Hi,
May be this helps:
m <- 0.01
T <- 90
t <- 0:T
set.seed(42)
res1 <- sapply(t,function(.t) runif(1,0,m*(T-.t)))
#or
set.seed(42)
res2 <- runif(91,rep(0,91),m*(rep(T,91)-t))
identical(res1,res2)
#[1] TRUE
A.K.
Hi,
I need to generate a sequence that consists of random numbers.
My first i
I want to map out a mostly flat area of land, 300 meters on a side.
I want to make (x,y,z) triples where x and y vary between -150 and 150 and
there is just one z value.
Eventually I will try to use graphics to actually draw this, but my first
problem is that I need to get 90601 values by i
Duncan,
Thanks. Why doesn't
coloursf2 <- factor(1:8, levels = 8:1)
give an ordering when you do str(coloursf2) like
"8"<"7"<"6" ...
Bill
On Mon, Dec 2, 2013 at 6:29 PM, Duncan Mackay wrote:
> Hi Bill
>
> eg
>
> > colours = 1:8
> > coloursf = factor(1:8)
> > colourso = ordered(1:8)
> > str(
Hi Bill
eg
> colours = 1:8
> coloursf = factor(1:8)
> colourso = ordered(1:8)
> str(coloursf)
Factor w/ 8 levels "1","2","3","4",..: 1 2 3 4 5 6 7 8
> str(colourso)
Ord.factor w/ 8 levels "1"<"2"<"3"<"4"<..: 1 2 3 4 5 6 7 8
coloursf2 <- factor(1:8, levels = 8:1)
str(coloursf2)
Duncan
Dunc
Duncan,
Your solution seems so much simpler. I was reading The Art of R Programming
and it says the following:
g=c("M","F","F","I","M","M","F",
ifelse (g == "M", 1, ifelse (g == "F", 2, 3))
[1] 1 2 2 3 1 1 2
What actually happens in that nested ifelse () ? Lets take a careful look.
First, for
Duncan,
Your solution seems so much simpler. I was reading The Art of R Programming
and it says the following:
g=c("M","F","F","I","M","M","F",
ifelse (g == "M", 1, ifelse (g == "F", 2, 3))
[1] 1 2 2 3 1 1 2
What actually happens in that nested ifelse () ? Lets take a careful look.
First, fo
Hi David,
You attach the dataset, which creates a copy of it. You set the contrasts
on the copy, but in your models, you reference the dataset explicitly. You
should set the contrasts directly in the data (and for your sake and your
students, would encourage not using attach() ).
contrasts(Eyse
I have been having trouble understanding the difference between
"contrasts(Group) <- contr.sum" and options(contrasts =
c("contr.sum","contr.poly"). They both seem to say that they have done
what I want, but only the latter works.
The reason why the question arises is tha,t using Fox's Anova,
i have a problem with big matix, in fact, after th matrix's creation many
rows ans colomuns are invisble because the big dimension of the matix
could you help me to get may complete matrix, have you any fonctions or any
solution to resolve this problem
--
View this message in context:
http://
Dear all
I have observations done in 4 different classes and the between classes
*variance* is too high that I decided to run a model without pooling the
*variance*. I used the following code first :
model<-lm(y~x+factor(class))
and got the following output:
Coefficients:
> It seems so inefficient.
But ifelse knows nothing about the expressions given
as its second and third arguments -- it only sees their
values after they are evaluated. Even if it could see the
expressions, it would not be able to assume that f(x[i])
is the same as f(x)[i] or things like
ifels
Hi. Thanks. Which part? I looked at the below but I don't think it clearly
addresses the nested ifelse situation.
ifelse {base}R DocumentationConditional Element Selection Description
ifelse returns a value with the same shape as test which is filled with
elements selected from either yes or no
On 13-12-02 7:49 PM, Bill wrote:
It seems so inefficient. I mean the whole first vector will be
evaluated. Then if the second if is run the whole vector will be
evaluated again. Then if the next if is run the whole vector will be
evaluted again. And so on. And this could be only to test the first
It seems so inefficient. I mean the whole first vector will be evaluated.
Then if the second if is run the whole vector will be evaluated again. Then
if the next if is run the whole vector will be evaluted again. And so on.
