I'm a little confused, because the sample code does something that
none of the suggestions does.
x1 <- c(116,0,115,137,127,0,0)
x2 <- c(0,159,0,0,0,159,127)
[You] want : xx <- c(116,115,137,127,159, 127)
Assuming that there should have been two copies of 159 in xx, this is
xx <- c(x1[x1 != 0], x2[
Thank you for the general code. Really appreciate it.
On Tue, Sep 5, 2023 at 7:59 PM Eric Berger wrote:
> As Duncan points out, ifelse() provides a more general approach than
> the specific pmax().
>
> Even more generally, you might want to consider the apply() function
> (and its relatives sap
Thank you for the general code. Really appreciate it.
On Tue, Sep 5, 2023 at 7:45 PM Duncan Murdoch
wrote:
> On 05/09/2023 4:55 a.m., roslinazairimah zakaria wrote:
> > Hi all,
> >
> > I have these data
> >
> > x1 <- c(116,0,115,137,127,0,0)
> > x2 <- c(0,159,0,0,0,159,127)
> >
> > I want : xx <
As Duncan points out, ifelse() provides a more general approach than
the specific pmax().
Even more generally, you might want to consider the apply() function
(and its relatives sapply(), lapply(), ...)
For example
apply(cbind(x1,x2), MAR=1, max)
In the above statement, x1 and x2 are combined i
On 05/09/2023 4:55 a.m., roslinazairimah zakaria wrote:
Hi all,
I have these data
x1 <- c(116,0,115,137,127,0,0)
x2 <- c(0,159,0,0,0,159,127)
I want : xx <- c(116,115,137,127,159, 127)
I would like to merge these data into one column. Whenever the data is '0'
it will be replaced by the value
Thank you very much for your help.
On Tue, Sep 5, 2023 at 6:39 PM Rui Barradas wrote:
> Às 09:55 de 05/09/2023, roslinazairimah zakaria escreveu:
> > Hi all,
> >
> > I have these data
> >
> > x1 <- c(116,0,115,137,127,0,0)
> > x2 <- c(0,159,0,0,0,159,127)
> >
> > I want : xx <- c(116,115,137,127
Thank you very much for your help.
On Tue, Sep 5, 2023 at 6:12 PM Eric Berger wrote:
> xx <- pmax(x1,x2)
>
> On Tue, Sep 5, 2023 at 11:56 AM roslinazairimah zakaria
> wrote:
> >
> > Hi all,
> >
> > I have these data
> >
> > x1 <- c(116,0,115,137,127,0,0)
> > x2 <- c(0,159,0,0,0,159,127)
> >
> >
Às 09:55 de 05/09/2023, roslinazairimah zakaria escreveu:
Hi all,
I have these data
x1 <- c(116,0,115,137,127,0,0)
x2 <- c(0,159,0,0,0,159,127)
I want : xx <- c(116,115,137,127,159, 127)
I would like to merge these data into one column. Whenever the data is '0'
it will be replaced by the valu
xx <- pmax(x1,x2)
On Tue, Sep 5, 2023 at 11:56 AM roslinazairimah zakaria
wrote:
>
> Hi all,
>
> I have these data
>
> x1 <- c(116,0,115,137,127,0,0)
> x2 <- c(0,159,0,0,0,159,127)
>
> I want : xx <- c(116,115,137,127,159, 127)
>
> I would like to merge these data into one column. Whenever the da
Thanks for your suggestion.
I have just returned from a vacation and started catching up on my emails.
Rolling join is an elegant and most suitable solution for my tasks. I invested
some time in learning data.table package. The vignette on secondary indices
and auto indexing refers to anoth
It sounds like you might want a rolling join, e.g.
https://dplyr.tidyverse.org/reference/join_by.html#rolling-joins.
(And data.table has similar functionality which inspired dplyr)
Hadley
On Mon, Aug 7, 2023 at 9:32 PM Naresh Gurbuxani
wrote:
>
>
> I have two dataframes, each with a column for
On Mon, 07 Aug 2023, Naresh Gurbuxani writes:
> I have two dataframes, each with a column for timestamp. I want to
> merge the two dataframes such that each row from first dataframe
> is matched with the row in the second dataframe with most recent but
> preceding timestamp. Here is an example.
