A.K. Wonderful - thank you. The last set of code using sapply(),
seq_along(), and get() worked like a charm and was what I was missing.
Brian
Brian S. Cade, PhD
U. S. Geological Survey
Fort Collins Science Center
2150 Centre Ave., Bldg. C
Fort Collins, CO 80526-8818
email: ca...@usgs.gov
t
Hi,
If the two pairs of matrices are in a list:
set.seed(42)
lst1 <- lapply(1:11, function(x) matrix(sample(40, 20, replace=TRUE), 5,4))
names(lst1) <- paste0("gs", paste0("4.",seq(0,100,by=10)))
set.seed(585)
lst2 <- lapply(1:11, function(x) matrix(sample(40, 20, replace=TRUE), 5,4))
names
So I know I must be missing something simple and obvious for the following
data manipulation where I have (in this example) 11 pairs of matrices
(gs4.0 to gs4.100 and ps1.0 to ps1.100) from some population simulations
(all with same dimensions) where I want to get some summary statistics on
the pro
abline(lm(Response1~Predictor,data=Site),col=colours[as.numeric(Site[1,1
])])
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf Of Tom Wilding
Sent: Montag, 27. Mai 2013 12:40
To: r-help@r-project.org
Subject: [R] Indexing within by
; -Original Message-
> From: tom.wild...@sams.ac.uk
> Sent: Mon, 27 May 2013 10:39:58 +
> To: r-help@r-project.org
> Subject: [R] Indexing within by statement - different coloured lines in
> abline wanted..
>
> Dear R-list
>
> I'm trying to get
Dear R-list
I'm trying to get each regression line, plotted using abline, to be of a
different colour as the following code illustrates. I'm hoping there is a
simple indexing solution. Many thanks.
## code from here
colours=c("black","red","blue","green","pink")
Mean=500;Sd=10;NosSites=5;Xaxi
j.brae...@uvt.nl
> Sent: Thu, 7 Feb 2013 16:41:49 +
> To: r-help@r-project.org
> Subject: [R] indexing operation based upon a sequence
>
> Dear R-list,
>
>
> We stumbled upon some weird problem when performing a simple indexing
> operation.
> Below some example code t
R-FAQ 7.31
Berend
On 07-02-2013, at 17:41, "J. Braeken" wrote:
> Dear R-list,
>
>
> We stumbled upon some weird problem when performing a simple indexing
> operation.
> Below some example code to illustrate the issue
>
> #FAILS TO FIND .55 Oo
> huh1 = seq(.1,.7,.15);huh1
> # [1] 0.10 0.25
Dear R-list,
We stumbled upon some weird problem when performing a simple indexing operation.
Below some example code to illustrate the issue
#FAILS TO FIND .55 Oo
huh1 = seq(.1,.7,.15);huh1
# [1] 0.10 0.25 0.40 0.55 0.70
huh1 == .25
# [1] FALSE TRUE FALSE FALSE FALSE
huh1 == .55
# [1] FALSE FA
07
#[8] 0.45993107 0.39590760
A.K.
- Original Message -
From: Jeffrey Stratford
To: Rui Barradas
Cc: r-help@r-project.org
Sent: Wednesday, November 14, 2012 1:01 PM
Subject: Re: [R] indexing for Wilcoxon test (take 2)
Hi everyone,
I've been asked to run a number of Wilcoxon tests on some
Hi everyone,
I've been asked to run a number of Wilcoxon tests on some behavioral data
(below is a sample). They want me to compare each trial to the first
trial, considered the control trial. I know I can use brackets to index and
select, for example, trial 1 vs. trial 4 (not exactly sure how to
Hello,
Thanks for the data example.
Try the following.
sp <- split(dat, dat$trial)
control <- sp[[1]]
lapply(sp[-1], function(x) wilcox.test(control[["removed"]],
x[["removed"]]))
Hope this helps,
Rui Barradas
Em 14-11-2012 18:01, Jeffrey Stratford escreveu:
Hi everyone,
I've been asked t
Hello,
Use ?dput to post data, like this it's unreadable. For instance,
dput(Mydata) # paste the output of this in a post.
