Re: [R] lattice: packet.number() versus panel.number()
On Aug 26, 2014, at 3:25 PM, Benjamin Tyner wrote: > Hi, > > According to > https://svn.r-project.org/R-packages/trunk/lattice/R/print.trellis.R, > >"[panel.number] is usually the same as, but can be different from > packet.number" > > and I had been under the impression that as long as the user is not > using a custom index.cond nor perm.cond, the panel.number would in fact > be the same as the packet.number. > > However, I have recently come across a case where the two are *not* the > same, even though I am not using a custom index.cond nor perm.cond. > > So my question is, what might be some other possible situations in which > the two would be expected to differ? The immediate hypothesis that leaps to mind is cases where there are multiple pages. On each page I suspect the upper left numbering restarts with 1, but I suspect the packet numbers are sequential increasing. -- David Winsemius Alameda, CA, USA __ R-help@r-project.org mailing list 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] NA's introduced by coercion
Hi Madhvi, First, please use "reply-all" when responding to emails form this list so that others can help (and benefit from) the discussion. Comment down below: On 26 Aug 2014, at 22:15, madhvi.gupta wrote: On 08/27/2014 10:42 AM, Steve Lianoglou wrote: Hi, On Tue, Aug 26, 2014 at 9:56 PM, madhvi.gupta wrote: Hi, I am applyin function as.numeric to a vector having many values as NA and it is giving : Warning message: NAs introduced by coercion Can anyone help me to know how to remove this warning and sor it out? Let's say that the vector you are calling `as.numeric` over is called `x`. If you could show us the output of the following command: R> head(x[is.na(as.numeric(x))]) You'll see why you are getting the warning. How you choose to sort it out probably depends on what you are trying to do with your data after you convert it to a "numeric" -steve Hi, I am having this error bacouse vector contains value NA but i want to convert that vector to numeric I don't quite follow what the problem is, then ... what is the end result that you want to happen? When you convert the vector to a numeric, the NA's that were in it originally, will remain as NAs (but they will be of a 'numeric' type). What would you like to do with the NA values? Do you just want to keep them, but want to silence the warning? If so, you can do: R> suppressWarnings(y <- as.numeric(x)) -steve -- Steve Lianoglou Computational Biologist Genentech __ R-help@r-project.org mailing list 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] NA's introduced by coercion
On Aug 26, 2014, at 9:56 PM, madhvi.gupta wrote:
> Hi,
>
> I am applyin function as.numeric to a vector having many values as NA and it
> is giving :
> Warning message:
> NAs introduced by coercion
>
> Can anyone help me to know how to remove this warning and sor it out?
You are the one that needs to identify the cause of hte warning: look at the
difference in console output for these two examples:
> as.numeric( c("a", 1, NA_character_) )
[1] NA 1 NA
Warning message:
NAs introduced by coercion
> as.numeric( c("2", 1, NA_character_) )
[1] 2 1 NA
So it is not the NA's in that character vector but rather values that were
coerce to NA because the conversion could not be accomplished.
--
David Winsemius
Alameda, CA, USA
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[R] working with matrices
hello i want to fill a matrix by its upper off diagonal elements specifically I want to take the first and second column of� the matrix and I apply a function to then that returns a single number which I want to place in the (1,2) entry of the matrix, then I want to take the first and third column of the matrix and apply the same function, get the single number and place it to (2,3) entry of the matrix and so on how can i do it? thanks anna [[alternative HTML version deleted]] __ R-help@r-project.org mailing list 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] NA's introduced by coercion
Hi, On Tue, Aug 26, 2014 at 9:56 PM, madhvi.gupta wrote: > Hi, > > I am applyin function as.numeric to a vector having many values as NA and it > is giving : > Warning message: > NAs introduced by coercion > > Can anyone help me to know how to remove this warning and sor it out? Let's say that the vector you are calling `as.numeric` over is called `x`. If you could show us the output of the following command: R> head(x[is.na(as.numeric(x))]) You'll see why you are getting the warning. How you choose to sort it out probably depends on what you are trying to do with your data after you convert it to a "numeric" -steve -- Steve Lianoglou Computational Biologist Genentech __ R-help@r-project.org mailing list 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] NA's introduced by coercion
Hi, I am applyin function as.numeric to a vector having many values as NA and it is giving : Warning message: NAs introduced by coercion Can anyone help me to know how to remove this warning and sor it out? Thanks Madhvi __ R-help@r-project.org mailing list 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] What the difference between .Golbalenv and package:base?
So, the decisive factor is whether the input string be on the search() name
list, and not related with the envir's name attribute.
When we using attach, it is becasue the name attribute just match the search()
name list(or say,search() name list just use the name attribute), so
as.environment() can work well.
Tks!