And this could be only to test the first element (if it is false for each
i
On 13-12-02 7:33 PM, Bill wrote:
ifelse ((day_of_week == "Monday"),1,
ifelse ((day_of_week == "Tuesday"),2,
ifelse ((day_of_week == "Wednesday"),3,
ifelse ((day_of_week == "Thursday"),4,
ifelse ((day_of_week == "Friday"),5,
ifelse ((day_of_week == "Saturday"),6,7)))
In cod
ifelse ((day_of_week == "Monday"),1,
ifelse ((day_of_week == "Tuesday"),2,
ifelse ((day_of_week == "Wednesday"),3,
ifelse ((day_of_week == "Thursday"),4,
ifelse ((day_of_week == "Friday"),5,
ifelse ((day_of_week == "Saturday"),6,7)))
In code like the above, day_of_week is a vector
Thanks to all, still a bit of a mystery. It is for the graphing I guess but
not sure why. I don't think there is too much sense to saying that Tuesday
precedes Wednesday but there could, as some of you suggest, be
circumstances where this is useful.
On Mon, Dec 2, 2013 at 12:00 PM, Bert Gunter w
> I ran a job to combine 2 dataframes using rbind.
> I received this error message that the names were not the same
> Error in match.names(clabs,names(xi)): names do not match previous names
The column names of the data.frames given to rbind must all be
permutations of one another. E.g.,
> r
Hi Julie,
On Mon, Dec 2, 2013 at 3:38 PM, Julie Royster wrote:
> Hello wise R folks,
>
> I ran a job to combine 2 dataframes using rbind.
> I received this error message that the names were not the same
>
> Error in match.names(clabs,names(xi)): names do not match previous names
>
> BUT when I en
Hi,
I couldn't reproduce the first part.
Lines1 <- readLines(textConnection("2 5 7 11
1 2 5
5 7 10 12 13"))
Max1 <- max(as.numeric(unlist(strsplit(Lines1," "
t(sapply(strsplit(Lines1," "), function(x) {x1<- as.numeric(x); x2 <-
numeric(Max1); x2[x1]<- 1; x2}))
#or
mat1<- as.matrix(rea
Hi,
I couldn't reproduce the first part.
Lines1 <- readLines(textConnection("2 5 7 11
1 2 5
5 7 10 12 13"))
Max1 <- max(as.numeric(unlist(strsplit(Lines1," "
t(sapply(strsplit(Lines1," "), function(x) {x1<- as.numeric(x); x2 <-
numeric(Max1); x2[x1]<- 1; x2}))
#or
mat1<- as.matrix(read.ta
Hello wise R folks,
I ran a job to combine 2 dataframes using rbind.
I received this error message that the names were not the same
Error in match.names(clabs,names(xi)): names do not match previous names
BUT when I entered this statement
Identical (names(data1[[1]]),names(data2[[2]]) )
R res
thanks andy
it's a real honour form me to get a reply by you;
I'm still a bit faraway from a proper grasp of the purpose of the plot...
may I ask you for a more technical (trivial) issue?
is it possible to add a legend in the MDS plot?
my problem is to link the color points in the chart to the fa
On Dec 2, 2013, at 1:01 PM, Duncan Murdoch wrote:
> On 02/12/2013 2:22 PM, Jacob Wegelin wrote:
>> I want to put the "plus or minus" symbol into a character variable, so that
>> this can be turned into a factor and be displayed in the "strip" of a
>> faceted ggplot2 plot.
>>
>> A very nice sol
On Dec 2, 2013, at 11:22 AM, Jacob Wegelin wrote:
>
> I want to put the "plus or minus" symbol into a character variable, so that
> this can be turned into a factor and be displayed in the "strip" of a faceted
> ggplot2 plot.
>
> A very nice solution, thanks to Professor Ripley's post of Nov
On 02/12/2013 2:22 PM, Jacob Wegelin wrote:
I want to put the "plus or minus" symbol into a character variable, so that this can be
turned into a factor and be displayed in the "strip" of a faceted ggplot2 plot.
A very nice solution, thanks to Professor Ripley's post of Nov 16, 2008;
3:13pm, v
Thank you David, it is exactly what I needed.
Regards,Phil
> From: dcarl...@tamu.edu
> To: pmassico...@hotmail.com; r-help@r-project.org
> Subject: RE: [R] legend position
> Date: Mon, 2 Dec 2013 14:29:06 -0600
>
> It is not straightforward unless you want the legend in the
> right or the bottom
It is not straightforward unless you want the legend in the
right or the bottom margins. To put the legend inside the plot
region it is simplest to use image() to plot the raster file and
then image.plot(legend.only=TRUE) to add the legend. In addition
to reading the help page for plot{raster}, you
On Dec 2, 2013, at 1:17 AM, Knut Krueger wrote:
> Am 29.11.2013 20:39, schrieb David Winsemius:
>>> Thats impossible, we are used to hit the comma
>> I don't know what that means.
> it is common here, that the decimal sign is commy
Believe me, I _do_ understand that in Europe it is common to use
Thank you, I'll try to work with lattice.
