:41
To: R help
Tárgy: [R] cbind question, please
Hello!
I have a cbind type question, please: Suppose I have the following:
dog - 1:3
cat - 2:4
tree - 5:7
and a character vector
big.char - c(dog,cat,tree)
I want to end up with a matrix that is a cbind of dog, cat, and tree.
This is a toy
On 24/04/15 10:41, Erin Hodgess wrote:
Hello!
I have a cbind type question, please: Suppose I have the following:
dog - 1:3
cat - 2:4
tree - 5:7
and a character vector
big.char - c(dog,cat,tree)
I want to end up with a matrix that is a cbind of dog, cat, and tree.
This is a toy example.
-Original Message-
From: erinm.hodg...@gmail.com
Sent: Thu, 23 Apr 2015 18:41:05 -0400
To: r-h...@stat.math.ethz.ch
Subject: [R] cbind question, please
Hello!
I have a cbind type question, please: Suppose I have the following:
dog - 1:3
cat - 2:4
tree - 5:7
To: R help
Subject: [R] cbind question, please
Hello!
I have a cbind type question, please: Suppose I have the following:
dog - 1:3
cat - 2:4
tree - 5:7
and a character vector
big.char - c(dog,cat,tree)
I want to end up with a matrix
, cat,
tree, big.char)))
John Kane
Kingston ON Canada
-Original Message-
From: erinm.hodg...@gmail.com
Sent: Thu, 23 Apr 2015 18:41:05 -0400
To: r-h...@stat.math.ethz.ch
Subject: [R] cbind question, please
Hello!
I have a cbind type question, please: Suppose I have
Is this what you're looking for?
dog - 1:3
bat - 2:4
tree - 5:7
big.char - c(dog,bat,tree)
do.call(cbind,lapply(big.char, get))
[,1] [,2] [,3]
[1,]125
[2,]236
[3,]347
On Thu, Apr 23, 2015 at 3:41 PM, Erin Hodgess erinm.hodg...@gmail.com wrote:
G'day Erin,
On Thu, 23 Apr 2015 20:51:18 -0400
Erin Hodgess erinm.hodg...@gmail.com wrote:
Here is the big picture. I have a character vector with all of the
names of the variables in it.
I want to cbind all of the variables to create a matrix.
Doing 3 is straightforward, but many, not
I am amazed at the number of rather obtuse misunderstandings of the
actual nature of Erin's question.
The suggestion that Erin should read the intro to R made me smile. Erin
is a long time and highly sophisticated user of R; she has no need to
read the intro. The person who made that
Hello Erin,
I think you have explain your goal more detailed. Maybe I am completely
lost but as far as I understand now you only need the command cbind:
m1 - cbind(dog, dat, tree)
dog cat tree
[1,] 1 25
[2,] 2 36
[3,] 3 47
But I can't imagine that is the solution
You could do something tricky like
do.call(cbind, lapply(big.char, as.name))
dog cat tree
[1,] 1 25
[2,] 2 36
[3,] 3 47
but you are usually better off creating these things as part of a list
and passing that to do.call(cbind, list).
There is a slight danger
Perhaps:
dog - 1:3
cat - 2:4
tree - 5:7
big.char - cbind(dog,cat,tree)
big.char
dog cat tree
[1,] 1 25
[2,] 2 36
[3,] 3 47
colnames(big.char)-c(dog,cat,tree)
big.char
dog cat tree
[1,] 1 25
[2,] 2 36
[3,] 3 47
Clint Bowman
Hi Erin,
Well, if I do this:
dog - 1:3
cat - 2:4
tree - 5:7
dct-cbind(dog,cat,tree)
I get this:
dct
dog cat tree
[1,] 1 25
[2,] 2 36
[3,] 3 47
If I assume that you want to include the character vector as well:
rownames(dct)-big.char
dct
Jim
On Fri, Apr 24, 2015
On Apr 23, 2015, at 5:41 PM, Erin Hodgess erinm.hodg...@gmail.com wrote:
Hello!
I have a cbind type question, please: Suppose I have the following:
dog - 1:3
cat - 2:4
tree - 5:7
and a character vector
big.char - c(dog,cat,tree)
I want to end up with a matrix that is a cbind of
: Friday, 24 April 2015 10:41a
To: R help
Subject: [R] cbind question, please
Hello!
