ldn't put the result into a R
> variable.
> Do you have any comments on this?
Hi.
Try the following.
writeLines(c("uno dos tres", "cuatro cinco", "seis"), "some_file.txt")
out <- system("wc some_file.txt", intern=TRUE)
out
1]
if (length(x) >= n) break
}
stopifnot(length(x) >= n)
x <- x[1:n]
Hope this helps.
Petr Savicky.
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occurrences of each protocol or
sum(all[, 2] == "UDP")
to get the number of UDP rows or
udp <- all[all[, 2] == "UDP", ]
to extract only UDP rows.
If you cannot change the export to .csv, you can use the function strsplit().
Petr Savicky.
_
On Fri, Nov 19, 2010 at 10:34:26AM -0800, wangwallace wrote:
>
> this is a simple question, but I wasn't able to figure it out myself.
>
> here is the data frame:
>
> M P Q
> 1 2 3
> 4 5 6
> 7 8 9
>
> M, P, Q each represent a variable
>
> I want to draw 2 random sample from each row separatel
On Fri, Nov 19, 2010 at 07:22:57PM -0800, wangwallace wrote:
> actually, what I meant is to draw two random numbers from each row
> separately to create a new data frame. for example, an example output could
> be:
>
> 1 3
> 4 5
> 9 8
This may be done, for example
X <- matrix(1:9, ncol = 3, byr
On Sun, Nov 21, 2010 at 10:56:14AM -0500, David Winsemius wrote:
> On Nov 21, 2010, at 10:43 AM, madr wrote:
> >Is there any way of suppressing that error, like in other programming
> >languages you can specifically invoke an error or simply exit,
>
> If you are in a function, then return()
>
> >
)]
col3 <- A[cbind(seq(nrow(A)), 3 + ind[, 3])]
B <- cbind(col1, col2, col3)
# or with a cycle over rows
C <- matrix(nrow=nrow(A), ncol=3)
for (i in seq(nrow(A))) {
C[i, 1] <- A[i, ind[i, 1]]
C[i, 2:3] <- A[i, 3 + ind[i, 2:3]]
}
Petr Savicky.
ize under Linux, one can use
Sys.getenv("COLUMNS"). I do not know, whether this applies also to MasOS.
Petr Savicky.
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and provide commented, minimal, self-contained, reproducible code.
s them in some situations, may be found
in the first section of
http://rwiki.sciviews.org/doku.php?id=misc:r_accuracy
and in
http://rwiki.sciviews.org/doku.php?id=misc:r_accuracy:decimal_numbers
Petr Savicky.
__
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ngle character
may be done using 0/1 instead of FALSE/TRUE.
Petr Savicky.
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and provide commente
000 0.400 0.375 0.3636364
M/M1
[,1] [,2] [,3] [,4]
[1,] 6 15 24 33
[2,]6 15 24 33
[3,]6 15 24 33
Alternatively, it is possible to use
sweep(M, 2, colSums(M), FUN="/")
Petr Savicky.
__
R-help@r
) {
out[i, seq(length(p[[i]]))] <- p[[i]]
}
out
[,1] [,2] [,3] [,4]
[1,]3 NA NA NA
[2,]25 NA NA
[3,]369 11
[4,]13 NA NA
Petr Savicky.
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h
ies in the table is 310660*17431. Using integer
type, this is 310660*17431*4 bytes, which is 20.17 GB. This probably
does not fit into RAM. Function table() produces a full matrix, not
a sparse one, even if there are empty cells.
Petr Savicky.
__
R-he
(rowSums((a - rep(x, each=nrow(a)))^2)))
xinit <- colMeans(a)
x <- optim(xinit, d, a=a)$par
plot(a)
points(rbind(x), col=2)
Is this, what you mean?
Function optim() has further parameters, which influence efficiency
and accuracy, and there are also
ng the end-nodes of the edges starting in each node.
In a sorted file, they form blocks of consecutive lines, so a simple text
processing with perl is sufficient.
Petr Savicky.
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3 4"
3 "2"
4 "1 3 5"
5 "2 4"
and to a text (with a possible file= argument)
cat(paste(names(out2), out2), sep="\n")
1 2 3 4 5
2 3 4
3 2
4 1 3 5
5 2 4
Petr Savicky.
