Note that
c(list(1,2,3), list(4, 5,6), list(7,8,9, 10))
is identical to
list(1, 2, 3, 4, 5, 6, 7, 8, 9, 10)
You want either
list(list(1,2,3), list(4,5,6), list(7,8,9,10)) # list of 3 lists of numeric
scalars
or
list(c(1,2,3), c(4,5,6), c(7,8,9,10)) # list of 3 numeric vectors
In any ca
Hi,
You could also try:
df1<-data.frame(name=c("a","b","c"),type=c(1,2,3),rtn=as.array(list(1:3,4:6,7:10)))
A.K.
- Original Message -
From: Kevin Zembower
To: r-help@r-project.org
Cc:
Sent: Friday, March 8, 2013 7:49 PM
Subject: [R] data.frame with variable-length list
Hello,
I'm
HI,
Try this:
df<-data.frame(name=c("a","b","c"),type=c(1,2,3),rtn=do.call(cbind,list(list(1:3,4:6,7:10
str(df)
#'data.frame': 3 obs. of 3 variables:
# $ name: Factor w/ 3 levels "a","b","c": 1 2 3
# $ type: num 1 2 3
# $ rtn :List of 3
# ..$ : int 1 2 3
#..$ : int 4 5 6
.#.$ : i
Hello,
I'm trying to create a data frame with three columns, one of which is a
variable-length list. I tried:
df <- data.frame(name = c("a", "b", "c"),
type=c(1, 2, 3),
rtn = c(list(1,2,3), list(4, 5,6), list(7,8,9, 10)
)
)
This would be useful,
Try using the combination
plot.new() ; par(new=TRUE)
to advance to the next position in the layout before querying par("pin").
Be sure to actually plot something after the par(new=TRUE).
E.g.,
> layout(matrix(c(1,2,3,4), nrow=2), width=c(1,3), height=c(1,3))
> plot.new() ; par(new=TRUE)
> par("
Hi R users,
I find par("pin") is kind of confusing (or maybe just me?). The manual said
it will give " The current plot dimensions, (width,height), in inches."
The word "current" is the key here. I thought it would give the dimensions
of the to-be plot, but it actually gives the dimension of the
Hi
I found the source of the problem. The R would not let me update the
"survival" and "KernSmooth" package at the R/library folder. I manually
delete them and then download the latest version of them to proceed with the
xcms installation. I work on a window 7 computer and R 2.15.3/2/1 (all
versi
Hello,
I had thought of something like that, but I'm not sure if the match must
be exact. If not, grep seems better. More complicated and slower but
more flexible.
Rui Barradas
Em 08-03-2013 21:32, arun escreveu:
Hi,
You can also try:
res2<-rowSums(x==word)
res1<-sapply(where,length)
On 03/08/2013 06:01 AM, Jan van der Laan wrote:
You could use the fact that scan reads the data rowwise, and the fact that
arrays are stored columnwise:
# generate a small example dataset
exampl <- array(letters[1:25], dim=c(5,5))
write.table(exampl, file="example.dat", row.names=FALSE. col.nam
Hi,
You can also try:
res2<-rowSums(x==word)
res1<-sapply(where,length)
res1[]<- sapply(res1,as.numeric)
identical(res1,res2)
#[1] TRUE
A.K.
- Original Message -
From: Rui Barradas
To: Sudip Chatterjee
Cc: r-help@r-project.org
Sent: Friday, March 8, 2013 4:26 PM
Subject: Re: [R]
This is even simpler
> aaa <- matrix(c("aa", "bb", "aa", "aa", "cc", "ee"), 2, 3,
dimnames=list(LETTERS[1:2], letters[3:5]))
> apply(aaa == "aa", 1, sum)
A B
2 1
On Fri, Mar 8, 2013 at 4:16 PM, Richard M. Heiberger wrote:
> > aaa <- matrix(c("aa", "bb", "aa", "aa", "cc", "ee"), 2, 3,
> dimnames
Hello,
I'm not sure I understand, but see if the following is an example of
counting occurences of a word in each row.
