> On Aug 23, 2017, at 10:06 PM, Dhivya Narayanasamy
> wrote:
>
> I have a vibration signal coming accelerometer. I converted this signal from*
> m/s^2* to *mm/s*. Now I am supposed to predict this vibration signal in R
> using historical data. (Please see the attached picture of vibration
> sig
> On Aug 22, 2017, at 11:31 PM, jagan krishnan via R-help
> wrote:
>
> Hi all,
> This is Jagan.i have been provided a task of analyzing sales data of a
> company in R programming...Just wanted to know,how can I pull Tally 9.1
> software data into R programming dataframe.
> Waiting eagerly for
I have a vibration signal coming accelerometer. I converted this signal from*
m/s^2* to *mm/s*. Now I am supposed to predict this vibration signal in R
using historical data. (Please see the attached picture of vibration
signal).
Can I use this vibration signal just like that to the prediction usi
## There is a bug in likert:::plot.likert.
## The centered argument is not handled correctly.
## In addition to centering, it also flips the order.
## Here it is:
likert:::plot.likert(results, type = "bar", centered = TRUE, ## correct order
group.order = c("Band 3", "Band 4")
Please don't start a new thread with the same question. My usual suggestion at
this point would be for you to respond to the answers that have already been
posted to your last question, but I think at this point that you need to
correspond directly with the maintainer of the likert package.
--
R- Help Forum
Working with the "likert" package and I can't figure out why my "bar" graphs
are backwards (see attached). The percentages are place correctly but the
bars are backwards.
#Sample code
# libraries
library(likert)
# create data
band <- c("Band 3","Band 3","Band 3","Band 3
Thanks.
I suspect that would help to track down the problem in the underlying
code, but I am not sufficiently motivated to do so. Perhaps the
maintainer will if he sees this thread.
Cheers,
Bert
Bert Gunter
"The trouble with having an open mind is that people keep coming along
and sticking thi
On 23/08/2017 6:25 PM, Bert Gunter wrote:
Doesn't sort by size of subgroup. I interpret the phrase I asterisked as:
You were fooled by Peter's tricky single negative.
Duncan Murdoch
Your code does the following:
First subsets of size 1 are given.
Then all subsets of size 2.
Then all subset
It seems that if you build the formula as a character string, and
postpone the "as.formula" into the lm call, it works.
instead of
frm1 <- as.formula(paste(trg,"~1"))
use
frm1a <- paste(trg,"~1")
and then
strt <- lm(as.formula(frm1a),dat)
regards,
Heinz
Stephen O'hagan wrote/hat geschrieben o
Note that the "doubled brackets" are not essential to this usage.
[:characterClass:] is recognized inside of square brackets as a shortcut
for listing a bunch of characters. You can mix it with other characters or
character classes inside a set of square brackets.
E.g., the following pattern matc
You need to learn how to access code for nonexported methods. See ? "::"
> methods(qda)
[1] qda.data.frame* qda.default*qda.formula*qda.matrix*
see '?methods' for accessing help and source code
Shows you that the methods are not exported from the namespace. Hence
you need to use the trip
On 24/08/17 02:46, Bert Gunter wrote:
Inline.
-- Bert
Bert Gunter
"The trouble with having an open mind is that people keep coming along
and sticking things into it."
-- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
On Wed, Aug 23, 2017 at 2:29 AM, Rolf Turner wrote:
On
Hello,
I am Souradeep Chattopadhyay and I am a graduate student at Iowa
State University Department of Statistics.
Can anyone please explain the mathematical formulation behind the scaling
matrix returned by the qda function in MASS package. I want to understand
how this scaling matrix
Doesn't sort by size of subgroup. I interpret the phrase I asterisked as:
Your code does the following:
First subsets of size 1 are given.
Then all subsets of size 2.
Then all subsets of size 3.
etc.
Your code does not do this (quite).
If you meant something else, then please clarify.
Cheers,
> On 23 Aug 2017, at 23:12 , Bert Gunter wrote:
>
>> This points to a different algorithm where you write 0:(2^n-1) as n-digit
>> binary numbers and chose items corresponding to the 1s. That won't give the
>> combinations **sorted by size of selected subgroup** though. Something like
>> this:
Inline.
-- Bert
Bert Gunter
"The trouble with having an open mind is that people keep coming along
and sticking things into it."
-- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
On Wed, Aug 23, 2017 at 1:58 PM, peter dalgaard wrote:
>
>> On 23 Aug 2017, at 20:51 , Ista Zahn
> On 23 Aug 2017, at 20:51 , Ista Zahn wrote:
>
> On Wed, Aug 23, 2017 at 12:35 PM, Bert Gunter wrote:
>> ummm, Ista, it's 2^n.
>
> ummm yes ug.
>
You didn't really say otherwise: sum(choose(n,0:n)) == 2^n by the binomial
expansion of (1+1)^n (but you knew that)
This points to a differ
On Wed, Aug 23, 2017 at 12:35 PM, Bert Gunter wrote:
> ummm, Ista, it's 2^n.
ummm yes ug.