--
PO SU
mail: desolato...@163.com
Majored in Statistics from SJTU
At 2014-08-27 00:02:07, "William Dunlap" wrote:
>as.environment(characterString) maps an entry from the output of
>search() to the environment at the named position in the search list.
>as.environment(number) maps an index into the output of search() to
>the the environment at that position in the search list. If
>'characterString' is not in the output of search() or 'number' is not
>in seq_along(search()) then as.environment throws an error. As far as
>I can tell, as.environment does not deal with the name of the
>environment at all. (When you attach an environment, attach will add
>a name attribute to the copied environment so the attached
>environment's name matches the name on the output of search(), but I
>don't think as.environment ever looks at that attribute.)
>
>Bill Dunlap
>TIBCO Software
>wdunlap tibco.com
>
>
>On Mon, Aug 25, 2014 at 9:07 PM, PO SU wrote:
>> First, sorry for my pool english expression which make you misunderstanding
>> of my original purpose.
>>
>> Sometimes, suppose a object in both stats and base, then i type the object
>> name, then after R search the search() list, R will use the object in stats,
>> is it right?( I just suppose, stats can be any package which libraried into
>> R.)
>> Then i know that, .GlobalEnv or globalenv() is the global environment
>> object, baseenv() returns the base environment object.
>> I also know that, i can convert the environment name into the real
>> environment object by using stats<-as.environment("package:stats"), And the
>> stats environment's name can be obtained using environmentName(stats), but
>> it returns "". (why?)
>> Then i use environmentName(.GlobalEnv) then i get "R_GlobalEnv", then i use
>> as.environment("R_GlobalEnv"), it does't work.(why?)
>>
>>
>> In one word, as.environment(x), x is somthing not the environment's name.
>>
>>
>> But, when i add a environment into the search() list, after i
>> attr(newenvir,"name")<-"new_name"
>> I can use the as.environment("new_name") to obtain the added environment.
>> (why?)
>>
>>
>> Hope you understand my meaning :)
>>
>>
>>
>>
>>
>>
>>
>>
>>
>>
>>
>>
>> --
>>
>> PO SU
>> mail: desolato...@163.com
>> Majored in Statistics from SJTU
>>
>>
>>
>> At 2014-08-26 02:51:54, "MacQueen, Don" wrote:
>>>Put simply,
>>> .GlobalEnvstores objects you create
>>> package:base contains functions and objects provided by R itself
>>>
>>>You don¹t need to use .GlobalEnv$a to use the variable named a. Just
>>>is ³a² by itself.
>>>
>>> a <- 4
>>> b <- 2*a
>>>print(a)
>>>print(b)
>>>
>>>Not necessary to use
>>> print(.GlobalEnv$a)
>>>
>>>Similarly, to find an object in the base package, just type its name.
>>>
>>>I don¹t know what you are trying to do, or why you think you have to use
>>>.GlobalEnv$a
>>>But in more than 20 years of using R for many different tasks, I have
>>>never had to do that.
>>>
>>>Furthermore, if you are new to R (which I would guess is the case), it
>>>seems unlikely to me that you need to work with environments or use
>>>attach() or assign(). In the vast majority of cases there are simpler ways
>>>that are easier to understand.
>>>
>>>You are aware, I hope, that
>>> > ls('.GlobalEnv')
>>> > ls(1)
>>> > ls()
>>>all return the same result?
>>>
>>>
>>>--
>>>Don MacQueen
>>>
>>>Lawrence Livermore National Laboratory
>>>7000 East Ave., L-627
>>>Livermore, CA 94550
>>>925-423-1062
>>>
>>>
>>>
>>>
>>>
>>>On 8/24/14, 11:07 PM, "PO SU" wrote:
>>>
Dear rusers,
As we know, there are a lot of environments in the search() path,
such as .Golbalenv and package:base .
And i can just use .Golbalenv$a ,.Golbalenv$b to use the virable, but
i must use as.envrionment("package:base") to find virable, i feel it not
very convenient.
For example, when i use the following codes to add a new env into the
search() path.
> tmp<-attach(NULL,name="new_name")
> assign("a",2,envir=as.environment("new_name"))
> a
[1] 2
> as.environment("new_name")$a
[1] 2
I must always convert the name to the environment, How can i just use
the following form:
> tmp<-attach(NULL,name="new_name")
> assign("a",2,envir=new_name) #like using .GlobalEnv
> a
[1] 2
> new_name$a
[1] 2
--
PO SU
mail: desolato...@163.com
Majored in Statistics from SJTU
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PLEASE do re
[R] Where to find source of C_pbinom?
Dear expeRts, I would like to find out how R computes pbinom(). A grep in the source code reveiled src/library/stats/R/distn.R:146: .External(C_pbinom, q, size, prob, lower.tail, log.p), so 'C_pbinom' refers to compiled C/C++ code loaded into R. Where can I find the source code of C_pbinom? Cheers, Marius __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Mixed sorting/ordering of strings acknowledging roman numerals?