Hope this helps,
Rui Barradas
Em 14-11-2012 15:58, Jeffrey Stratford escreveu:
Hi everyone,
I've been asked to run a number of Wilcoxon tests on some behavioral data
(b
Hi everyone,
I've been asked to run a number of Wilcoxon tests on some behavioral data
(below is a sample). They want me to compare each trial to the first
trial, considered the control trial. I know I can use brackets to index and
select, for example, trial 1 vs. trial 4 (not exactly sure how to
ware
> wdunlap tibco.com
>
>
>> -Original Message-
>> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
>> Behalf
>> Of R. Michael Weylandt
>> Sent: Thursday, August 09, 2012 4:22 PM
>> To: arun
>> Cc: R help;
tibco.com
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
> Behalf
> Of R. Michael Weylandt
> Sent: Thursday, August 09, 2012 4:22 PM
> To: arun
> Cc: R help; jimi adams
> Subject: Re: [R] indexing in data frames
>
On Thu, Aug 9, 2012 at 5:30 PM, arun wrote:
>
> lapply(1:length(a$c),function(x) a$b[x]-a$c[[x]])
Arun,
I've seen you use this idiom a few times lately and I'd just like to note that
seq_along()
is an (underutilized) primitive and a safer and faster alternative
(avoiding the pathological leng
ab <- data.frame(ab, diff=ab$year-ab$d)
>> new <- split(ab$diff, ab$group)
>> new
> $G1988
> [1] 3 6 4
>
> $G1989
> [1] 1 9
>
> --
> David L Carlson
> Associate Professor of Anthropology
> Texas A&M Univer
-
> From: jimi adams
> To: r-help@r-project.org
> Cc:
> Sent: Thursday, August 9, 2012 4:42 PM
> Subject: [R] indexing in data frames
>
> I'm still not fully understanding exactly how R is handling data frames, but
> am getting closer. Any help with this one will likel
HI,
In the reply I sent, I forgot to add,
anew<-list()#before,
for(i in 1:length(b1)){
anew[[i]]<-list()
anew[[i]]<-b1[[i]]-c[[i]]
}
A.K.
- Original Message -
From: jimi adams
To: r-help@r-project.org
Cc:
Sent: Thursday, August 9, 2012 4:42 PM
Subject: [R] indexing in da
-2 -6
A.K.
- Original Message -
From: jimi adams
To: r-help@r-project.org
Cc:
Sent: Thursday, August 9, 2012 4:42 PM
Subject: [R] indexing in data frames
I'm still not fully understanding exactly how R is handling data frames, but am
getting closer. Any help with this one will likely
ge -
From: jimi adams
To: arun
Cc: R help
Sent: Thursday, August 9, 2012 5:42 PM
Subject: Re: [R] indexing in data frames
Thanks. Yes, I got it to work with loops for small data. I was just hoping,
given the size of the data.frame (hundreds of thousands) and the length of the
lists (varying
> -Original Message-
> From: David Winsemius [mailto:dwinsem...@comcast.net]
> Sent: Thursday, August 09, 2012 5:17 PM
> To: dcarl...@tamu.edu
> Cc: 'jimi adams'; r-help@r-project.org
> Subject: Re: [R] indexing in data frames
>
>
> On Aug 9, 2012, at 2:43 PM, D
$year-ab$d)
new <- split(ab$diff, ab$group)
new
$G1988
[1] 3 6 4
$G1989
[1] 1 9
--
David L Carlson
Associate Professor of Anthropology
Texas A&M University
College Station, TX 77843-4352
-Original Message-
From: r-help-boun...@r-project
p)
> new
$G1988
[1] 3 6 4
$G1989
[1] 1 9
--
David L Carlson
Associate Professor of Anthropology
Texas A&M University
College Station, TX 77843-4352
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
>
I'm still not fully understanding exactly how R is handling data frames, but am
getting closer. Any help with this one will likely go a long way in getting me
there. Let's say I have a data frame, let's call it "a". Within that data frame
i have two variables, let's call them "b" and "c", where
2012 01:20
To: Duncan Murdoch
Cc: r-help@r-project.org
Subject: Re: [R] Indexing matrices from the Matrix package with [i, j] seems to
be very slow. Are there "faster alternatives"?