>
I was able to adapt your solution using packages with which I am more familiar.
myres2 <- merge(option.trades, stock.trades, by = "timestamp", all =
TRUE)
myres2[,"stock.timestamp"] <- ifelse(is.na(myres2$stock.price), NA,
myres2$timestamp)
myres2$stock.timestamp <- as.POSIXct(myres2$stock.time
Hi Naresh,
Perhaps the below is faster than your approach
library(dplyr)
library(tidyr)
merge(option.trades, stock.trades, by="timestamp", all=TRUE) |>
dplyr::arrange(timestamp) |>
dplyr::mutate(stock.timestamp =
as.POSIXct(ifelse(is.na(option.price), timestamp, NA))) |>
tidyr::fill(stock.pr
Hi Jim and all,
All functions worked beautifully. I really appreciate your help.
*Thank you and best regards.*
On Fri, Nov 19, 2021 at 4:26 AM Jim Lemon wrote:
> Hi RosalinaZakaria,
> Talk about using a sledgehammer to crack a nut. In your example the
> two objects are character vectors. Ho
Hi RosalinaZakaria,
Talk about using a sledgehammer to crack a nut. In your example the
two objects are character vectors. How about:
dt_comb1gd <-paste0(dtpaigd,dtpmgd)
Jim
On Fri, Nov 19, 2021 at 2:15 AM ROSLINAZAIRIMAH BINTI ZAKARIA .
wrote:
>
> Dear all,
>
> I try to merge two columns consi
Hello,
Use an index giving the "" positions.
i <- nchar(dtpmgd) == 0L
dtpmgd[i] <- dtpaigd[i]
Or, in one line,
dtpmgd[nchar(dtpmgd) == 0L] <- dtpaigd[nchar(dtpmgd) == 0L]
Hope this helps,
Rui Barradas
Às 07:02 de 18/11/21, ROSLINAZAIRIMAH BINTI ZAKARIA . escreveu:
Dear all,
I try t
On 18/11/2021 09:02, ROSLINAZAIRIMAH BINTI ZAKARIA . wrote:
Dear all,
I try to merge two columns consisting of characters using the 'coalesce'
function from dplyr package. However, two data still have not merged, data
no. 124 1nd 143. Any help is very much appreciated. I provide the data as
fo
A and B) and not(B).
David C
-Original Message-
From: Jeff Newmiller [mailto:jdnew...@dcn.davis.ca.us]
Sent: Monday, January 7, 2019 11:04 AM
To: Priya Arasu ; Priya Arasu via R-help
; David L Carlson ; David Winsemius
; r-help@r-project.org
Subject: Re: [R] Merge the data from multiple
quot;(A and C)" "not(D)"
>
>$C
>[1] "D"
>
>> TF.and <- lapply(TF.list, paste, collapse=" and ")
>> TF.final <- lapply(names(TF.and), function(x) paste(x, "=",
>TF.and[[x]]))
>> TF.final <- do.call(rbind, TF.final)
>>
t;
[2,] "B = (A and C) and not(D)"
[3,] "C = D"
> write(TF.final, file="TF.output.txt")
The text file "TF.output.txt" contains the three lines.
--
David L. Carlson
Department of Anthropology
Texas A&M Unive
[1,] "A = (D and E) and not(B or C)"
[2,] "B = (A and C) and not(D)"
[3,] "C = D"
> write(TF.final, file="TF.output.txt")
The text file "TF.output.txt" contains the three lines.
--
David L. Ca
On 1/5/19 7:28 AM, Priya Arasu via R-help wrote:
I have multiple text files, where each file has Boolean rules.
Example of my text file 1 and 2
Text file 1:
A = not(B or C)
B = A and C
C = D
Text file 2:
A = D and E
B = not(D)
I want to merge the contents in text file as follows
A = not(B or C
HI All,
I found that other R users experienced the same problem when using sqldf
package. An expert provided solution, adding "method="raw" at the end as below,
one or more of the columns in my data.frame are of class "AsIs" ( I don't know
what it is) . When sqldf retrieves the result from SQL
Have you tried 'foverlaps' in the data.table package?
