This is as much a mathematics as an R question, in the this should be easy but I don't
see it category.
Assume I have a full rank p by p matrix V (aside: V = (X'X)^{-1} for a particular setup),
a p by k matrix B, and I want to complete an orthagonal basis for the space with distance
function
Hi Terry,
maybe I'm missing something, but why not define a matrix BB = V'B;
then t(B) %*% V = t(BB), then your problem reduces to finding A such
that t(BB) %*% A = 0?
Peter
On Thu, Jul 16, 2015 at 10:28 AM, Therneau, Terry M., Ph.D.
thern...@mayo.edu wrote:
This is as much a mathematics as an
Yes it is obvious --- once someone else pointed it out.
Thanks for the hint.
Terry T.
On 07/16/2015 12:52 PM, Peter Langfelder wrote:
Hi Terry,
maybe I'm missing something, but why not define a matrix BB = V'B;
then t(B) %*% V = t(BB), then your problem reduces to finding A such
that t(BB)
On Jul 4, 2015, at 3:09 AM, Alex Kim dumboisveryd...@gmail.com wrote:
Hi guys,
Suppose I have an extremely large data frame with 2 columns and .5 mil
rows. For example, the last 6 rows may look like this:
.
..
...
89 100
93 120
95 125
101NA
115
Hi guys,
Suppose I have an extremely large data frame with 2 columns and .5 mil
rows. For example, the last 6 rows may look like this:
.
..
...
89 100
93 120
95 125
101NA
115NA
123NA
124NA
I would like to manipulate this data frame to
Hi guys,
Suppose I have an extremely large data frame with 2 columns and .5 mil
rows. For example, the last 6 rows may look like this:
.
..
...
89 100
93 120
95 125
101NA
115NA
123NA
124NA
I would like to manipulate this data frame to
Hi guys,
Suppose I have an extremely large data frame with 2 columns and .5 mil
rows. For example, the last 6 rows may look like this:
.
..
...
89 100
93 120
95 125
101NA
115NA
123NA
124NA
I would like to manipulate this data frame to
Many thanks,
Stéphane
Le 30 mars 2015 à 10:42, peter dalgaard pda...@gmail.com a écrit :
On 30 Mar 2015, at 09:59 , Stéphane Adamowicz
stephane.adamow...@avignon.inra.fr wrote:
However, in order to help me understand, would you be so kind as to give me
a matrix or data.frame example
On 30-03-2015, at 09:59, Stéphane Adamowicz
stephane.adamow...@avignon.inra.fr wrote:
Le 27 mars 2015 à 18:01, David Winsemius dwinsem...@comcast.net a écrit :
On Mar 27, 2015, at 3:41 AM, Stéphane Adamowicz wrote:
Well, it seems to work with me.
No one is doubting that it
On 30 Mar 2015, at 09:59 , Stéphane Adamowicz
stephane.adamow...@avignon.inra.fr wrote:
However, in order to help me understand, would you be so kind as to give me a
matrix or data.frame example where « complete.cases(X)== T » or «
complete.cases(X)== TRUE » would give some unwanted
Le 27 mars 2015 � 18:01, David Winsemius dwinsem...@comcast.net a �crit :
On Mar 27, 2015, at 3:41 AM, St�phane Adamowicz wrote:
Well, it seems to work with me.
No one is doubting that it worked for you in this instance. What Peter D. was
criticizing was the construction :
On 27 Mar 2015, at 09:58 , Stéphane Adamowicz
stephane.adamow...@avignon.inra.fr wrote:
data_no_NA - data[, complete.cases(t(data))==T]
Ouch! logical == TRUE is bad, logical == T is worse:
data[, complete.cases(t(data))]
--
Peter Dalgaard, Professor,
Center for Statistics, Copenhagen
: [R] matrix manipulation question
Well, it seems to work with me.
Y - as.matrix(airquality)
head(Y, n=8)
Ozone Solar.R Wind Temp Month Day
[1,]41 190 7.4 67 5 1
[2,]36 118 8.0 72 5 2
[3,]12 149 12.6 74 5 3
[4,]18 313 11.5 62 5
Well, it seems to work with me.