I have a cbind type question, please: Suppose I have the following:
dog - 1:3
cat - 2:4
tree - 5:7
and a character vector
big.char - c(dog,cat,tree)
I want to end up with a matrix that is a cbind of dog, cat
These are great! Thank you!
On Thu, Apr 23, 2015 at 7:14 PM, William Dunlap wdun...@tibco.com wrote:
You could do something tricky like
do.call(cbind, lapply(big.char, as.name))
dog cat tree
[1,] 1 25
[2,] 2 36
[3,] 3 47
but you are usually better
Hello!
I have a cbind type question, please: Suppose I have the following:
dog - 1:3
cat - 2:4
tree - 5:7
and a character vector
big.char - c(dog,cat,tree)
I want to end up with a matrix that is a cbind of dog, cat, and tree.
This is a toy example. There will be a bunch of variables.
I
Two solutions proposed -- not entirely orthogonal, but both do the
trick. Instead of nesting cbin in a loop (as I did originally -- OP,
below),
1\ do.call(cbind, lapply(mat_list, as.vector))
or
2\ sapply(mat_list,function(x) as.vector(x))
Both work fine. Thanks to Jeff Laake (2) +
Subject: Re: [R] cbind in a loop...better way? | summary
Two solutions proposed -- not entirely orthogonal, but both do the
trick. Instead of nesting cbin in a loop (as I did originally -- OP,
below),
1\ do.call(cbind, lapply(mat_list, as.vector))
or
2\ sapply(mat_list,function(x) as.vector(x
]
On Behalf Of Evan Cooch
Sent: Thursday, October 9, 2014 7:37 AM
To: Evan Cooch; r-help@r-project.org
Subject: Re: [R] cbind in a loop...better way? | summary
Two solutions proposed -- not entirely orthogonal, but both do the
trick. Instead of nesting cbin in a loop (as I did originally -- OP
, 2014 7:37 AM
To: Evan Cooch; r-help@r-project.org
Subject: Re: [R] cbind in a loop...better way? | summary
Two solutions proposed -- not entirely orthogonal, but both do the
trick. Instead of nesting cbin in a loop (as I did originally -- OP,
below),
1\ do.call(cbind, lapply(mat_list, as.vector
...or some such. I'm trying to work up a function wherein the user
passes a list of matrices to the function, which then (1) takes each
matrix, (2) performs an operation to 'vectorize' the matrix (i.e., given
an (m x n) matrix x, this produces the vector Y of length m*n that
contains the
of Anthropology
Texas AM University
College Station, TX 77840-4352
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Evan Cooch
Sent: Wednesday, October 8, 2014 2:13 PM
To: r-help@r-project.org
Subject: [R] cbind in a loop...better way
[mailto:r-help-boun...@r-project.org] On
Behalf Of Evan Cooch
Sent: Wednesday, October 8, 2014 2:13 PM
To: r-help@r-project.org
Subject: [R] cbind in a loop...better way?
...or some such. I'm trying to work up a function wherein the user
passes a list of matrices to the function, which then (1
Hi,
The rownames part is not clear as your expected output and input files didn't
show them as rownames.
##Suppose you have all the files in a folder
##here I am creating the names of those files
files1 - paste0(sample, rep(1:777, each=29),chr,1:29,.txt)
length(files1)
#[1] 22533
lst1 -
Hi All,
I have create a matrix using cbind() function as follows:
a=c(1,2,3)
b=c('a','b','c')
c=c(ee,tt,rr)
k=cbind(a,b,c)
Problem: when we print the matrix k,
k
a b c
[1,] 1 a ee
[2,] 2 b tt
[3,] 3 c rr
we can see that rows are represented by [1,] , [2,] and [3,].
Hi,
Try:
k[,a]
#[1] 1 2 3
k[,b]
#[1] a b c
k[,c]
#[1] ee tt rr
A.K.
