In '?rep' find out about the 'each' argument.
Also there is the function 'gl' which creates a factor and offers a shorter
syntax for your problem.
If n equals 5 use one of:
rep(seq(5), each = 4)
gl(5,4)
On 19 April 2015 at 15:44, John Sorkin wrote:
> Windows 7 64-bit
> R 3.1.3
> RStudio 0.98.1
Hi Dimitri,
str_replace_all is not in the base libraries, you could use 'gsub' as well,
for example:
a = "What a nice day today! - Story of happiness: Part 2."
b = "What a nice day today: Story of happiness (Part 2)"
sa = gsub("[^A-Za-z0-9]", "", a)
sb = gsub("[^A-Za-z0-9]", "", b)
a==b
# [1] FAL
Hi Karim,
you should learn ?Map to iterate along the list and supply mutliple list
arguments (there is also parallel:::mcMap for multicore).
The magic of the color code generation you figure out yourself, I guess...
Here 'i' intends to be the value, 'n' the name, e.g.
# returns color by charact
Another solution:
CaseID <- c("1015285", "1005317", "1012281", "1015285", "1015285", "1007183",
"1008833", "1015315", "1015322", "1015285")
Primary.Viol.Type <- c("AS.Age", "HS.Hours", "HS.Hours", "HS.Hours",
"RK.Records_CL",
"OT.Overtime", "OT.Overtime", "OT.Overtime", "V.Poster_Other",
"V.Poster
you can also define 'na.rm' in sum() by 'NA state' of x (where x is
your vector holding the data):
sum(x, na.rm=!all(is.na(x)))
On 26 January 2015 at 13:45, Martin Maechler
wrote:
>> Jim Lemon
>> on Mon, 26 Jan 2015 11:21:03 +1100 writes:
>
> > Hi Allen, How about this:
>
>
Maybe this is due to the usage of rep() in ifelse():
f.rep <- function(ans){ans <- rep(ans,1);return(ans)}
f <- function(ans){return(ans)}
f(a <- 123) # no print here
f.rep(a <- 123) # prints:
# [1] 123
On 27 January 2015 at 11:54, Bert Gunter wrote:
> Huh??
>
>> ifelse(TRUE, a <- 2L, a <- 3L)
Make use of the plyr and reshape2 package (both on CRAN):
library(plyr)
d<-adply(ArrayDiseaseCor, 1:2)
# adply calls function identity by default
d<-melt(d)
d<-subset(d,value>.5)
head(d)
You will have to rename columns, or adjust arguments in melt/adply.
Note: use set.seed before sampling for rep
see inline
On 12 February 2015 at 09:10, Sven E. Templer wrote:
> Make use of the plyr and reshape2 package (both on CRAN):
>
I forgot:
library(reshape2)
> library(plyr)
> d<-adply(ArrayDiseaseCor, 1:2)
> # adply calls function identity by default
> d<-melt(d)
> d&l
Without (example) code it is hard to follow... use ?dput to present
some data (subset).
But if it is data.frames you are dealing with (for sure with read.csv,
but not so sure at all with raster maps), give this a try:
?merge
On 19 February 2015 at 17:44, Simon Tarr wrote:
> Hello everyone,
>
> I
Hi,
you didn't specify values in A, and you first wanted to compare bi
with Aij, but then also which bi is less/equal to zero.
For the first case, with
A <- matrix(0:3,2)
b <- seq(-1,5)
and a comparison function for bi less/equal to Aij like
f <- function (bi) {as.integer(bi<=A)}
you can iter
> Q3: any other recommendations?
You might be interested in the very easy to use R markdown, see:
http://rmarkdown.rstudio.com/
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://st
On 30 March 2015 at 16:47, Sarah Goslee wrote:
> colnames(e) <- paste0('pop',1:12)
>
> isn't a function and doesn't return anything.
>
But
function(e){colnames(e) <- paste0('pop', 1:2)}
is a function and it returns something (the last evaluated expression! -
here the paste0 return):
> mylist2 <
ct (the data.frames) with each rowSums
output.
So, use cbind within your first lapply.
p.s. Is it a standard convention to always copy the reply to the last
> person who responded?
>
I guess it depends on which answer you refer to.
>
> On Mon, Mar 30, 2015 at 10:56 AM, S
y not wisdom."
> Clifford Stoll
>
>
>
>
> On Mon, Mar 30, 2015 at 7:56 AM, Sven E. Templer
> wrote:
> > On 30 March 2015 at 16:47, Sarah Goslee wrote:
> >
> >> colnames(e) <- paste0('pop',1:12)
> >>
> >> isn't a fun
On 30 March 2015 at 17:50, Sarah Goslee wrote:
> On Mon, Mar 30, 2015 at 11:43 AM, Sven E. Templer
> wrote:
> >
> >
> > On 30 March 2015 at 17:31, Bert Gunter wrote:
> >>
> >> Sarah's statement is correct.
