Hi Arnold,
I like the "Data and Story" library a lot
and have already used it in my own courses:
http://lib.stat.cmu.edu/DASL/
BW
Bernd
On Tue, 12 Feb 2013 10:51:04 -0800 (PST)
David Arnold wrote:
> All,
>
> If you have any good links for sites that contain data sets that can easily
> be a
Hi,
For the dataset A,
Assuming that this is what you meant: the maximum value of "lig" in each 2 rows
excluding the first row:
#dataset: A
A[-1,]<-do.call(rbind,lapply(split(A[-1,],((1:nrow(A[-1,])-1)%/%2)+1),function(x){x[4]<-max(x[4]);x}))
A
# ok time secs lig geo
Dear all mailing listers,
I failed in install source packages in Windows system. I have installed
Rtools in order to install the source package based on the discussion
available on the following link:
http://r.789695.n4.nabble.com/Installing-a-Package-tar-gz-in-windows-td3411393.html.
By the way,
Hi,
Sorry, in my previous reply, I omitted one of the columns.
Lines1 <- readLines(textConnection("Date:10.09.19 Time:21:39:05 Lat:N62.37.18
Long:E018.07.32
-0010 | 28| 28
0010-0020| 302| 302
0020-0030| 42| 42
0030-0040| 2| 2
0040-0050| 1| 1
0060-0070| 1| 1
_
Date
Hi Dimitri,
neededcolumns<- c(1,2,0)
mapply(`[`,mylist,lapply(neededcolumns,function(x) x),SIMPLIFY=FALSE)
#[[1]]
# a
#1 1
#2 2
#
#[[2]]
# b
#1 5
#2 6
#[[3]]
#data frame with 0 columns and 2 rows
A.K.
From: Dimitri Liakhovitski
To: arun
Cc: R help
Sen
Hi all,
I am searching for a way to recover results from lapply() when one or more
values returns an error. I have a written a function that uses tryCatch()
to recover from errors. Here is a very simple example of such a function.
divideBy2<-function(X){
result<-tryCatch(X/2, error = function(
Hi,
mapply(`[`,mylist,list(1,2,0),SIMPLIFY=FALSE)
#[[1]]
# a
#1 1
#2 2
#[[2]]
# b
#1 5
#2 6
#[[3]]
#data frame with 0 columns and 2 rows
A.K.
- Original Message -
From: Dimitri Liakhovitski
To: r-help
Cc:
Sent: Tuesday, February 12, 2013 4:33 PM
Subject: [R] grabbing from eleme
Hello All,
I 'm trying to plot R graph in landscape orientation in a .rtf file.
i 'm using library(rtf) for this.
do we have any option in addPlot() function to rotate the orientation of
the rtf file?
Thank you,
Deepthi BM
PGDB-10-10-04
Institute of Bioinformatics and Applied Biotechnology
B
Dear R users,
writing an R package I wonder, if there is a place to put executable R
scripts into. So that when the package is installed one can use the
script from the command line.
Say we have e new package called FunkyR and wanted to deliver with it
an executable called FunkyCold.R, so that fr
Hi,
sorry for not useable data...
When I tried to use the dput function to the dataframe let's call it A, had
some problems with the time and clock variables... however I attach an
other dataframe B, which is created by merging A with other dataframe. The
B dataframe has a little different struct
One could try using unlist(mylist, recursive=FALSE) to make a list of all the
columns
of the data.frames in mylist and subscripting from that. E.g., here is some
barely tested
code:
> nCols <- vapply(mylist, ncol, 0L)
> neededcolumns<-c(1,2,0) # I assume 0 means no column wanted from 3rd df
Than you so much for your suggestions, everyone!
Arun's solution is elegant and it works - but it takes the same amount of
time as my original code.
Bill's code works too, it's a tiny-bit faster (for 10,000 iterations of the
same code - it took about 16.2 secs vs. 16.6 sec for Arun's solution and
m
Maybe I am mistaken, but my code seems to work with a 0 index. In my final
do.call I don't need anything if the index was 0.
I've tested Bert's solution vs. my original one. It's about 25% faster. So,
thanks a lot, Bert.
Dimitri
On Tue, Feb 12, 2013 at 5:52 PM, Bert Gunter wrote:
> The answer is
Thank you, David. I missed the key detail.
Then Arun's option doesn't work as I need to parametrise needed columns
(they will be a vector).