On Tuesday, October 22, 2013 11:37 PM, Vivek Singh vksingh.ii...@gmail.com
wrote:
Hi All,
I have create a matrix using cbind() function as follows:
a=c(1,2,3)
b=c('a','b','c')
c=c(ee,tt,rr)
k=cbind(a,b,c)
Hard to say, not sure what you want to do. But the columns are not denoted by
[a], [b] or [c]. You should learn to use the str function to understand what
various expressions really are, and return to the Introduction to R document
that comes with the software. There is a distinct difference
In addition to what the others have told you, it looks like you might
be confusing
matrices with data.frames. Please see
?data.frame
I think what you are looking for is
b - c('a','b','c')
c - c(ee,tt,rr)
k - cbind(a,b,c)
K - data.frame(a, b, c)
K
a b c
1 1 a ee
2 2 b tt
3 3 c rr
I
Hi,
You can save it in file. I copy and paste:
Subtype,Gender,Expression
A,m,-0.54
A,f,-0.8
B,f,-1.03
C,m,-0.41
on the gedit and save it as data1.csv. You might be able to do the same
with notepad.
x - read.csv(data1.csv,header=T,sep=,)
x2 - read.csv(data2N.csv,header=T,sep=,)
x3 -
Hi,
I can't seem to get this to work:
http://www.endmemo.com/program/R/cbind.php
Do I save the data as data1.csv in note pad and pull in the file? Do I type
data1.csv-Subtype,Gender,Expression,A,m,-0.54,A,f,-0.8,B,f,-1.03,C,m,-0.41??
I can do a simple matrix. But, I want to have headers and
Hello all,
I posted a question about cbind and an error when specifying the check.names
argument here:
http://stackoverflow.com/questions/17810470/cbind-error-with-check-names
I was advised to take this to R-devel, but before that, I'd like to get the
opinions of people on this list.
Thanks for the explanations.
Wouldn't the following bit of checking in do.call() make it easier to figure
such things out in the future?
my.call - function(what,args,...) {
## Get the name of the function to call.
if (!is.character(what))
whatStr - deparse(substitute(what))
On Tue, Apr 9, 2013 at 7:31 AM, Harry Mamaysky h.mamay...@gmail.com wrote:
Thanks for the explanations.
Wouldn't the following bit of checking in do.call() make it easier to figure
such things out in the future?
Sure, it would have helped you figure out your issue, but you don't
want a
That's true. So perhaps there should be a flag that turns on this error
checking. Often args is just a list that gets generated automatically and you
don't know what all of its elements are. It just leads to a bit of
non-deterministic behavior. It would be useful to have the option of flagging
On Tue, Apr 9, 2013 at 9:15 AM, Harry Mamaysky h.mamay...@gmail.com wrote:
That's true. So perhaps there should be a flag that turns on this error
checking. Often args is just a list that gets generated automatically and
you don't know what all of its elements are. It just leads to a bit of
That's a nice solution. Thanks.
Sent from my iPhone
On Apr 9, 2013, at 10:00 AM, Gabor Grothendieck ggrothendi...@gmail.com wrote:
On Tue, Apr 9, 2013 at 9:15 AM, Harry Mamaysky h.mamay...@gmail.com wrote:
That's true. So perhaps there should be a flag that turns on this error
checking.
Can someone explain why this happens when one of the list elements is named
'all'?
zz - list( zoo(1:10,1:10), zoo(101:110,1:10), zoo(201:210,1:10) )
names(zz)-c('test','bar','foo')
do.call(cbind,zz)
test bar foo
1 1 101 201
2 2 102 202
3 3 103 203
4 4 104 204
5 5 105
Because 'all' is the name of one of the arguments to cbind.zoo:
R args(cbind.zoo)
function (..., all = TRUE, fill = NA, suffixes = NULL, drop = FALSE)
NULL
do.call constructs a call somewhat like:
R cbind(test=zz$test, all=zz$all, foo=zz$foo)
The same thing would happen for list elements named
On Mon, Apr 8, 2013 at 3:54 PM, Harry Mamaysky h.mamay...@gmail.com wrote:
Can someone explain why this happens when one of the list elements is named
'all'?
zz - list( zoo(1:10,1:10), zoo(101:110,1:10), zoo(201:210,1:10) )
names(zz)-c('test','bar','foo')
do.call(cbind,zz)
test bar foo
## Hi, I'm having trouble understanding how the cbind function decides what
method to apply to its arguments. Easy cut and paste code below.
## create two columns
c1 - c(A,A,B,B)
c2 - 1:4
## cbind outputs a matrix with elements that are characters, seems
reasonable
out -
Short answer: there is a default method used here which returns a matrix. It's
defined at the C-level for speed so you don't see it with methods()
Longer: cbind() isn't a regular S3 generic since it has no UseMethod(). Look at
WRE and R-Internals (internal generics) for more info.
Best (and
Hi, we have a problem concerning the use and forecasting on the GARCH (1, 1) on
financial data.