set.seed(1855)
x <- matrix(sample(LETTERS[1:5], 400, replace = TRUE), ncol = 4)
word <- "A"
where <- apply(x, 1, function(.x) grep(word, .x))
sapply(where, length) # count t
> aaa <- matrix(c("aa", "bb", "aa", "aa", "cc", "ee"), 2, 3,
dimnames=list(LETTERS[1:2], letters[3:5]))
> aaa
cde
A "aa" "aa" "cc"
B "bb" "aa" "ee"
> apply(aaa, 1, function(x, word) sum(x==word), word="aa")
A B
2 1
>
On Fri, Mar 8, 2013 at 11:04 AM, Sudip Chatterjee wrote:
> Hi All,
>
>
On Mar 8, 2013, at 9:05 AM, Jing Lu wrote:
> Hi everyone,
>
> I hope this question is beyond "read the manual". My task is simple, just
> to plot the following, but the plot in the middle should be a
> filled.contour plot:
> http://gallery.r-enthusiasts.com/graph/Scatterplot_with_marginal_histog
Hi,
Try this:
z1[seq(which(time(z1)==as.POSIXct("01.01.2012 02:00:00",format="%d.%m.%Y
%H:%M:%S")),which(time(z1)==as.POSIXct("01.01.2013 03:00:00",format="%d.%m.%Y
%H:%M:%S"))),]
# Value1 Value2
#2012-01-01 02:00:00 4.8 5.3
#2012-02-01 02:30:00 4.9 5.2
#2012-08
Hello,
I was wondering if there is a way to pass a custom command line argument to
slave processes using the parallel package. I need to do this because I have
some logic in my .Rprofile that checks what arguments it was called with using
commandArgs() and performs certain actions only if it wa
... the row.names will not interfere with any merging operation ...
Row names do not interfere with merge, but they cause other problems. In the
example, I want to test whether rows have the same entries (in some or all
columns). identical fails because of the row names, and all( == ) can fail
if
Hi everyone,
I hope this question is beyond "read the manual". My task is simple, just
to plot the following, but the plot in the middle should be a
filled.contour plot:
http://gallery.r-enthusiasts.com/graph/Scatterplot_with_marginal_histograms_78
Background: I prefer filled.contour rather than
Hi All,
I am wondering if there is any examples where you can count your
interested "word" in each row. For an example if you have data with *'ID*'
and '*write-up*' for 100 rows, how would I calculate the word frequency for
each row ?
Thank you for all your time.
[[alternative HTML v
Funnily, the bivpois package is no longer available too...
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__
R-help@
Hi,
Using your code:
sub("[.]$",".1",sim.code)
#[1] "1.1234.1a.1" "1.1234.1a.2" "1.3245.2c.5" "4.6743.3c.1" "4.3254.6b.4"
#[6] "3.5463.2a.1"
A.K.
- Original Message -
From: chris201
To: r-help@r-project.org
Cc:
Sent: Friday, March 8, 2013 5:23 AM
Subject: [R] Substitute value
Hi,
I h
Yes, it works, but I wonder if I can make the y title horizontal just setting
any "hidden" parameter...
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_
Hi,
I have the following bwplots, and want to add a label with the mean value at
each dote.
Can you give me suggestions?
Also, if I wanted to add another bwplot below the first, can I use the
par(mfrow=c(2,1)) function?
This is my code:
Code
#Data
x<-c('Small','Large','Large','Medium','Medium
Hi Group,
I'm trying to build a model to predict a product's sale price. I'm
researching the dlm package. Looks like I should use dlmModPoly, dlmMLE,
dlmFilter, dlmSmooth, and finally dlmForecast. I'm looking at the Nile
River example and I have a few questions:
1.
If I only want to predic
Hi,
try this:
sim.df<-data.frame(sim.code,sim.val,stringsAsFactors=FALSE)
sim.df[,1][-grep("\\d+$",sim.df[,1])]<-
paste(sim.df[,1][-grep("\\d+$",sim.df[,1])] , 1,sep="")
sim.df
# sim.code sim.val
#1 1.1234.1a.1 4
#2 1.1234.1a.2 5
#3 1.3245.2c.5 3
#4 4.6743.3c.1
Hi Jakob,
dat1<-read.table(text="
TIME, Value1, Value2
01.08.2011 02:30:00, 4.4, 4.7
01.09.2011 03:00:00, 4.2, 4.3
01.11.2011 01:00:00, 3.5, 4.3
01.12.2011 01:40:00, 3.4, 4.5
01.01.2012 02:00:00, 4.8, 5.3
01.02.2012 02:30:00, 4.9, 5.2
01.08.2012 02:30:00, 4.1, 4.7
01.12.2012 03:00:00, 4.