__
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> some of the basic function will cause old code to break?
I think that this is an important part of the reason.
Petr Savicky.
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On Wed, Dec 15, 2010 at 11:08:06AM -0200, Henrique Dallazuanna wrote:
> Try this:
>
> gsub("[^0-9]", "", "AB15E9SDF654VKBN?dvb.65")
Consider also
strsplit("AB15E9SDF654VKBN?dvb.65", "[^.0-9][^.0-9]*")
[[1]]
[1] """15" "9" "654" ".65"
PS.
> On Wed, Dec 15, 2010 at 6:55 AM, Luis Fe
N"
> > pullchar("AB15E9SDF654VKBN?dvb.65", "[0-9]")
> [1] "15965465"
>
> Still learning regex so if there is a "positive" strategy I'm all
> ears. ...er, eyes?
One of the suggestions in this thread was to use an external program.
On Thu, Dec 16, 2010 at 06:17:45AM -0800, Dieter Menne wrote:
> Petr Savicky wrote:
> >
> > One of the suggestions in this thread was to use an external program.
> > A possible solution without negation in Perl is
> >
> > @a = ("AB15E9SDF654VKBN?dvb.
)
cars1_diff <- cars2_plus[ - seq(nrow(cars2_set)), ]
cars2_diff <- cars1_plus[ - seq(nrow(cars1_set)), ]
all(cars1_unique == cars1_diff) # [1] TRUE
all(cars2_unique == cars2_diff) # [1] TRUE
Petr Savicky.
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h
"9 Ladies Dancing"
> [5] "8 Maids-a-Milking""7 Swans-a-Swimming"
Additionally, if you know the exact character positions, which have
to be changed, then substr() can be used.
x <- "123456789"
substr(x, 5, 7) &l
On Fri, Dec 17, 2010 at 07:39:46AM -0500, Gabor Grothendieck wrote:
> On Thu, Dec 16, 2010 at 11:42 AM, Petr Savicky wrote:
[...]
> > Can something similar be done in R either specifically for numbers or
> > for a general regular expression?
>
> Dieter's first p
006 for row 1, 2008 for row 2 and 2008 for row 3.
If the pattern is always c("0","1"), the number of rows is large
and the number of years is relatively small, then this may
computed also using matrix calculations. For example
M <-
matrix(c("0","0&
Hi Mark:
> However, if the dataframe contains non-unique rows (two rows with
> exactly the same values in each column) then the unique function will
> delete one of them and that may not be desirable.
In order to get information about equal rows between two dataframes
without removing duplicated
ding errors
in simple situations.
Petr Savicky.
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
x[i-2]), f(x[i-1])),
col=2)
}
points(x[i], 0)
aux <- readline("press Enter to continue")
}
x[i]
}
regulafalsi(function(x) x^(1/2)+3*log(x)-5,1,10)
regulafalsi(function(x) x^(1/2)+3*log(x)-5,1,100)
then it may be seen that
://cran.at.r-project.org/web/views/gR.html
Petr Savicky.
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, se
der). Which
of these two was in your actual R code? In a formula, the ASCII tilda is
required.
Petr Savicky.
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
"Subscribing to R-help" and follow the description.
Petr Savicky.
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
sample(which()) samples from 1:i.
However, with the parameters 34 and 40, your code uses sample() to vectors
of length at least 35 or at least 40 - 34.
If you want to keep all cases and only reassign the groups, you can either
modify df$mar.y (and not the
0) > 34) {
df$mar.y[sample(ind0, length(ind0) - 34)] <- 1
} else {
df$mar.y[sample(ind1, 34 - length(ind0))] <- 0
}
table(old, new=df$mar.y) # just to check the change
Does this work in your situation?
Petr Savicky.
__
R-help@
0.661
[1] 0.868
[1] 1.079
[1] 1.29
[1] 1.507
[1] 1.724
[1] 1.947
[1] 2.172
Petr Savicky.
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
like the following?
xx <- c("abc", "abcd", "abcde")
z1 <- rep("0", times=length(xx))
z2 <- substr(z1, 1, 9 - nchar(xx))
yy <- paste(z2, xx, sep="")
cbind(yy)
# yy
#[1,] "00abc"
#[2,]
le of the identifiers would be helpful.