We would like to make predictions on future returns in a loop which moves the
window of estimation forward one unit for each iteration. Our code is as
follows:
BJORN -
Complete newbie to R -- struggling with something which should be pretty
basic. Trying to create a simple data set (which I gather R refers to as a
data.frame). So
a - c(1,2,3,4,5);
b - c(0.3,0.4,0.5,0,6,0.7);
Stick the two together into a data frame (call test) using cbind
test -
Don't use cbind() -- it forces everything into a single type, here
string, which in turn becomes factor.
Simply,
data.frame(a, b, c)
Like David mentioned a few days ago, I have no idea who is promoting
this data.frame(cbind(...)) idiom, but it's a terrible idea (albeit
one that seems to be very
Still didn't work for me without cbind , although you really don't need it ;)
worked after i set options(stringsAsFactors=F).
options(stringsAsFactors=F)
df-data.frame(intVec,chaVec)
df
intVec chaVec
1 1 a
2 2 b
3 3 c
df$chaVec
[1] a b c
documentation of
cbind() works as well, but only if c is attached to the existing test variable:
tst - cbind( test, c )
tst
ab c
Sorry, I missed that the OP's real question was in character/factor,
not in the why are these all factors bit...good catch.
Rant about cbind() still stands though. :-) [Your way with cbind()
would give him all characters, not some characters and some numerics
since cbind() gives a matrix by
On Apr 10, 2012, at 11:58 AM, Rainer Schuermann wrote:
cbind() works as well, but only if c is attached to the existing
test variable:
tst - cbind( test, c )
tst
ab c
1 1 0.3 y1
2 2 0.4 y2
3 3 0.5 y3
4 4 0.6 y4
5 5 0.7 y5
str( tst )
'data.frame': 5 obs. of 3
On Apr 10, 2012, at 12:19 PM, David Winsemius wrote:
On Apr 10, 2012, at 11:58 AM, Rainer Schuermann wrote:
cbind() works as well, but only if c is attached to the existing
test variable:
tst - cbind( test, c )
tst
ab c
1 1 0.3 y1
2 2 0.4 y2
3 3 0.5 y3
4 4 0.6 y4
5
I have two one dimensional list of elements and want to perform cbind and
then write into a file. The number of entries are more than a million in
both lists. R is taking a lot of time performing this operation.
Is there any alternate way to perform cbind?
x = table1[1:100,1]
y =
, 2012 12:43 PM
To: r-help@r-project.org
Subject: [R] cbind alternate
I have two one dimensional list of elements and want to perform cbind
and then write into a file. The number of entries are more than a
million in both lists. R is taking a lot of time performing this
operation.
Is there any
On Jan 6, 2012, at 11:43 AM, Mary Kindall wrote:
I have two one dimensional list of elements and want to perform cbind and
then write into a file. The number of entries are more than a million in
both lists. R is taking a lot of time performing this operation.
Is there any alternate way to
On Jan 6, 2012, at 12:43 PM, Mary Kindall wrote:
I have two one dimensional list of elements and want to perform
cbind and
then write into a file. The number of entries are more than a
million in
both lists. R is taking a lot of time performing this operation.
Is there any alternate way
On Jan 6, 2012, at 12:39 PM, Marc Schwartz wrote:
On Jan 6, 2012, at 11:43 AM, Mary Kindall wrote:
I have two one dimensional list of elements and want to perform cbind and
then write into a file. The number of entries are more than a million in
both lists. R is taking a lot of time
Hello,
I believe this function can handle a problem of that size, or bigger.
It does NOT create the full matrix, just writes it to a file, a certain
number of lines at a time.
write.big.matrix - function(x, y, outfile, nmax=1000){
if(file.exists(outfile)) unlink(outfile)
testf
Sorry Mary,
My function would write the remainder twice, I had only tested it
with multiples of the chunk size.
(And without looking at the lenghty output correctly.)
Now checked:
write.big.matrix - function(x, y, outfile, nmax=1000){
if(file.exists(outfile)) unlink(outfile)
testf -
What is it you want to do with the data after you save it? Are you
just going to read it back into R? If so, consider using save/load.
