[R] Archer-Lemeshow Goodness of Fit Test for Survey Data with Log. Regression
?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-Complex-Survey-Logistic-Regression-td4668233.html
Everything is going well until I get to the point where I have to add the
objects 'r' and 'g' as variables to the data frame by either using the
transform function or the update function to update the svrepdesign object.
The log. regression model involved uses a subset of data and some of the values
in the data frame are NA, so that is affecting my ability to add 'r' and 'g' as
variables; I am getting an error because I only have 8397 rows for the new
variables and 16197 in the data frame and svrepdesign object. I am not sure
how to overcome this error.
The following is a MRE:
##Archer Lemeshow Goodness of Fit Test for Complex Survey Data with Logistic
Regression
library(RCurl)
library(survey)
data <-
getURL("https://raw.githubusercontent.com/cbenjamin1821/careertech-ed/master/elsq1adj.csv";)
elsq1ch <- read.csv(text = data)
#Specifying the svyrepdesign object which applies the BRR weights
elsq1ch_brr<-svrepdesign(variables = elsq1ch[,1:16], repweights =
elsq1ch[,18:217], weights = elsq1ch[,17], combined.weights = TRUE, type = "BRR")
elsq1ch_brr
##Resetting baseline levels for predictors
elsq1ch_brr <- update( elsq1ch_brr , F1HIMATH = relevel(F1HIMATH,"PreAlg or
Less") )
elsq1ch_brr <- update( elsq1ch_brr , BYINCOME = relevel(BYINCOME,"0-25K") )
elsq1ch_brr <- update( elsq1ch_brr , F1RACE = relevel(F1RACE,"White") )
elsq1ch_brr <- update( elsq1ch_brr , F1SEX = relevel(F1SEX,"Male") )
elsq1ch_brr <- update( elsq1ch_brr , F1RTRCC = relevel(F1RTRCC,"Academic") )
#Log. Reg. model-all curric. concentrations including F1RTRCC as a predictor
allCC <-
svyglm(formula=F3ATTAINB~F1PARED+BYINCOME+F1RACE+F1SEX+F1RGPP2+F1HIMATH+F1RTRCC,family="binomial",design=elsq1ch_brr,subset=BYSCTRL==1&G10COHRT==1,na.action=na.omit)
summary(allCC)
#Recommendations from Lumley (from R Help Archive) on implementing the Archer
Lemeshow GOF test
r <- residuals(allCC, type="response")
f<-fitted(allCC)
g<- cut(f, c(-Inf, quantile(f, (1:9)/10, Inf)))
# now create a new design object with r and g added as variables
#This is the area where I am having problems as my model involves a subset and
some values are NA as well
#I am also not sure if I am naming/specifying the new variables of r and g
properly
transform(elsq1ch,r=r,g=g)
elsq1ch_brr <- update(elsq1ch_brr,tag=g,tag=r)
#then:
decilemodel<- svyglm(r~g, design=newdesign)
regTermTest(decilemodel, ~g)
#is the F-adjusted mean residual test from the Archer Lemeshow paper
Thank you,
Courtney
?
