?Hello R Experts,
I am trying to implement the Archer-Lemeshow GOF Test for survey data on a
logistic regression model using the survey package based upon an R Help Archive
post that I found where Dr. Thomas Lumley advised how to do it:
http://r.789695.n4.nabble.com/Goodness-of-t-tests-for-Comp
> On Nov 16, 2016, at 4:53 PM, Nelly Reduan wrote:
>
> Thank you very much for your help !
>
>
> I 'm trying to use the package "rPithon" but I obtain this error message:
Are you sure you are not just misspelling rPython? If that's not the issue than
you need to say where you got rPithon,
Thank you very much for your help !
I 'm trying to use the package "rPithon" but I obtain this error message:
> if (pithon.available())
+ {
+ nRow <-50
+ nCol <-50
+ h <- 0.75
+
+ # this file contains the definition of function concat
+ pithon.load("C:/Users/Anaconda2/Lib/site-
take a look at rpython or rPithon package
On Wed, Nov 16, 2016 at 4:53 PM, Nelly Reduan wrote:
> Hello,
>
>
> How can I run this Python code from R ?
>
>
import nlmpy
nlm = nlmpy.mpd(nRow=50, nCol=50, h=0.75)
nlmpy.exportASCIIGrid("raster.asc", nlm)
>
>
> Nlmpy is a Python package
Hello,
How can I run this Python code from R ?
>>> import nlmpy
>>> nlm = nlmpy.mpd(nRow=50, nCol=50, h=0.75)
>>> nlmpy.exportASCIIGrid("raster.asc", nlm)
Nlmpy is a Python package to build neutral landscape models
https://pypi.python.org/pypi/nlmpy . The example comes from this website. I
Thank you, Peter!
On Wed, Nov 16, 2016 at 4:21 PM, peter dalgaard wrote:
>
>> On 16 Nov 2016, at 21:58 , Dimitri Liakhovitski
>> wrote:
>>
>> Hello!
>>
>> I need to calculate the maximum of each row of a data frame.
>> This works:
>>
>> x <- data.frame(a = 1:5, b=11:15, c=111:115)
>> x
>> do
> On 16 Nov 2016, at 21:58 , Dimitri Liakhovitski
> wrote:
>
> Hello!
>
> I need to calculate the maximum of each row of a data frame.
> This works:
>
> x <- data.frame(a = 1:5, b=11:15, c=111:115)
> x
> do.call(pmax, x)
> [1] 111 112 113 114 115
>
> However, how should I modify it if my
Thanks a lot, Sarah.
I just had no idea where to put na.rm = T in the do.call call.
Appreciate it!
Dimitri
On Wed, Nov 16, 2016 at 4:06 PM, Sarah Goslee wrote:
> pmax has a na.rm argument. Why not just use that?
>
> x <- data.frame(a = c(1:5), b=11:15, c=c(111:114,NA))
>
>> do.call(pmax, c(x, na.
pmax has a na.rm argument. Why not just use that?
x <- data.frame(a = c(1:5), b=11:15, c=c(111:114,NA))
> do.call(pmax, c(x, na.rm=TRUE))
[1] 111 112 113 114 15
On Wed, Nov 16, 2016 at 3:58 PM, Dimitri Liakhovitski
wrote:
> Hello!
>
> I need to calculate the maximum of each row of a data frame
To clarify:
I know I could do:
apply(x, 1, max, na.rm = T)
But I was wondering if one can modify the pmax one...
On Wed, Nov 16, 2016 at 3:58 PM, Dimitri Liakhovitski
wrote:
> Hello!
>
> I need to calculate the maximum of each row of a data frame.
> This works:
>
> x <- data.frame(a = 1:5, b=1
Hello!
I need to calculate the maximum of each row of a data frame.
This works:
x <- data.frame(a = 1:5, b=11:15, c=111:115)
x
do.call(pmax, x)
[1] 111 112 113 114 115
However, how should I modify it if my data frame has NAs?
I'd like it to ignore NAs and return the maximum of all non-NAs
Hello,
I am a novice in Generalized Additive Models (GAMs) and I would need some
advice on these models. From capture data, I would like to assess the effect of
longitudinal changes in proportion of forests on abundance of skunks. To test
this, I built this GAM where the dependent variable is th
Dear list members,
I am very glad to announce the introductory course: "Statistical Data
Analysis with R".
I have running this course for 10 years and previous participants have
had a nice time learning R at the LinuxHotel.
You are invited to bring their own data to analyze during the course.
This looks like one of those 'please talk to a statistician' questions ...
You appear to have requested a 12-run placket-burman experiment, which is a
design that requires up to 11 two-level factors. You then fitted (I think)
simulated data to that design, using those factors converted to their
Hi,
As I sit and learn how to work with time series, I've run into a problem
that is eluding a quick easy answer (and googling for answers seems to
really slow the process...)
Question #1--
In a simple example on R 3.3.1 (sorry my employer hasn't upgraded to 3.3.2
yet):
x=rnorm(26,0,1)
x.ts<-ts(
> On Nov 16, 2016, at 8:43 AM, Jeff Newmiller wrote:
>
> I will start by admitting I don't know the answer to your question.
>
> However, I am responding because I think this should not be an issue in real
> life use of R. Data frames are lists of distinct vectors, each of which has
> its own
I will start by admitting I don't know the answer to your question.
However, I am responding because I think this should not be an issue in real
life use of R. Data frames are lists of distinct vectors, each of which has its
own reason for being present in the data, and normally each has its own
Hi All,
I build an empty dataframe to fill it will values later. I did the
following:
-- cut --
matrix(NA, 2, 2)
[,1] [,2]
[1,] NA NA
[2,] NA NA
> data.frame(matrix(NA, 2, 2))
X1 X2
1 NA NA
2 NA NA
> as.data.frame(matrix(NA, 2, 2))
V1 V2
1 NA NA
2 NA NA
-- cut --
Why does data.
> -Original Message-
>
> You can check for an empty vector as follows:
> ...
> vals <- apply(plabor[c("colA","colB","colC")],1,function(x)
> length(na.omit(x)))
> vals # [1] 3 0 3 2
> <- ifelse(vals>0, plabor$colD, NA)
>
plabor
A slightly more compact variant that avoids the interme
> I have data set contains one variable "*Description*"
>
> *Description** Category*
>
> 1. i want ice cream food
> 2. i like banana very much fruit
> 3. tomorrow i will eat chick
You say you are coming back to R after a pause. I think the key word in
Sarah's response is re-reading. Why not start with the material you used
before and re-read it? If you have never read the Introduction to R
which comes with your installation that is worth a read too.
On 15/11/2016 20:47,
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