Filtering out different types of delimiters may be done as
a preprocessing step, for example, using gsub()
s <- c("ab cd", "ab cd", "a b cd")
gsub(" ", "", s) # [1] "abcd" "abcd" &qu
> b b b b
> c c c c
> d d d d
>
> But what I really want is:
>
> a b c d
> b c d a
> c d a b
> d a b c
>
> How can I do this?
Try the following
A <- c("a","b","c","d")
B <- matrix(
to the previous ones
immediately.
nro <- as.numeric(readline("no of teams "))
teams <- rep(NA, times=nro)
for (i in seq(length=nro)) {
repeat {
current <- readline(paste("team", i, ""))
if (curren
lution needs to be applied many times, so I need something quick -- I
> was hoping a base function would do it, but I'm drawing a blank.
If the matrix can have different number of columns, then
also the following can be used
combs <- as.matrix(expand.grid(c(0,1),c(0,1),c(0,1)))
x <
4 4
> 4
> [75] 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4
Let me suggest a slightly simpler code, which produces the same
output, if the input has length at most 9.
xx <- c("abc", "abcd", "abcde")
xx <- paste(&quo
e sum of numbers, which have very different magnitudes, may
be approximated by their maximum, so max(logx) is an
approximation of the required logarithm. A more accurate
calculation can be done, for example, as follows
maxlog + log(sum(exp(logx - maxlog)))
# [1] 5675.977
Petr Savick
calculate the ratios of the
summands
to the largest of them or to a close approximation of the largest.
Petr Savicky.
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l=4)
a[, ] <- v[row(a) + col(a) - 1]
a
[,1] [,2] [,3] [,4]
[1,] "a" "b" "c" "d"
[2,] "b" "c" "d" "a"
[3,] "c" "d" "a" "b"
[4,] "d" "a" "b" "c"
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
dimensions represent the columns.
This does not reorder the elements of the array, only changes the
dimension information.
A1 <- matrix(A, nrow=3*3, ncol=2)
Compute the means
B1 <- c(B)
C1 <- rowsum(A1, group=B1)/c(table(B1))
Put the means to the required positions
for(a in
f the residues in the columns are periodic, like in the
example above, then also the vector x is periodic. If this
really occurs in the application, generating the new column
using this fact may also be useful. The "length.out" argument
of rep() can be used.
y <- rep(rowSums(a[
Sums(a %% 13 == 0) != 0)
b <- cbind(a, x)
or
x <- (rowSums(a %% 13 == 0) != 0) + 0
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"Resp", "other")
Stat <- Vals[c(1, 3, 1, 2, 3, 2, 1, 3, 2)]
ind <- which(Stat %in% c("MagDwn", "Resp"))
Reduced <- Stat[ind]
ind[which(diff(Reduced == "Resp") == 1) + 1]
# [1] 4 9
The positions of the corresponding MagDwn are
ind[w
NCE MUCH OF THE TIME THIS WILL NOT BE THE CASE, THIS SCRIPT DEALS WITH
> THE REMAINDER...
> for(i in 1:length(positions)){
If "positions" may, in some situations, be of length 0, then it
is better to use
for(i in seq(along=positions))
or
for(i
when needed. There is an R Wiki
page with some tips concerning factors at
http://rwiki.sciviews.org/doku.php?id=tips:data-factors:factors
Petr Savicky.
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PLEASE do read the posting
uot;id", "diagnosis"), class = "data.frame", row.names = c(NA,
-10L))
tab <- table(df$id, df$diag)
Then, for example, the data cases for "2. Patients with ah but no ihd"
may be obtained
sel <- tab[, "ah"] != 0 &
))
# AUC diff
# [1,] 0.73920
The difference is not always exactly zero, but is at the level
of the machine rounding error.
Petr Savicky.
>
>
> On Thu, Jan 20, 2011 at 3:04 PM, He, Yulei wrote:
>
> > Hi, there.
> >
> > Suppose I already have sensitivit
tisfy some property, which allows to find a unique solution, then
the algorithm depends on what is known about the original matrices.
Hope this helps.