Dear Duncan
Thanks for your suggestion, but I really need sparse matrices: I have
implemented various gr
),
columns=c("test", "replications", "elapsed", "relative"),
replications=5)
----
-Original Message-
From: Duncan Murdoch [mailto:murdoch.dun...@gmail.com]
Sent: 25. juni 2012 11:27
To: Søren Hø
On 12-06-24 4:50 PM, Søren Højsgaard wrote:
Dear all,
Indexing matrices from the Matrix package with [i,j] seems to be very slow. For
example:
library(rbenchmark)
library(Matrix)
mm<- matrix(c(1,0,0,0,0,0,0,0), nr=20, nc=20)
MM<- as(mm, "Matrix")
lookup<- function(mat){
for (i in
Dear all,
Indexing matrices from the Matrix package with [i,j] seems to be very slow. For
example:
library(rbenchmark)
library(Matrix)
mm <- matrix(c(1,0,0,0,0,0,0,0), nr=20, nc=20)
MM <- as(mm, "Matrix")
lookup <- function(mat){
for (i in 1:nrow(mat)){
for (j in 1:ncol(mat)){
> Behalf
> Of Peter Maclean
> Sent: Wednesday, June 13, 2012 7:15 AM
> To: R mailing list
> Subject: Re: [R] Indexing Grouped Data
>
> I need help in indexing grouped data. In this excample (df1 data), the first
> child had a
> first immunization at age 2. The second c
I need help in indexing grouped data. In this excample (df1 data), the first
child had a first immunization at age 2. The second child had the first, second
and third immunization at age 5,10, and 12, the third child had first and
second immunization at age 4 and 6 and the fourth child had the f
I'm trying to use a self-written function with the summaryBy function (doBy
package).
I have lots of data from Monte Carlo experiments comparing different
estimators across different (combinations of) parameter values, similar to
the following form:
colnames(mydata) <- c("X", "b0", "b1", # parame
Josh
Apologies I haven't responded earlier. This looks great - I ended up doing
what I needed done piece-by-piece because of a looming deadline, but
understanding this code and your suggestions below will be a weekend
project.
Many thanks for all your help.
*
*
Best
Ben
On Tue, Apr 3, 2012 at 8
Le jeudi 05 avril 2012 à 12:40 -0700, Peter Meilstrup a écrit :
> Consider the data.frame:
>
> df <- data.frame(A = c(1,4,2,6,7,3,6), B= c(3,7,2,7,3,5,4), C =
> c(2,7,5,2,7,4,5), index = c("A","B","A","C","B","B","C"))
>
> I want to select the column specified in 'index' for every row of 'df', to
On Thu, Apr 5, 2012 at 1:40 PM, Peter Meilstrup
wrote:
> Consider the data.frame:
>
> df <- data.frame(A = c(1,4,2,6,7,3,6), B= c(3,7,2,7,3,5,4), C =
> c(2,7,5,2,7,4,5), index = c("A","B","A","C","B","B","C"))
>
> I want to select the column specified in 'index' for every row of 'df', to
> get
>
>
Consider the data.frame:
df <- data.frame(A = c(1,4,2,6,7,3,6), B= c(3,7,2,7,3,5,4), C =
c(2,7,5,2,7,4,5), index = c("A","B","A","C","B","B","C"))
I want to select the column specified in 'index' for every row of 'df', to
get
goal <- c(1, 7, 2, 2, 3, 5, 5)
This sounds a lot like the indexing-by
Hi Benjamin,
You seem to have the right basic ideas, but a lot of your code had
typos and some logic flaws that I guess came from trying to move from
just code to in a function. I attached the changes I made. What I
would strongly encourage you to do, is work through each of the little
functions
Petr,
You can get an example dataset at
http://dl.dropbox.com/u/3845242/torhelp.csv. I'll look at the rest of your
suggestions in the morning, thanks for taking a look at it.
Ben
*
*
On Mon, Apr 2, 2012 at 10:33 PM, Petr PIKAL wrote:
> Hi
>
> There is some mismatch in curly braces in your plo
Hi
There is some mismatch in curly braces in your plotter function but when I
try to use it I get
>
plotter(10,3,fram=rwb,framvec=rwb$prcnt.char.depth,obj=prcnt.char.depth,form1=
+ post.f.crwn.length~shigo.av,form2=post.f.crwn.length~shigo.av-1,
+ form3=leaf.area~(1/exp(shigo.av*x))*n,type=2,xl
Josh,
Many thanks - here's a subset of the data and a couple examples:
plotter(10,3,fram=rwb,framvec=rwb$prcnt.char.depth,obj=prcnt.char.depth,form1=
post.f.crwn.length~shigo.av,form2=post.f.crwn.length~shigo.av-1,
form3=leaf.area~(1/exp(shigo.av*x))*n,type=2,xlm=70,ylm=35)
plotter(10,3,fram=rwb
Hi,
Glancing through your code it was not immediately obvious to me why it
does not work, but I can see a lot of things that could be simplified.