On Aug 26, 2014, at 5:24 PM, Henrik Bengtsson wrote:
> Hi,
>
> does anyone know of an implementation/function that sorts strings that
> *contain* roman numerals (I, II, III, IV, V, ...) which are treated as
> numbers. In 'gtools' there is mixedsort() which does this for strings
> that contains (decimal) numbers. I'm looking for a "mixedsortroman()"
> function that does the same but with roman numbers, e.g.
It's pretty easy to sort something you know to be congruent with the existing
roman class:
romanC <- as.character( as.roman(1:3899) )
match(c("I", "II", "III","X","V"), romanC)
#[1] 1 2 3 10 5
But I guess you already know that, so you want a regex approach to parsing.
Looking at the path taken by Warnes, it would involve doing something like his
regex based insertion of a delimiter for "Roman numeral" but simpler because he
needed to deal with decimal points and signs and exponent notation, none of
which you appear to need. If you only need to consider character and Roman,
then this hack of Warnes tools succeeds:
mixedorderRoman <- function (x)
{
if (length(x) < 1)
return(NULL)
else if (length(x) == 1)
return(1)
if (is.numeric(x))
return(order(x))
delim = "\\$\\@\\$"
roman <- function(x) {
suppressWarnings(match(x, romanC))
}
nonnumeric <- function(x) {
suppressWarnings(ifelse(is.na(as.numeric(x)), toupper(x),
NA))
}
x <- as.character(x)
which.nas <- which(is.na(x))
which.blanks <- which(x == "")
if (length(which.blanks) > 0)
x[which.blanks] <- -Inf
if (length(which.nas) > 0)
x[which.nas] <- Inf
delimited <- gsub("([IVXCL]+)",
paste(delim, "\\1", delim, sep = ""), x)
step1 <- strsplit(delimited, delim)
step1 <- lapply(step1, function(x) x[x > ""])
step1.roman <- lapply(step1, roman)
step1.character <- lapply(step1, nonnumeric)
maxelem <- max(sapply(step1, length))
step1.roman.t <- lapply(1:maxelem, function(i) sapply(step1.roman,
function(x) x[i]))
step1.character.t <- lapply(1:maxelem, function(i) sapply(step1.character,
function(x) x[i]))
rank.roman <- sapply(step1.roman.t, rank)
rank.character <- sapply(step1.character.t, function(x)
as.numeric(factor(x)))
rank.roman[!is.na(rank.character)] <- 0
rank.character <- t(t(rank.character) + apply(matrix(rank.roman),
2, max, na.rm = TRUE))
rank.overall <- ifelse(is.na(rank.character), rank.numeric,
rank.character)
order.frame <- as.data.frame(rank.overall)
if (length(which.nas) > 0)
order.frame[which.nas, ] <- Inf
retval <- do.call("order", order.frame)
return(retval)
}
y[mixedorderRoman(y)]
[1] "chr I""chr II" "chr III" "chr IV" "chr IX"
[6] "chr V""chr VI" "chr VII" "chr VIII" "chr X"
[11] "chr XI" "chr XII"
--
David.
>
> ## DECIMAL NUMBERS
>> x <- sprintf("chr %d", 12:1)
>> x
> [1] "chr 12" "chr 11" "chr 10" "chr 9" "chr 8"
> [6] "chr 7" "chr 6" "chr 5" "chr 4" "chr 3"
> [11] "chr 2" "chr 1"
>
>> sort(x)
> [1] "chr 1" "chr 10" "chr 11" "chr 12" "chr 2"
> [6] "chr 3" "chr 4" "chr 5" "chr 6" "chr 7"
> [11] "chr 8" "chr 9"
>
>> gtools::mixedsort(x)
> [1] "chr 1" "chr 2" "chr 3" "chr 4" "chr 5"
> [6] "chr 6" "chr 7" "chr 8" "chr 9" "chr 10"
> [11] "chr 11" "chr 12"
>
>
> ## ROMAN NUMBERS
>> y <- sprintf("chr %s", as.roman(12:1))
>> y
> [1] "chr XII" "chr XI" "chr X""chr IX"
> [5] "chr VIII" "chr VII" "chr VI" "chr V"
> [9] "chr IV" "chr III" "chr II" "chr I"
>
>> sort(y)
> [1] "chr I""chr II" "chr III" "chr IV"
> [5] "chr IX" "chr V""chr VI" "chr VII"
> [9] "chr VIII" "chr X""chr XI" "chr XII"
>
>> mixedsortroman(y)
> [1] "chr I""chr II" "chr III" "chr IV"
> [5] "chr V""chr VI" "chr VII" "chr VIII"
> [9] "chr IX" "chr X""chr XI" "chr XII"
>
> The latter is what I'm looking for.