Jim Holtman
Data Munger Guru
What is the problem that you are trying to solve?
Tell me what you want to do, not how you want to do it.
On Mon, Sep 4, 2017 at 8:31 AM, Mohammad Tanvir Ahamed via R-help <
r-help@r-project.org> wrote:
> Hi,
>
> On Apr 13, 2017, at 7:56 AM, Mohammad Tanvir Ahamed via R-help
> wrote:
>
> Hi,
> I have a list like
> kk<- list (a = 1:5, b = 6:10, c = 4:11)
>
> Now i want to merger (Union) the list element "a" and "c" by name .
>
> My expected outcome is
> kk1<- list(a_c = 1:11, b = 6:10)
>
>
> I c
list(a_c = 1:11, b = 6:10) just to show by
> expected outcome. But i am expecting the resulting code will naming "a_c" by
> itself also.
> Hope i can make clear about problem.
>
>
>
>
> Tanvir Ahamed
> Göteborg, Sweden | mashra...@yahoo.com
>
>
>
>
rsday, 13 April 2017, 18:37
Subject: Re: [R] Merge selected list element by name
Hello,
There's no need to send the same question twice, we've got it at the
first try.
Maybe I don't understand but is this it?
kk1 <- list(a_c = union(kk$a, kk$c), b = kk$b)
kk1
$a_c
[1] 1 2
Hello,
There's no need to send the same question twice, we've got it at the
first try.
Maybe I don't understand but is this it?
kk1 <- list(a_c = union(kk$a, kk$c), b = kk$b)
kk1
$a_c
[1] 1 2 3 4 5 6 7 8 9 10 11
$b
[1] 6 7 8 9 10
Hope this helps,
Rui Barradas
Em 13-04-2017 1
Dear Milu,
If your objective is to match the places from one table to the nearest
place in the second table, you can generally use knn algorithm for 1
nearest neighbourhood.
But please, check what David suggests first.
Best regards,
Michal
2016-10-16 19:24 GMT+02:00 David Winsemius :
>
> > On Oc
> On Oct 16, 2016, at 6:32 AM, Miluji Sb wrote:
>
> Dear all,
>
> I have two dataframe 1 by latitude and longitude but they always do not
> match. Is it possible to merge them (e.g. nearest distance)?
>
> # Dataframe 1
> structure(list(lat = c(54L, 55L, 51L, 54L, 53L, 50L, 47L, 51L,
> 49L, 54L
Hi Lily,
I think below codes can work:
f<- list.files("D:/output/test/your folde
rname",full.names=TRUE,recursive=TRUE)
files<- grep(".csv", f)
files_merge<- data.frame()
for (i in 1:length(f[files])){
data<- read.csv(file=f[files][i],header=TRUE, sep=",")
files_merge<- rbind(files_merge,
Hello,
Maybe something like this.
fls <- list.files(pattern = "*.csv")
dat.list <- lapply(fls, read.csv)
dat <- do.call(rbind, dat.list)
Hope this helps,
Rui Barradas
Citando lily li :
> Hi R users,
>
> I'd like to ask that how to merge several datasets into one in R? I put
> these csv file
Lily Li wrote :
I think you are going to have to give us some more detail. What commands did
you execute? what are the names of the .csv files in your directory? Can you
read one of them as asingle read.csv?
> Hi R users,
>
> I'd like to ask that how to merge several datasets into one in R? I
Lily:
If you mean that you have several csv files in a directory/folder on
your computer and you are using lapply() to do something with them,
then you do not have a clue about how R works and you need to go
through some tutorials to learn. There are many good ones on the web.
Some recommendations
On 20/04/2016 7:38 AM, Gaston wrote:
I indeed used is.na() to check length, as I was not sure weather
lenght() was a simple query or would go through the whole vector to
count the elements.
length() is a simple query, and is very fast. The other problem in your
approach (which may not be a pr
I indeed used is.na() to check length, as I was not sure weather
lenght() was a simple query or would go through the whole vector to
count the elements.