Courtney Benjamin
Broome-Tioga BOCES
Automotive Technology II Teacher
Located at Gault Toyota
Doctoral Candidate-Educational Theory & Practice
State University of New York at Binghamton
cbenj...@btboces.org
607-763-8633
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Run a Python code from R
> 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,
>
>
>> 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-packages/nlmpy/nlmpy.py")
>
> + pithon.call( "mpd", nRow, nCol, h)
>
> +
>
> + } else {
>
> + print("Unable to execute python")
>
> + }
>
> Show Traceback
>
>
>
> Rerun with Debug
>
>
>
> Error in pithon.get("_r_call_return", instance.name = instname) :
>
> Couldn't retrieve variable: Traceback (most recent call last):
>
> File "C:/Users/Documents/R/win-library/3.3/rPithon/pythonwrapperscript.py",
> line 110, in
>
>reallyReallyLongAndUnnecessaryPrefix.data =
> json.dumps([eval(reallyReallyLongAndUnnecessaryPrefix.argData)])
>
> File "C:\Users\ANACON~1\lib\json\__init__.py", line 244, in dumps
>
>return _default_encoder.encode(obj)
>
> File "C:\Users\ANACON~1\lib\json\encoder.py", line 207, in encode
>
>chunks = self.iterencode(o, _one_shot=True)
>
> File "C:\Users\ANACON~1\lib\json\encoder.py", line 270, in iterencode
>
>return _iterencode(o, 0)
>
> File "C:\Users\ANACON~1\lib\json\encoder.py", line 184, in default
>
>raise TypeError(repr(o) + " is not JSON serializable")
>
> TypeError: array([[ 0.36534654, 0.31962481, 0.44229946, ..., 0.11513079,
>
> 0.07156331, 0.00286971],
>
> [ 0.41534291, 0.41333479, 0.48118995, ..., 0.19203674,
>
> 0.04192771, 0.03679473],
>
> [ 0.5188
>
>
>
>
>
> Nell
>
>
>
> De : Wensui Liu
> Envoy� : mercredi 16 novembre 2016 16:00:03
> � : Nelly Reduan
> Cc : r-help@r-project.org
> Objet : Re: [R] Run a Python code from R
>
> 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 to build neutral landscape models
>>
>> https://pypi.python.org/pypi/nlmpy . The example comes from this website. I
>> tried to use the function system2 but I don't know how to use it.
>>
>>
>> path_script_python <- "C:/Users/Anaconda2/Lib/site-packages/nlmpy/nlmpy.py"
>>
>> test <- system2("python", args = c(path_script_python, as.character(nRow),
>> as.character(nCol), as.character(h)))
>>
>> Thanks a lot for your help.
>> Nell
>>
>>
>> nlmpy 0.1.3 : Python Package Index
>> pypi.python.org
>> NLMpy. NLMpy is a Python package for the creation of neutral landscape
>> models that are widely used in the modelling of ecological patterns and
>> processes across ...
>>
>>
>>
>>[[alternative HTML version deleted]]
>>
>> __
>> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
>> https://stat.ethz.ch/mailman/listinfo/r-help
>> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>
> [[alternative HTML version deleted]]
>
> __
> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
David Winsemius
Alameda, CA, USA
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Run a Python code from R
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-packages/nlmpy/nlmpy.py")
+ pithon.call( "mpd", nRow, nCol, h)
+
+ } else {
+ print("Unable to execute python")
+ }
Show Traceback
Rerun with Debug
Error in pithon.get("_r_call_return", instance.name = instname) :
Couldn't retrieve variable: Traceback (most recent call last):
File "C:/Users/Documents/R/win-library/3.3/rPithon/pythonwrapperscript.py",
line 110, in
reallyReallyLongAndUnnecessaryPrefix.data =
json.dumps([eval(reallyReallyLongAndUnnecessaryPrefix.argData)])
File "C:\Users\ANACON~1\lib\json\__init__.py", line 244, in dumps
return _default_encoder.encode(obj)
File "C:\Users\ANACON~1\lib\json\encoder.py", line 207, in encode
chunks = self.iterencode(o, _one_shot=True)
File "C:\Users\ANACON~1\lib\json\encoder.py", line 270, in iterencode
return _iterencode(o, 0)
File "C:\Users\ANACON~1\lib\json\encoder.py", line 184, in default
raise TypeError(repr(o) + " is not JSON serializable")
TypeError: array([[ 0.36534654, 0.31962481, 0.44229946, ..., 0.11513079,
0.07156331, 0.00286971],
[ 0.41534291, 0.41333479, 0.48118995, ..., 0.19203674,
0.04192771, 0.03679473],
[ 0.5188
Nell
De : Wensui Liu
Envoy� : mercredi 16 novembre 2016 16:00:03
� : Nelly Reduan
Cc : r-help@r-project.org
Objet : Re: [R] Run a Python code from R
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 to build neutral landscape models
>
> https://pypi.python.org/pypi/nlmpy . The example comes from this website. I
> tried to use the function system2 but I don't know how to use it.