> >>
> >> So is yours.
If you don't mind an extra column, you could use something similar to:
data.frame(r=seq(8),foo=NA,bar=NA)
If you do, here is another approach (see function body):
empty.frame <- function (r = 1, n = 1, fill = NA_real_) {
data.frame(setNames(lapply(rep(fill, length(n)), rep, times=r), n))
}
emp
One way I know to do this is (in bash) to use a dummy variable and make the
comment a multiline character string:
dummy <- c("
This is my multiline
comment or code block.
")
or if printing does not disturb you, just use:
"
...
"
Use ' if you have " in the block.
Other workarounds are here, whic
see inline for another vectorized example.
On 25 September 2014 23:05, David L Carlson wrote:
> Another approach
>
> fun <- function(i, dat=x) {
> grp <- rep(1:(nrow(dat)/i), each=i)
> aggregate(dat[1:length(grp),]~grp, FUN=sum)
> }
>
> lapply(2:6, fun, dat=TT)
>
>
> ---
in ?which read about arr.ind
following jims assumption (column instead of row indices is what you
want) this also works:
m <- matrix(1:20,4)
unique(which(m>11, arr.ind = T)[,"col"])
On 27 September 2014 12:23, Jim Lemon wrote:
> On Fri, 26 Sep 2014 10:15:14 PM Fix Ace wrote:
>> Hello, there,
>>
Hello,
how can I open (by an R command) the index page in html mode, as obtained by:
options(browser="firefox") # or any other
options(help_type="html")
?help
# and then following the html reference on the page bottom named "Index"
In text mode I know library(help='utils') to open the utils pack
Dear Barry,
some thoughts:
1) e in your function status_fnc is a vector when applied on a matrix
like object, but you index it as a matrix (e[,i] should be e[i]).
2) You can simplify the if statement by using the function any
(replacing all the OR statements) on the vector, so use any(e=='Y')
her
follow instructions on
https://stat.ethz.ch/mailman/listinfo/r-help
at
"To unsubscribe from R-help, get a password reminder, or change your
subscription options enter your subscription email address: "
...
On 10 October 2014 16:16, Tasnuva Tabassum wrote:
> I want to get rid of this thread. what
use:
which(p<=.05)
this will not yield logical, but integer indices without NA
On 14 October 2014 11:51, Rainer M Krug wrote:
> Hi
>
> I want to evaluate NA and NaN to FALSE (for indexing) so I would like to
> have the result as indicated here:
>
> ,
> | > p <- c(1:10/100, NA, NaN)
> | > p
Prevent graphic menues with:
options(menu.graphics = FALSE)
or and define repositories:
options(repos = c(CRAN = "http://cran.r-project.org";))
On 14 October 2014 17:00, wrote:
> Subscribers,
>
> A version of R is installed in a virtual machine, which has complete
> internet access via the host.
;
[4] " Copyright (C) 2007 Free Software Foundation, Inc. <http://fsf.org/>"
[5] " Everyone is permitted to copy and distribute verbatim copies"
[6] " of this license document, but changing it is not allowed."
On 15 October 2014 10:51, wrote:
> On 2014-10
Hi.
You can't.
But using a second file where you first write your header and then
append the original file is a solution. ?cat and ?write.table with a
focus on the 'append' argument should help. you can then use ?unlink
to delete the original file and ?file.rename to rename the second, if
desired
> On Oct 22, 2014 12:33 AM, "Sven E. Templer" wrote:
>>
>> Hi.
>>
>> You can't.
>>
>> But using a second file where you first write your header and then
>> append the original file is a solution. ?cat and ?write.table with a
>> focus
With melt and rep you are close. If you combine them it works:
library(reshape)
# your data:
df1 <- data.frame(area=c(1,2),group1=c(2,3),group2=c(1,5),group3=c(4,0))
df2<-data.frame(person_id=seq(1:15),area=c(rep(1,7),rep(2,8)),group_num=c(1,1,2,3,3,3,3,1,1,1,2,2,2,2,2))
# first melt
d <- melt(df1
seems like a transpose, so use
?t
t(your.data.frame)
On 22 October 2014 11:34, Matthias Weber wrote:
> Hello together,
>
> i have a little problem. Maybe anyone can help me.
>
> I have a data. frame which look like this one:
> 1000 1001 10021003
> 15 6 12
# fixed formula part:
f <- dat ~ a0 * exp(-S*(x - 250)) + K
# convert to character
f <- as.character(f)
# component:
C <- "(p0*exp(-0.5*((x-p1)/p2)^2))"
# number of components (defined randomly):
n <- sample(1:3, 1)
C <- rep(C, n)
# collapse:
C <- paste(C, collapse = "+")
# combine
f <- paste(f[2],
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