Rodrigo Díaz wrote
Hi. I have a matrix like this:
cycle=c(rep(1,3),rep(2,3),rep(3,3),rep(4,3))col=c(rep(blue,2),rep(green,2),rep(blue,2),rep(green,2),rep(blue,2),rep(green,2))values=c(1:12)data.frame(cycle,col,values)
# cycle col values#1 1 blue 1#2 1 blue 2#3 1
dimnik wrote
thank you for your answer.Yes,that sounds right.I thought the same thing
but the problem is how can i generalize the command for every vector of
numbers not only for the specific example?not only for c(1,2),c(0.1,0.8).
2015-01-04 0:45 GMT+00:00 Pete Brecknock [via R]
ml-node
dimnik wrote
i want to find a functionthattakes in two vectors of numbers
thathave
the same
length.The output should be a listof vectors, where each vector
is a
sequence of
randomly generated Poisson variableswhere the
Lee wrote
Hi,
I am struggling with this issue and need some helps. The data set 'DF'
includes persons' IDs and other variables A, B, and C. Each Person has
multiple values in A, B, and C. What I am trying to do is 1) selecting a
maximum value of B within same ID, and 2) making a new data
slavia wrote
Hi,
I have some values that I need to represente in the same plot.
For exemple, if I have,
c-c(200,205,210,215,220,225,230,235)
a-c(0.032,0.44,0.86,0.65,0.53,0.213,0.46,0.231)
b-c(0.325,0.657,0.784,0.236,0.798,0.287,0,748,0.785)
Pete Brecknock wrote
Hi
The code below plots a stacked barchart.
I would like to overlay on this chart a circular plotting character at the
sum of the bars for each month. The plotted characters should be joined
with a line.
So, for 1/1/2014, I would like to see a point at 200 (-1000
Krishia wrote
Hello,
I am pretty new to R and would like to transform my 272x12 matrix into a
3264X1. I'm trying to have the setup change from:
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12
13,14,15,16,17,18,19,20,21,22, 23, 24
etc.
to
1
2
3
4
5
6
7
8
9
10
11
12
etc.
Any
eric wrote
Is there an easy way to get the midpoint between two dates in a data frame
? If I have a dataframe that looks like this :
head(x)
instDay remDay exp.time mpy
1 2006-02-02 2006-04-03 60 0.2
2 2006-04-17 2006-08-17 122 0.3
4 2006-08-17 2006-10-23 67
wudadan wrote
Dear R users,
I wonder if there is a way that I can plot a time series data which is in
a
wide format like this:
CITY_NAME 2000Q12000Q2 2000Q32000Q4 2001Q1
2001Q2 2001Q3 2001Q4 2002Q1 2002Q2
CITY1100.5210
Hi
I am trying to add the contents of the list myList to a new column z in
the data frame myDataframe
myList - list(c(A1,B1), c(A2,B2,C2), c(A3,B3))
myDataframe - data.frame(x=c(1,2,3), y=c(R,S,T))
Would like to produce
x y z
1 R A1,B1
2 S A2,B2,C2
3 T A3,B3
where z is a character
Thanks Brian. Perfect.
Brian Diggs wrote
On 7/19/2013 12:54 PM, Pete Brecknock wrote:
Hi
I am trying to add the contents of the list myList to a new column z
in
the data frame myDataframe
myList - list(c(A1,B1), c(A2,B2,C2), c(A3,B3))
myDataframe - data.frame(x=c(1,2,3), y=c(R,S,T
fitz_ra wrote
I know this is posted a lot, I've been through about 40 messages reading
how to do this so let me apologize in advance because I can't get this
operation to work unlike the many examples shown.
I have a 2 row matrix
temp
[,1] [,2] [,3] [,4] [,5]
sunny0 wrote
I'd like to integrate vectors 't' and 'w' for log(w)/(1-t)^2 where i can
vary the upper limit of the integral to change with each value of 't' and
'w', and then put the output into another vector.
So, something like this...
w=c(.33,.34,.56)
t=c(.2,.5,.1)
k-c(.3,.4,.5)
cdouglass wrote
Hello all,
Totally new to this and I'm just doing a frequency distribution analysis
on T-shirt sales by size. I have a .csv with 60 orders. I read in the
data using read.csv. If I look at the summary() or table() of the data it
looks fine, except that the shirt sizes are
tasnuvat wrote
I have a vaiable named NAM having value : 1,2,3,4,5,6,7,8,9. I want to
make an indicator variable that will take value 1 if NAM=7 or NAM=8 or
NAM=9. How can I do that?
