Hi
> -Original Message-
> From: Dr. Debasis Ghosh [mailto:d...@debasis.in]
> Sent: Tuesday, September 20, 2016 3:59 PM
> To: PIKAL Petr ; R-help@r-project.org
> Subject: RE: [R] Issue on LGP solving
>
> Thanks Petr!! However, I found in the goalprog package I found
> "achievements" as "a
Thank you for the answer but I had already tried that way; when I introduce
weights in the glm appears the error:
Warning: non-integer #successes in a binomial glm!
I tried to run the glm regression using the family quasibinomial:
eq <- glm(pip ~ men + age_pr + age_c + I(age_pr^2) + I(age_c^2),
Dear
Although I can install the new version of the R, I can not open the soft.
How do I do it?
Kyuzi Kamoi, MD.
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.e
If a data.frame (r4) does not exist in my R environment, I would like to create
it before I move on to the next step in my script. How do I make that happen?
Here is what I want to do from a code perspective:
if (exists(r4))
{
is.data.frame(get(r4))
}
else
{
a <- 0, b <- 0, c <- 0, d <- "x", r4
Hello DRUGs,
I'm new to R and would appreciate some expert advice on prepping files for,
and running, PCA...
My data set consists of aquatic invertebrate and zooplankton count data and
physicochemical measurements from an ecotoxicology study. Four chemical
treatments were applied to mesocosm tanks
You can use the latter IF you know there are no problems with the input data.
If you need to troubleshoot then you need separate columns so you can compare
them.
--
Sent from my phone. Please excuse my brevity.
On September 20, 2016 4:22:41 PM PDT, lily li wrote:
>Thanks. The former method wo
Thanks. The former method works. I confused character with factor.
Besides, I should use: dta$DischargeNum <- as.numeric( dta$Discharge )
instead of: dta$Discharge <- as.numeric( dta$Discharge )
On Tue, Sep 20, 2016 at 5:18 PM, Jeff Newmiller
wrote:
> Which means it avoided converting to facto
Which means it avoided converting to factor... Success!
Note that the column apparently has garbage characters in one or more of the
rows, which should be evident when you LOOK AT THE CHARACTERS in the column.
They should all be numeric symbols, plus or minus, and perhaps decimal points.
If the
Yes, I tried to add this statement when reading the dataset.
But when I use summary(df), it shows:
Discharge
Length:
Class :character
Mode :character
On Tue, Sep 20, 2016 at 5:06 PM, Joe Ceradini wrote:
> read.csv("your_data.csv", stringsAsFactors=FALSE)
> (I'm just reiterating Jianling said.
Find the offending data. One approach is to look at the input data with your
image sensors and neural pattern processor (eyes and brain). One way to reduce
the load on those told is to read in the data with the stringsAsFactors=TRUE
argument and try manually converting the resulting character st
read.csv("your_data.csv", stringsAsFactors=FALSE)
(I'm just reiterating Jianling said...)
Joe
On Tue, Sep 20, 2016 at 4:56 PM, lily li wrote:
> Is there a function in read.csv that I can use to avoid converting numeric
> to factor? Thanks a lot.
>
>
>
> On Tue, Sep 20, 2016 at 4:42 PM, lily li
Is there a function in read.csv that I can use to avoid converting numeric
to factor? Thanks a lot.
On Tue, Sep 20, 2016 at 4:42 PM, lily li wrote:
> Thanks. Then what should I do to solve the problem?
>
> On Tue, Sep 20, 2016 at 4:30 PM, Jeff Newmiller
> wrote:
>
>> I suppose you can do what
Thanks. Then what should I do to solve the problem?
On Tue, Sep 20, 2016 at 4:30 PM, Jeff Newmiller
wrote:
> I suppose you can do what works for your data, but I wouldn't recommend
> na.rm=TRUE because it hides problems rather than clarifying them.