On Mar 8, 2013, at 10:59 AM, David Winsemius wrote:
>
> On Mar 8, 2013, at 9:31 AM, David Winsemius wrote:
>
>>
>> On Mar 8, 2013, at 6:01 AM, Jan van der Laan wrote:
>>
>>>
>>> You could use the fact that scan reads the data rowwise, and the fact that
>>> arrays are stored columnwise:
>>>
On Mar 8, 2013, at 9:31 AM, David Winsemius wrote:
>
> On Mar 8, 2013, at 6:01 AM, Jan van der Laan wrote:
>
>>
>> You could use the fact that scan reads the data rowwise, and the fact that
>> arrays are stored columnwise:
>>
>> # generate a small example dataset
>> exampl <- array(letters[1
On 3/8/2013 12:09 PM, John Kane wrote:
donner<-read.csv("http://www.ling.upenn.edu/~joseff/data/donner.csv";)
ggplot(donner, aes(AGE, NFATE, colour = GENDER))+
geom_point(position = position_jitter(height = 0.02, width = 0)) +
stat_smooth(method = "glm", family = binomial, for
On Mar 8, 2013, at 6:01 AM, Jan van der Laan wrote:
>
> You could use the fact that scan reads the data rowwise, and the fact that
> arrays are stored columnwise:
>
> # generate a small example dataset
> exampl <- array(letters[1:25], dim=c(5,5))
> write.table(exampl, file="example.dat", row.n
Is this roughly what you want? Shamelessly stolen , I mean , adapted from
http://docs.ggplot2.org/0.9.3/stat_smooth.html
donner<-read.csv("http://www.ling.upenn.edu/~joseff/data/donner.csv";)
ggplot(donner, aes(AGE, NFATE, colour = GENDER))+
geom_point(position = position_jitter(height
In the example below, from
http://www.ling.upenn.edu/~joseff/rstudy/summer2010_ggplot2_intro.html
I'd like to make (a) the fitted line thicker and (b) change the
background fill color for the confidence
envelope around each fitted line to a low-alpha transparent version of
the same color used
fo
Okay, I think I see what you want. I had though of that earlier and then
decided that I was wrong. The png got through nicely. Data set dd slightly
revised as we don't need that dummy x variable.
dd <- structure(list(abnr2 = c(11425, 11425, 11555, 11888),
time = c(2, 1, 1, 2), cat =
Hello together
There is another try as a png file. Hope you can see it now, what i want to do
with my bar chart.
Your example with ggplot2 works, but it wont help to convert my data like this
one:
1 2 3 4
abnr2
Hi
some brute force
aggregate(test$Value, list(rep(1:61, each=7)[1:422]), mean)
or
aggregate(test$Value, list(findInterval(1:nrow(test), seq(1,422,by=7)),
test$Data), mean)
gives you aggregated values for each week.
or
lll <- split(test, test$Data)
lapply(lll, function(x) aggregate(x, list(fin
Thanks for your replies !
The lapply hack does the job in my context, so I'll stick to it (and
actually in any case where I expect variable length results).
About your quote from ?apply : I read it some time ago actually - but with
my recent use in "variable length returning FUN", I got fooled...
Hello,
Something like this?
sp <- split(test, test$Data)
res <- do.call(rbind, lapply(sp, function(x){
Week <- (seq_len(nrow(x)) %/% 7) + 1
aggregate(Value ~ Data + Week, data = x, FUN = mean)}))
rownames(res) <- seq_len(nrow(res))
res
Hope this helps,
Rui Barradas
Em 08-03-
HI Irucka,
The situation is slightly different here. I was under the assumption that the
list elements sometimes had one two columns or three columns. Here, all the
columns are present, but some with entire rows of missing values. Also, there
was a mistake in the code when I updated the code
Hi
Some issues:
1 do not use HTML mail
2 what OS
3 which R version
4 when and how it stopped working - exact error message (if any)
5 did you ask xcms maintainer for help? (it is not in standard CRAN packages)
6 try to send reproducible example for others to enable them to test the problem
Regar
Dear all, I have a big data matrix and I want to convert those data into
weekly basis which means 7 days needs to be avaraged and aggregate a single
value
> dput(test)
structure(list(locid = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L,
On Fri, Mar 8, 2013 at 5:23 AM, chris201 wrote:
> Hi,
> I have a large data frame and within this there is one column which contains
> individual codes (eg. 1.1234.2a.2). I am splitting these codes into their 4
> components using strsplit (eg. "1", "1234", "2a", "2"). However there are
> some in
Hi
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of chris201
> Sent: Friday, March 08, 2013 11:23 AM
> To: r-help@r-project.org
> Subject: [R] Substitute value
>
> Hi,
> I have a large data frame and within this there is one c
https://github.com/hadley/devtools/wiki/Reproducibility
Is this what your matrix looks like?