Dimitri
On Tue, Feb 12, 2013 at 6:04 PM, David Winsemius wrote:
>
> On Feb 12, 2013, at 3:01 PM, Dimitri Liakhovitski wrote:
>
> > Arun, do you know why the
On Feb 12, 2013, at 3:01 PM, Dimitri Liakhovitski wrote:
> Arun, do you know why the first line works and the second doesn't?
> newlist<-mapply(`[`,mylist,list(1,2,0),SIMPLIFY=FALSE) # it works
> neededcolumns<-c(1,2,0) # it does not work
> newlist<-mapply(`[`,mylist,list(needed
Arun, do you know why the first line works and the second doesn't?
newlist<-mapply(`[`,mylist,list(1,2,0),SIMPLIFY=FALSE) # it works
neededcolumns<-c(1,2,0) # it does not work
newlist<-mapply(`[`,mylist,list(neededcolumns),SIMPLIFY=FALSE)
Dimitri
On Tue, Feb 12, 2013 at 5:08 PM, a
On Feb 12, 2013, at 2:51 PM, Lopez, Dan wrote:
> Hi,
>
> Is there a way to write a dataframe or table to the clipboard as an image? I
> want to paste into an email. I use snagit or print screen to do this today
> but want to know if there is a way to do this through R.
>
> For example I know
The answer is essentially no, a loop is required (as best I can see),
although it can be in the form of an apply type call instead. Also,
your code will fail with a 0 index. Something like this should work:
newlist <- lapply(1:3,function(i)if(!neededcolumns[i])NULL else
mylist[[c(i,neededcolumns[
Hi,
Is there a way to write a dataframe or table to the clipboard as an image? I
want to paste into an email. I use snagit or print screen to do this today but
want to know if there is a way to do this through R.
For example I know how to write to clipboard for copy and pasting into Excel:
> w
I apologize for the noise. I did not read the original question carefully
enough.
Dan
Daniel J. Nordlund
Washington State Department of Social and Health Services
Planning, Performance, and Accountability
Research and Data Analysis Division
Olympia, WA 98504-5204
> -Original Message-
>
Anindya,
Completely understand why you wouldn't want to look into the code. I don't
understand your suggestion well enough to implement it, I lost you at
"unlist your function variable . . ."; I don't understand why one would
want to do that. Are there examples you'd suggest? If not I'll take a lo
I was wondering if anyone knows of a random forest implementation (or way
of tweaking the standard randomForest library) that allows one to specify
some sort of variable importance a* *priori.
For example, say that I know that some variables/factors could be more
informative than others for class
That was the problem. Thanks David!
James
On Tue, Feb 12, 2013 at 2:08 PM, David Winsemius wrote:
>
> On Feb 12, 2013, at 10:53 AM, James Jong wrote:
>
> > When I try to crate a grid of parameters for training with caret I get
> > various errors:
> >
I'm sure the OP has reasons to reorder the months as shown but his data
seems to argue that he shouldn't split the wet season (Oct - Apr) but
should, instead, tack the 1972 months (Jan - Aug after the 1971 months
(Sep - Dec).
Clint
Clint BowmanINTERNET: cl...@ecy.wa.
Hi,
Lines1 <- readLines(textConnection("Date:10.09.19 Time:21:39:05 Lat:N62.37.18
Long:E018.07.32
-0010 | 28| 28
0010-0020| 302| 302
0020-0030| 42| 42
0030-0040| 2| 2
0040-0050| 1| 1
0060-0070| 1| 1
_
Date:10.09.19 Time:21:44:52 Lat:N62.38.00 Long:E018.09.07
Take a look at:
>get("%in%")
to see why this (i.e. which(y %in% x) ) is not a good idea.
-- Bert
On Tue, Feb 12, 2013 at 9:48 AM, Nordlund, Dan (DSHS/RDA)
wrote:
> Another option is
>
> which(y %in% x)
>
>
> Dan
>
> Daniel J. Nordlund
> Washington State Department of Social and Health Service
On Feb 12, 2013, at 10:53 AM, James Jong wrote:
> When I try to crate a grid of parameters for training with caret I get
> various errors:
>
>> my_grid <- createGrid("rf")
> Error in if (p <= len) { : argument is of length zero
>
>> my
When I try to crate a grid of parameters for training with caret I get
various errors:
> my_grid <- createGrid("rf")
Error in if (p <= len) { : argument is of length zero
> my_grid <- createGrid("rf", 4)
Error in if (p <= len) { : argume
All,
If you have any good links for sites that contain data sets that can easily
be accessed and copied (or downloaded) by students in introductory
statistics, could you please share them?