>
> If in fact your data includes true NA value
I suppose you can do what works for your data, but I wouldn't recommend
na.rm=TRUE because it hides problems rather than clarifying them.
If in fact your data includes true NA values (the letters NA or simply nothing
between the commas are typical ways this information may be indicated), then
I reread the data, and use 'na.rm = T' when reading the data. This time it
has no such problem. It seems that the existence of NAs convert the integer
to factor. Thanks for your help.
On Tue, Sep 20, 2016 at 4:09 PM, Jianling Fan wrote:
> Add the "stringsAsFactors = F" when you read the data,
Add the "stringsAsFactors = F" when you read the data, and then
convert them to numeric.
On 20 September 2016 at 16:00, lily li wrote:
> Yes, it is stored as factor. I can't check out any problem in the original
> data. Reread data doesn't help either. I use read.csv to read in the data,
> do yo
Yes, it is stored as factor. I can't check out any problem in the original
data. Reread data doesn't help either. I use read.csv to read in the data,
do you think it is better to use read.table? Thanks again.
On Tue, Sep 20, 2016 at 3:55 PM, Greg Snow <538...@gmail.com> wrote:
> This indicates th
This indicates that your Discharge column has been stored/converted as
a factor (run str(df) to verify and check other columns). This
usually happens when functions like read.table are left to try to
figure out what each column is and it finds something in that column
that cannot be converted to a
Hi R users,
I have a problem in reading data.
For example, part of my dataframe is like this:
df
month day year Discharge
31 20106.4
32 2010 7.58
33 2010 6.82
34 2010 8.63
3
Sorry, I've made a stupid mistake.
It's obviously the other way around.
ix <- which(rownames(data) %in% c("601", "604"))
clean <- data[-ix, ]
Rui Barradas
Citando ruipbarra...@sapo.pt:
Hello,
Try something like the following.
ix <- which(c("601", "604") %in% rownames(data))
clean <- data[
Hello,
Try something like the following.
ix <- which(c("601", "604") %in% rownames(data))
clean <- data[-ix, ]
Hope this helps,
Rui Barradas
Citando Pauline Laïlle :
Dear all,
I built a dataframe with read.csv2(). Initially, row names are integers
(order of answers to a survey). They a
You might consider the Predict.Plot and TkPredict functions in the
TeachingDemos package. These help you explore multiple linear
regression models by plotting the "line" relating the response to one
of the predictors at given values of the other predictors. These
lines can be combined in a single
Hint: "601" is not 601.
Have you gone through any R tutorials?
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 Tue, Sep 20, 2016 at 5:42 AM, Paul
Dear all,
I built a dataframe with read.csv2(). Initially, row names are integers
(order of answers to a survey). They are listed in the csv's first column.
The import works well and my dataframe looks like I wanted it to look.
Row names go as follows :
[1] "6" "29" "31" "32" "52" "55" "6
Hi all
I am using the bam function of the mgcv package to model behavioral data of
a learning experiment. To model individual variation in learning rate, I am
testing models with (a) by-participant random intercepts of trial, (b)
by-participant random slopes and random intercepts of trial, and (c)
If you want your records to be weighted by the survey weights during the
analysis, then use the weights= argument of the glm() function.
Jean
On Tue, Sep 20, 2016 at 5:04 AM, laura roncaglia
wrote:
> I am a beginner user of R. I am using a national survey to test what
> variables influence the
There are many good R tutorials on the web. Some recommendations can
be found here:
https://www.rstudio.com/online-learning/#R
Please spend some time learning fundamental R constructs and
functionality before posting what appear to be very basic questions
here.
Cheers,
Bert
Bert Gunter
"The t
You need statistical help, which is generally off topic here. I
suggest you post to a statistcal site like stats.stackexchange.com
instead. Better yet, find a local statistical expert with whom you can
consult.
