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/
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,
> 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
Did you mean this:
> n <- c('a', 'b')
> structure(list(1, 2), names = n)
$a
[1] 1
$b
[1] 2
On Fri, Mar 12, 2010 at 5:28 PM, Rune Schjellerup Philosof
wrote:
> I often find myself making lists similar to this
> list(var1=var1, var2=var2)
>
> It doesn't seem list has an option, to make it use th
Amazing! I haven't seen usage of calling `names<-` like this before.
Thanks so much!
On Thu, Mar 11, 2010 at 9:50 PM, Henrique Dallazuanna wrote:
> Yes, just in the list.
> If they want change the name in Environment GlobalEnv:
>
> for(i in ls(pattern = "DF[0-9]"))
> assign(i, `names<-`(ge
It seems that the names of original data frames have not changed in
this way. I guess textConnection() could help, like this:
for (name in objects(pattern = "df[0-9]"))
eval(parse(textConnection(paste('names(', name, ') <-
column_names'
On Thu, Mar 11, 2010 at 9:25 PM, Henrique Dallazuanna w
> DF
V1 V2 V3
1 10:03:13 3.4 1002
2 10:03:14 5.6 1001
3 10:05:27 7.2 999
4 10:05:33 8.2 998
> DF2 <- t(sapply(split(DF[,-1], gsub('(.{5}).*', '\\1:00', DF$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.
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 <
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
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
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
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,
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)
> }
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
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
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 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
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.
>
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,
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>
(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]]
>
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
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
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 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) :
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
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,
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)
> ___
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
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,
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:
> 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,]
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
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
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 !!!
>
>
>
>
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
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
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
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 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
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,
>
> 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
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
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
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
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
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
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
Did you mean this:
> write.table(t, eol=",\n", row.names=FALSE, col.names=FALSE)
"",
"01001001011011101100",
"1001001011010101",
"1101110100000011",
"000100100101001001011001",
"000101101101101001101001",
> t <- c(
+ "",
+ "01001001011011101100",
+ "1001001011010101",
+ "1101110100000011",
+ "000100100101001001011001",
+ "000101101101101001101001")
> {
+ cat ('rom_array := (\n');
+ for (i in 1:length(t)) {
+
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
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
> 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
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
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
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
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
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)
>>
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] [,
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
60 matches
Mail list logo