I usually do: Var001- ifelse(NAM==7,1,0) for the simplest case.
[[alternative HTML version deleted]]
bradleyd wrote
Excuse the request from an R novice! I have a data frame (DATA) that has
two numeric columns (YEAR and DAY) and 4000 rows. For each YEAR I need to
determine the 10% and 90% quantiles of DAY. I'm sure this is easy enough,
but I am a new to this.
quantile(DATA$DAY,c(0.1,0.9))
bradleyd wrote
Thanks for your help Pete. I can almost get it to work with;
by(day,year,quantile)
but this only gives me 0% 25% 50% 75% 100%, not the ones I'm looking
for, 10% and 90%.
I have tried;
by(day,year,quantile(c(0.1, 0.9))) but this is rejected by
Error in FUN(X[[1L]],
jdbaba wrote
Hi ,
I am trying to convert the date as factor to date using as.date function
in R. I have the date in the following format
2008-01-01 02:30
I tried to use the following command :
as.Date(mydata$Date, format=%y-%m-%d )
Can somebody help me with this
bradleyd wrote
That does it, thanks. Do you think you help me a little bit further?
I actually have 4 columns, YEAR, DAY, TEMP , and IBI. They are all
numeric. I need to calculate the average TEMP and IBI values between the
10% and 90% quantiles for each YEAR.
The code
*
bradleyd wrote
Thanks Pete. The TRIM argument in the MEAN function tells me how to trim
off decimal points, but I am lost as to how to append the mean values of
TEMP and IBI between the 10% and 90% quantiles of DAY in each YEAR.
DAY is the julian date that an event occurred in certain
monicamir88 wrote
Hello!
I have a doubt with the R software. I have this function:
results - function(bCODBOD, BOD, VSS, COD, sBOD, sCOD, TSS, TKNpa, T,
NH3Ne, DO, Q, TKN, MLSS, NO3Ne, RAS, tanoxic1, tanoxic2, rbCOD, SDNR1,
SDNR2, tdanoxic, tanaerobic, IRp, P) {
bCOD - bCODBOD*BOD
nbCOD -
simonj16 wrote
Consider an urn that contains 10 tickets, labelled: 1,1,1,1,2,5,5,10,10,10
I want to draw with replacement n=40 tickets. I am interested in the sum,
Y, of the 40 ticket values that I draw
Write an R function named urn.model that simulates this experiement. What
I have below
malaka wrote
Hi,
I want to assign the ar1 , ma 1 and the intercept estimated by the
following code to three variables a, b and c respectively.
Can anyone help me with this please?
code:
a0 = 0.05; a1 = 0.1; b1 = 0.85
nu = rnorm(2500)
epsi = rep(0, 2500)
h = rep(0, 2500)
for (i in
domcastro wrote
Hi
I'm trying to convert a column of strings (nominal types) to a set of
boolean / binary / logical values. For example, in the column there is
red, blue, green and yellow. There are 100 rows and each has a colour. I
want to convert the column to 4 columns: red, blue,
Any recommendations for how I can embed my title below in a single red
strip/box across the plot area in the outer margin?
I would like to avoid the color appearing in any other area defined by the
oma.
# Example Plot
par(mfrow=c(2,2),mar=c(4,4,2,2), oma = c(1, 1, 3, 1))
plot(rnorm(100),1:100)
David Winsemius wrote
On Jan 15, 2013, at 2:49 PM, Pete Brecknock wrote:
Any recommendations for how I can embed my title below in a single red
strip/box across the plot area in the outer margin?
I would like to avoid the color appearing in any other area defined by
the
oma.
The code
David Winsemius wrote
On Jan 15, 2013, at 3:25 PM, Pete Brecknock wrote:
David Winsemius wrote
On Jan 15, 2013, at 2:49 PM, Pete Brecknock wrote:
Any recommendations for how I can embed my title below in a single red
strip/box across the plot area in the outer margin?
I would like
Yao He wrote
Dear All:
I want to calculate the mean and sd for each column in a data.frame.