Thanks.
David
--
View this message in context:
http://r.789695.n4.nabble.com/Data-sets-online-for-stude
On Feb 11, 2013, at 10:09 PM, nick pardikes wrote:
> I have data that looks like below and I would like to re-order the values
> within the "date" column. I would like to have each year organized like so:
>
> Sep-71
> Oct-71
> Nov-71
> Dec-71
> Jan-71
> Feb-71
> Mar-71
> Apr-71
> May-71
> Jun-7
On Feb 12, 2013, at 5:10 AM, Mat wrote:
> hello togehter,
>
> how can i match a complete data.frame to another one?
>
> I know i can match one column with that code:
>
> A<-match(data$PJ_ID,project$PROJ_ID)
> data$Name<-proj$Name[A]
How about:
data <- data[ match(data$PJ_ID,project$PROJ_ID)
Hi,
I'm trying to fit a not linear model with the nls function to some data.
So far this is the best fitting that i found.
download.file("http://dl.dropbox.com/u/29337496/data"; , destfile="./data")
load("data")
plot (x06Veg,y06Pop)
nlmod=nls( y06Pop ~ B + A * log(x06Veg) , start = list(A = 1,
Hello,
Your data example is a mess. Can't you please use ?dput to post it?
Supposing your data is named 'dat', use
dput(head(dat, 20)) # Paste the output of this in a post
Hope this helps,
Rui Barradas
Em 12-02-2013 15:30, zuzana zajkova escreveu:
Hello,
I would like to ask you for help
Hi Ozgul,
the interesting part is the "select" parameter in subset:
subset(bg,select=c_bg[,1])
cheers
Am 12.02.2013 15:29, schrieb Ozgul Inceoglu:
> Hello,
>
> I have a very data matrix and I have a file which has the names that I need
> to subset. However I cannot manage to subset the main f
Another option is
which(y %in% x)
Dan
Daniel J. Nordlund
Washington State Department of Social and Health Services
Planning, Performance, and Accountability
Research and Data Analysis Division
Olympia, WA 98504-5204
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-he
Thanks. That works!
min() function didn't vectorized but "abs( )", "pmin" or "+ /-" did.
On Tue, Feb 12, 2013 at 12:18 PM, William Dunlap wrote:
> integrate() want a vectorized function, one such that
> length(f(x))==length(x).
> Use pmin or pmax instead of min or max.
>
> Bill Dunlap
> Spotfir
integrate() want a vectorized function, one such that length(f(x))==length(x).
Use pmin or pmax instead of min or max.
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
On Feb 12, 2013, at 12:38 AM, Rainer M Krug wrote:
> -BEGIN PGP SIGNED MESSAGE-
> Hash: SHA1
>
> On 12/02/13 08:30, Ian Renner wrote:
>> Hi,
>>
>> I am trying to write a function which defines some arguments, then uses
>> those arguments as
>> arguments of other function calls. It's a
use pmin instead of min.
hth
Matthias
On 12.02.2013 16:41, dan wang wrote:
Hi All,
Can any one help to explain why min and max function couldn't work in the
integrate function directly.
For example, if issue following into R:
integrand <- function(x) {min(1-x, x^2)}
integrate(integrand, lower
Dear all,
I am currently running an ENFA in R using Adehabitat pkg to assess species
distribution and I am really stuck and need some help to finalize my
project.