>
>
> path_script_python <- "C:/Users/Anaconda2/Lib/site-packages/nlmpy/nlmpy.py"
>
> test <- system2("python", args = c(path_script_python, as.character(nRow),
> as.character(nCol), as.character(h)))
>
> Thanks a lot for your help.
> Nell
>
>
> nlmpy 0.1.3 : Python Package Index
> pypi.python.org
> NLMpy. NLMpy is a Python package for the creation of neutral landscape models
> that are widely used in the modelling of ecological patterns and processes
> across ...
>
>
>
> [[alternative HTML version deleted]]
>
> __
> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Run a Python code from R
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 to build neutral landscape models
>
> https://pypi.python.org/pypi/nlmpy . The example comes from this website. I
> tried to use the function system2 but I don't know how to use it.
>
>
> path_script_python <- "C:/Users/Anaconda2/Lib/site-packages/nlmpy/nlmpy.py"
>
> test <- system2("python", args = c(path_script_python, as.character(nRow),
> as.character(nCol), as.character(h)))
>
> Thanks a lot for your help.
> Nell
>
>
> nlmpy 0.1.3 : Python Package Index
> pypi.python.org
> NLMpy. NLMpy is a Python package for the creation of neutral landscape models
> that are widely used in the modelling of ecological patterns and processes
> across ...
>
>
>
> [[alternative HTML version deleted]]
>
> __
> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] Run a Python code from R
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
tried to use the function system2 but I don't know how to use it.
path_script_python <- "C:/Users/Anaconda2/Lib/site-packages/nlmpy/nlmpy.py"
test <- system2("python", args = c(path_script_python, as.character(nRow),
as.character(nCol), as.character(h)))
Thanks a lot for your help.
Nell
nlmpy 0.1.3 : Python Package Index
pypi.python.org
NLMpy. NLMpy is a Python package for the creation of neutral landscape models
that are widely used in the modelling of ecological patterns and processes
across ...
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] using pmax in presence of NAs
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.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 in each row: >> >> x <- data.frame(a = c(1:5), b=11:15, c=c(111:114,NA)) >> x >> I'd like it to return: >> [1] 111 112 113 114 15 >> >> Thanks a lot! > > The first thing to notice is that pmax allows na.rm=TRUE, so it works fine to > do > >> pmax(a = c(1:5), b=11:15, c=c(111:114,NA), na.rm=TRUE) > [1] 111 112 113 114 15 > > I.e., it is your desire to use do.call() that is the challenge. The 2nd > argument should be a list containing the arguments as in the above call, so > this works too > >> do.call(pmax, list(a = c(1:5), b=11:15, c=c(111:114,NA), na.rm=TRUE)) > [1] 111 112 113 114 15 > > The form do.call(pmax,x) uses the fact that a data.frame is a kind of list. > What you need to do is to tack on the element na.rm=TRUE. This works > >> do.call(pmax, c(x, list(na.rm=TRUE))) > [1] 111 112 113 114 15 > > and it even works without the list() > >> do.call(pmax, c(x, na.rm=TRUE)) > [1] 111 112 113 114 15 > > -pd > > >> >> -- >> Dimitri Liakhovitski >> >> __ >> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see >> https://stat.ethz.ch/mailman/listinfo/r-help >> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html >> and provide commented, minimal, self-contained, reproducible code. > > -- > Peter Dalgaard, Professor, > Center for Statistics, Copenhagen Business School > Solbjerg Plads 3, 2000 Frederiksberg, Denmark > Phone: (+45)38153501 > Office: A 4.23 > Email: pd@cbs.dk Priv: pda...@gmail.com > > > > > > > > > -- Dimitri Liakhovitski __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] using pmax in presence of NAs