It would really help if you could give us a reproducible example.
Find/upload/create (in R) some data, and examples of how you would use
the function.
Hello,
I've written a small function that's supposed to save me some time, and
it's ending up killing it- the intention is to iteratively subset a dataset
fram on framevec, fit a model (either lm or nls depending on type) and
return the r2 or AIC from the model, respectively. Although as far as I
Thanks, Dear Petr Savick.
Your help is enough to solve my problem. With your help I've dealt with the
problem.
Many thanks for your effort,
Sincerely,
Helin
--
View this message in context:
http://r.789695.n4.nabble.com/indexing-tp4428210p4431168.html
Sent from the R help mailing list archiv
On Tue, Feb 28, 2012 at 11:42:32AM -0800, helin_susam wrote:
> Dear Petr Savicky,
>
> Actually, this is based on jackknife after bootstrap algorithm. In summary,
>
> I have a data set, and I want to compute some values by using this
> algorithm.
>
> Firstly, using bootstrap, I create some bootst
Dear Petr Savicky,
Actually, this is based on jackknife after bootstrap algorithm. In summary,
I have a data set, and I want to compute some values by using this
algorithm.
Firstly, using bootstrap, I create some bootstrap re-samples. This step O.K.
Then, for each data point within these re-samp
On Tue, Feb 28, 2012 at 08:50:45AM -0800, helin_susam wrote:
> Dear Petr Pikal and Petr Savicky thank you for your replies..
>
> If the y vector contains different elements my algorithm gives this result;
> y <- c(1,2,3,4,5,6,7,8,9,10)
> x <- c(1,0,0,1,1,0,0,1,1,0)
>
> n <- length(x)
>
> t <-
On Tue, Feb 28, 2012 at 08:50:45AM -0800, helin_susam wrote:
> Dear Petr Pikal and Petr Savicky thank you for your replies..
>
> If the y vector contains different elements my algorithm gives this result;
> y <- c(1,2,3,4,5,6,7,8,9,10)
> x <- c(1,0,0,1,1,0,0,1,1,0)
>
> n <- length(x)
>
> t <-
Dear Petr Pikal and Petr Savicky thank you for your replies..
If the y vector contains different elements my algorithm gives this result;
y <- c(1,2,3,4,5,6,7,8,9,10)
x <- c(1,0,0,1,1,0,0,1,1,0)
n <- length(x)
t <- matrix(cbind(y,x), ncol=2)
z = x+y
for(j in 1:length(x)) {
out <- vector(
On Tue, Feb 28, 2012 at 05:59:24AM -0800, helin_susam wrote:
> Hello All,
>
> My algorithm as follows;
> y <- c(1,1,1,0,0,1,0,1,0,0)
> x <- c(1,0,0,1,1,0,0,1,1,0)
>
> n <- length(x)
>
> t <- matrix(cbind(y,x), ncol=2)
>
> z = x+y
>
> for(j in 1:length(x)) {
> out <- vector("list", )
>
> for(i
Hi
>
> My algorithm as follows;
> y <- c(1,1,1,0,0,1,0,1,0,0)
> x <- c(1,0,0,1,1,0,0,1,1,0)
>
> n <- length(x)
>
> t <- matrix(cbind(y,x), ncol=2)
Do not use t, it is a function for transposing matrix and after you
redefine it you can get nasty surprise in future.
tt <- cbind(y,x)
is enough
Hello All,
My algorithm as follows;
y <- c(1,1,1,0,0,1,0,1,0,0)
x <- c(1,0,0,1,1,0,0,1,1,0)
n <- length(x)
t <- matrix(cbind(y,x), ncol=2)
z = x+y
for(j in 1:length(x)) {
out <- vector("list", )
for(i in 1:10) {
t.s <- t[sample(n,n,replace=T),]
y.s <- t.s[,1]
x.s <- t.s[,2]
z.s <- y.s+x.s
help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of Ista Zahn
> Sent: Wednesday, February 01, 2012 5:45 AM
> To: r-help@r-project.org
> Subject: Re: [R] indexing by empty string (was RE: Error in
> predict.randomForest ... subscript out of bounds with NULL
Hi Andy,
On Tuesday, January 31, 2012 08:44:13 AM Liaw, Andy wrote:
> I'm not exactly sure if this is a problem with indexing by name; i.e., is
> the following behavior by design? The problem is that names or dimnames
> that are empty seem to be treated differently, and one can't index by them:
>
I'm not exactly sure if this is a problem with indexing by name; i.e., is the
following behavior by design? The problem is that names or dimnames that are
empty seem to be treated differently, and one can't index by them:
R> junk = 1:3
R> names(junk) = c("a", "b", "")
R> junk
a b
1 2 3
R> j
>From: William Dunlap [mailto:wdun...@tibco.com]
>You can build the 2nd argument to do.call with alist() instead.