Regards,Phil
> Date: Mon, 2 Dec 2013 12:06:50 -0800
> From: c...@witthoft.com
> To: r-help@r-project.org
> Subject: Re: [R] legend position
>
> It occurs to me that perhaps you're referring to the 'color bar' on the right
> of the plot. AFAIK you cannot
It occurs to me that perhaps you're referring to the 'color bar' on the right
of the plot. AFAIK you cannot get at that from the raster::plot method.
However lattice::levelplot does allow you to manipulate or remove that
colorbar.
--
View this message in context:
http://r.789695.n4.nabble.
Did you not see Mark Leeds's post?
The OP apparently did not really mean R's "ordered factors" as
produced by the R ordered() constructor; rather, he meant "factors
with levels ordered differently than the default", for which Rich's
answer was apropos. Mine -- and now yours -- in which we wrongly
Thank you for reply.
If I'm not wrong, legend(...) will works for discrete elements. I'm not sure
hot to use it for a colorbar legend sur as the one in the example bellow.
Phil
> Date: Mon, 2 Dec 2013 11:49:19 -0800
> From: c...@witthoft.com
> To: r-help@r-project.org
> Subject: Re: [R] legend po
On 12/2/2013 9:35 AM, Bert Gunter wrote:
> Not true, Rich.
The point about alphabetical ordering explains why the author likely
explicitly set the levels for the factor, though.
As to why ordered factors, we may never know, but one possible
explanation is that at some point he was going to use
See ?legend . you can add a legend directly to an existing plot. An
example:
legend('topright',c('hot','cold'),lty=1,col=c('red','green'),bg='white')
Now if you're trying to place the legend outside the plot area (i.e. in some
other part of the window),
you'll need to invoke par(xpd=TRUE) . Se
What is the value of
diag(c1) - 1
?
(Or, use digits=16 when printing c1.)
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 Dan Abner
> Sent: Monday, December 02, 201
I want to put the "plus or minus" symbol into a character variable, so that this can be
turned into a factor and be displayed in the "strip" of a faceted ggplot2 plot.
A very nice solution, thanks to Professor Ripley's post of Nov 16, 2008;
3:13pm, visible at
http://r.789695.n4.nabble.com/Sym
Hi all.
I'm ploting a raster and I can't find the proper way to move the legend. For
example,
r = raster(system.file("external/test.grd", package="raster"))plot(r)
How can I put the legend at the desired position?
Thank in advance,Phil
[[alternative HT
On 12/2/2013 10:12 AM, Duncan Murdoch wrote:
dbinom can take vector inputs for the parameters, so this would be a
bit simpler:
x <- seq(0,12)
x <- rep(x, 4)
p <- rep(c(1/6, 1/3, 1/2, 2/3), each=13)
bin.df <- data.frame(x, prob = dbinom(x, 12, p), p)
Thanks, Duncan
What I was missing was how
On 02/12/2013 2:08 PM, Dan Abner wrote:
Hi all,
Can anyone explain what is happening with element 4,4 of c1? ifelse()
is not recongizing it as value 1:
FAQ 7.31.
Duncan Murdoch
> c1
q1q2q3q4
q1 1.000 0.6668711 0.6948419 0.5758860
q2 0.6668711 1.
Hi all,
Can anyone explain what is happening with element 4,4 of c1? ifelse()
is not recongizing it as value 1:
> c1
q1q2q3q4
q1 1.000 0.6668711 0.6948419 0.5758860
q2 0.6668711 1.000 0.6040746 0.4917447
q3 0.6948419 0.6040746 1.000 0.4730732
q4 0.57
Thank you. I can't believe I didn't notice that. What a relief.
From: David Carlson
Sent: Monday, December 2, 2013 1:47 PM
To: White, William Patrick; 'Pascal Oettli'
Cc: 'r-help@R-project.org'
Subject: RE: [R] Days to solstice calculation
They are a day a
They are a day apart. Summer solstice is day 172 in both cases
so the calendar dates should be one day apart and they are (June
22 in 2007 and June 21 in 2008):
> strptime("2007-06-22", format="%Y-%m-%d")$yday
[1] 172
> strptime("2008-06-21", format="%Y-%m-%d")$yday
[1] 172
Your Daylength() funct
On 02/12/2013 11:59 AM, Federico Calboli wrote:
Hi All,
together with colleagues we are planning to submit a 2.0 version of a package
we have on CRAN. Because the package deals with high throughput genomic data
we though it would be nice to have some sort of guidance for the users. This
sho
Yes, that's part of the intention anyway. One can also use them to do
clustering.