> 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 data frame has NAs? > I'd like it to ignore NAs and return the maximum of all non-NAs in each row: > > x <- data.frame(a = c(1:5), b=11:15, c=c(111:114,NA)) > x > I'd like it to return: > [1] 111 112 113 114 15 > > Thanks a lot! The first thing to notice is that pmax allows na.rm=TRUE, so it works fine to do > pmax(a = c(1:5), b=11:15, c=c(111:114,NA), na.rm=TRUE) [1] 111 112 113 114 15 I.e., it is your desire to use do.call() that is the challenge. The 2nd argument should be a list containing the arguments as in the above call, so this works too > do.call(pmax, list(a = c(1:5), b=11:15, c=c(111:114,NA), na.rm=TRUE)) [1] 111 112 113 114 15 The form do.call(pmax,x) uses the fact that a data.frame is a kind of list. What you need to do is to tack on the element na.rm=TRUE. This works > do.call(pmax, c(x, list(na.rm=TRUE))) [1] 111 112 113 114 15 and it even works without the list() > do.call(pmax, c(x, na.rm=TRUE)) [1] 111 112 113 114 15 -pd > > -- > Dimitri Liakhovitski > > __ > R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. -- Peter Dalgaard, Professor, Center for Statistics, Copenhagen Business School Solbjerg Plads 3, 2000 Frederiksberg, Denmark Phone: (+45)38153501 Office: A 4.23 Email: pd@cbs.dk Priv: pda...@gmail.com __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] using pmax in presence of NAs
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.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. >> 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 in each row: >> >> x <- data.frame(a = c(1:5), b=11:15, c=c(111:114,NA)) >> x >> I'd like it to return: >> [1] 111 112 113 114 15 >> >> Thanks a lot! >> >> -- >> Dimitri Liakhovitski >> -- Dimitri Liakhovitski __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] using pmax in presence of NAs
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. > 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 in each row: > > x <- data.frame(a = c(1:5), b=11:15, c=c(111:114,NA)) > x > I'd like it to return: > [1] 111 112 113 114 15 > > Thanks a lot! > > -- > Dimitri Liakhovitski > __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] using pmax in presence of NAs
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=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 in each row: > > x <- data.frame(a = c(1:5), b=11:15, c=c(111:114,NA)) > x > I'd like it to return: > [1] 111 112 113 114 15 > > Thanks a lot! > > -- > Dimitri Liakhovitski -- Dimitri Liakhovitski __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] using pmax in presence of NAs
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 in each row: x <- data.frame(a = c(1:5), b=11:15, c=c(111:114,NA)) x I'd like it to return: [1] 111 112 113 114 15 Thanks a lot! -- Dimitri Liakhovitski __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] GAMs: test of simple effects following a significant interaction
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 the number of unique skunks and the independent variables are the X coordinates of the centroids of trapping sites (called "X" in the GAM) and the proportion of forests within the trapping sites (called "prop_forest" in the GAM): mod <- gam(nb_unique ~ s(x,prop_forest), offset=log_trap_eff, family=nb(theta=NULL, link="log"), data=succ_capt_skunk, method = "REML", select = TRUE) summary(mod) Family: Negative Binomial(13.446) Link function: log Formula: nb_unique ~ s(x, prop_forest) Parametric coefficients: Estimate Std. Error z value Pr(>|z|) (Intercept) -2.020950.03896 -51.87 <2e-16 *** --- Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 Approximate significance of smooth terms: edf Ref.df Chi.sq p-value s(x,prop_forest) 3.182 29 17.76 0.000102 *** --- Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 R-sq.(adj) = 0.37 Deviance explained = 49% -REML = 268.61 Scale est. = 1 n = 58 Should I include the simple effects of independent variables "X" and "prop_forest" into the GAM when the interaction is significant? I ask this question because the longitude and latitude are often included as an interaction term in a GAM (i.e., s(X,Y)) without the simple effects (however, I tested for the simple effects and they were not significant in my case). Is it correct to include the interaction between X and proportion of forests when my objective is to test longitudinal changes in proportion of forests? Thanks a lot for your time. Have a nice day. Marine [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Course: "Statistical Data Analysis with R"