>alist() produces a list that you can use c() and subscripting on
>to add or modify arguments. It is usually better to encapsulate
>this sort of thing in a function like extract() tha
nction like extract() that has a convenient
interface.
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
> Behalf Of davavra
> Sent: Thursday, December 22, 2011 10:31 AM
>
>From help("[", package="base"): "An index value of NULL is treated as
> if it were integer(0).".
Yeah, I should have read it better.
>I don't think there is an easy way to achieve:
> y[,2:3,1,drop=FALSE]
> using do.call("[") without explicitly specify the indices for that
> "missing" dimensio
> (This does imply you knew the number of dimensions was 3.)
Yes, at run time.
>It looks as though the Nulls became 0's. So if you wanted to use
>do.call(`[` then this succeeds:
> do.call(`[`, list(tbl, x, 1:dim(tbl)[2], 1:dim(tbl)[3]) )
...
>As does this using the "empty comma" approach:
>
On Dec 21, 2011, at 10:34 PM, David A Vavra wrote:
I want to take slices of a multi-dimensional table (or array) without
knowing the number of dimensions in advance.
As a test I tried using (in this example a 3d table):
do.call(`[`, list(tbl, x,NULL,NULL)]
Surely that was meant to be:
On Thu, Dec 22, 2011 at 4:34 AM, David A Vavra wrote:
> I want to take slices of a multi-dimensional table (or array) without
> knowing the number of dimensions in advance.
>
>
>
> As a test I tried using (in this example a 3d table):
>
>
>
> do.call(`[`, list(tbl, x,NULL,NULL)]
>
>
>
> where
I want to take slices of a multi-dimensional table (or array) without
knowing the number of dimensions in advance.
As a test I tried using (in this example a 3d table):
do.call(`[`, list(tbl, x,NULL,NULL)]
where I built the list on the fly. It works great as long as I only want th
R-help -
This code iterates over a function with 2 free parameters to find a list of
values (which are the number of incorrect predictions for a computational
model). I want to find the values of i,e when there is the minimum number
of incorrect predictions. In other words, the value of "i" and
Thanks Joshua and Berton for your pointers. Yes it was typo as I wanted to mean
to select few columns.
-Original Message-
From: Joshua Wiley [mailto:jwiley.ps...@gmail.com]
Sent: 22 July 2011 22:02
To: Bogaso Christofer
Cc: r-help@r-project.org
Subject: Re: [R] Indexing problem with
See the "drop" argument (and link) in ?"["
-- Bert
On Fri, Jul 22, 2011 at 9:46 AM, Bogaso Christofer
wrote:
> Dear all, assume I have a matrix with just 1 row. Now suppose I want to
> fetch 1st few rows from that matrix, however resulting object becomes
> vector. Here is 1 such example:
>
>
>
On Fri, Jul 22, 2011 at 9:46 AM, Bogaso Christofer
wrote:
> Dear all, assume I have a matrix with just 1 row. Now suppose I want to
> fetch 1st few rows from that matrix, however resulting object becomes
I'm assuming you mean the first few columns. The place to look is
?"[" where you would find
Dear all, assume I have a matrix with just 1 row. Now suppose I want to
fetch 1st few rows from that matrix, however resulting object becomes
vector. Here is 1 such example:
> matrix(1:5, 1)
[,1] [,2] [,3] [,4] [,5]
[1,]12345
>
> matrix(1:5, 1)[,-1]
[1] 2 3 4 5
Hi Daisy,
You've got a conceptual problem and a couple of practical ones, I think.
On Tue, Jun 28, 2011 at 9:29 PM, Daisy Englert Duursma
wrote:
> Hello,
>
> I think this is a simple problem but I am not coming up with a simple
> solution. I think it just an indexing problem.
>
> I can easily re
Thanks, that worked like a charm.