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PLEA
619691, 0.018334730, -0.009747171)
x <- numeric(length(y))
for (i in 1:length(y)) {
x[i] <- ifelse(i==1, 1*(1+y[i]), (1+y[i])*x[i-1])
}
z <- 1*cumprod(1 + y)
max(abs(x - z))
# [1] 1.818989e-12
Petr Savicky.
__
R
, digits=2)
yields the correct comparison
round(t, 2) == round(tt, 2)
# [1] TRUE
athough 0.2 is also not exactly representable. Both sides are rounded
to the same representable number.
See also
http://rwiki.sciviews.org/doku.php?id=misc:r_accuracy
for other examples.
Petr Savicky.
___
nfluence the next iteration of the loop.
For example, the following loop always makes m*n repetitions, although
using the same variable in nested loops is definitely not suggested.
m <- 3
n <- 5
for (i in seq(length=m)) {
for (i in seq(length=n)) {
cat("*")
}
, if k is 0, since 1:0 is a vector of length 2.
If k may be 0, then it is better to use
for (n in seq(length=k))
since seq(length=0) has length 0.
Hope this helps.
Petr Savicky.
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On Tue, Jan 25, 2011 at 09:05:03AM +0100, Petr Savicky wrote:
[...]
> to foreach loop in Perl. If v is a vector, then
>
> for (n in v)
>
> first creates the vector v and then always performs length(v) iterations.
I forgot that ‘break’ may stop the loop. See ?"for" fo
(-1, 1, length=5)
xints <- data.frame(
x1=cut(x[, 1], breaks=breaks),
x2=cut(x[, 2], breaks=breaks),
x3=cut(x[, 3], breaks=breaks))
table(complete.cases(xints))
xtabs(~ ., xints)
Hope this helps.
Petr Savicky.
__
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nd faster, as others already
mentioned. Using replicate(), i obtained on my computer a speed up by a
factor between 5 and 7 for k <= 20 and there is a remarkable speed up
also for larger k. The function seq.int() is more general than the other
two. In particular, it can ge
e, then also the data frame will not
contain the unwanted level.
a<-matrix(c("ww","ww","xx","yy","ww","yy","xx","yy","NA"),
ncol=3, byrow=TRUE)
a[a[,3]=="NA",3]<-NA
d<-data.fr
A <- matrix(nrow=n+1, ncol=n)
for(i in 1:n){
A[i, seq.int(along=x)] <- x
x <- diff(x)
}
M <- matrix(A, nrow=n, ncol=n)
M[upper.tri(M)] <- t(M)[upper.tri(M)]
return(M)
}
Reorganizing an (n+1)
root(f, c(0, 10), x=1, y=3)$root
[1] 5
Hope this helps.
Petr Savicky.
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide com
logical result, then use
all(A == B)
for exact equality and
all(abs(A - B) <= eps)
for approximate equality of all entries.
See also ?all.equal, which uses the relative error, not absolute
difference.
Hope this helps.
Petr Savicky.
__
a b
> 1 1 1 3 3 5 6
>
> So the desired result is:
>
> id name
> 8 e
> 17f
> 20g
> 4c
> 11c
> 19c
> 6d
> 9d
> 10d
> 1a
> 5a
> 12a
> 14a
> 15a
> 2b
> 3b
>
colMeans() and variance using
var(). See also ?var.
Hope this helps.
Petr Savicky.
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and
f a vector of
> strings without looping -- I have to think not.
Try the following
x <- c("this is a string", "this is a numeric")
reassemble <- function(x, ind) paste(x[ind], collapse=" ")
vapply(strsplit(x," "), reassemble, "chara
Col2 = with(x, ave(H, paste(Site, Prof),
FUN=function(y)y-min(y
Petr Savicky.
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
mory, but it is more
efficient than read.table(), since it does no parsing of
the file as a whole.
x <- readLines("file")
strsplit(x[length(x)], " +")[[1]][3]
Hope this helps.
Petr Savicky.