Cheers,
Bert
Bert Gunter
"The trouble with having an open mind is that people keep com
On 9/19/2016 10:37 PM, John wrote:
Hi,
I have the following dataframe:
temp<-data.frame(a=c(1,1,2), b=2:4, c=1:3)
row.names(temp)<-c("D", "E", "F")
temp
a b c
D 1 2 1
E 1 3 2
F 2 4 3
I would like R to tell me which rows has value "a" equal to 1. The
answer is the first row and t
Thanks Petr!! However, I found in the goalprog package I found "achievements"
as "a data frame with the deviation variables for each objective together with
the
priority level". I defined
> p1<-c(2,0,0,0,0,0)
> p2 <- c(0,0,0,0,1,0)
> p3<- c(0,0,0,0,0,1)
> achievement <- data.frame(p1,p2,p3)
He
First answer is that R is not the proper environment for such a
question. There a re many free package for image analysis or even GIS.
Try for example Q-GIS
David
Il 19/09/2016 22:34, GwanSeon Kim ha scritto:
Hi, all
I am just beginner to use R.
I am working with TIF image file, and the info
> On 20 Sep 2016, at 11:34, Michael Haenlein wrote:
>
> Dear all,
>
> I am trying to estimate a lm model with one continuous dependent variable
> and 11 independent variables that are all categorical, some of which have
> many categories (several dozens in some cases).
If I’m not wrong, ( I as
Hi
Just a wild guess. Achievement in the goalprog package is data frame with four
named columns (objective, priority, p and n).
Your achievement is 3 column data.frame with names p1, p2 and p3.
Maybe data frame with defined structure is required.
Cheers
Petr
> -Original Message-
> Fr
I am a beginner user of R. I am using a national survey to test what
variables influence the partecipation in complementary pensions (the
partecipation in complementary pension is voluntary in my country).
Since the dependent variable is a dummy (1 if the person partecipate and 0
otherwise) I want
I was solving a LGP problem which is very basic.
Find x0 = [x1; x2], n0 = [n1; n2; n3] and p0 = [p1; p2; p3] that minimize a
= [(2p1); (n2); (n3)]
The objectives are as follows
10x1 + 15x2 + n1 - p1 = 40
100x1 + 100x2 + n2 - p2 = 1000
x2 + n3 - p3 = 7
x; n; p >= 0
The solution is x' = [4
Hi
see in line
> -Original Message-
> From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of mviljamaa
> Sent: Tuesday, September 20, 2016 10:01 AM
> To: r-help@r-project.org
> Subject: [R] Using lm's subset parameter results in Error in xj[i] : invalid
> subscript type 'list'
>
Dear all,
I am trying to estimate a lm model with one continuous dependent variable
and 11 independent variables that are all categorical, some of which have
many categories (several dozens in some cases).
I am not interested in statistical inference to a larger population. The
objective of my mo
By subsetting the rows in the firstkids data frame, you've already
subsetted your data.
Try specifying firstkids as your data instead of a subset in the lm call.
Also, eliminate the kidmomhsage prefix from all of your variables since
you're running the linear model on a different data frame(firstk
On Tue, 20 Sep 2016, 15:55 GwanSeon Kim wrote:
> Hi, all
> I am just beginner to use R.
> I am working with TIF image file, and the information about the raster is
> following:
>
> class : RasterLayer
> dimensions : 11150, 21808, 243159200 (nrow, ncol, ncell)
> resolution : 30, 30 (x, y
I'm trying to take lm on a subset of my dataset and to do this I believe
I need to pass my subset of the data as the subset parameter of lm.
So I do my subsetting:
firstkids <- kidmomhsage[0:234,], i.e. the first 234 rows of the data
frame.
Then construct the model:
fit4 <- lm(kidmomhsage$k
Mike Wojnowicz writes:
> I have successfully installed R on my AWS EC2 r3.8 box running Linux with
>>sudo yum install -y R
>
> However, I cannot find R anywhere (which I want for the sake of
> tar'ing it up and decompressing to make future installations easier.)
> For example,
>
>> rpm -ql R
>
>
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