frame(matrix(1:10,2))
> df
X1 X2 X3 X4 X5
1 1 3 5 7 9
2 2 4 6 8 10
> df[,-2]
X1 X3 X4 X5
1 1 5 7 9
2 2 6 8 10
> df[,-which(names(df)=="X2")]
X1 X3 X4 X5
1 1 5 7 9
2 2 6 8 10
On Wed, May 27, 2009 at 6:37 PM, Zeljko Vrba wrote:
> Given an
0.2
Any comment is welcome! ;)
On Wed, May 27, 2009 at 11:04 PM, Linlin Yan wrote:
>> m <- matrix( c(2, 1, 1, 3, .5, .3, .5, .2), 4)
>> m
> [,1] [,2]
> [1,] 2 0.5
> [2,] 1 0.3
> [3,] 1 0.5
> [4,] 3 0.2
>> m[unlist(sapply(sort(unique(m[,1
On Sat, May 30, 2009 at 2:48 AM, Grześ wrote:
>
> I have a vector like this:
> h <- c(4, 6, NA, 12)
> and I create the secound logical vector like this:
> g <- c(TRUE, TRUE, FALSE, TRUE)
Why don't you create vector g like this:
g <- ! is.na(h)
>
> And my problem is that I would like to get a new
e.g.
dat[ order(dat$a), ]
On Sun, May 31, 2009 at 2:34 PM, Угодай n/a wrote:
> I have a data frame, for exampe
>
>> dat <- data.frame(a=rnorm(5),b=rnorm(5),c=rnorm(5))
> a b c
> 1 -0.1731141 0.002453991 0.1180976
> 2 1.2142024 -0.413897606 0.7617472
> 3 -0.942848
I think you can use readLines(n=1) in loop to skip unwanted rows.
On Mon, Jun 1, 2009 at 12:56 AM, wrote:
> Thanks, Juliet.
> It works for filtering columns.
> I am also wondering if there is a way to filter rows.
> Thanks again.
> -james
>
>> One can use colClasses to set which columns get read
> c(x[1], x[-length(x)]) != x
[1] FALSE FALSE FALSE TRUE TRUE FALSE TRUE
On Mon, Jun 1, 2009 at 11:57 PM, liujb wrote:
>
> Hello,
>
> I have a vector:
> x <- c("A", "A", "A", "B", "A", "A", "C")
>
> I'd like to compare each of elements of vector x from its previous element
> (except for the 1s
What's wrong with the copyright? And how could that trouble your linking?
On Wed, Jun 3, 2009 at 2:53 PM, RamaR Guru wrote:
>
> From: techzone2...@hotmail.com
> To: r-h...@stat.math.ethz.ch; r-help-requ...@stat.math.ethz.ch
> Subject: R.dll Reg.,
> Date: Wed, 3 Jun 2009 11:57:52 +0530
>
> Sir,
>
How about this:
> "%==%" <- function(x, y) {
if (length(x) > 1) {
sapply(x, function(z) isTRUE(all.equal(z, y)));
} else {
sapply(y, function(z) isTRUE(all.equal(z, x)));
}
}
> seq(0, 1, by=0.1) %==% 0.1
[1] FALSE TRUE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALS
Try this:
for (i in 1:dim(ALLRESULTS)[1]) {
ALLRESULTS[i,23] <- length(ALLRESULTS[i,][ALLRESULTS[i,16:22] <= 0.05])
}
On Wed, Jun 10, 2009 at 12:17 AM, Amit Patel wrote:
>
> Hi
> I am trying to create a column in a data frame which gives a sigificane score
> from 0-7. It should read values from
How about like this:
> t1 <- data.frame(row.names=c('c1','c2','c3','c4'), mk1=c(1,1,0,0),
> mk2=c(0,0,0,1), mk3=c(1,1,1,1), mk4=c(0,0,0,0), mk5=c(0,0,0,1), S=c(4,5,3,2))
> t1
mk1 mk2 mk3 mk4 mk5 S
c1 1 0 1 0 0 4
c2 1 0 1 0 0 5
c3 0 0 1 0 0 3
c4 0 1 1 0 1
Function parameters in R are passed by value, not by reference. In
order to resolve it, just remove "clusters" from the parameter list,
and use "clusters[i] <<- ..." to change the value of global variable.
On Wed, Jun 17, 2009 at 7:52 PM, Nick Angelou wrote:
>
> Hi,
>
> I have a problem with dynam
I guess that the matrix dimension changed because matrix in R are
filled by columns. Since you try:
apply(b, 1, function(y) sort(y, na.last=F))
The second parameter make it scan matrix b row by row but store result
by columns, which make the result be a matrix transposed.
If you try:
apply(b, 2, fu
> d <- 1:4
> f <- c(2,3,3,1)
> rep(d,f)
[1] 1 1 2 2 2 3 3 3 4
On Sat, Jul 10, 2010 at 10:55 PM, nn roh wrote:
> Hi,
>
> I have a question relating to R package
>
> If i have the frequencies of data how i can get the data as following :
>
> 1 2 3 4 data
> 2 3 3 1 frequency
>
> Whic
If I am not wrong, it seems that you want to get factor counts in the
whole scale of data.raw. Maybe you can do that just like this:
table(data.raw)
On Sat, Jul 24, 2010 at 1:44 AM, Marcus Liu wrote:
> Hi everyone,
>
> Is there any command for updating table withing a loop? For instance, at i,
try to use difftime() instead of as.difftime().
On Thu, Sep 2, 2010 at 10:32 PM, Dunia Scheid wrote:
> Hello all,
>
> I've 2 strings that representing the start and end values of a date and
> time.
> For example,
> time1 <- c("21/04/2005","23/05/2005","11/04/2005")
> time2 <- c("15/07/2009", "03/
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