Taking data(iris) for example:
I tried sapply(iris[,-5],mean,na.rm=T) or sapply(iris[,-5],sd,na.rm=T)
to calculate the mean and sd .But sapply() transfer a function per
time. How to transfer two
kiotoqq wrote
I want to add a shorter column to my dataset with the function merge,
it
should be filled with NAs wo be as long as the other colums, like this:
idage
946
856
6 52
5 NA
4 NA
3 NA
1 NA
i did this:
pa1 - merge(pa1, an1, by=mergeid)
and it says
jholtman wrote
What do you want to do with the samples after you generate them? What
are the parameters for the normal distribution? You left a lot of
information out. You can generate 500,000 numbers and then store them
in a 1x50 matrix quite easily.
On Sat, Nov 24, 2012 at 5:03 PM,
Jasmin wrote
I try to use hansen-hurwitz and horvitz-thompson estimator.So I should
generate samples which come from normal distribution (mu=50,sigma=3).
I have taken the liberty of scaling the problem down to something more
digestible and have changed lines 5 and 7 in your code
nsamples=10
frespider wrote
Hi,
it is possible. but don't you think it will slow the code if you convert
to data.frame?
Thanks
Date: Thu, 22 Nov 2012 18:31:35 -0800
From:
ml-node+s789695n4650500h51@.nabble
To:
frespider@
Subject: RE: Summary statistics for matrix columns
TheRealJimShady wrote
Hi Peter,
Yes, I did miss an e from the first 'not' in the brackets at the end
of the message, sorry.
Thanks for that code, but when I use it, it creates a new column
called csum which simply contains the values of the variable x . i.e.
it just duplicates the values
frespider wrote
Hi,
is there a way I can calculate a summary statistics for a columns matrix
let say we have this matrix
x - matrix(sample(1:8000),nrow=100)
colnames(x)- paste(Col,1:ncol(x),sep=)
if I used summary
summary(x)
i get the output for each column but I need the output
Peterso wrote
Uwe:
I was actually trying to stack one table on top of the other. All column
names are the same except for the Part1 and Part 2. My final table should
look like the table below. Maybe it is possible to change the names of
Part1 and Part 2 to Part?
A B C Part
1 0 1 550
:\\temp\\psw09.xls, :
cannot open URL 'http://ir.eia.gov/wpsr/psw09.xls'
In addition: Warning message:
In download.file(http://ir.eia.gov/wpsr/psw09.xls;, c:\\temp\\psw09.xls,
:
InternetOpenUrl failed: 'The operation timed out'
Thanks for any insights.
Pete Brecknock
--
View this message
maris478 wrote
Good afternoon,
I've encountered a little bit of a problem, would appreciate any help
here.
I made a small vector consisting of ones and zeros.
Something like this x - c(0,1,0,1,0,0,1,0), and all I need is to count
how many times 0 becomes 1.
Tried various, of what I
galemago wrote
Dear Forum,
I just recently started to work with R, I´d like to know if there is a way
to give instructions/do operations related to values with different
subscripts within a vector.
Let´s assume I have a vector like this:
A=368369370371 393394395
How about
# Read Data
nb10 - read.table(http://www.adjoint-functors.net/su/web/314/R/NB10;)
# Calculate Stats
total = length(nb10[,1])
mean = mean(nb10[,1])
sd = sd(nb10[,1])
# Function ... nSD is the number of SD you are looking at
pData - function(nSD){
lo = mean - nSD/2*sd
hi =
How about using the legend function ...
plot(rnorm(100))
legend(60,2,100 Random Normal Draws,cex=.8,text.col=blue,
box.col=red,bg=yellow)
You can customize my effort to fit your specific needs
HTH
Pete
Henry wrote
New to R - rookie question.
I'm a mechanical engineer and enjoying using
Hi Valerie
One way would be to use the match function.
# Your Data
u =data.frame(coe=c(0,0,0,0,0,0,0,0),
name=c(Time,Poten,AdvExp,Share,Change,Accounts,Work,Rating))
v = data.frame(coeff=c(0.7272727,0.322,0.0500123),
enter=c(Accounts,Time,Poten))
# Match
of time required to traverse the link, its name etc. That said, my
thought was that the situation was too simple to fire up a full-blown
object system beyond what R provides natively. I guess it's like making
a data frame that has some 3-d elements.