Y - as.matrix(airquality)
head(Y, n=8)
Ozone Solar.R Wind Temp Month Day
[1,]41 190 7.4 67 5 1
[2,]36 118 8.0 72 5 2
[3,]12 149 12.6 74 5 3
[4,]18 313 11.5 62 5 4
[5,]NA NA 14.3 56
Why not use complete.cases() ?
data_no_NA - data[, complete.cases(t(data))==T]
Le 27 mars 2015 à 06:13, Jatin Kala jatin.kala...@gmail.com a écrit :
Hi,
I've got a rather large matrix of about 800 rows and 60 columns.
Each column is a time-series 800 long.
Out of these 60 time
Le 27 mars 2015 à 12:34, PIKAL Petr petr.pi...@precheza.cz a écrit :
Very, very, very bad solution.
as.matrix can change silently your data to unwanted format,
complete.cases()==T is silly as Peter already pointed out.
Perhaps, but it happens that in the original message, the
Hi
-Original Message-
From: Stéphane Adamowicz [mailto:stephane.adamow...@avignon.inra.fr]
Sent: Friday, March 27, 2015 1:26 PM
To: PIKAL Petr
Cc: peter dalgaard; r-help@r-project.org
Subject: Re: [R] matrix manipulation question
Le 27 mars 2015 à 12:34, PIKAL Petr petr.pi
example. Furthermore in my example no unwanted format occurred. You can
Yes because data.frame was (luckily) numeric.
Luck has nothing to do with this. I Chose this example on purpose …
Stéphane
__
R-help@r-project.org mailing list -- To
On Mar 27, 2015, at 3:41 AM, Stéphane Adamowicz wrote:
Well, it seems to work with me.
No one is doubting that it worked for you in this instance. What Peter D. was
criticizing was the construction :
complete.cases(t(Y))==T
... and it was on two bases that it is wrong. The first is that
On 2015-03-27 11:41, Stéphane Adamowicz wrote:
Well, it seems to work with me.
Y - as.matrix(airquality)
head(Y, n=8)
Ozone Solar.R Wind Temp Month Day
[1,]41 190 7.4 67 5 1
[2,]36 118 8.0 72 5 2
[3,]12 149 12.6 74 5 3
[4,]18 313
Thanks Richard,
This works, rather obvious now that i think of it!
=)
On 27/03/2015 4:30 pm, Richard M. Heiberger wrote:
just reverse what you did before.
newdata - data
newdata[] - NA
newdata[,!apply(is.na(data), 2, any)] - myfunction(data_no_NA)
On Fri, Mar 27, 2015 at 1:13 AM, Jatin Kala
Hi,
I've got a rather large matrix of about 800 rows and 60 columns.
Each column is a time-series 800 long.
Out of these 60 time series, some have missing values (NA).
I want to strip out all columns that have one or more NA values, i.e.,
only want full time series.
This should do the
just reverse what you did before.
newdata - data
newdata[] - NA
newdata[,!apply(is.na(data), 2, any)] - myfunction(data_no_NA)
On Fri, Mar 27, 2015 at 1:13 AM, Jatin Kala jatin.kala...@gmail.com wrote:
Hi,
I've got a rather large matrix of about 800 rows and 60 columns.
Each column is a
Dear R users,
I have a question about matrix manipulation with its rows.
Plz see the simple example below
sample - list(matrix(1:6, nr=2,nc=3), matrix(7:12, nr=2,nc=3),
matrix(13:18,nr=2,nc=3))
sample
[[1]]
[,1] [,2] [,3]
[1,]135
[2,]246
[[2]]
[,1] [,2]
Of Kathryn Lord
Sent: Wednesday, January 16, 2013 9:00 AM
To: r-help@r-project.org
Subject: [R] matrix manipulation with its rows
Dear R users,
I have a question about matrix manipulation with its rows.