I run the ENFA in both original location data and pseudo-absence points
(the latter randomly generated using dismo pkg), and now I w
Hi,
bg<- read.table(text="
Otu00022 Otu00029 Otu00039 Otu00042 Otu00101 Otu00105 Otu00125 Otu00131
Otu00137 Otu00155 Otu00158 Otu00172 Otu00181 Otu00185 Otu00190 Otu00209 Otu00218
Gi20Jun11 0.001217 0 0.001217 0 0.00 0 0
0 0.001217 0 0
Hello,
I would like to ask you for help. I have quite a big dataframe (119 313
rows), this is a part of it:
"jul" "ind" "time" "secs" "geo" "act" "lig" "date.x" "clock" "h" "m" "s"
"d" "mo" "y" "dtime" "land" "date.y" "sriseIC" "ssetIC" "dssetIC" "dsriseIC"
"1207" 14628 5500519 2010-01-19 15:14:5
Duncan,
thanks for the response. stopifnot() looks almost right. it's very
useful if assertion failures issue an error message pointing at the
source file/line (such as error produces); it might also be nice (but
not necessary) to allow the caller to pass in a "higher level" message
to be append
On Feb 12, 2013, at 2:56 AM, Niklas Larson wrote:
> Hi R users,
> Wonder if somebody could give me help on how to reshape this type of data:
>
#Adding a bit of code to read this into R:
Lines <- readLines(textConnection("Date:10.09.19 Time:21:39:05 Lat:N62.37.18
Long:E018.07.32
-0010 | 2
Hello,
Read the help page ?subset more carefully, it's argument 'select' you
should be using:
subset(bg, select = c_bg)
Hope this helps,
Rui Barradas
Em 12-02-2013 14:29, Ozgul Inceoglu escreveu:
Hello,
I have a very data matrix and I have a file which has the names that I need to
subse
Hi,
Monogram Biosciences is recruiting a bioinformatics scientist for our
bioinformatics/biostatistics group. If you are interested in this
position, please apply directly using our career website at:
https://www3.apply2jobs.com/LabCorp/ProfExt/index.cfm?fuseaction=mExternal.showJob&RID=17724&Cur
On Feb 12, 2013, at 10:12 , Sarah d'Apollonia wrote:
> please remove me from the R-help mailing list
Instructions are found in the footer.
>
> thank you!
>
> [[alternative HTML version deleted]]
>
> __
> R-help@r-project.org mailing list
> ht
Abu Naser hotmail.com> writes:
> I have been trying to fit my data (only right censored)
> with gumbel distribution using fitdistrplus. I am
> getting very high standard error. I have been wondering why.
> The followings are the outputs:
>
> fit1=fitdistcens(dr0, "gumbel", start=list(a=99, b=0.
Rolf, many thanks for these suggestions, I will definitely give these a go.
I ended up resorting to locator() to ascertain the x y coordinates I required,
but obviously this isn't optimal.
Thanks again,Oli
> Date: Thu, 7 Feb 2013 12:38:22 +1300
> From: rolf.tur...@xtra.co.nz
> To: oti...@hotmail
Dear all,
I have been trying to fit my data (only right censored) with gumbel
distribution using fitdistrplus. I am getting very high standard error. I have
been wondering why.
The followings are the outputs:
fit1=fitdistcens(dr0, "gumbel", start=list(a=99, b=0.6), optim.method=
"L-B
On 12/02/2013 9:11 AM, Greg Minshall wrote:
hi. i've looked for (and not found) an assert() function. needing one,
i created the following (from stop()). i'm posting it in case 1)
someone sees a problem with this; and 2) someone else has a need.
stopifnot() might do what you want.
The trick
Hello,
I have a very data matrix and I have a file which has the names that I need to
subset. However I cannot manage to subset the main file. ANy idea?
bg <- read.table (file.choose(), header=T, row.names)
bg
Otu00022 Otu00029 Otu00039 Otu00042 Otu00101 Otu00105 Otu00125
Otu00131 Otu
Hi,
You could try:
dat1<- read.table(text="
site date year precipitation temp_max temp_min
1 castlepeak Jan-71 1971 26.2903226 38.29032 18.06452
2 castlepeak Feb-71 1971 9.1071429 39.60714 17.5
3 castlepeak Mar-71 1971 36.3548387 38.87097 17.77419
4 castlepeak Apr
Dear all,
I am currently running an ENFA in R using Adehabitat pkg to assess species
distribution and I am really stuck and need some help to finalize my
project.
I run the ENFA in both original location data and pseudo-absence points
(the latter randomly generated using dismo pkg), and now I w
hello togehter,
how can i match a complete data.frame to another one?
I know i can match one column with that code:
A<-match(data$PJ_ID,project$PROJ_ID)
data$Name<-proj$Name[A]
but how can i match the total data.frame from project to data?
indivduel matching for each column is not very producti
i found a way :-)
data<-merge(data,project,by.x="PJ_ID",by.y="PROJ_ID",all.x=TRUE)
perhaps this can help someone
Mat
--
View this message in context:
http://r.789695.n4.nabble.com/match-a-complete-data-frame-tp4658250p4658252.html
Sent from the R help mailing list archive at Nabble.com.