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. Below you find more information about this course. If you have any question don't hesitate to contact me. Best regards, Pablo ++ Course: Introduction to Statistical Data Analysis with R Where: Linux Hotel, Essen-Horst, Germany When: 15.12 - 17.12.2016 week 50 Instructor: PD Dr. rer. nat. Pablo Emilio Verde ++ *Target audience* This course is for data analyst who are familiar with classical statistics software like SPSS, SAS, etc. and they want to get a working knowledge in R. This is a 3 days intensive training course with 8 hours per day including lecturing and exercises. The course presentation is practical with many worked examples. To attend the course you do NOT need experience with R. Lectures are given in English. Discussions can be in English, German or Spanish. ++ Day 1 * Introduction to concepts in R * Data structures, objects and classes * Data management with R (indexing and other advanced techniques) * Graphics with R (classical functions, lattice plots and ggplot2) Day 2 * Introduction to statistical functions in R * Bootstrap methods and Monte Carlo simulation inference in R * Linear models and regression techniques in R * Model checking and model diagnostics Day 3 * Logistic regression, Lognormal models and GLM with R * Exploratory multivariate analysis * Exploratory regression analysis with Classification Trees and Random Forest * Own projects ++ Costs: Public sector and commercial: 3 days, 998.00 € + 19% vat = 1,187.62€ (three days course, full eight clock hours per day, complete set of literature, WiFi, complimentary notebook, full board, drinks, pastries, homemade cakes, sauna, social program) Additionally acommodation on demand: 63,13 € shared double room per night (incl. VAT) or 138,03 € single room per night (inc. VAT). Student: 675 € three days course Some of the courses are frequently fully booked. So please notice that you may have to try several times, until you get a spare place. For more information, please contact: i...@linuxhotel.de ++ [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Variable 'A' is not a factor Error message
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 integer
representation - which is a completely different thing in model matrix terms.
So your predictors are still two-level factors, and your linear model has,
after building a model matrix from the default contrasts, probably got one
coefficient for each upper level of your two level factors and one for the
intercept. (I'm assuming default contrasts here).
You then decided to redefine your predictors as numeric variables with a very
large number of levels given by rnorm, none of which (except by very rare
coincidence) were in your original design. So your model cannot possibly
predict the output - it has no coefficients for all those new levels. To avoid
that, R quite accurately told you that you can't do that.
If you want to fit a linear model with continuous variables, you need to set up
your DOE data frame with (meaningful) numeric predictors, not factors. You will
then get a numerical gradient for each factor instead of a single offset for
each upper level. That isn't really what Placket and Burman had in mind, so I
would not normally start with a P-B design if I wanted to do that. Consider a
response surface model instead.
S Ellison
> -Original Message-
> From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of ahmed
> meftah
> Sent: 11 November 2016 17:52
> To: r-help@r-project.org
> Subject: [R] Variable 'A' is not a factor Error message
>
> I am running a DOE with the following code library(Rcmdr)
> library(RcmdrMisc)
> library(RcmdrPlugin.DoE)
> # Define plackett burman experiment
> PB.DOE <- pb(nruns= 12 ,n12.taguchi= FALSE ,nfactors= 12 -1, ncenter= 0 ,
> replications= 1 ,repeat.only= FALSE ,randomize= TRUE ,seed=
> 27241 ,
> factor.names=list(
> A=c(100,1000),B=c(100,200),C=c(1,3),D=c(1,1.7),
> E=c(1000,1500),G=c(-2,2) ) )
>
> as.numeric2 <- function(x) as.numeric(as.character(x))
>
> # Calculate response column
> IP <- with(PB.DOE,(as.numeric2(A)*as.numeric2(B)*(5000-
> 3000))/(141.2*as.numeric2(C)*as.numeric2(D)*(log(as.numeric2(E)/0.25)-
> (1/2)+as.numeric2(G
> # Combine response column with exp design table
> final_set <- within(PB.DOE, {
> IP<- ((as.numeric2(A)*as.numeric2(B)*(5000-
> 3000))/(141.2*as.numeric2(C)*as.numeric2(D)*(log(as.numeric2(E)/0.25)-
> (1/2)+as.numeric2(G
> })I then ran a regression as follows:LinearModel.1 <- lm(IP ~ A + B + C +
> D +
> E + G,