My point is, if the number of groups is large, check it before hand.
If you can check it without embarrassing yourself in public like I did
that's even better.
Best,
Ista
>
> Cheers,
> Bert
>
>
> Bert
most likely you have a factor with default levels which are not
ordered the way you want them to be ordered.
You can reorder the levels with the factor function. If you still
need help after adjusting the order of the levels, then
please send the data to the list by including the output of
dput(da
Any responds?!
On Wednesday, August 23, 2017 5:50 AM, Elahe chalabi via R-help
wrote:
Hi all,
I would like to do cross validation in random forest using rfcv function. As
the documentation for this package says:
rfcv(trainx, trainy, cv.fold=5, scale="log", step=0.5, mtry=function(p) ma
When I try to use the following code, I get the error message shown. This
is quite confusing to me, insofar as family is a recognized argument for
MCMCglmm. Can anyone spot an obvious glitch?
model1 <-MCMCglmm(fixed = NoRRVpos ~ Year, random = ~County,
family="zipoisson", data=Rabies_Project_Init
: Thanks. But when I apply your codes I get all NA instead of TRUE and FALSE
You need to show a self-contained example of your problem, one that others
can copy and paste into their R sessions. E.g.,
data <- read.table(header=TRUE, stringsAsFactors=FALSE, text="
COL1 COL2
6/1/14 5/1/15
On 2017-08-23 11:35 AM, Bert Gunter wrote:
ummm, Ista, it's 2^n.
or (2^n-1) if the empty set is not considered as a "combination"
;-) spencer
Cheers,
Bert
Bert Gunter
"The trouble with having an open mind is that people keep coming along
and sticking things into it."
-- Opus (ak
ummm, Ista, it's 2^n.
Cheers,
Bert
Bert Gunter
"The trouble with having an open mind is that people keep coming along
and sticking things into it."
-- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
On Wed, Aug 23, 2017 at 8:52 AM, Ista Zahn wrote:
> On Wed, Aug 23, 2017 at 1
Thanks. But when I apply your codes I get all NA instead of TRUE and FALSE
From: PIKAL Petr
Sent: Wednesday, August 23, 2017 11:20:00 AM
To: Patrick Casimir; r-help@r-project.org
Subject: RE: Comparing 2 dale columns
Hi
your code is wrong.
I get
> test<-read.t
Hi Mark,
Instead of 1 and 0. It generates NA
NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA
From: Mark Sharp
Sent: Wednesday, August 23, 2017 11:28:39 AM
To: Patrick Casimir
Cc: r-help@r-project.org
Subject: Re: [R] Comparing 2 dale columns
Pat
On Wed, Aug 23, 2017 at 11:33 AM, Christofer Bogaso
wrote:
> Hi again,
>
> I am exploring if R can help me to get all possible combinations of
> members in a group.
>
> Let say I have a group with 5 members : A, B, C, D, E
>
> Now I want to generate all possible unique combinations with all
> poss
Please at least do a basic internet search before posting here.
Searching on "R combinations" brought up the combn() function, which
you can use in a simple loop to get your answer:
> lapply(seq_len(5), FUN = function(x)combn(LETTERS[1:5], x))
## You can use a for() loop if you prefer
[[1]]
lapply(1:5, function(x) combn(groups, x))
or perhaps
unlist(lapply(1:5, function(x) combn(groups, x, FUN = paste, collapse = ", ")))
Best,
Ista
On Wed, Aug 23, 2017 at 11:33 AM, Christofer Bogaso
wrote:
> Hi again,
>
> I am exploring if R can help me to get all possible combinations of
> membe
Hi again,
I am exploring if R can help me to get all possible combinations of
members in a group.
Let say I have a group with 5 members : A, B, C, D, E
Now I want to generate all possible unique combinations with all
possible lengths from that group e.g.
1st combination : A
2nd combination : B
Patrick,
## Run the following script an notice the different values of the dataframe
"data" in each instance.
# I understand you have done something like the following:
data <- data.frame(COL1 = c("6/1/14", "7/1/14"),
COL2 = c("5/1/15", "5/1/15"), stringsAsFactors = FALSE)
dat
Hi
your code is wrong.
I get
> test<-read.table("clipboard", header=T)
> str(test)
'data.frame': 2 obs. of 2 variables:
$ COL1: Factor w/ 2 levels "6/1/14","7/1/14": 1 2
$ COL2: Factor w/ 1 level "5/1/15": 1 1
> test$COL2<- as.Date(as.character(test$COL2, format="%y/%m/%d"))
> test$COL1<- as
Hi
Well, yes I tried it about two weeks ago but my post did not get through as it
still awaits moderator approval. I could check which column is offending but
actually it is only minor nuisance, I can live with selection of columns before
fitting a model. What seems to me strange is that both f
Dear R fellows,
I created a new column Date_flag to compare the dates of COL1 and COL2 using
the code
below. But it showed that 5/1/15 is greater than 6/1/2014 and 5/1/2015 greater
than
7/1/2014 despite the year is greater. How do I fix that? I did try to format as
%y/%m/%d
but it does not f
Better posted on r-sig-mixed-models , no?