___
Hi Ian,
The remark of Rainer is correct, but I think the solution is very simpe:
f1 = function(a)
{
b = a + 1
b
}
f2 = function(x, z)
{
y = x*z(x) -2
y
}
> f2(x = 3, z = f1)
[1] 10
Or are you intending something else?
Best wishes,
Frans
-Oorspronkelijk bericht-
Van: r-help-boun...
please remove me from the R-help mailing list
thank you!
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-gu
hi. i've looked for (and not found) an assert() function. needing one,
i created the following (from stop()). i'm posting it in case 1)
someone sees a problem with this; and 2) someone else has a need.
cheers, Greg
## an assert mechanism...
assert <- function (shouldbe, ...) {
if (!shoul
I'm sorry, no clue how I did not see that. Thank you!
On 12 February 2013 15:21, Uwe Ligges wrote:
>
>
> On 12.02.2013 15:15, Torvon wrote:
>
>> The code is quite long because I am running a WLS regression instead of an
>> OLS regression (due to heteroscedasticity). First, I get mean structure,
On 12.02.2013 15:15, Torvon wrote:
The code is quite long because I am running a WLS regression instead of an
OLS regression (due to heteroscedasticity). First, I get mean structure,
then get mean/SD relationship, then improve the variance structure by using
weights proportional to 1/variance.
The code is quite long because I am running a WLS regression instead of an
OLS regression (due to heteroscedasticity). First, I get mean structure,
then get mean/SD relationship, then improve the variance structure by using
weights proportional to 1/variance.
I am quite sure this is not relevant,
On 12.02.2013 14:44, Torvon wrote:
Thank you, Uwe.
summary(m1) gives me p-value estimates of:
(Intercept) 2e-16
x1 6.9e-15
x2 1.9e-07
x3 2.7e-09
While coef(summary(m1))[,4] gives me:
(Intercept) 3.0e-23
x1 5.7e-13
x2 2.6e-07
x3 1.7e-17
While the first one confirms my suspicion (-23 instead
Torvon gmail.com> writes:
>
> Thank you, Uwe.
>
> summary(m1) gives me p-value estimates of:
> (Intercept) 2e-16
> x1 6.9e-15
> x2 1.9e-07
> x3 2.7e-09
>
> While coef(summary(m1))[,4] gives me:
> (Intercept) 3.0e-23
> x1 5.7e-13
> x2 2.6e-07
> x3 1.7e-17
>
> While the first one confirms my su
Thank you, Uwe.
summary(m1) gives me p-value estimates of:
(Intercept) 2e-16
x1 6.9e-15
x2 1.9e-07
x3 2.7e-09
While coef(summary(m1))[,4] gives me:
(Intercept) 3.0e-23
x1 5.7e-13
x2 2.6e-07
x3 1.7e-17
While the first one confirms my suspicion (-23 instead of -16), the latter
one vary drastically
Hello,
Maybe there's some date trick, but I think the following will do the job.
dat <- read.table(text = "
site date year precipitation temp_max temp_min
1 castlepeak Jan-71 197126.2903226 38.29032 18.06452
2 castlepeak Feb-71 1971 9.1071429 39.60714 17.5
3 castlepea
On 12.02.2013 13:37, Torvon wrote:
I need to report exact p-values in my dissertation. Looking at my lm()
results of many regressions with huge datasets I have the feeling that
p-values are rounded to the smallest value of "2e-16", because this p-value
is very common.
Is that true or just chan
Hi Jon,
zoo is great for tasks like this, not just for na.approx. :)
I would approach the problem like this:
library(zoo)
# put lightdata into a zoo object
z <- with(lightdata, zoo(light,
as.POSIXct(paste(date, time), format="%m/%d/%y %H:%M:%S")))
# merge the above zoo object with an "empty" z
Dear All,
I encounter some discrepancies when comparing the deviance of a weighted and
unweigthed model with the AIC values. A general example (from 'nls'):
DNase1 <- subset(DNase, Run == 1)
fm1DNase1 <- nls(density ~ SSlogis(log(conc), Asym, xmid, scal), DNase1)
Now for a weighted fit:
fm2D
I need to report exact p-values in my dissertation. Looking at my lm()
results of many regressions with huge datasets I have the feeling that
p-values are rounded to the smallest value of "2e-16", because this p-value
is very common.