> data=final_set)
> summary(LinearModel.1)Following this i wanted to run a predict using
> specified values as predictors in a Monte Carlo:n = 1 # Define probability
> distributions of predictors A = rnorm(n,450,100) hist(A,col = "blue",breaks =
> 50)
>
> B = rnorm(n, 150,10)
> hist(B,col = "blue",breaks = 50)
>
> C = rnorm(n, 1.5, 0.5)
> hist(C,col = "blue",breaks = 50)
>
> D = runif(n,1.2,1.7)
> hist(D,col = "blue",breaks = 50)
>
> E = rnorm(n,1250,50)
> hist(E,col = "blue",breaks = 50)
>
> G = rnorm(n,0,0.5)
> hist(G,col = "blue",breaks = 50)
>
> MCtable <- data.frame(A=A,B=B,C=C,D=D,E=E,G=G)
>
> for (n in 1:n) {
> N=predict(LinearModel.1,MCtable)
> }
>
> hist(N,col = "yellow",breaks = 10)I end up getting this error:"Warning in
> model.frame.default(Terms, newdata, na.action = na.action, xlev =
> object$xlevels) :
> variable 'A' is not a factor"Using str() to get some info on the
> LinearModel.1
> and from what I understand seems to indicates that since the predictors
> A,B,C etc are factors with 2 levels I have to convert my data.frame table to
> factors aswell. Is that correct?Doing this would mean I would also need to
> specify the number of levels which would mean that since I have set my n to
> 1 would mean 1 levels for each factor. How would I go about doing
> this? Is there a better solution? Any help would be appreciated.
> [[alternative HTML version deleted]]
>
> __
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[R] Some basic time series questions
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(x,start=c(2014,9),frequency=12) inputting x.ts at the prompt gives me a table with the rooms denoted by year and columns denoted by months and everything lines up wonderfully. Now my problem comes when I type at the prompt plot(x.ts) or plot(x.ts, xlab="") or plot.ts(x.ts,xlab="") I get a plot of the values, but my x-axis labels are 2015.0, 2015.5, 2016.0, and 2016.5 . January 2015 is coming out as 2015.0... Is there a way of getting a more intelligible x-axis labeling? Even 2015.1 for Janaury, etc. would work, or even getting an index (either Septemebr 2014 representing 0 or 1 and it incrementally increasing each month). Question #2-- If I have a time series of decadal events, how best should I set the frequency. It is historical data, in the form of say AD 610-619 5 events, AD 620-629 7 events, etc. Sorry for such a basic questions. Any advice would be appreciated. Thanks in advance, MEH Mark E. Hall, PhD Assistant Field Manager Black Rock Field Office Winnemucca District Office 775-623-1529. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Different results when converting a matrix to a data.frame
> 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 reason for being present in the data, and normally each has its own
> storage mode. Your use of a matrix as a short cut way to create many columns
> at once does not change this fundamental difference between data frames and
> matrices. You should not be surprised that putting the finishing touches on
> this transformation takes some personal attention.
>
> Normally you should give explicit names to each column using the argument
> names in the data.frame function. When using a matrix as a shortcut, you
> should either immediately follow the creation of the data frame with a
> names(DF)<- assignment, or wrap it in a setNames function call.
>
> setNames( data.frame(matrix(NA, 2, 2)), c( "ColA", "ColB" ) )
>
> Note that using a matrix to create many columns is memory inefficient,
> because you start by setting aside a single block of memory (the matrix) and
> then you move that data column at a time to separate vectors for use in the
> data frame. If working with large data you might want to consider allocating
> each column separately from the beginning.
>
> N <- 2
> nms <- c( "A", "B" )
> as.data.frame( setNames( lapply( nms, function(n){ rep( NA, 2 ) } ), nms ) )
>
> which is not as convenient, but illustrates that data frames are truly
> different than matrices.
> --
> Sent from my phone. Please excuse my brevity.
>
> On November 16, 2016 7:20:38 AM PST, g.maub...@weinwolf.de wrote:
>> 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.frame deliver different results than as.data.frame with
>> regard to the variable names (V instead of X)?