Cheers,
Bert
Bert Gunter
"The trouble with having an open mind is that people keep coming along
and sticking things into it."
-- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
On Wed, Aug 23, 2017 at 5:17 AM, PIKAL Petr wrote:
> Dear
Inline.
-- Bert
Bert Gunter
"The trouble with having an open mind is that people keep coming along
and sticking things into it."
-- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
On Wed, Aug 23, 2017 at 2:29 AM, Rolf Turner wrote:
>
> On 23/08/17 18:33, Stefan Evert wrote:
>
First of all I want to mention the _warmup_ parameter to _control_argument of
microbenchmark function. Default value is 2 and function runs the code 2 times
before count the time intervals.
See ?microbenchmark
> However, the first iteration always takes the longest. I'm wondering if I
> should
Hi all,
I would like to do cross validation in random forest using rfcv function. As
the documentation for this package says:
rfcv(trainx, trainy, cv.fold=5, scale="log", step=0.5, mtry=function(p) max(1,
floor(sqrt(p))), recursive=FALSE, ...)
however I don't know how to build trianx and tr
Hi all,
I would like to do cross validation in random forest using rfcv function. As
the documentation for this package says:
rfcv(trainx, trainy, cv.fold=5, scale="log", step=0.5, mtry=function(p) max(1,
floor(sqrt(p))), recursive=FALSE, ...)
however I don't know how to build trianx and train
Hi there
Thanks for your answers. I didn't expect that this would be so complex.
Honestly, I don't understand everything you wrote since I'm not an IT
specialist. But I read something that reading *.rds files is faster than
loading *.Rdata and I wanted to proof that for my system and R version.
Hi Bill
Thanks for your answer and the explanations. I tried to use garbage collection
but I'm still not satisfied with the result. Maybe the question was not stated
clear enough. I want to test the speed of reading/loading of data into R when a
'fresh' R session is started (or even after a new
Dear all
I encountered strange behaviour of augPred with virtually the same data
First I made groupedData object.
> mar.g<-groupedData(rutilizace~doba|int, data=mar)
When I perform nlme on complete dataset I get an error with augPred
> fit<-nlsList(rutilizace~SSasymp(doba, Asym, R0, lrc), data=
As far as I can tell from the docs, strt & frm1 should be the simplest model,
'y1~1', since we’re doing stepAIC in the forward direction.
I found that using 'y1~.' for frm2 doesn't always work, hence the explicit
enumeration of the xvars into a full formula. I had forgotten that you don't
need
Until I get a fix that works, a work-around would be to rename the 'y1' column,
used a fixed formula, and rename it back afterwards.
Thanks for your help.
SGO.
-Original Message-
From: Bert Gunter [mailto:bgunter.4...@gmail.com]
Sent: 22 August 2017 20:38
To: Stephen O'hagan
Cc: r-help
On 23/08/17 18:41, PIKAL Petr wrote:
Hi Rolf
I am not at all an expert in regex but
gsub("[[:alpha:]]","",x)
Works as you expected. Do not ask me why.
Thanks Petr. Stefan Evert's explanation clarified the issue. Which I
must say *needed* some clarification!
cheers,
Rolf
--
Technical E
On 23/08/17 18:33, Stefan Evert wrote:
On 23 Aug 2017, at 07:45, Rolf Turner wrote:
My reading of ?regex led me to believe that
gsub("[:alpha:]","",x)
should give the result that I want.
That's looking for any of the characters a, l, p, h, : .
OK. I see that now. I don't think t
Le 22/08/2017 à 16:26, niharika singhal a écrit :
Hello I have a vector
v=c(0.0886,0.1744455,0.1379778,0.1209769,0.1573065,0.1134463,0.2074027)
when i do
sum(v)
or
0.0886+0.1744455+0.1379778+0.1209769+0.1573065+0.1134463+0.2074027
i am getting output as 1
But if i add them manually i get
If this is not docuemnted that way, write to the package maintainer.
Best,
Uwe Ligges
On 23.08.2017 02:51, Jeff Reichman wrote:
R- Help Forum
Working with the "likert" package and find that my "bar" graphs are
backwards (see attached)
summary(results)
Item low neutral
Hi all,
This is Jagan.i have been provided a task of analyzing sales data of a company
in R programming...Just wanted to know,how can I pull Tally 9.1 software data
into R programming dataframe.
Waiting eagerly for your inputs.
With Regards,Jagannathan Krishnan
Sent from Yahoo Mail on Android
I know there are ways around the 'can't allocate a vector of size x GB' errors,
but I'm stumped.
So my raw data has >7 million rows and eight columns. That's not a problem
itself.
Using the confreq package (for configural frequency analysis), I take my data
and run it through the package's dat2
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