On Fri, Jan 6, 2012 at 12:43 PM, Mary Kindall mary.kind...@gmail.com wrote:
I have two one dimensional list of elements and want to perform cbind and
then
Note that this issue was raised on StackOverflow recently.
http://stackoverflow.com/questions/7678090/xts-merge-odd-behaviour
Here's the solution:
index(a) - index(a)
index(b) - index(b)
merge(a,b)
ZWD.UGX SCHB.Close
2010-03-31 NA 28.02
2010-04-01 7.6343 NA
I have two xts objects, call them a and b, and am trying to merge them...
class(a)
[1] xts zoo
class(b)
[1] xts zoo
head(a)
2010-04-01 7.6343
2010-04-02 7.6343
2010-04-03 7.5458
2010-04-04 7.4532
2010-04-05 7.4040
2010-04-06 7.3317
head(b)
2010-04-01 568.80
2010-04-05 571.01
Hmm, that's quite puzzling. I don't know but I'd willing to guess the
time/date stamps on a,b are more different than the output is leading us to
believe. My experience is that there's always more to a time/date object to
trick us up than one would expect.
If you could, dput() them so we can see
On 26 August 2011 03:37, R. Michael Weylandt michael.weyla...@gmail.com wrote:
If you could, dput() them so we can see everything about them. You also
might see if merge() gives you more expected behavior
Ok...
dput(a)
structure(c(7.6343, 7.6343, 7.5458, 7.4532, 7.404, 7.3317), class =
Hi
On 26 August 2011 03:37, R. Michael Weylandt
michael.weyla...@gmail.com wrote:
If you could, dput() them so we can see everything about them. You
also
might see if merge() gives you more expected behavior
Ok...
dput(a)
structure(c(7.6343, 7.6343, 7.5458, 7.4532, 7.404,
Hi:
Try this:
require('xts')
merge.zoo(zoo(a), zoo(b), all = c(TRUE, TRUE))
ZWD.UGX SCHB.Close
2010-03-31 NA 28.02
2010-04-01 7.6343 NA
2010-04-02 7.6343 NA
2010-04-03 7.5458 NA
2010-04-04 7.4532 28.30
2010-04-05 7.4040 28.38
I was rather too quick
It has probably something to do with versions of zoo and xts
after updating to zoo 1.7.4 and xts 0.8.2 I got with your examples
merge(a,b)
ZWD.UGX SCHB.Close
2010-04-01 NA 28.02
2010-04-01 7.6343 NA
2010-04-02 7.6343 NA
2010-04-03
This seems to be the easiest way to handle the problem:
a = xts(coredata(a), time(a))
b = xts(coredata(b), time(b))
merge(a,b)
ZWD.UGX SCHB.Close
2010-03-31 NA 28.02
2010-04-01 7.6343 NA
2010-04-02 7.6343 NA
2010-04-03 7.5458 NA
2010-04-04
For a little lateral thinking, consider the use of . on the LHS. That could
play out as follows:
myvars - c(Ozone,Wind)
f - . ~ Month
j - union(all.vars(f[[3]]), myvars)
aggregate(. ~ Month, data=airquality[j], mean, na.rm=T)
MonthOzone Wind
1 5 23.61538 11.457692
2 6
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 Dimitri Liakhovitski
Sent: Thursday, July 14, 2011 1:45 PM
To: David Winsemius
Cc: r-help
Subject: Re: [R] cbind in aggregate
On Jul 15, 2011, at 15:06 , peter dalgaard wrote:
For a little lateral thinking, consider the use of . on the LHS. That could
play out as follows:
myvars - c(Ozone,Wind)
f - . ~ Month
j - union(all.vars(f[[3]]), myvars)
aggregate(. ~ Month, data=airquality[j], mean, na.rm=T)
Month
Hello!
I am aggregating using a formula in aggregate - of the type:
aggregate(cbind(var1,var2,var3)~factor1+factor2,sum,data=mydata)
However, I actually have an object (vector of my variables to be aggregated):
myvars-c(var1,var2,var3)
I'd like my aggregate formula (its cbind part) to be able
On Jul 14, 2011, at 3:05 PM, Dimitri Liakhovitski wrote:
Hello!
I am aggregating using a formula in aggregate - of the type:
aggregate(cbind(var1,var2,var3)~factor1+factor2,sum,data=mydata)
However, I actually have an object (vector of my variables to be
aggregated):
Thank you, David, it does work.
Could you please explain why? What exactly does changing it to as matrix do?
Thank you!