__
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On Wed, May 04, 2011 at 08:52:07AM -0700, William Dunlap wrote:
> > -Original Message-
> > From: r-help-boun...@r-project.org
> > [mailto:r-help-boun...@r-project.org] On Behalf Of Petr Savicky
> > Sent: Wednesday, May 04, 2011 12:51 AM
> > To: r-help@r-p
t;train" is a subset of "master".
master <- data.frame(ID=2001:2011)
train <- data.frame(ID=2004:2006)
valid <- master[! (master[, 1] %in% train[ ,1]), , drop=FALSE]
Hope this helps.
Petr Savicky.
__
R-help@r-project.org m
[1] 6 6 6 6 6 6 6 7 7 7 7 7 7 6 6 6 6 6 6 6 7 7 7 7 7 7
The input vector may be obtained using c() from a matrix. The
output vector may be reformatted using matrix(). However, for
a matrix solution, a more precise description of the question
is needed.
Hope this helps.
Petr Savicky.
_
mes
more elements than the input, since 1040/160/1 = 6.5. This
corresponds to the understanding that odd elements should repeat 7
times and even elements 6 times. However, it is not clear, what
the dimension of the output matrix should be.
Hope this helps.
Petr Savicky.
__
the standard arithmetic and includes explicit tests like
abs(x) < 1e-20.
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and pr
the second datset.
I am not sure, whether you can consider also other types
of subsets to increase the number of different samples.
For example, the following selects 16 rows at random
a[sort(sample(1:32, 16)), ]
Hope this helps.
Petr Savicky.
__
56
Hi.
Try the following
a <- matrix(c(0, 0, 2, 0, 4, 0, 1, 8, 0, 56), ncol=2)
a[rowSums(a != 0) != 0, ]
Hope this helps.
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PLEASE do read the p
the desired covariance matrices would
> be appreciated.
Hello.
Let me suggest the following procedure.
1. Generate a symmetric matrix A with the desired distribution of the
non-diagonal elements and with zeros on the diagonal.
2. Compute the smallest eigenvalu
On Fri, Jun 03, 2011 at 01:54:33PM -0700, Ned Dochtermann wrote:
> Petr,
> This is the code I used for your suggestion:
>
> k<-6;kk<-(k*(k-1))/2
> x<-matrix(0,5000,kk)
> for(i in 1:5000){
> A.1<-matrix(0,k,k)
> rs<-runif(kk,min=-1,max=1)
> A.1[lower.tri(A.1)]<-r
, 0.60330865, 0.61832829)
x <- matrix(rnorm(36), nrow=6, ncol=6) %*% diag(w)
x <- x/sqrt(rowSums(x^2))
a <- x %*% t(x)
Hope this helps.
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PLEASE do
ot;identification",
which contains seeds, a critical function used in the simulation
and also a package and R version. The last two things may be obtained
for example as
R.version.string
library(help=mvtnorm)[[3]][[1]][3]
Up to now, i did not really needed these
e negative eigenvalues. For example,
if all components in sigma[1:20] are 4, which is in
the range used for sigma, then we have a matrix, whose
diagonal elements are 4 and nondiagonal elements are
0.3*4^2 = 4.8 > 4. This matrix has negative eigenvalues,
so it is not a covarian
jumping ahead in the sequence
without generating all intermediate numbers, but i do
not know about an efficient available implementation.
Petr Savicky.
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PLEASE do r
squared test for given probabilities
data: x
X-squared = 25, df = 4, p-value = 5.031e-05
It is not clear, whether this is suitable for your application.
If you generate the values in a different way, then another
test may be needed. Can you specify more detail on how the
numbers are generated
er of species, which are contained
in a sum of a random selection of k rows may be computed easily,
since we can consider the columns (species) individually and
for each column, the probability to get a nonzero sum may be
computed without actually constructing all the subsets.
If you need a parame
On Thu, Jan 27, 2011 at 05:30:15PM +0100, Petr Savicky wrote:
> On Thu, Jan 27, 2011 at 11:30:37AM +0100, Serena Corezzola wrote:
> > Hello everybody!
> >
> >
> >
> > I?m trying to define the optimal number of surveys to detect the highest
> > numbe
0.04101562 0.03222656 0.0625 -0.05468750 0.04687500
matrix(apply(expand.grid(x, y), 1, FUN=FUN), nrow=length(x), ncol=length(y))
[,1] [,2]
[1,] 0.05468750 0.0625000
[2,] -0.04101562 -0.0546875
[3,] 0.03222656 0.0468750
Hope this helps.