So to sum up, function calls are expensive, therefore recursion should
be avoided, and growing the size of a vector (which is probably
reass
On 19/04/2016 3:39 PM, Gaston wrote:
Hello everyone,
I am learning R since recently, and as a small exercise I wanted to
write a recursive mergesort. I was extremely surprised to discover that
my sorting, although operational, is deeply inefficient in time. Here is
my code :
merge <- function(
This is not a well defined question, until your notions of "small" and
"nearest" are defined. In your specific example
rect.hclust(hc, k = 3, border = 2:5)
... will do what you are asking for. This is not likely to work in the general
case - imagine that your cluster of size two only meets t
The root of your problems lie in your assumption that xts variables act like
data frames. Instead they are matrices with an index attribute. All values in a
matrix must be of the same storage mode.
You might want to investigate the data.table package. It is not a time series
object but you can
c"
2015-09-06 "def"
2015-09-07 "abc"
-Original Message-
From: "Joshua Ulrich" [josh.m.ulr...@gmail.com]
Date: 09/03/2015 09:43 PM
To: "ce"
CC: "R-Help"
Subject: Re: [R] merge xts objects with different data types ?
On Thu, Se
On Thu, Sep 3, 2015 at 7:40 PM, ce wrote:
>
> Hello
>
> Let's say some questions about merging xts variables :
>
> a<- xts("abc", Sys.Date())
> b <- xts("def", Sys.Date())
> c <- xts(1, Sys.Date())
>
>> merge(a,b)
>a b
> 2015-09-03 "abc" "def"
>> merge(a,b,c)
> a b c
Thank you Jeff,
Your solutions have two great aspects: a) you provide a different
approach by using reshape2 syntax / tidyr, and b) the concern that it is
better to update x.HHu.map with y.HHo.map, without overwriting x.HHu.map
with NA from y.HHo.map, thus keeping intact the old value(s). That i
Thank you Ista,
Your solution is smart, by sub-setting from x.HHu.map data only "HHid",
"position" as indices (because they are unique) for the merge, and any
extra columns in x.HHu.map that are not present in y.HHo,map, thus when
the merge is done with option all=T, will work among the two sets
I get confused by your use of the position map table. If I follow your
description toward your desired result, I take a different route that
makes sense to me. Perhaps it will make sense to you as well. The key idea
is to make individual comparisons of the values for each combination of
HHid an
I think this does what you want:
## find idiv coloumns in x.HHu.map that don't exist in y.HHo.map
x.HHu.map <- x.HHu.map[
c("HHid",
"position",
names(x.HHu.map)[
!names(x.HHu.map)
%in% names(y.HHo.map)]
)]
## merge, adding extra column from x.HHu
Exactly what I thought too the first time I read ?merge. R sometimes has its
own approach.
John Kane
Kingston ON Canada
> -Original Message-
> From: r-help@r-project.org
> Sent: Mon, 1 Jun 2015 14:47:07 + (UTC)
> To: li...@dewey.myzen.co.uk, r-help@r-project.org
>
You do not appear to understand what merge() does. Go through the worked
examples in ?merge so that you do.
FWIW, I would agree that the Help file is cryptic and difficult to
understand. Perhaps going through a tutorial on database "join" operations
might help.
Cheers,
Bert
Bert Gunter
"Data is
I understood that by would take the intersection of names(x) and names(y),
names(x) being the column names of x and names(y), column names of y.
if x has 5 col and the col names of x are col1, col2... col5 and y has 3 col
and their names are col1, col2, col3, I thought that the merged data set wi
On 01/06/2015 14:46, carol white via R-help wrote:
Hi,By default the merge function should take the intersection of column names
(if this is understood from by = intersect(names(x), names(y)),
Dear Carol
The by parameter specifies which columns are used to merge by. Did you
understand it t
> Sent: Mon, 1 Jun 2015 06:29:41 -0800
> To: wht_...@yahoo.com, r-help@r-project.org
> Subject: RE: [R] merge function
>
> As Burt says it is not exactly clear what you want but is something like
> this what you are looking for?