Dimitri
On Thu, Jul 14, 2011 at 3:25 PM, David Winsemius dwinsem...@comcast.net wrote:
On Jul 14, 2011, at 3:05 PM, Dimitri Liakhovitski wrote:
Hello!
I am aggregating
Dmitri:
Look at my vars from myvars-c(value1,value2)
It's just a character vector of length 2!
You can't cbind a character vector of length 2! These are not
references/pointers.
It's not at all clear to me what you ultimately want to do, but IF
it's: pass a character vector of names to be used
Dmitri:
as.matrix makes a matrix out of the dataframe that is passed to it.
As a further note I attempted and failed for reasons that are unclear
to me to construct a formula that would (I hoped) preserve the column
names which are being mangle in the posted effort:
form -
Thanks a lot!
actually, what I tried to do is very simple - just passing tons of
variable names into the formula. Maybe that get thing suggested by
Bert would work...
Dimitri
On Thu, Jul 14, 2011 at 4:01 PM, David Winsemius dwinsem...@comcast.net wrote:
Dmitri:
as.matrix makes a matrix out
David - I tried exactly the thing you did (and after that asked my
question to the forum):
form - as.formula(paste(
cbind(,
paste( myvars, collapse=,),
) ~ group+mydate,
sep= ) )
And it did not work - although
To: David Winsemius
Cc: r-help
Subject: Re: [R] cbind in aggregate formula - based on an existing object
(vector)
Thanks a lot!
actually, what I tried to do is very simple - just passing tons of
variable names into the formula. Maybe that get thing suggested by
Bert would work...
Dimitri
On Thu, Jul
Hi:
I think Bill's got the right idea for your problem, but for the fun of
it, here's how Bert's suggestion would play out:
# Kind of works, but only for the first variable in myvars...
aggregate(get(myvars) ~ group + mydate, FUN = sum, data = example)
group mydate get(myvars)
1 group1
How can I cbind three or more matrices like A,B and C. This does not work:
cbind(A,B,C)
--
Thanks,
Jim.
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read
do.call(cbind, list(A, B, C))
On Sat, Jun 4, 2011 at 7:14 PM, Jim Silverton jim.silver...@gmail.com wrote:
How can I cbind three or more matrices like A,B and C. This does not work:
cbind(A,B,C)
--
Thanks,
Jim.
--
Sarah Goslee
http://www.functionaldiversity.org
A, B, C should have the same number of rows.
mlist = replicate(3, matrix(rnorm(6), 2), simplify=FALSE)
names(mlist) = LETTERS[seq_along(mlist)]
with(mlist, cbind(A,B,C))
or,
do.call(cbind, mlist)
HTH,
baptiste
On 5 June 2011 11:14, Jim Silverton jim.silver...@gmail.com wrote:
How can I
Jim -
In what sense does cbind(A,B,C) not work?
A = matrix(rnorm(10),5,2)
B = matrix(rnorm(15),5,3)
C = matrix(rnorm(20),5,4)
cbind(A,B,C)
[,1] [,2] [,3] [,4] [,5][,6]
[1,] -0.54194873 -1.1105170 -0.479010 0.619911163 0.1610162 0.49028633
[2,]
Hello jholtman,
Thanks very much for the suggestion.
I tried using as.is=TRUE and it worked as the way I expected. Sorry for
not being clear about the problem in my mail.
The characters are very much needed, cos I am trying to create a signaling
network using Rgraphviz.
Thanks again.
Regards,
I am new to R and request your kind help.
I have a table like the one below,
one two
1 apple fruit
2 ball game
3 chair wood
4 wood plain
5 fruitbanana
6 cloth silk
Note: duplicate entries are there
the task is to create relations to each
This comes about since when using read.table (which I assume you did,
but you did not show us what commands you were using), characters are
converted to factors. If you are not using factors, then you probably
want the data read in as characters. You should understand the use of
'str' to look at
Is there a way to rename the columns to something like A and B in the cbind
function?
x - rnorm(n = 10, mean = 0, sd = 1)
y - rnorm(n = 10, mean = 0, sd = 1)
cbind(x,y)
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https://stat.ethz.ch/mailman/listinfo/r-help
cbind(A=x, B=y)
On Sat, Aug 21, 2010 at 6:53 PM, r.ookie r.oo...@live.com wrote:
Is there a way to rename the columns to something like A and B in the cbind
function?
x - rnorm(n = 10, mean = 0, sd = 1)
y - rnorm(n = 10, mean = 0, sd = 1)
cbind(x,y)
On 2010-08-21 16:53, r.ookie wrote:
Is there a way to rename the columns to something like A and B in the cbind
function?
x- rnorm(n = 10, mean = 0, sd = 1)
y- rnorm(n = 10, mean = 0, sd = 1)
cbind(x,y)
Unless I completely misunderstand your query, ?cbind tells you:
... vectors or
Thanks.