mat1 <- structure(c(11425, 11425, 11555, 11888, 2, 1, 1, 2, 1, 2, 1,
2), .Dim = c(4L, 3L), .Dimnames = list(NULL, c("abnr2", "time",
"cat")))
It is good practice to use dput() to supply sample data.
I
I am trying to create a very big matrix with big.matrix from package
big.memory in order to apply dbscan afterwards.
I have two problems:
I need to create 4 matrices with 12 rows X 12 columns. I have
tested various R packages for big data (particularly bigmemory and
ff). Since ff cannot c
Perhaps you could process this with a unix/Linux utility "Awk", before reading
the file into R.
-Sohail
From: r-help-boun...@r-project.org [r-help-boun...@r-project.org] On Behalf Of
peter dalgaard [pda...@gmail.com]
Sent: Friday, March 08, 2013 5:08 AM
T
Hi,
I have a large data frame and within this there is one column which contains
individual codes (eg. 1.1234.2a.2). I am splitting these codes into their 4
components using strsplit (eg. "1", "1234", "2a", "2"). However there are
some individual codes which do not have a last component (eg. 2.43
Hi
I have sent couple of days ago an email describing an issue in regards to R
software when I use the xcms package. I haven't got any response whether my
post has been made public to receive any help for other users. Is my email
again treated as spam or considered as inappropriate to be publis
if i can sort my list as follows, i can create the bar chart.
But how can i sort my list from:
[,1] [,2] [,3]
[,4]
abnr2 11425 11425 11555 11888
TIME 21 1
Hello together,
perhabs anyone of you, has an ideal, how i can do this:
I have a matrix, like this one:
[,1] [,2] [,3]
[,4]
abnr2 11425 11425 11555 11888
TIME 21
Dear All
I am kind of stuck up with a code a part of which seems to be causing a
problem, or at least I think so. May be the community can help me. Its
simple but I suppose I am missing something.
I generate a data matrix X, say of order n*p, where n represents
independent row-vectors and p corr
It works for me. Or at least I got something to download. What is another
matter. I am just trying to learn how to do this and really don't have any good
advice.
Perhaps the site was down or overloaded? Any firewalls to worry about at your
end?
John Kane
Kingston ON Canada
> -Original M
-- begin included message --
I have a competing risk data where a patient may die from either AIDS or
Cancer. I want to compare the cox model for each of the event of interest
with a competing risk model. In the competing risk model the cumulative
incidence function is used directly.
-end inclusi
On 3/6/2013 11:50 PM, Charles Determan Jr wrote:
Generic question... I am familiar with generic power calculations in R,
however a lot of the data I primarily work with is multivariate. Is there
any package/function that you would recommend to conduct such power
analysis? Any recommendations wo
You could use the fact that scan reads the data rowwise, and the fact
that arrays are stored columnwise:
# generate a small example dataset
exampl <- array(letters[1:25], dim=c(5,5))
write.table(exampl, file="example.dat", row.names=FALSE. col.names=FALSE,
sep="\t", quote=FALSE)
# and re
Hi,
You can try:
mat1<- do.call(rbind,x)
lapply(seq_len(ncol(mat1)),function(i) mat1[,i])
#[[1]]
#[1] 12.10 3.44
#[[2]]
#[1] 0.1 3.0
#[[3]]
#[1] 12.0 33.1
#[[4]]
#[1] 1.1 23.0
A.K.