On 2/3/2012 9:32 PM, Pete Brecknock wrote
Not entirely sure why you would want a data.frame that has multiple entries
in one of the columns (Connect.down) but leaving that aside is the following
of any use?
nn=list()
nn[[1]] = list(Node = 1, Connect.up = c(NULL), Connect.down = c(2,3))
nn[[2]] = list(Node = 2, Connect.up = c(1),
Axel Urbiz wrote
Dear List,
I'll appreciate your help on this. I'm trying to create a variable as in
cumsum_y.cond1 below, which should compute the cumulative sum of y
conditional on the value of cond==1.
set.seed(1)
d - data.frame(y= sample(c(0,1), 10, replace= T),
Erin Hodgess-2 wrote
Dear R People:
Short of doing a series of ablines, is there a way to produce graph
paper in R please?
Thanks,
Erin
--
Erin Hodgess
Associate Professor
Department of Computer and Mathematical Sciences
University of Houston - Downtown
mailto:
Dom Pazzula wrote
All,
I'm attempting to run this:
table.CAPM(series[,Strat.Return,drop=FALSE],series[,spy.Return,drop=FALSE])
and getting this error
Error in as.vector(data[, i]) : subscript out of bounds
I've searched around and cannot find a solution to the problem. I've used
jdog76 wrote
I am positive this problem has a very simple solution, but I have been
unable to find it, so I am asking for your help. I need to know how to
look something up in one data frame and add it as a column in another. If
I have a data frame that looks like this:
frame1
ID
eigenvalet wrote
How can I display a number as currency including a $ sign and commas?
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I am trying to rename column names in a dataframe within a function. I am
seeing an error (listed below) that I don't understand.
Would be grateful of an explanation of what I am doing wrong and how I
should rewrite the function to allow me to be able to rename my variables.
Thanks.
# Test
Aurélien PHILIPPOT wrote
Dear R-experts,
I am struggling with the following problem, and I am looking for advice
from more experienced R-users: I have a data frame with 2 identifying
variables (comn and mi), and an output variable (x). comn is a variable
for
a company and mi is a variable
alaios wrote:
Dear all,
I have a function that recognizes the following format for timestamps
%Y-%m-%d %H:%M:%S
my function takes two input arguments the TimeStart and TimeEnd
I would like to help me create the right list with pairs of TimeStart and
TimeEnd which I can feed to lapply
kaallen wrote:
Hi,
I am working on a data set which looks like this:
head(temp)
Day Month Year PW ROW
1 1 1 1959 NA 6.40
2 2 1 1959 6.65 6.35
3 3 1 1959 2.50 3.60
4 4 1 1959 0.60 2.25
5 5 1 1959 0.85 0.30
6 6 1 1959 0.00 2.20
I am
Vining, Kelly wrote:
Dear UseRs,
I have a data frame that looks like this:
head(test2)
attributes start end StemExplant Callus RegenPlant
1 LTR_Unknown 120 535 3.198 1.931 1.927
3 LTR_Unknown 2955 3218 0.541 0.103 0.613
6 LTR_Unknown 6210 6423
feargalr wrote:
I am currently trying to create 3 histograms from 3 sets of data and in
order to compare them I need them all to have a common scale, the Y axis
is the only place its a problem, as one histogram only goes up to 4,
another 5, and another 7 on the Y axis and obviously they all
Erik Svensson wrote:
Hello,
In a data frame I want to identify ALL duplicate IDs in the example to be
able to examine OS and time.
(df-data.frame(ID=c(userA, userB, userA, userC),
OS=c(Win,OSX,Win, Win64),
time=c(12:22,23:22,04:44,12:28)))
IDOS time
1 userA Win 12:22
wizykid wrote:
Hi there.
I went through the manual but I couldn't find a solution for my problem.
I have list like this one :
lst1
[[1]]
[1] 0 1 2 3
[[2]]
[1] 0 1 5
[[3]]
[1] 2 3 4
and I want to save it as Matrix in Matlab mat format like :
0 1 2 3
0 1 5 0
2 3 4 0
DimmestLemming wrote:
I am interning in a computer science lab and I'm very new to R. The
language basics are clear, but this particular problem isn't:
I have a very large dataframe called data which holds data from Halo
matches. I'm trying to analyze a certain window such that
Lisa wrote:
Dear all,
I have two sets of numbers that look like
a - c(1, 2, 3, 3, 4, 4, 5, 6, 1, 2, 2, 3, 1, 2, 1, 2, 3, 3, 4, 5, 1, 2,
3, 4)
b - c(1, 5, 8, 9, 14, 20, 3, 10, 12, 6, 16, 7, 11, 13, 17, 18, 2, 4, 15,
19)
I just want to use “b” to encode “a” so that “a” looks like
I have a zoo object that contains 2 time series named A-B and V1.