They are two different functions:
It's fairly easy to see by looking at the code:
as.data.frame.matrix uses: names(value) <- paste0("V", ic) when there are no
column names and data.frame calls make.names which prepends an "X" as the first
letter of invalid or missing names.
As to why the authors did it this way, I'm unable to comment.
>>
>> Kind regards
>>
>> Georg
>>
>> [[alternative HTML version deleted]]
David Winsemius
Alameda, CA, USA
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Re: [R] Different results when converting a matrix to a data.frame
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 storage
mode. Your use of a matrix as a short cut way to create many columns at once
does not change this fundamental difference between data frames and matrices.
You should not be surprised that putting the finishing touches on this
transformation takes some personal attention.
Normally you should give explicit names to each column using the argument names
in the data.frame function. When using a matrix as a shortcut, you should
either immediately follow the creation of the data frame with a names(DF)<-
assignment, or wrap it in a setNames function call.
setNames( data.frame(matrix(NA, 2, 2)), c( "ColA", "ColB" ) )
Note that using a matrix to create many columns is memory inefficient, because
you start by setting aside a single block of memory (the matrix) and then you
move that data column at a time to separate vectors for use in the data frame.
If working with large data you might want to consider allocating each column
separately from the beginning.
N <- 2
nms <- c( "A", "B" )
as.data.frame( setNames( lapply( nms, function(n){ rep( NA, 2 ) } ), nms ) )
which is not as convenient, but illustrates that data frames are truly
different than matrices.
--
Sent from my phone. Please excuse my brevity.
On November 16, 2016 7:20:38 AM PST, g.maub...@weinwolf.de wrote:
>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.frame deliver different results than as.data.frame with
>regard to the variable names (V instead of X)?
>
>Kind regards
>
>Georg
>
> [[alternative HTML version deleted]]
>
>__
>R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
>https://stat.ethz.ch/mailman/listinfo/r-help
>PLEASE do read the posting guide
>http://www.R-project.org/posting-guide.html
>and provide commented, minimal, self-contained, reproducible code.
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[R] Different results when converting a matrix to a data.frame
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.frame deliver different results than as.data.frame with regard to the variable names (V instead of X)? Kind regards Georg [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Issues with the way Apply handled NA's
> -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 intermediate 'vals' variable is
to apply an anonymous function that does the check internally:
plabor$colD <- apply(plabor, 1, function(x) if(all(is.na(x))) NA else prod(x,
na.rm=TRUE))
S Ellison
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Re: [R] Text categories based on the sentences
> 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 chicken food
> 4. yesterday i went to birthday partyfestival
> 5. i lost my mobile last week mobile
>
> Please remember that i have only "*Description*" Variables only.How can i
> get the categories column based on the sentences of *Description *column.
You could look at something like ReadMe (http://gking.harvard.edu/readme) to
generate a classifier based on a suitable subsample of your data that then
classifies the rest of your data set.
Alternatively you could do it the hard way; use a natural language parser to
extract all the noun phrases (or just split oout the words), list the unique
noun phrases, manually classify all of them using your own criteria to give a
pair list (phrase->category), and then match classes back to the rows that
contain each noun phrase.
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Re: [R] extracting all different items in a matrix colum or vector
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, bgnumis bgnum wrote: Many Thanks Sarah Really I´m going to do it. If you can suggest me one complete and didactic I will be very gratefull. Anyway many thanks for your answer. 2016-11-15 18:56 GMT+01:00 Sarah Goslee : Your question strongly suggests that you need to reread at least one introductory guide to R. But the answer to your specific question is the unique() function. Sarah On Tue, Nov 15, 2016 at 7:33 AM, bgnumis bgnum wrote: Hi all, >From many time ago, I have return to R, I have a matrix with this values. C(jose, pepe, jose, luis, pepe, raul) I want to "read" this matrix or element and extract all different values, so the output matrix (that I want to download is: c(jose, pepe, luis, raul). There is a function to do this? Many thanks in advance. -- Sarah Goslee http://www.functionaldiversity.org [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Michael http://www.dewey.myzen.co.uk/home.html __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