>
> Before hacking together something myself (e.g. identify roman numerals
> substrings, translate them to decimal numbers, use gtools::mixedsort()
> to sort them and then translate them back to roman numbers), I'd like
> to hear if someone already has this implemented/know of a package that
> does this.
>
> Thanks,
>
> Henrik
>
> __
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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.
David Winsemius
Alameda, CA, USA
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Re: [R] Clip smaller domain from large domain netCDF file
Using the ncdf4 library requires some knowledge of netcdf files and how they
work. However, if you can provide the following information I may be able to
provide some pointers. I am assuming your file is named "myFile.nc". Where
you see that replace with the actual name.
library(ncdf4)
myFile<-nc_open('myFile.nc')
str(myFile)
The output of the last command will show what is basically a dump of metadata
content of the file, showing its structure.From the bounds I assume this is
a Canadian dataset?
-Roy
On Aug 26, 2014, at 4:46 PM, Aseem Sharma wrote:
> Hi,
> I have this huge ( ~30GB) .nc file (NC_FORMAT_NETCDF4_CLASSIC)) for the
> whole country 141.00 to 52.00 W, 41.00 to 84.00 N".
> I am trying to clip this big dataset for a small region specific domain
> (120.00 to 130.00 W, 50.00 to 60.00 N).
> I am trying to do using netCDF4 r package but could not figure out how to
> do so.
> Kindly please suggest me how should i proceed.
>
>
> Thank you,
>
> [[alternative HTML version deleted]]
>
> __
> R-help@r-project.org mailing list
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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.
**
"The contents of this message do not reflect any position of the U.S.
Government or NOAA."
**
Roy Mendelssohn
Supervisory Operations Research Analyst
NOAA/NMFS
Environmental Research Division
Southwest Fisheries Science Center
***Note new address and phone***
110 Shaffer Road
Santa Cruz, CA 95060
Phone: (831)-420-3666
Fax: (831) 420-3980
e-mail: roy.mendelss...@noaa.gov www: http://www.pfeg.noaa.gov/
"Old age and treachery will overcome youth and skill."
"From those who have been given much, much will be expected"
"the arc of the moral universe is long, but it bends toward justice" -MLK Jr.
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[R] Mixed sorting/ordering of strings acknowledging roman numerals?
Hi,
does anyone know of an implementation/function that sorts strings that
*contain* roman numerals (I, II, III, IV, V, ...) which are treated as
numbers. In 'gtools' there is mixedsort() which does this for strings
that contains (decimal) numbers. I'm looking for a "mixedsortroman()"
function that does the same but with roman numbers, e.g.
## DECIMAL NUMBERS
> x <- sprintf("chr %d", 12:1)
> x
[1] "chr 12" "chr 11" "chr 10" "chr 9" "chr 8"
[6] "chr 7" "chr 6" "chr 5" "chr 4" "chr 3"
[11] "chr 2" "chr 1"
> sort(x)
[1] "chr 1" "chr 10" "chr 11" "chr 12" "chr 2"
[6] "chr 3" "chr 4" "chr 5" "chr 6" "chr 7"
[11] "chr 8" "chr 9"
> gtools::mixedsort(x)
[1] "chr 1" "chr 2" "chr 3" "chr 4" "chr 5"
[6] "chr 6" "chr 7" "chr 8" "chr 9" "chr 10"
[11] "chr 11" "chr 12"
## ROMAN NUMBERS
> y <- sprintf("chr %s", as.roman(12:1))
> y
[1] "chr XII" "chr XI" "chr X""chr IX"
[5] "chr VIII" "chr VII" "chr VI" "chr V"
[9] "chr IV" "chr III" "chr II" "chr I"
> sort(y)
[1] "chr I""chr II" "chr III" "chr IV"
[5] "chr IX" "chr V""chr VI" "chr VII"
[9] "chr VIII" "chr X""chr XI" "chr XII"
> mixedsortroman(y)
[1] "chr I""chr II" "chr III" "chr IV"
[5] "chr V""chr VI" "chr VII" "chr VIII"
[9] "chr IX" "chr X""chr XI" "chr XII"
The latter is what I'm looking for.
Before hacking together something myself (e.g. identify roman numerals
substrings, translate them to decimal numbers, use gtools::mixedsort()
to sort them and then translate them back to roman numbers), I'd like
to hear if someone already has this implemented/know of a package that
does this.
Thanks,
Henrik
__
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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.