>
> dat1 <- data.frame(aa = c
As Burt says it is not exactly clear what you want but is something like this
what you are looking for?
dat1 <- data.frame(aa = c("a", "b", "c"), bb = 1:3)
dat2 <- data.frame(xx = c("b", "c", "d"), yy = 3:1)
merge(dat1, dat2, by.x = "aa", by.y = "xx")
For further reference here are some sugg
1. Please read and follow the posting guide.
2. Reproducible example? (... at least I don't understand what you mean)
3. Plain text, not HTML.
Cheers,
Bert
Bert Gunter
"Data is not information. Information is not knowledge. And knowledge is
certainly not wisdom."
-- Clifford Stoll
On Mon,
This would be a perfect time for you to use best practices to convey what you
have to us [1]... most specifically, posting using plain text will keep your
code from getting munged up, and using dput to provide an unambiguous form we
can put into our R sessions.
[1]
http://stackoverflow.com/que
On 13/11/2014 14:02, Matthias Weber wrote:
Hello togehter,
i have a little problem. Maybe anyone can help me.
I think you might find
?merge
enlightening
Indeed given that the word merge occurs in your subject line and your
text it is surprising you have not already found it.
I have 2 d
merge(df1, df2, all=TRUE)
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Thu, Nov 13, 2014 at 6:02 AM, Matthias Weber <
matthias.we...@fntsoftware.com> wrote:
> Hello togehter,
>
> i have a little problem. Maybe anyone can help me.
>
> I have 2 data.frames, which look like as follows:
> First:
Hello,
See ?merge, in particular the argument 'all'.
dat1 <- read.table(text = "
NAMEMONTH BONUS
1 Andy 2014-10 100
2 Pete 2014-10200
3 Marc2014-10300
4 Andy2014-11400
", header = TRUE, stringsAsFactors = FALSE)
On 10/28/2014 12:13 PM, John Fox wrote:
Hi Michael,
How about this?
That's perfect! And more general than I had hoped for. You probably
used mindreader() :-)
-Michael
coef.glmlist <- function(object, result=c("list", "matrix", "data.frame"),
...){
result <- match.arg(result)
coe
Hi Michael,
How about this?
coef.glmlist <- function(object, result=c("list", "matrix", "data.frame"),
...){
result <- match.arg(result)
coefs <- lapply(object, coef)
if (result == "list") return(coefs)
coef.names <- unique(unlist(lapply(coefs, names)))
n.mods <- length(objec
Hello,
Try
aggregate(Rain ~ Year + Month, data = dat, FUN = sum)
Hope this helps,
Rui Barradas
Em 23-10-2014 01:29, Hafizuddin Arshad escreveu:
Dear R users,
Can someone help me on this? I would like to find the sum of the Rain if
the Month appears more than once. For example in row 3 and
Thank you Jean, Petr, Terry, William and everyone else who thought about
my problem.
It is sooo good that this mailing list exists!
I solved my problem using Petr's suggestion, that didn't seem so
complicated and worked fine for me.
Thanks again and have a great weekend,
Dagmar
Am 02.10.2014
Hi Terry,
Some of that combination of sort() and approx() can be done by
findInterval(), which may be quick enough that you don't need the
'thinning' part of the code.
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Fri, Oct 3, 2014 at 6:05 AM, Therneau, Terry M., Ph.D.
wrote:
> I've attached
our mydata and use na.locf to fill
gaps you should get desired result.
Regards
Petr
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of Dagmar
> Sent: Thursday, October 02, 2014 11:25 PM
> Cc: R help
> Subject: Re: [R]
e.g. by
?complete.cases
Regards
Petr
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of Adams, Jean
> Sent: Thursday, October 02, 2014 11:38 PM
> To: Dagmar
> Cc: R help
> Subject: Re: [R] merge by time, certain va
Thanks, Dagmar.
So, shouldn't row 3 with a time of 09:51:01 be "low" and not "high"?
