try to use difftime() instead of as.difftime().
On Thu, Sep 2, 2010 at 10:32 PM, Dunia Scheid dunia.sch...@gmail.com 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 -
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 marcusliu...@yahoo.com wrote:
Hi everyone,
Is there any command for updating table withing a loop?
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 nn.r...@gmail.com 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
Did you mean this:
n - c('a', 'b')
structure(list(1, 2), names = n)
$a
[1] 1
$b
[1] 2
On Fri, Mar 12, 2010 at 5:28 PM, Rune Schjellerup Philosof
rphilo...@health.sdu.dk wrote:
I often find myself making lists similar to this
list(var1=var1, var2=var2)
It doesn't seem list has an option,
DF
V1 V2 V3
1 10:03:13 3.4 1002
2 10:03:14 5.6 1001
3 10:05:27 7.2 999
4 10:05:33 8.2 998
DF2 - t(sapply(split(DF[,-1], gsub('(.{5}).*', '\\1:00', DF$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 15.4
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
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 www...@gmail.com 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,
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 Petti
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 jholt...@gmail.com wrote:
Try
Try this:
df[!duplicated(df[,'x']),]
On Sun, Feb 28, 2010 at 8:56 AM, Juliet Ndukum jpnts...@yahoo.com 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
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, khaz...@ceremade.dauphine.fr wrote:
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 rprojecth...@gmail.com 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){
format(x, nsmall = 2)
On Mon, Feb 15, 2010 at 5:41 PM, t.wun...@stud.uni-heidelberg.de 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
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
I have saw it now. Thank you for your excellent works.
On Fri, Feb 12, 2010 at 9:15 PM, Tal Galili tal.gal...@gmail.com 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:
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,
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 diver...@univecom.ch wrote:
Mark, not sure
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 xiang@gmail.com wrote:
Is there a function in R that could find the significant digit of a specific
number? Such as
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 pengyu...@gmail.com wrote:
a=1:10
b=a
a=1:10
tracemem(a)# I assume the following is address 'a' points to
[1]
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 mudiver1...@yahoo.com wrote:
Dear List,
I have a list
There is no year() function. Maybe you can try format() instead.
On Sun, Nov 29, 2009 at 8:44 PM, DispersionMap frenc...@btinternet.com 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
3 C 3
4 D 4
5 E 5
On Sun, Nov 29, 2009 at 3:43 AM, Linlin Yan yanlinli...@gmail.com wrote:
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
On Tue, Nov 24, 2009 at 1:01 AM, Henrique Dallazuanna www...@gmail.com 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 ramya.vict...@gmail.com
wrote:
Hi There,
I have a named List
Try this:
gsub([?], , x)
On Mon, Nov 23, 2009 at 7:01 AM, Steven Kang stochastick...@gmail.com 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
Regular expression needs double the '\' again, so try this:
gsub('/','',string)
On Mon, Nov 16, 2009 at 7:35 AM, Peng Yu pengyu...@gmail.com 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, any))
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 johannesr...@gmail.com 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
Try this:
x[, colnames(x) != 'a']
[1] 3 4
On Tue, Nov 3, 2009 at 9:31 AM, Peng Yu pengyu...@gmail.com 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
How about using operator ==
On Sat, Oct 31, 2009 at 5:00 AM, bamsel benam...@gmail.com 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.
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 brag...@poleis.eu 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,] 2
I think NppToR may be a good choice.
http://sourceforge.net/projects/npptor/
On Tue, Aug 11, 2009 at 6:37 AM, Farley, Robertfarl...@metro.net 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.
rep(A, each=2)
On Thu, Jul 30, 2009 at 12:15 AM, Inchallah
Yarabinchallahya...@yahoo.fr 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,
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 Malbertolo...@gmail.com wrote:
Hi Everybody
I have the following problem
suppose that we
a-c(uno,dos,tres)
I am working with
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 Angelounikola...@yahoo.com wrote:
Hi,
I have a problem
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 2
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 Patelamitrh...@yahoo.co.uk wrote:
Hi
I am trying to create a column in a data frame which gives a sigificane score
from 0-7. It should
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 FALSE
e.g.
dat[ order(dat$a), ]
On Sun, May 31, 2009 at 2:34 PM, Угодай n/a ugo...@gmail.com 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
I think you can use readLines(n=1) in loop to skip unwanted rows.
On Mon, Jun 1, 2009 at 12:56 AM, g...@ucalgary.ca 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
On Sat, May 30, 2009 at 2:48 AM, Grześ gregori...@gmail.com 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
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
maithili_sh...@yahoo.com 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,
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 zv...@ifi.uio.no wrote:
,]3 0.2
m[order.matrix(m),]
[,1] [,2]
[1,]1 0.3
[2,]1 0.5
[3,]2 0.5
[4,]3 0.2
m[order.matrix(m, c(1=FALSE, 2=TRUE)),]
[,1] [,2]
[1,]1 0.5
[2,]1 0.3
[3,]2 0.5
[4,]3 0.2
Any comment is welcome! ;)
On Wed, May 27, 2009 at 11:04 PM, Linlin Yan
Did you mean this:
write.table(t, eol=,\n, row.names=FALSE, col.names=FALSE)
,
01001001011011101100,
1001001011010101,
1101110100000011,
000100100101001001011001,
000101101101101001101001,
Try
t - c(
+ ,
+ 01001001011011101100,
+ 1001001011010101,
+ 1101110100000011,
+ 000100100101001001011001,
+ 000101101101101001101001)
{
+ cat ('rom_array := (\n');
+ for (i in 1:length(t)) {
+ cat('',
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 zrou...@gmail.com wrote:
Dear R-users,
in a minimal example exists() gives FALSE on an object which obviously does
exist. How
z - paste(x, y, sep = '')
z
[1] A1 B2 C3 D4 E5 F6
On Mon, May 18, 2009 at 7:09 PM, Henning Wildhagen hwildha...@gmx.de 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 want to combine them into a single
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 simon.pick...@bto.org wrote:
z-c(x,y)
cheers, Simon.
- Original Message - From: Henning Wildhagen hwildha...@gmx.de
To: r-help@r-project.org
Sent: Monday,
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 absolute value of the original numeral. Am I right? I am
afraid that there may not be any single function could work
On Sat, May 16, 2009 at 12:05 PM, Debbie Zhang debbie0...@hotmail.com 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
How about ceiling(x), which return the smallest integer not less than x?
On Sun, May 17, 2009 at 2:49 AM, Thomas Mang thomas.m...@fiwi.at 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
On Thu, May 14, 2009 at 2:16 PM, christiaan pauw cjp...@gmail.com 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]
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 andrej.ble...@nib.si wrote:
Try this
x%in%x[which(y)]
From your example
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
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 debbie0...@hotmail.com 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
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