[R] Clip smaller domain from large domain netCDF file
Hi, I have this huge ( ~30GB) .nc file (NC_FORMAT_NETCDF4_CLASSIC)) for the whole country 141.00 to 52.00 W, 41.00 to 84.00 N". I am trying to clip this big dataset for a small region specific domain (120.00 to 130.00 W, 50.00 to 60.00 N). I am trying to do using netCDF4 r package but could not figure out how to do so. Kindly please suggest me how should i proceed. Thank you, [[alternative HTML version deleted]] __ R-help@r-project.org mailing list 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] lattice: packet.number() versus panel.number()
Hi, According to https://svn.r-project.org/R-packages/trunk/lattice/R/print.trellis.R, "[panel.number] is usually the same as, but can be different from packet.number" and I had been under the impression that as long as the user is not using a custom index.cond nor perm.cond, the panel.number would in fact be the same as the packet.number. However, I have recently come across a case where the two are *not* the same, even though I am not using a custom index.cond nor perm.cond. So my question is, what might be some other possible situations in which the two would be expected to differ? Regards, Ben __ R-help@r-project.org mailing list 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] plot for "survreg" and "intcox" (rewritten)
On Aug 26, 2014, at 2:33 PM, Silong Liao wrote: > Dear R users, > > I'm trying to plot survival probability against time(in years) using > "survreg" and "intcox". Please can you help me with this problem? (I have > rewritten using plain text.) I tried to use "curve" function but have no clue. > I suspect you want survfit (in the survial package which is where I suspect survreg is coming from. It returns an object that has a plot method. You could also scroll through help(pack=survival) to see other plotting functions. You could also use survest in the rms package. > For survreg, >> mod.reg1=survreg(s_new~type+sex+eye+preopiop+preopva,dist="weibull") >> summary(mod.reg1) > Call: > survreg(formula = s_new ~ type + sex + eye + preopiop + preopva, dist = > "weibull") > Value Std. Error z p > (Intercept) 40.539 20.582 1.970 4.89e-02 > typeTrab -6.606 4.279 -1.544 1.23e-01 > sexM -1.055 3.765 -0.280 7.79e-01 > eyeR -2.112 3.587 -0.589 5.56e-01 > preopiop -0.308 0.269 -1.147 2.52e-01 > preopva -0.461 1.771 -0.260 7.95e-01 > Log(scale) 2.058 0.285 7.222 5.12e-13 > > Scale= 7.83 > Weibull distribution > Loglik(model)= -78.7 Loglik(intercept only)= -81.4 > Chisq= 5.37 on 5 degrees of freedom, p= 0.37 > Number of Newton-Raphson Iterations: 10 > n= 339 > > For intcox, You are asked to provide the package name for functions that are not in the base or default packages. I have quite a few packages loaded including survival_2.37-7 , coxme_2.2-3, and rms_4.2-0 but I get: > ?intcox No documentation for ‘intcox’ in specified packages and libraries: you could try ‘??intcox’ > -- David. >> cox.fit=intcox(s_new~type+eye+sex+age+preopiop+preopva,data=glaucoma_new) >> summary(cox.fit) > Call: > intcox(formula = s_new ~ type + eye + sex + age + preopiop +preopva, data = > glaucoma_new) > > n= 339 > > coef exp(coef) se(coef) z Pr(>|z|) > typeTrab 0.59391 1.81106 NA NA NA > eyeR 0.28419 1.32868 NA NA NA > sexM -0.11597 0.89050 NA NA NA > age -0.06556 0.93655 NA NA NA > preopiop 0.03903 1.03980 NA NA NA > preopva -0.05517 0.94632 NA NA NA > > exp(coef) exp(-coef) lower .95 upper .95 > typeTrab 1.8111 0.5522 NA NA > eyeR 1.3287 0.7526 NA NA > sexM 0.8905 1.1230 NA NA > age 0.9365 1.0678 NA NA > preopiop 1.0398 0.9617 NA NA > preopva 0.9463 1.0567 NA NA > > Rsquare= NA (max possible= 0.327 ) > Likelihood ratio test= NA on 6 df, p=NA > Wald test = NA on 6 df, p=NA > Score (logrank) test = NA on 6 df, p=NA > > __ > R-help@r-project.org mailing list > 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 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] plot for "survreg" and "intcox" (rewritten)