Plz see the simple example below
sample - list(matrix(1:6, nr=2,nc=3), matrix(7:12, nr
Not a great solution, I don't think, but:
kronecker(diag(2), matrix(1:6, 2, byrow=TRUE))[c(1,4),]
[,1] [,2] [,3] [,4] [,5] [,6]
[1,]123000
[2,]000456
So using a function that does this in 'lapply'
should solve the problem you state. I'm
:
Sent: Wednesday, January 16, 2013 2:59 AM
Subject: [R] matrix manipulation with its rows
Dear R users,
I have a question about matrix manipulation with its rows.
Plz see the simple example below
sample - list(matrix(1:6, nr=2,nc=3), matrix(7:12, nr=2,nc=3),
matrix(13:18,nr=2,nc=3))
sample
Dear Rlisters,
I'm writing to ask how to manipulate a matrix or dataframe in a specific way.
To elaborate, let's consider an example. Assume we have the following
3 by 4 matrix A with elements either 0 or 1,
0 1 1 0
1 0 1 1
0 0 0 1
From the original matrix A, I'd like to generate a new
On Thu, Jun 14, 2012 at 01:11:45PM +, G. Dai wrote:
Dear Rlisters,
I'm writing to ask how to manipulate a matrix or dataframe in a specific way.
To elaborate, let's consider an example. Assume we have the following
3 by 4 matrix A with elements either 0 or 1,
0 1 1 0
1 0 1 1
0
thank you, Petr.
This is exactly what I'm looking for in my post.
An related question can be how to get an arbitrary weight, say if row1
and row 2 have 1 common value 1, then assign a weight 10, if row 1 and
row 2 have 2 common value 1, then assign a weight 12. I'm not so sure
how to expand your
On Thu, Jun 14, 2012 at 02:24:20PM -0400, cowboy wrote:
thank you, Petr.
This is exactly what I'm looking for in my post.
An related question can be how to get an arbitrary weight, say if row1
and row 2 have 1 common value 1, then assign a weight 10, if row 1 and
row 2 have 2 common value 1,
Hello everyone,
I have a 2 x 5 matrix: say
0.2 0.3 1 -1 3
0.2. 0.4 5 0.5 -1
I want to replace all the values greater than or equal to 1 with 1 and those
less than or equal to 0 with 0. So I should end up with a mtrix looking
like:
0.2 0.3 1 0 1
0.2. 0.4 1 0.5 0
It's very easy to do in two steps:
testmat - matrix(c(.2, .3, 1, -1, 3, .2, .4, 5, .5, -1), byrow=TRUE, nrow=2)
testmat
[,1] [,2] [,3] [,4] [,5]
[1,] 0.2 0.31 -1.03
[2,] 0.2 0.45 0.5 -1
testmat[testmat = 1] - 1
testmat[testmat 0] - 0
testmat
[,1] [,2] [,3] [,4]
Hi all,
I have a problem with getting my code to do what I want!
This is the code I have:
create.means.one.size-function(nsample,var,nboot){
mat.x-matrix(0,nrow=nboot,ncol=nsample)
for(i in 1:nboot){
mat.x[i,]-sample(var,nsample,replace=T)
}
mean.mat-rep(0,nboot)
for(i in
hello
I think if you try this:
for(j in 1: length(nsample)){
MEANS[,]-create.means.one.size(j,var,nboot)
}
it will work
--
View this message in context:
http://r.789695.n4.nabble.com/Matrix-manipulation-in-for-loop-tp3525849p3525888.html
Sent from the R help mailing list
HI,
here is another solution:
int - sample(1:20,10)
int
[1] 10 4 5 2 14 17 9 11 16 13
mat-matrix(11:30,ncol=4)
mat
[,1] [,2] [,3] [,4]
[1,] 11 16 21 26
[2,] 12 17 22 27
[3,] 13 18 23 28
[4,] 14 19 24 29
[5,] 15 20 25 30
mat[apply(mat,1,
Hi all!
I have a vector, let's say for example int - sample(1:20,10);
for now:
now I have a matrix...
M = m x n
where the first column is a feature column and most likely shares at least
one of the int (interesting) numbers.