Daisy
On Wed, Jun 29, 2011 at 11:47 AM, David Winsemius
wrote:
>
> On Jun 28, 2011, at 9:18 PM, Daisy Englert Duursma wrote:
>
>> Hello,
>>
>> I think this is a simple problem but I am not coming up with a simple
>> solution. I think it just an indexing problem.
A) Relying on floating point numbers for predictable indexing is a recipe for
failure. Convert your floating point numbers to integers before using them to
index.
B) Row and column labels are strings, and you are giving numerics to the
brackets. You would need to use as.character() on ddr$x and
On Jun 28, 2011, at 9:18 PM, Daisy Englert Duursma wrote:
Hello,
I think this is a simple problem but I am not coming up with a simple
solution. I think it just an indexing problem.
I can easily replace values in a matrix from a dataframe when the
dataframe has row and column numbers. In the
Hello,
I think this is a simple problem but I am not coming up with a simple
solution. I think it just an indexing problem.
I can easily replace values in a matrix from a dataframe when the
dataframe has row and column numbers. In the example below I use row
and column names and I can not get it
Hello,
I think this is a simple problem but I am not coming up with a simple
solution. I think it just an indexing problem.
I can easily replace values in a matrix from a dataframe when the
dataframe has row and column numbers. In the example below I use row
and column names and I can not get it
I have a data frame in which missing values exist, and I need to
recode the string "missing" to a missing value. For the example, let's
assume I cannot do this while reading it in. Even though this has been
discussed extensively, I'm still a little confused about when to index
with "which" and when
Dear expeRts,
I'm struggling a bit with arrays of lists. The nice thing about arrays is that
one can easily access subsets. For example:
arr. <- array(1:24, dim=c(2,3,4), dimnames=list(a=c("a1","a2"),
b=c("b1","b2","b3"),
c=c("c1","c2","c3","c4")))
arr.[,,4]
me
On Apr 30, 2011, at 4:18 PM, Alice Wines wrote:
Hello all,
I have a quandry I have been scratching my head about for a
while. I've searched the manual and the web and have not been able to
find an acceptable result, so I am hoping for some help.
I have two data frames and I want to in
Hi:
Here are two possibilities:
df1 <- data.frame(rows=c("A","B","C", "B", "C", "A"),
columns=c("21_2", "22_2", "23_2", "21_2", "22_2", "23_2"),
values=c(3.3, 2.5, 67.2, 44.3, 53, 66))
with(df1, xtabs(values ~ rows + columns))
columns
rows 21_2 22_2 23_2
A 3.3 0.0 66.0
B 44.3 2.
Alice Wines wrote:
>
> Hello all,
>
> I have a quandry I have been scratching my head about for a
> while. I've searched the manual and the web and have not been able to
> find an acceptable result, so I am hoping for some help.
>
> I have two data frames and I want to index into the
Hello all,
I have a quandry I have been scratching my head about for a
while. I've searched the manual and the web and have not been able to
find an acceptable result, so I am hoping for some help.
I have two data frames and I want to index into the first using
the second, and replace t
Dear Simon,
Try any of the following:
sapply(r, function(l) l[,2] / l[1, 2])
lapply(r, function(l) l[,2] / l[1, 2])
HTH,
Jorge
On Fri, Apr 22, 2011 at 5:52 PM, Simon Kiss <> wrote:
> Dear colleagues,
> I have a list that looks like what the code below produces. I need a
> function to go thr
Dear colleagues,
I have a list that looks like what the code below produces. I need a function
to go through each list element and work on the second column of each list
element (the first column is irrelevant to me...if the proposed function works
on the first column as a consequence of a wri
That's kind of hard to follow without example data, but have
you looked at merge() ?
On Wed, Feb 2, 2011 at 1:31 PM, Ross Dunne wrote:
> Hello, thank you all for your patience and time
>
> I am essentially trying to get disorganised data into long form for linear
> modelling.
>
> I have 2 datafr
Hello, thank you all for your patience and time
I am essentially trying to get disorganised data into long form for linear
modelling.
I have 2 dataframes "rec" and "book"
Each row in "book" needs to be pasted onto the end of several of the rows of
"rec" according to two variables in the row:"
Hard to know exactly without seeing the structrure of rec and book,
but I would start with ?merge
Best,
Ista
On Wed, Feb 2, 2011 at 2:06 PM, dunner wrote:
>
> Hello, thank you all for your patience and time
>
> I am essentially trying to get disorganised data into long form for linear
> modellin
Hello, thank you all for your patience and time
I am essentially trying to get disorganised data into long form for linear
modelling.