Dear R users, I'm trying to plot survival probability against time(in years) using "survreg" and "intcox". Please can you help me with this problem? (I have rewritten using plain text.) I tried to use "curve" function but have no clue. For survreg, >mod.reg1=survreg(s_new~type+sex+eye+preopiop+preopva,dist="weibull") >summary(mod.reg1) Call: survreg(formula = s_new ~ type + sex + eye + preopiop + preopva, dist = "weibull") Value Std. Error z p (Intercept) 40.539 20.582 1.970 4.89e-02 typeTrab -6.606 4.279 -1.544 1.23e-01 sexM -1.055 3.765 -0.280 7.79e-01 eyeR -2.112 3.587 -0.589 5.56e-01 preopiop -0.308 0.269 -1.147 2.52e-01 preopva -0.461 1.771 -0.260 7.95e-01 Log(scale) 2.058 0.285 7.222 5.12e-13 Scale= 7.83 Weibull distribution Loglik(model)= -78.7 Loglik(intercept only)= -81.4 Chisq= 5.37 on 5 degrees of freedom, p= 0.37 Number of Newton-Raphson Iterations: 10 n= 339 For intcox, >cox.fit=intcox(s_new~type+eye+sex+age+preopiop+preopva,data=glaucoma_new) >summary(cox.fit) Call: intcox(formula = s_new ~ type + eye + sex + age + preopiop +preopva, data = glaucoma_new) n= 339 coef exp(coef) se(coef) z Pr(>|z|) typeTrab 0.59391 1.81106 NA NA NA eyeR 0.28419 1.32868 NA NA NA sexM -0.11597 0.89050 NA NA NA age -0.06556 0.93655 NA NA NA preopiop 0.03903 1.03980 NA NA NA preopva -0.05517 0.94632 NA NA NA exp(coef) exp(-coef) lower .95 upper .95 typeTrab 1.8111 0.5522 NA NA eyeR 1.3287 0.7526 NA NA sexM 0.8905 1.1230 NA NA age 0.9365 1.0678 NA NA preopiop 1.0398 0.9617 NA NA preopva 0.9463 1.0567 NA NA Rsquare= NA (max possible= 0.327 ) Likelihood ratio test= NA on 6 df, p=NA Wald test = NA on 6 df, p=NA Score (logrank) test = NA on 6 df, p=NA __ R-help@r-project.org mailing list 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] buit a package using Rstudio and existing R files
I would suggest building your first package with as few functions as you can
(one?). Once you get that working, try adding more functions and keep
rebuilding as you go so the error messages don't drown you. In any event, your
immediate problem is that CanceR-package.Rd has not been properly edited.
You should take the advice given in your error message and read the Writing R
Extensions manual. While there are some shortcuts such as roxygen to creating
the required files, when things go wrong the errors will generally be about
files that are documented in the WRE document.
You should also read the Posting Guide which among other things mentions that
this is a plain text mailing list. We are more likely to be able to decipher
your email if you don't let your email program format it as HTML.
---
Jeff NewmillerThe . . Go Live...
DCN:Basics: ##.#. ##.#. Live Go...
Live: OO#.. Dead: OO#.. Playing
Research Engineer (Solar/BatteriesO.O#. #.O#. with
/Software/Embedded Controllers) .OO#. .OO#. rocks...1k
---
Sent from my phone. Please excuse my brevity.
On August 26, 2014 11:45:07 AM PDT, Karim Mezhoud wrote:
>Dear All,
>I am trying to built for the first time a package. I am following this
>tutorial:
>https://support.rstudio.com/hc/en-us/articles/200486488-Developing-Packages-with-RStudio
>
>I create a new Project (package) and I added 44 .R sources files
>(functions).
>After, I got error message when I built and reload the project (please,
>see
>bellow).
>
>Any suggestion?
>
>I need to fill the doc informations (title, manual, vignette,
>description).
>Can I do these directly from Rd files located in Man folder?
>Thanks!
>
>Karim
>
>#00install.out
>* installing *source* package ‘CanceR’ ...
>** R
>** preparing package for lazy loading
>** help
>Warning:
>/home/mezhoud/CGDS-R/CanceR.Rcheck/00_pkg_src/CanceR/man/CanceR-package.Rd:33:
>All text must be in a section
>Warning:
>/home/mezhoud/CGDS-R/CanceR.Rcheck/00_pkg_src/CanceR/man/CanceR-package.Rd:34:
>All text must be in a section
>Warning:
>/home/mezhoud/CGDS-R/CanceR.Rcheck/00_pkg_src/CanceR/man/CanceR-package.Rd:35:
>All text must be in a section
>*** installing help indices
>Error in Rd_info(db[[i]]) :
> missing/empty \title field in
>'/home/mezhoud/CGDS-R/CanceR.Rcheck/00_pkg_src/CanceR/man/CanceR.Rd'
>Rd files must have a non-empty \title.
>See chapter 'Writing R documentation' in manual 'Writing R Extensions'.