Jean
On Thu, Oct 2, 2014 at 4:25 PM, Dagmar wrote:
> Dear Jean and all,
>
> I want all lines to be "low", but during times 9:55 - 10:05 a.m (i.e. a
> timespan of 10 min) I want them to be "high".
> In my real
Dear Jean and all,
I want all lines to be "low", but during times 9:55 - 10:05 a.m (i.e. a
timespan of 10 min) I want them to be "high".
In my real data "low" and "high" refer to "lowtide" and "hightide" in
the waddensea and I want to assign the location of my animal at the time
it was taken to
Dagmar,
Can you explain more fully why rows 1, 2, and 5 in your result are "low"
and rows 3 and 4 are "high"? It is not clear to me from the information
you have provided.
> result[c(1, 2, 5), ]
Timestamp location Event
1 24.09.2012 09:05:011 low
2 24.09.2012 09:49:50
Cautionary note on the solution below. Be sure the 'sndr' is either
factor or character because, if sndr is numeric, as the list is
populated, R will fill in non-adjacent list items with NULLs, leaving a
list with many empty entries. So, the modified line is
newdat[[as.character(vnam)]]<
This is a (very) slightly modified version of Jim's reply that takes the
sender's email our of the list element and uses it as the name so it can
be accessed as newdat$'senders email' or newdat[['senders email']]
newdat<-list()
for(sndr in unique(rdvdf$sender)) {
newvec<-
as.character(unique
On Fri, 11 Jul 2014 12:19:39 PM Ryan de Vera wrote:
> Hello all,
>
> I have a data frame filled with senders and recipients. Some of the
senders
> have multiple rows with different recipients and I want to merge
those
> rows. For example I have
>
> a...@email.com b...@email.com
> a...@email
Hi John,
I don't think your x2 is right, but who knows?
One possible approach would be:
R> lapply(split(x2$recipients, unique(x2$sender)), paste, collapse=", ")
$`a...@email.com`
[1] "b...@email.com, f...@email.com"
$`r...@email.com`
[1] "c(\"c...@email.com\", \"d...@email.com\"), h...@email.co
Hi Ryan,
We can't tell from your example what structure your original data are
in, nor what your output is intended to look like, or for that matter
how you got from one to the other. Please don't post in HTML because
it gets mangled!
Using dput() to provide your example data is the best thing, b
Rolf,
I hear you.
But, after reflection, ie I looked at my situation again, it is great :-)-O
el
Sent from Dr Lisse's iPad mini
> On Jun 30, 2014, at 0:48, Rolf Turner wrote:
>
>
>> On 30/06/14 10:32, Dr Eberhard W Lisse wrote:
>>
>> Thanks,
>>
>> I then set NA to 0, and can do the sutr
Thank you very much.
el
On 2014-06-30, 00:48 , Rolf Turner wrote:
>
> On 30/06/14 10:32, Dr Eberhard W Lisse wrote:
>
>> Thanks,
>>
>> I then set NA to 0, and can do the sutraction,
>>
>> great.
>
> Not so great. I haven't gone through the issues underlying this post,
> but replacing NA by 0
On 30/06/14 10:32, Dr Eberhard W Lisse wrote:
Thanks,
I then set NA to 0, and can do the sutraction,
great.
Not so great. I haven't gone through the issues underlying this post,
but replacing NA by 0 will almost surely yield nonsense. "Missing" is
***not*** the same thing as "zero". Pr
Thanks,
I then set NA to 0, and can do the sutraction,
great.
el
On 2014-06-29, 22:32 , Michael Peng wrote:
> you can get a new data frame by
> merge(qpiso, qplegit, all.x = TRUE, all.y = TRUE, by = "iso" )
> Take the subtraction on the new data frame.
>
>
>
>
> 2014-06-29 11:24 GMT-05:00
you can get a new data frame by
merge(qpiso, qplegit, all.x = TRUE, all.y = TRUE, by = "iso" )
Take the subtraction on the new data frame.