Is that true or just chance? If it is true, how do I obtain the
In one particular situation predict.coxph gives an error message. Namely:
stratified data, predict='expected', new data, se=TRUE. I think I found the
error but I'll leave that to you to decide.
Thanks,
Chris
CODE
library(survival)
set.seed(20121221)
nn <- 10 # sample size in each
On 13-02-12 5:34 AM, Jan T Kim wrote:
On Tue, Feb 12, 2013 at 09:38:19AM +0100, Rainer M Krug wrote:
-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1
On 12/02/13 08:30, Ian Renner wrote:
Hi,
I am trying to write a function which defines some arguments, then uses those
arguments as
arguments of
On 13-02-12 5:18 AM, Robert Latest wrote:
Hi guys,
like so often, the answert came to me minutes after posting. pmatch()
does exactly what I need. match() gives the values of the elements,
but not their positions.
I think you should read the docs more closely. match() does what you
need; pma
Hi R users,
Wonder if somebody could give me help on how to reshape this type of data:
---
Date:10.09.19 Time:21:39:05 Lat:N62.37.18 Long:E018.07.32
-0010 | 28| 28
0010-0020
Hi
Why you do not change date and time to POSIX object? It is simple and saves you
a lot of frustration when merging two data frames.
If you changed lightdata date and time to new column
lightdata$newdate <- strptime(paste(lightdata$date, lightdata$time, sep=" "),
format = "%d/%m/%y %H:%M:%S")
On Tue, Feb 12, 2013 at 09:38:19AM +0100, Rainer M Krug wrote:
> -BEGIN PGP SIGNED MESSAGE-
> Hash: SHA1
>
> On 12/02/13 08:30, Ian Renner wrote:
> > Hi,
> >
> > I am trying to write a function which defines some arguments, then uses
> > those arguments as
> > arguments of other functio
Hello,
?match
> x <- c(4,5,6)
> y <- c(10,1,5,12,4,13,14)
> match(x,y)
[1] 5 3 NA
Hope this helps,
Pascal
Le 12/02/2013 19:09, Robert Latest a écrit :
Hello all,
given two vectors X and Y I'd like to receive a vector Z which
contains, for each element of X, the index of the corresponding
Hi guys,
like so often, the answert came to me minutes after posting. pmatch()
does exactly what I need. match() gives the values of the elements,
but not their positions.
Thanks,
robert
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman
On 12-02-2013, at 11:09, Robert Latest wrote:
> Hello all,
>
> given two vectors X and Y I'd like to receive a vector Z which
> contains, for each element of X, the index of the corresponding
> element in Y (or NA if that element isn't in Y).
>
> Example:
>
> x <- c(4,5,6)
> y <- c(10,1,5,12,
Am 12.02.2013 11:09 (UTC+1) schrieb Robert Latest:
> x <- c(4,5,6)
> y <- c(10,1,5,12,4,13,14)
Please try
#match(x,y)
[1] 5 3 NA
HTH,
Rainer Hurling
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PLEASE do read the
Robert,
Try
match(x, y)
HTH,
Jorge.-
On Tue, Feb 12, 2013 at 9:09 PM, Robert Latest wrote:
> Hello all,
>
> given two vectors X and Y I'd like to receive a vector Z which
> contains, for each element of X, the index of the corresponding
> element in Y (or NA if that element isn't in Y).
>
>
Hello all,
given two vectors X and Y I'd like to receive a vector Z which
contains, for each element of X, the index of the corresponding
element in Y (or NA if that element isn't in Y).
Example:
x <- c(4,5,6)
y <- c(10,1,5,12,4,13,14)
z <- findIndexIn(x, y)
z
[1] 5 3 NA
1st element of z is 5,
On Tue, Feb 12, 2013 at 7:30 AM, Ian Renner wrote:
> Where I am running into trouble is when I want to call function f1 within
> function f2:
>
> f2(x = 3, z = f1(x))
>
> This returns the error:
>
> "Error in f1(x) : object 'x' not found"
>
> I'm not sure how to define environments within the fu
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On 12/02/13 08:30, Ian Renner wrote:
> Hi,
>
> I am trying to write a function which defines some arguments, then uses those
> arguments as
> arguments of other function calls. It's a bit tricky to explain, so a simple
> example will have
> to suff
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