>* removing ‘/home/mezhoud/CGDS-R/CanceR.Rcheck/CanceR’
>
>###CanceR-package.Rd #
>
>\name{CanceR-package}
>\alias{CanceR-package}
>\alias{CanceR}
>\docType{package}
>\title{
>What the package does (short line)
>~~ package title ~~
>}
>\description{
>More about what it does (maybe more than one line)
>~~ A concise (1-5 lines) description of the package ~~
>}
>\details{
>\tabular{ll}{
>Package: \tab CanceR\cr
>Type: \tab Package\cr
>Version: \tab 1.0\cr
>Date: \tab 2014-08-26\cr
>License: \tab What license is it under?\cr
>}
>~~ An overview of how to use the package, ~~
>~~ including the most important functions ~~
>}
>\author{
>Who wrote it
>
>Maintainer: Who to complain to
>~~ The author and/or maintainer of the package ~~
>}
>\references{
>~~ Literature or other references for background information ~~
>}
>~~ Optionally other standard keywords, ~~
>~~ one per line, from file KEYWORDS in ~~
>~~ the R documentation directory ~~
>\keyword{ package }
>\seealso{
>~~ Optional links to other man pages, e.g. ~~
>~~ \code{\link[:-package]{}} ~~
>}
>\examples{
>~~ simple examples of the most important functions ~~
>}
>
> [[alternative HTML version deleted]]
>
>__
>R-help@r-project.org mailing list
>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
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] Help with lsmeans
Dan Dillon gmail.com> writes: > > Colleagues: > [snip] > > My data are from a behavioral experiment in which two groups of subjects > complete 200+ trials of a task with two conditions. Each subject is tested > in one of four separate locations. I record accuracy (0 or 1) and response > time (RT) on each trial--these are the DVs for the two regressions. Thus, > my dataframe has columns "location", "group", "subject", "trial", > "condition", "accuracy", and "RT". > > The regression model for accuracy looks like this: > > acc.fm = glmer(accuracy ~ location + group*condition + (1|subject), > family=binomial, data=my_data) > > The results look as expected and I'm using lsmeans to do some follow-up > analyses. For example, to compare accuracy by group and condition, I'm > doing this: > > acc.lsm <- lsmeans(acc.fm, ~group|condition) > > pairs(acc.lsm) > > [snip] > Here is my model for the RT data > (RT is a continuous variable so no logistic regression here): > > rt.fm = lmer(rt ~ location + group*condition*accuracy + (1|subject), > data=my_data) > > The results from this regression look fine, but if I try this . . . > > rt.lsm <- lsmeans(rt.fm ~ group|condition) > > . . . or if I try to specify a reference grid like this . . . > > rt.rg <- ref.grid(rt.fm) > > . . . my machine hangs. > [snip] It's a little hard to say without a reproducible example, and this question would probably be slightly more appropriate for r-sig-mixed-mod...@r-project.org (although I can't actually tell for sure whether it is an lme4-specific problem or a more general ls.means::ref.grid question), but: how big a reference is ref.grid() trying to construct? Is it fairly high-resolution/high-dimensional? I would probably try some experiments with small subsets of your data to see how the results scale. __ R-help@r-project.org mailing list 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] buit a package using Rstudio and existing R files
Dear All,
I am trying to built for the first time a package. I am following this
tutorial:
https://support.rstudio.com/hc/en-us/articles/200486488-Developing-Packages-with-RStudio
I create a new Project (package) and I added 44 .R sources files
(functions).
After, I got error message when I built and reload the project (please, see
bellow).
Any suggestion?
I need to fill the doc informations (title, manual, vignette, description).
Can I do these directly from Rd files located in Man folder?
Thanks!
Karim
#00install.out
* installing *source* package ‘CanceR’ ...
** R
** preparing package for lazy loading
** help
Warning:
/home/mezhoud/CGDS-R/CanceR.Rcheck/00_pkg_src/CanceR/man/CanceR-package.Rd:33:
All text must be in a section
Warning:
/home/mezhoud/CGDS-R/CanceR.Rcheck/00_pkg_src/CanceR/man/CanceR-package.Rd:34:
All text must be in a section
Warning:
/home/mezhoud/CGDS-R/CanceR.Rcheck/00_pkg_src/CanceR/man/CanceR-package.Rd:35:
All text must be in a section
*** installing help indices
Error in Rd_info(db[[i]]) :
missing/empty \title field in
'/home/mezhoud/CGDS-R/CanceR.Rcheck/00_pkg_src/CanceR/man/CanceR.Rd'
Rd files must have a non-empty \title.
See chapter 'Writing R documentation' in manual 'Writing R Extensions'.