On Aug 21, 2010, at 4:01 PM, RICHARD M. HEIBERGER wrote:
cbind(A=x, B=y)
On Sat, Aug 21, 2010 at 6:53 PM, r.ookie r.oo...@live.com wrote:
Is there a way to rename the columns to something like A and B in the cbind
function?
x - rnorm(n = 10, mean = 0, sd = 1)
y - rnorm(n = 10, mean
Hi there,
I need to merge/bind two time series objects (from RPackage: timeSeries) by
column. The theory is laid out nicely, even for overlapping indices.
In my example, I have overlapping indices (01.01.2001), where in one time
series I have one data point and in the other redundant data.
I have 30 files in the current directories, i would like to perform the
cbind(fil1,file2,file3,file4file30)
how could i do this in a for loop:
such as:
file2 - list.files(pattern=.out3$)
for (j in file2) {
cbind(j)...how to implement cbind here
}
Thanks.
Hi,
Assuming that you have read the files into R,
and that their names (in R) are held in some object
(e.g., 'file2'), then this works
do.call(what = cbind, args = mget(x = file2, envir = .GlobalEnv)
Here is a reproducible example:
x1 - data.frame(x = 1:10)
x2 - data.frame(y = 1:10)
file.names
On Jul 12, 2010, at 2:32 AM, Joshua Wiley wrote:
Hi,
Assuming that you have read the files into R,
and that their names (in R) are held in some object
(e.g., 'file2'), then this works
do.call(what = cbind, args = mget(x = file2, envir = .GlobalEnv)
Here is a reproducible example:
x1 -
Try this:
do.call(cbind, lapply(dir(pattern = '.out3$'), read.table))
On Mon, Jul 12, 2010 at 1:08 AM, jd6688 jdsignat...@gmail.com wrote:
I have 30 files in the current directories, i would like to perform the
cbind(fil1,file2,file3,file4file30)
how could i do this in a for loop:
I wrote a function that cbinds vectors of different lengths. Basically, the
function adds NAs to shorter vectors and cbinds in the end. I'm attaching
the code for guidance.
# This function takes in a list() of vectors and cbinds them into a
data.frame.
timerMelt - function(x, write.down = FALSE,
Hello R help
I have a dataframe, with 71 samples (rows) and 30 variables. I got linear
models for some of the variables, and I want to join fitted and residuals of
these models to the data frame. Sometimes, these vectors have the same length
of the dependant variable, but in a few cases, NA
On 01.05.2010 21:09, Giovanni Azua wrote:
Hello,
I have three method types and 100 generalization errors for each, all in the
range [0.65,0.81]. I would like to make a stacked histogram plot using ggplot2
with this data ...
Therefore I need a data frame of the form e.g.
Method
Hello,
I have three method types and 100 generalization errors for each, all in the
range [0.65,0.81]. I would like to make a stacked histogram plot using ggplot2
with this data ...
Therefore I need a data frame of the form e.g.
Method GE
-- --
Hello,
I read the help as well as the examples, but I can not figure out why
the following code does not produce the *given* row names, x and y:
x - 1:20
y - 21:40
rbind(
x=cbind(N=length(x), M=mean(x), SD=sd(x)),
y=cbind(N=length(y), M=mean(y), SD=sd(y))
)
Could you please help?
Thank
This gives what you want:
rbind.data.frame(
x=cbind(N=length(x), M=mean(x), SD=sd(x)),
y=cbind(N=length(y), M=mean(y), SD=sd(y))
)
On Fri, Jan 29, 2010 at 8:49 AM, soeren.vo...@eawag.ch wrote:
Hello,
I read the help as well as the examples, but I can not figure out why the
following
Hi!
29.01.2010 12:49, soeren.vo...@eawag.ch wrote:
Hello,
I read the help as well as the examples, but I can not figure out why
the following code does not produce the *given* row names, x and y:
x - 1:20
y - 21:40
rbind(
x=cbind(N=length(x), M=mean(x), SD=sd(x)),
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