Petr Savicky.
__
2 chr1 70 80
3 chr2 90 110
4 chr2 130 140
5 chr3 190 230
The column tb1$index contains for each row the index of the interval [V2, V3]
in tb2, which contains the values V2, V3 from tb1. For example, line
10 chr2 130 131 1 4
of tb1 contains index 4, because the interval
4 chr
tb2. If you need to test a nonempty
intersection, it is slightly more complicated, but not much.
Are you from the same research team?
Petr Savicky.
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PLEASE do read t
227.7373
If r1, r2 should be the names of the columns, then use named
arguments in the call of the function data.frame()
rr<-data.frame(r1=rnorm(1000,10,5),r2=rnorm(1000,220,5))
rr[1:3, ]
r1 r2
1 12.3274362 224.7632
2 13.1347464 214.3805
3 0.7495177 219.6179
n <- 33
y <- cumsum(runif(n))
plot(y)
# restarting indices
ind <- 1:n - (1:n) %% 10
ind[ind == 0] <- 1
plot(y - y[ind])
Is this close to what you want?
If not, then i suggest to send the loop solution as a part of the description.
Petr Savicky.
manage to find the
> answer.
Hi.
The reason is that a number, which has a finite expansion as a decimal
number, need not have a finite expansion as a binary number. Besides FAQ 7.31,
see also
http://rwiki.sciviews.org/doku.php?id=misc:r_accuracy
for further examples and some hints.
Petr Savicky
density function over that region. If the
region is a single point, then this integral is zero.
Functions related to a multivariate normal distribution may
be computed using package
http://cran.at.r-project.org/web/packages/mvtnorm/index.html
Petr Savicky.
_
on of the points into a finite
number of regions and keeps the information needed for any of the
tasks, which you mention.
Hope this helps.
Petr Savicky.
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PLEASE do read the
1 0
3 0 2 0 0
Does this approach work for your data?
Petr Savicky.
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
]
rownames(expand) <- NULL
expand
animal N
1 a 2
2 a 2
3 b 1
4 c 3
5 c 3
6 c 3
Hope this helps.
Petr Savicky.
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PLEA
e numbers, consider
also the function lfactorial(), which computes the logarithm
in the standard numeric type.
lfactorial(1000)
[1] 5912.128
Hope this helps.
Petr Savicky.
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eal variables, which has
a measurable difference of expected value on Mersenne Twister numbers and
truly random ones, then this is likely to be an interesting mathematical
discovery.
Petr Savicky.
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https://stat.ethz.ch/mai
t;- rbind(x %o% y, zero)
k <- m + n - 1
b <- matrix(c(a)[1:(n*k)], nrow=k, ncol=n)
rowSums(b)
}
Testing this on computing the product of the polynomials (1+t)^4 (1+t)^3
yields
x <- choose(4, 0:4)
y <- choose(3, 0:3)
convolution(x, y)
[1] 1 7 21
391843
5.803329 5.803329
6.289873 6.28988
6.876084 6.876084
7.5992 7.599201
8.518665 8.518683
9.736212 9.736212
11.44329 11.44328
14.05345 14.05345
18.68155 18.68155
29.97659 29.9766
219.3156 219.3155
Hope this helps.
Petr Savicky.
This also suggests that the same distribution on the random assignments
is obtained, if area is created already sorted and only the second
column of "ass" is random
ass <- as.data.frame(cbind("area"=1:7, "strategy"=sample(1:7, 7)))
Whether creating only thi
;- data.frame(V1=1:3, V2="CC", V3=seq(3.1, 3.3, by=0.1))
for (j in 1:3) {
# here any command using tab[[j]] may be used
# using print() for simplicity
print(tab[[j]])
}
See chapter 6 Lists and data frames of R-intro.pdf available at
http://cran.r-project.org/
.
>
> How do I do it in R ?
Hi.
Let me use a small example
df <- data.frame(a=1:7, b=11:17, row.names=letters[1:7])
df
a b
a 1 11
b 2 12
c 3 13
d 4 14
e 5 15
f 6 16
g 7 17
Is the following what you are asking for in terms of this small example?
df[c("
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