When I create a third series V2, the name of the A-B series is changed
to A.B.
Although I could recreate the names for the 3 series I am wondering if there
is a way of preventing the name change from happening ( ... maybe an
Gabor Grothendieck wrote:
On Thu, May 26, 2011 at 7:35 PM, Pete Brecknock
lt;peter.breckn...@bp.comgt; wrote:
I have a zoo object that contains 2 time series named A-B and V1.
When I create a third series V2, the name of the A-B series is
changed
to A.B.
Although I could recreate
boki2b wrote:
Hello,Could somebody tell me what is the difference between theese 3
calls of functionsarma(x,order=c(1,0)), arima(x,order=c(1,0,0))
ar(x,order=1)?I expected same residuals of theese three models,but
unexpectably for the first two R requiredinitial value of something
tornanddesperate wrote:
Hi its me again
I don't mean to get on your nerves, but the use of R proofs to be a bit
more complicated than envisaged.
I would like to calculate the mean of a group of values, here
wage_accepted. The group is determined by the stage and period, so in
the end
... is the apply function what you are looking for?
A=matrix(1,2,4)
apply(A,1,sum)
HTH
Pete
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You could try the urca package
Also, I would maybe have a look a the CRAN Task View on computational
econometrics at http://cran.r-project.org/web/views/Econometrics.html
HTH
Pete
--
View this message in context:
Here's one way .
Lines-x1 x2 x3 x4 x5 x6
NA NA 3 4 NA NA
5 3 4 NA NA NA
7 3 4 4 NA NA
11 3 4 5 NA NA
67 4 4 NA NA NA
d - read.table(textConnection(Lines), header = TRUE,
colClasses=c(integer))
closeAllConnections()
res = t(apply(d, 1, function(x) ave(x,is.na(x),FUN=cumsum)))
print(res)
Doesn't use apply but maybe the following will work for you?
temp1- data.frame(ID=c(Herb,Shrub),stat=c(4,5),pvalue=c(.03,.04))
temp2- data.frame(ID=c(Herb,Shrub,
Tree),stat=c(12,15,13),pvalue=c(.2,0.4,.3))
L-list(a=temp1,b=temp2)
d.f = do.call(rbind,L)
d.f$tableName = substring(rownames(x),1,1)
Hi
I would like to be able to add reference lines to a series of plots that are
built using the Grid graphics package. These lines should coincide with tick
marks which are different on each plot.
I can add the lines manually using the grid.lines() function but would like
to understand how to
Apologies
I forgot to include that the reference lines should be for the y axis only.
Thanks.
Pete
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Thanks Gabor. I owe you again.
Kind regards
Pete
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__
For 5 years
set.seed = 1
d=data.frame(year=rep(2007:2011,each=12), month=rep(1:12,5), meanTemp =
rnorm(60,10,5))
meanByMonth = ave(d$meanTemp, d$month, FUN = mean)[7:9]
HTH
Pete
--
View this message in context:
You might want to take a look at Bob Muenchen's book R for SAS and SPSS
Users
There is an 80 page preview at .. http://rforsasandspssusers.com/
Additionally, there is lots of documentation for getting started with R on
the CRAN website http://cran.r-project.org/
HTH
Pete
--
1. Using the quantmod package.
Look at ?getSymbols
e.g. getSymbols(^GSPC,src=yahoo)
then use the to.monthly function
2. Using the tseries package .
Look at ?get.hist.quote
e.g. get.hist.quote(instrument = ^GSPC, start = as.yearmon(Jan 1950)
,compression = m, quote = Close)
There
If I have understood correctly, then as an example for 4 columns how about
d=data.frame(C1=c(1,1,1,1,1),C2=c(2,2,2,2,2),C3=c(3,3,3,3,3),C4=c(4,4,4,4,4))
combs=combn(paste(C,seq_len(4),sep=),2,simplify=FALSE)
one = sapply(combs,function(x) d[x[1]])
two = sapply(combs,function(x) d[x[2]])
try
subset(D, D$x 5|D$y 5)
HTH
Pete
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__
R-help@r-project.org mailing
Terry
Any good?