2014-06-29 11:24 GMT-05:00 Dr Eberhard Lisse :
> I have two data frames like so
>
> > qpiso
> iso requests
> 1A1 20
> 2A2 199
> 3
If you really prefer solution code, then provide reproducible example code as
the footer requests.
Use ?merge with the all.x=TRUE parameter, and then perform your calculations on
the resulting combined data frame, using ?ifelse and ?is.na as needed.
--
Many thank for all your answers. it has helped me save time
--
View this message in context:
http://r.789695.n4.nabble.com/Merge-two-vectors-into-one-tp4687361p4687447.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.
Just to satisfy my curiosity:
> library(microbenchmark)
>
> a <- 1:10
> b <- 101:110
>
> microbenchmark(
+ m1=as.vector( rbind(a,b) ),
+ m2=c( rbind(a,b) ),
+ m3=as.vector( matrix(c(a,b), nrow=2, byrow=TRUE) ),
+ m4={x <- integer(length(a)*2); x[c(TRUE,FALSE)] <- a;
x[c(FALSE,TRUE)] <- b; x},
+ m5
On Mar 22, 2014, at 3:22 PM, Tham Tran wrote:
> Dear R users,
>
> Given two vectors x and y
> a=1 2 3
> b=4 5 6
>
> i want to combine them into a single vector z as 1 4 2 5 3 6
>
> Thanks for your help
Searching Stackoverflow for [r] interleave produced this idea fron @Arun, which
I think i
He wants a[1] b[1] a[2] b[2] a[3] b[3]
I think you can do:
x = as.vector(rbind(a, b))
On Mon, Mar 24, 2014 at 2:39 PM, Frans Marcelissen <
fransiepansiekever...@gmail.com> wrote:
> Why not simply
>
> > a<-1:3
> > b<-4:5
> > c(a,b)
> [1] 1 2 3 4 5
>
>
> 2014-03-22 23:22 GMT+01:00 Tham Tran :
>
Why not simply
> a<-1:3
> b<-4:5
> c(a,b)
[1] 1 2 3 4 5
2014-03-22 23:22 GMT+01:00 Tham Tran :
> Dear R users,
>
> Given two vectors x and y
> a=1 2 3
> b=4 5 6
>
> i want to combine them into a single vector z as 1 4 2 5 3 6
>
> Thanks for your help
>
> Tham
>
>
>
> --
> View this message in c
On 24/03/14 08:37, David Winsemius wrote:
On Mar 22, 2014, at 3:22 PM, Tham Tran wrote:
Dear R users,
Given two vectors x and y
a=1 2 3
b=4 5 6
i want to combine them into a single vector z as 1 4 2 5 3 6
One way:
c( matrix(c(a,b), nrow=2, byrow=TRUE) )
It is more perspicuous to use
On Mar 22, 2014, at 3:22 PM, Tham Tran wrote:
> Dear R users,
>
> Given two vectors x and y
> a=1 2 3
> b=4 5 6
>
> i want to combine them into a single vector z as 1 4 2 5 3 6
One way:
c( matrix(c(a,b), nrow=2, byrow=TRUE) )
Leaving in the usual footer because Nabble's interface usually o
Hi,
May be this helps:
a <- 1:3
b <- 4:6
z <- as.vector(rbind(a,b))
z
#[1] 1 4 2 5 3 6
#or
z1 <- setNames(c(a,b),rep(seq_along(a),2))
z1 <- as.vector(z1[order(names(z1))])
z1
#[1] 1 4 2 5 3 6
A.K.
Dear R users,
Given two vectors x and y
a=1 2 3
b=4 5 6
i want to combine them into a sin
Hi,You haven't provided any example dataset.
set.seed(42)
dat1 <- data.frame(dates=seq(as.POSIXct("2009-01-01 00:00:00",format="%Y-%m-%d
%H:%M:%S"),by= '12 hour', length=12),Field1=rnorm(12),Field2=LETTERS[1:12])
set.seed(395)
dat2 <- data.frame(dates=seq(as.POSIXct("2009-01-01 00:00:00",for
Hi,
It would be better to post a reproducible example.