I want to extract the rows where int[] = M[,1]
I thought:
Hi:
Here' s one approach:
int - sample(1:20,10)
m - matrix(sample(1:40, 20), nrow = 10)
int
[1] 7 12 4 6 1 19 17 20 15 5
m
[,1] [,2]
[1,]9 15
[2,] 23 32
[3,] 40 14
[4,] 19 38
[5,] 286
[6,] 26 18
[7,] 34 22
[8,]7 35
[9,] 213
Hi,
Is there a quick way to go from this matrix:
A
[,1] [,2] [,3]
[1,]111
[2,]222
[3,]333
[4,]444
[5,]5 NA5
[6,] NA NA6
[7,] NA NA NA
to this matrix:
B
[,1] [,2] [,3]
[1,]1 NA NA
[2,]2 NA1
[3,]
try this:
x
V2 V3 V4
[1,] 1 1 1
[2,] 2 2 2
[3,] 3 3 3
[4,] 4 4 4
[5,] 5 NA 5
[6,] NA NA 6
[7,] NA NA NA
offset - c(0,2,1)
# add the control to the data and make two copies so we can offset
x.new - rbind(offset, x, x)
result - apply(x.new, 2, function(.col){
+
Many thanks-its worked a treat :-)
Emma
--
View this message in context:
http://r.789695.n4.nabble.com/Matrix-Manipulation-tp3027266p3027307.html
Sent from the R help mailing list archive at Nabble.com.
__
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Dear fellow R-users,
Say we have a matrix x, defined as follows
set.seed(50)
x - matrix(rbinom(100*5,1, p=0.75),nrow=100, ncol=5)
Now the interpretation of x is that each for of x is actually a sequence of
length 5, and i would like to transform x in such a way that I can describe the
On Tue, 1 Sep 2009, Gregory Gentlemen wrote:
Dear fellow R-users,
Say we have a matrix x, defined as follows
set.seed(50)
x - matrix(rbinom(100*5,1, p=0.75),nrow=100, ncol=5)
Now the interpretation of x is that each for of x is actually a sequence
of length 5, and i would like to transform
Hello everyone,
I have a couple of fairly simple questions (I hope) the answers to which I
cannot find through the documentation at the moment.
1. I would like to delete the a row from a matrix if a certain elimination
criterion is met. I am familiar with x - x[-7,] (to remove row 7, for
Try this:
For the first and the second question:
transform(subset(d, row.names(d) != 2), row.names=NULL)
On Thu, Jun 11, 2009 at 3:53 PM, Payam Minoofar payam.minoo...@meissner.com
wrote:
Hello everyone,
I have a couple of fairly simple questions (I hope) the answers to which I
cannot
On Jun 11, 2009, at 2:53 PM, Payam Minoofar wrote:
Hello everyone,
I have a couple of fairly simple questions (I hope) the answers to
which I cannot find through the documentation at the moment.
1. I would like to delete the a row from a matrix if a certain
elimination criterion is
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of David Winsemius
Sent: Thursday, June 11, 2009 1:49 PM
To: Payam Minoofar
Cc: r-help@r-project.org
Subject: Re: [R] Matrix manipulation
On Jun 11, 2009, at 2:53 PM, Payam
Or perhaps:
M10[rowSums(M10 == 63) == 0, ]
On Thu, Jun 11, 2009 at 4:49 PM, David Winsemiusdwinsem...@comcast.net wrote:
On Jun 11, 2009, at 2:53 PM, Payam Minoofar wrote:
Hello everyone,
I have a couple of fairly simple questions (I hope) the answers to which I
cannot find through the
From: metal_lical...@live.com
To: r-help@r-project.org
Subject: About Matrix
Date: Tue, 28 Apr 2009 11:28:43 +0300
Hi, Dear R users,
I have a question:
I have A matrix which is 11519X14
and B matrix which is 1764X14,
How do I get C matrix which is The remaining matrix after
use the '-' feature.
mat - matrix(rnorm(100), nrow = 10)
#snip the second row
mat[-2,]
#snip the third column
mat[,-3]
#snip rows 5 and 7
mat[-c(5,7),]
cheers
tc
On 10/23/07, [EMAIL PROTECTED] [EMAIL PROTECTED] wrote:
Hi everyone,
suppose I have a 2D matrix, is there a command to snip out
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