list -A B C N1 N2 N3 N4 N5 N6 N7 N8 N9 N10
Apples Bananas Peaches 1 2 3 4 5 6 7 8 9 10
Oranges Kumquats Plums 1 1 3 5 5 5 7 6 6 12
Pears Kiwis Grapes 2 4 6 8 5 6 7 6 9 1
d = read.table(textConnection(list), header=TRUE)
Nrowsums =
Jessica
Any good?
lines -DateTime, Q
2004-12-09 15:30:01, 2
2004-12-09 15:30:01, 1
2004-12-09 15:30:06, 1
2004-12-09 15:30:14, 5
2004-12-09 15:30:21, 1
2004-12-09 15:30:22, 11
2004-12-09 15:30:24, 4
2004-12-09 15:30:32, 1
2004-12-09 15:30:32, 1
2004-12-09 15:30:32, 3
2004-12-09 15:30:41, 4
d =
Dieter is correct, the lengths of the 2 series are different
Try
s = merge(s1,s2)
corr = cor(s[,Close.s1],s[,Close.s2],use=pairwise.complete.obs)
print(corr)
HTH
Pete
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Typing ? (no quotes) followed by a topic of interest will throw up the R
Help documentation.
1. Have a look at ?runif
2. Try ?subset and ?cumsum
3. Look at ?rle. Use in conjunction with cumsum and maybe ifelse.
HTH
Pete
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One way is to convert your data into a zoo object and use the rollapply
function
# Your data
lines = Date Close
2011-01-28 56.42
2011-01-27 57.37
2011-01-26 56.48
2011-01-25 56.39
2011-01-24 55.74
2011-01-21 55.46
d = read.table(textConnection(lines), header = TRUE)
# create zoo object
how about ...
j=NULL
for(i in 4: length(xyz$Close)) {
j[i] = sd(xyz$Close[i-3:i])
}
print(j)
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I believe there are two reasons why your code doesn't work
1. You should replace
sd(Close[i]:Close[(i-3)])
with
sd(Close[(i-3):i]).
This will ensure you select the appropriate obsevations to feed in the sd
function.
2. Per Ray's point above, you need to output the calculated value of sd
In addition to what has already been suggested you could use ..
mapply(function(x,y) x-y, z,means)
which returns
V1 V2
[1,] 0.333 -2.7142857
[2,] NA 7.2857143
[3,] -0.667 -3.7142857
[4,] -6.667 NA
[5,] NA -0.7142857
[6,]
Eric
Your problem lies in the way cumprod deals with NAs
If you look at ?cumprod you will see
An NA value in x causes the corresponding and following elements of the
return value to be NA
Not sure what behaviour you want to see on encountering an NA (ignore it,
restart the cumprod process,
Amy
It would have been helpful if you had sent your R code of how you
constructed the sab object.
If you have a data.frame, the subset command you are having trouble with
should work fine. See below.
# Working Example
sab = data.frame(group=c('Group A', 'Group A', 'Group C', 'Group B', 'Group
or
d = data.frame(Col1=c(1,2,3),Col2=c(2,3,4),Col3=c(3,4,5))
names(d)
names(d)[1] = NewName1
names(d)
HTH
Pete
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If I have understood your question correctly, how about the following ...
m = matrix(c(7,11,15,17,10,19,4,18,18), nrow = 3, ncol=3)
sum_m = sum(m)
new_m = summ-m
HTH
Pete
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typo ...
should have been
m = matrix(c(7,11,15,17,10,19,4,18,18), nrow = 3, ncol=3)
sum_m = sum(m)
new_m = sum_m-m
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try ...
new_m = m[c(2,7,8),c(1,4,6,7)]
HTH
Pete
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__
I believe you want to select a subset of rows and subset of columns of your
original matrix m.
If you had wanted only the first row of m, you could have used m[1,]
Alternatively, if you had wanted only the second column of m then you could
have used m[,2]
m[1,2] would give you the element at
or
rm(list=ls(pattern=^NY)) if you only want those objects that begin with
NY ...
HTH
Pete
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