set.seed(42)
df1 <-
cbind(as.data.frame(matrix(sample(60,40,replace=TRUE),ncol=4)),date=seq(as.Date("2013-05-10"),length.out=10,by=1))
colnames(df1)[-5] <- paste0("mod",1:4)
set.seed(14)
df2 <-
data.frame(obs=1:20,wy=sample(12,20,replac
Thanks
Simone
> Il giorno 15/nov/2013, alle ore 15:30, Joshua Ulrich
> ha scritto:
>
>> On Fri, Nov 15, 2013 at 6:32 AM, Simone Medori wrote:
>> Hi,
>> I'm triyng to merge two xts time series objects, one of them is from
>> Return.portfolio (PerformanceAnalytics).
>>
>> Despite the merging
On Fri, Nov 15, 2013 at 6:32 AM, Simone Medori wrote:
> Hi,
> I'm triyng to merge two xts time series objects, one of them is from
> Return.portfolio (PerformanceAnalytics).
>
> Despite the merging xts objects have the same indexes, the merged object
> shows extra lines at the day before of ever
I would suggest you post this on the r-sig-ecology or r-sig-geo lists.
These seem both more relevant (r-help is for general R programming
questions, mostly) and more likely to have participants with suitable
expertise.
Cheers,
Bert
On Wed, Sep 11, 2013 at 12:32 PM, Maria Fernanda Bonetti <
ferna
?do.call
## as in
do.call(rbind, yourlist)
Cheers,
Bert
On Tue, Sep 10, 2013 at 4:00 AM, Rosario Garcia Gil wrote:
> Hello
>
> I have done resampling from a matrix data with two variables (A,B). After
> resampling a get a set of lists (see below example for resample size 2). In
> both lists
Hello
I have done resampling from a matrix data with two variables (A,B). After
resampling a get a set of lists (see below example for resample size 2). In
both lists I have the same two variables (A,B).
I want to export this data to an excel file where I have only two columns, one
per variabl
D0407 D0409 D0410 D0462 D0463 D0473 D0475 D0488 D0489 D0492
#Conti_Australia 0 0 1 0 0 0 0 0 0 0
#Conti_Korea 0 0 0 0 0 0 0 0 0 0
#Conti_Malay 0 0 0 1 0 0 0 0 0 0
#
0 1 0 0 0 0
#Conti_Malay 0 0 0 0 0 1 0 0 0 0
#Island_Sumatra 0 0 0 0 0 0 0 0 0 1
# D0493 D0504 D0536
#Conti_Australia 0 0 1
#Conti_Malay 1 0 1
#Island_Sumatra
10 D0462 D0463 D0473 D0475 D0488 D0489 D0492 D0493 D0504
#IslandA 0 0 0 0 0 0 0 0 0 0 0 0
#IslandB 0 0 0 0 0 1 0 0 0 0 1 0
# D0536
#IslandA 0
#IslandB 1
A.K.
__
t;> requested you to dput() the data to reduce these confusions.
>>
>> mat1<-as.matrix(read.table(text="
>> GID D0989 D9820 D5629 D4327 D2134
>> 1 100 1 0
>> 2 011 0 0
>> 4
dA","IslandB"),function(x) {x1<-
mat1[match(gsub(".*\\s+","",get(x)),mat1[,1]),-1];(!!colSums(x1))*1}))
# D0989 D9820 D5629 D4327 D2134
#IslandA 1 1 0 1 0
#IslandB 0 1 1 0 1
A.K.
__
10 0 0
> 7 010 0 1
> ",sep="",header=TRUE))
> IslandA<-c("GID 1", "GID 5")
> IslandB<- c("GID 2", "GID 4", "GID
> 7")t(sapply(c("IslandA","
Sums(x1)}))
> #D0989 D9820 D5629 D4327 D2134
> #IslandA FALSE FALSE TRUE FALSE TRUE
> #IslandB TRUE FALSE FALSE TRUE FALSE
>
> t(sapply(c("IslandA","IslandB"),function(x) {x1<-
> mat1[match(gsub(".*\\s+","",get(x)),row.names(m
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