I have 2 dataframes "rec" and "book"
Each row in "book" needs to be pasted onto the end of several of the rows of
"rec" according to two variables in the row:" M
On Nov 15, 2010, at 7:58 PM, Chris Carleton wrote:
Thanks for the suggestions, but 'cat' is not causing name space
conflicts for me
install.packages(
library(fortunes)
fortune("dog")
and since I'm not packaging the code for anyone else to use, I'm
less than concerned about potential confl
Thanks for the suggestions, but 'cat' is not causing name space conflicts
for me and since I'm not packaging the code for anyone else to use, I'm less
than concerned about potential conflicts. I did type that too quickly, and I
have resolved my problem using a workaround that does not involve findi
On Nov 15, 2010, at 5:07 PM, Chris Carleton wrote:
Thanks for the suggestions. The issue for me is that the top level
index is
also like a database key so it might be a bit annoying to coerce it to
char() so that I can reference it with a $ and then I would have to
still be
able to find out
Thanks for the suggestions. The issue for me is that the top level index is
also like a database key so it might be a bit annoying to coerce it to
char() so that I can reference it with a $ and then I would have to still be
able to find out what the name was automatically. I've got a function right
On Nov 15, 2010, at 4:24 PM, Chris Carleton wrote:
Hi List,
I'm trying to work out how to use which(), or another function, to
find the
top-level index of a list item based on a condition. An example will
clarify
my question.
a <- list(c(1,2),c(3,4))
a
[[1]]
[1] 1 2
[[2]]
[1] 3 4
I wan
Hi Chris,
Does this do what you're after? It just compares each element of a
(i.e., a[[1]] and a[[2]]) to c(1, 2) and determines if they are
identical or not.
which(sapply(a, identical, y = c(1, 2)))
There were too many 1s floating around for me to figure out if you
wanted to find elements of a
Chris,
Well, the 'answer' could be:
which(sapply(a, function(x) all(x == c(1,2
But I wonder how these elements of 'a' in your
actual application are coming to be? If you're
constructing them, you can give the elements of
the list names, and then it doesn't matter what
numerical index they
Hi List,
I'm trying to work out how to use which(), or another function, to find the
top-level index of a list item based on a condition. An example will clarify
my question.
a <- list(c(1,2),c(3,4))
a
[[1]]
[1] 1 2
[[2]]
[1] 3 4
I want to find the top level index of c(1,2), which should return
Hi:
Is this what you were looking for?
frd <- rep('fred', 5)
bigfred <- lapply(frd, get)
On Tue, Sep 21, 2010 at 5:04 AM, Alaios wrote:
> I would like to thank you very much for your reply.
> Actually I would like to ask you if there is
> a small list called fred:
> fred <- list(happy = 1:10,
I would like to thank you very much for your reply.
Actually I would like to ask you if there is
a small list called fred:
fred <- list(happy = 1:10, name = "squash")
and a big list called bigfred that included fred list 5 times
bigfred <- rep(fred,5)
Is it possible somehow to index all these sub
t;Z"
However, given the long history of S/R, there must be logical or
philosophical obstacles to this...?
best,
Kingsford Jones
>> -Original Message-
>> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
>> Behalf Of Christian Raschke
>&g
d it all happens, no questions asled. (I can see a generic
function emerging here, perhaps...)
W.
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org
] On Behalf Of Christian Raschke
Sent: Tuesday, 20 July 2010 9:16 AM
To: r-help@r-project.org
Subj
appens, no questions asled. (I can see a generic function
> emerging here, perhaps...)
>
> W.
>
>
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
> Behalf Of Christian Raschke
> Sent: Tuesday, 20 July 2010 9:
On Mon, 2010-07-19 at 19:46 -0400, David Winsemius wrote:
> On Jul 19, 2010, at 7:16 PM, Christian Raschke wrote:
>
> > Dear R-Listers,
> >
> > My question concerns indexing vectors by logical vectors that are
> > based on the original vector. Consider the following simple example
> > to hopef
questions asled. (I can see a generic function emerging
here, perhaps...)
W.
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Christian Raschke
Sent: Tuesday, 20 July 2010 9:16 AM
To: r-help@r-project.org
Subject: [R] Indexing b
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