Best,
Andy
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Massimo Bressan
Sent: Monday, December 02, 2013 6:34 AM
To: r-help@r-project.org
Subject
Bert,
the issue is the sort order of the levels. Time series graphs in the
alphabetical sort
order will be uninterpretable. I show the three sets of contrasts for
factors, factors
with specified levels, and ordered factors.
week <-
c("Sunday","Monday","Tuesday","Wednesday","Thursday","Friday","
On 02 Dec 2013, at 16:35 , Bert Gunter wrote:
> Not true, Rich.
>
>> z <-factor(letters[1:3],lev=letters[3:1])
>> sort(z)
> [1] c b a
> Levels: c b a
>
> What you say is true only for the **default** sort order.
>
> (Although maybe the code author didn't realize this either)
The coding is ce
Hi All,
together with colleagues we are planning to submit a 2.0 version of a package
we have on CRAN. Because the package deals with high throughput genomic data
we though it would be nice to have some sort of guidance for the users. This
should ideally mean a 'vignette', but as the time of
#2 can be done simply with predict(fmi, type="prob"). See the help page for
predict.randomForest().
Best,
Andy
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of arun
Sent: Tuesday, November 26, 2013 6:57 PM
To: R help
Subject: Re:
Thank you for your response. I looked at the insol package and it does seem to
contain a daylength function, but it would be more informative and aid in my
growth as an R user to trace the source of the current problem rather than use
a package as a work around. Understanding that an alternate
Hi: I asked Bert privately and he recommended posting what I asked/said to
him to the list.
My comment/question was that I looked at the code and didn't actually
see an ordered factor being created. So my guess is that there is
a confusion with the use of the term "ordered". I'm not clear on wheth
You can use recycling to simplify things:
> set.seed(42)
> x <- seq(0,12)
> bin.df <- as.data.frame(
+ rbind( cbind(x, prob=dbinom(x,12,1/6), p=1/6),
+cbind(x, prob=dbinom(x,12,1/3), p=1/3),
+cbind(x, prob=dbinom(x,12,1/2), p=1/2),
+cbind(x, prob=dbinom(x,12
Not true, Rich.
> z <-factor(letters[1:3],lev=letters[3:1])
> sort(z)
[1] c b a
Levels: c b a
What you say is true only for the **default** sort order.
(Although maybe the code author didn't realize this either)
-- Bert
On Mon, Dec 2, 2013 at 7:24 AM, Richard M. Heiberger wrote:
> If days of
If days of the week is not an Ordered Factor, then it will be sorted
alphabetically.
Fr Mo Sa Su Th Tu We
Rich
On Mon, Dec 2, 2013 at 6:24 AM, Bill wrote:
> I am reading the code below. It acts on a csv file called dodgers.csv with
> the following variables.
>
>
>> print(str(dodgers)) # check t
"BIll" :
(Sorry -- Doubt that this will be helpful, but I couln't resist)
"I don't understand why the author of the code decided to make the factor
days_of_week into an ordered factor. Anyone know why this should be done?"
A definitive answer would require either psychic abilities or asking
the
On 02/12/2013 9:47 AM, Michael Friendly wrote:
I want to generate a collection of probability distributions in a data
frame, with
varying parameters. There must be some simpler way than what I have below
(avoiding rbind and cbind), but I can't quite see it.
x <- seq(0,12)
bin.df <- as.data.fram
I want to generate a collection of probability distributions in a data
frame, with
varying parameters. There must be some simpler way than what I have below
(avoiding rbind and cbind), but I can't quite see it.
x <- seq(0,12)
bin.df <- as.data.frame(
rbind( cbind(x, prob=dbinom(x,12,1/6), p
Given this general example:
set.seed(1)
data(iris)
iris.rf <- randomForest(Species ~ ., iris, proximity=TRUE, keep.forest=TRUE)
#varImpPlot(iris.rf)
#varUsed(iris.rf)
MDSplot(iris.rf, iris$Species)
I’ve been reading the documentation about random forest (at best of my -
poor - knowledge) b
I am reading the code below. It acts on a csv file called dodgers.csv with
the following variables.
> print(str(dodgers)) # check the structure of the data frame
'data.frame': 81 obs. of 12 variables:
$ month : Factor w/ 7 levels "APR","AUG","JUL",..: 1 1 1 1 1 1 1 1 1
1 ...
$ day
Hi to
everyone, I have a big data set where rows are observations and columns are
variables. It contains a lot of missing values. I have used multiple imputation
with library mice and I get an exact prediction of each missing value. Now, I
would like to know the error I can commit or the confid
Am 29.11.2013 20:39, schrieb David Winsemius:
Thats impossible, we are used to hit the comma
I don't know what that means.
it is common here, that the decimal sign is commy
All computer in the cip-pools are using the "comma" ( an I think 99.9%
of all other computers here)
Can you imagine what
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