- Original Message -
From: ishi soichi
To: PIKAL Petr
Cc: r-help
Sent: Friday, March 8, 2013 5:06 A
Le lundi 04 mars 2013 à 14:43 -0600, Joanna Papakonstantinou a écrit :
>
> Thank you.
> I actually ended up using:
> > CrossTable(mdt)
>
>
>Cell Contents
> |-|
> | N |
> | Chi-square contribution |
> | N / Row Total |
> | N
On 08/03/2013 11:37, Thomas wrote:
I realise this isn't exactly an R query, but does anyone know what ports
I need to open in order to get the install package function working?
It's blocked by our University firewall.
You mean install.packages() ?
It depends what you are connecting to. Most C
I realise this isn't exactly an R query, but does anyone know what
ports I need to open in order to get the install package function
working? It's blocked by our University firewall.
I have been asked for the following info and I don't know where to
find it or how to get it. I guess my comp
One option would be
x <- list(c(12.1, 0.1, 12, 1.1), c(3.44, 3, 33.1, 23))
do.call(c, apply(do.call(rbind, x), 2, list))
HTH,
Jorge.-
On Fri, Mar 8, 2013 at 9:06 PM, ishi soichi wrote:
> Thanks. The result should be a list of lists like...
>
> > x
> [[1]]
> [1] 12.10 3.44
>
> [[2]]
> [1] 0.
Thanks! Certainly they do help!
ishida
2013/3/8 D. Rizopoulos
> two possibilities are:
>
> lis <- list(c(12.1,0.1,12.0,1.1), c(3.44,3.00,33.10,23.00))
>
> # 1st
> m <- do.call(rbind, lis)
> split(m, col(m))
>
> # 2nd
> lapply(seq_along(lis[[1]]),
> function (i) sapply(lis, "[", i))
>
>
>
two possibilities are:
lis <- list(c(12.1,0.1,12.0,1.1), c(3.44,3.00,33.10,23.00))
# 1st
m <- do.call(rbind, lis)
split(m, col(m))
# 2nd
lapply(seq_along(lis[[1]]),
function (i) sapply(lis, "[", i))
I hope it helps.
Best,
Dimitris
On 3/8/2013 11:06 AM, ishi soichi wrote:
> Thanks. The
On Mar 7, 2013, at 01:18 , Yao He wrote:
> Dear all:
>
> I have a big data file of 6 columns and 6 rows like that:
>
> AA AC AA AA ...AT
> CC CC CT CT...TC
> ..
> .
>
> I want to transpose it and the output is a new like that
> AA
Thanks. The result should be a list of lists like...
> x
[[1]]
[1] 12.10 3.44
[[2]]
[1] 0.1 3.0
[[3]]
[1] 12.0 33.1
[[4]]
[1] 1.1 23.0
lapply(x, t) doesn't do the job, I think.
ishida
2013/3/8 PIKAL Petr
> Hi
>
> > -Original Message-
> > From: r-help-boun...@r-project.org [mailt
Hi
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of ishi soichi
> Sent: Friday, March 08, 2013 10:50 AM
> To: r-help
> Subject: [R] transpose lists
>
> Can you think of a function that transposes a list like
What shall be the
Can you think of a function that transposes a list like
> x
[[1]]
[1] 12.1 0.1 12.0 1.1
[[2]]
[1] 3.44 3.00 33.10 23.00
?
ishida
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listi
This is nice fodder for 'The R Inferno' -- thanks.
As Milan said, 'which' will suffice as the function.
Here is a specialized function that only returns a
list and is only implemented to work with matrices.
It should solve your current dilemma.
applyL <-
function (X, MARGIN, FUN, ...)
{
Le vendredi 08 mars 2013 à 09:29 +0100, Pierrick Bruneau a écrit :
> Hello everyone,
>
> Considering the following code sample :
>
>
> indexes <- function(vec) {
> vec <- which(vec==TRUE)
> return(vec)
> }
This is essentially which(), what did you write such a convoluted
function to
Hello everyone,
Considering the following code sample :
indexes <- function(vec) {
vec <- which(vec==TRUE)
return(vec)
}
mat <- matrix(FALSE, nrow=10, ncol=10)
mat[1,3] <- mat[3,1] <- TRUE
Issuing apply(mat, 1, indexes) returns a 10-cell list, as expected.
Now if I do:
ma
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