* removing ‘/home/mezhoud/CGDS-R/CanceR.Rcheck/CanceR’
###CanceR-package.Rd #
\name{CanceR-package}
\alias{CanceR-package}
\alias{CanceR}
\docType{package}
\title{
What the package does (short line)
~~ package title ~~
}
\description{
More about what it does (maybe more than one line)
~~ A concise (1-5 lines) description of the package ~~
}
\details{
\tabular{ll}{
Package: \tab CanceR\cr
Type: \tab Package\cr
Version: \tab 1.0\cr
Date: \tab 2014-08-26\cr
License: \tab What license is it under?\cr
}
~~ An overview of how to use the package, ~~
~~ including the most important functions ~~
}
\author{
Who wrote it
Maintainer: Who to complain to
~~ The author and/or maintainer of the package ~~
}
\references{
~~ Literature or other references for background information ~~
}
~~ Optionally other standard keywords, ~~
~~ one per line, from file KEYWORDS in ~~
~~ the R documentation directory ~~
\keyword{ package }
\seealso{
~~ Optional links to other man pages, e.g. ~~
~~ \code{\link[:-package]{}} ~~
}
\examples{
~~ simple examples of the most important functions ~~
}
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
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] Plot survreg and intcox
On Aug 26, 2014, at 4:17 AM, Silong Liao wrote: > Dear R users, > I'm trying to plot survival probability against time(in years) using > "survreg" and "intcox". Please can you help me with this problem? That line at the bottom of the message delivered to the list subscribers saying that the HTL version was deleted explains why this is such a mess. Please read the Posting Guide and post in plain text: -- David. > My code shows > below:>mod.reg1=survreg(s_new~type+sex+eye+preopiop+preopva,dist="weibull")>summary(mod.reg1) > Call:survreg(formula = s_new ~ type + sex + eye + preopiop + preopva, > dist = "weibull") Value Std. Error zp(Intercept) > 40.539 20.582 1.970 4.89e-02typeTrab-6.606 4.279 -1.544 > 1.23e-01sexM-1.055 3.765 -0.280 7.79e-01eyeR-2.112 > 3.587 -0.589 5.56e-01preopiop-0.308 0.269 -1.147 2.52e-01preopva > -0.461 1.771 -0.260 7.95e-01Log(scale) 2.058 0.285 7.222 5.12e-13 > Scale= 7.83 > Weibull distributionLoglik(model)= -78.7 Loglik(intercept only)= -81.4 > Chisq= 5.37 on 5 degrees of freedom, p= 0.37 Number of Newton-Raphson > Iterations: 10 n= 339 > ->cox.fit=intcox(s_new~type+eye+sex+age+preopiop+preopva,data=glaucoma_new)>summary(cox.fit)Call:intcox(formula > = s_new ~ type + eye + sex + age + preopiop + preopva, data = > glaucoma_new) > n= 339 > coef exp(coef) se(coef) z Pr(>|z|)typeTrab 0.59391 1.81106 >NA NA NAeyeR 0.28419 1.32868 NA NA NAsexM > -0.11597 0.89050 NA NA NAage -0.06556 0.93655 NA > NA NApreopiop 0.03903 1.03980 NA NA NApreopva -0.05517 > 0.94632 NA NA NA > exp(coef) exp(-coef) lower .95 upper .95typeTrab1.8111 0.5522 >NANAeyeR1.3287 0.7526NANAsexM > 0.8905 1.1230NANAage 0.9365 1.0678 > NANApreopiop1.0398 0.9617NANApreopva > 0.9463 1.0567NANA > Rsquare= NA (max possible= 0.327 )Likelihood ratio test= NA on 6 df, > p=NAWald test= NA on 6 df, p=NAScore (logrank) test = NA on 6 > df, p=NA > > > Kind regards,Sid > > > [[alternative HTML version deleted]] > > __ > R-help@r-project.org mailing list > 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 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] What the difference between .Golbalenv and package:base?
as.environment(characterString) maps an entry from the output of
search() to the environment at the named position in the search list.
as.environment(number) maps an index into the output of search() to
the the environment at that position in the search list. If
'characterString' is not in the output of search() or 'number' is not
in seq_along(search()) then as.environment throws an error. As far as
I can tell, as.environment does not deal with the name of the
environment at all. (When you attach an environment, attach will add
a name attribute to the copied environment so the attached
environment's name matches the name on the output of search(), but I
don't think as.environment ever looks at that attribute.)
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Mon, Aug 25, 2014 at 9:07 PM, PO SU wrote:
> First, sorry for my pool english expression which make you misunderstanding
> of my original purpose.
>
> Sometimes, suppose a object in both stats and base, then i type the object
> name, then after R search the search() list, R will use the object in stats,
> is it right?( I just suppose, stats can be any package which libraried into
> R.)
> Then i know that, .GlobalEnv or globalenv() is the global environment object,
> baseenv() returns the base environment object.
> I also know that, i can convert the environment name into the real
> environment object by using stats<-as.environment("package:stats"), And the
> stats environment's name can be obtained using environmentName(stats), but it
> returns "". (why?)
> Then i use environmentName(.GlobalEnv) then i get "R_GlobalEnv", then i use
> as.environment("R_GlobalEnv"), it does't work.(why?)
>
>
> In one word, as.environment(x), x is somthing not the environment's name.
>
>
> But, when i add a environment into the search() list, after i
> attr(newenvir,"name")<-"new_name"
> I can use the as.environment("new_name") to obtain the added environment.
> (why?)
>
>
> Hope you understand my meaning :)
>
>
>
>
>
>
>
>
>
>
>
>
> --
>
> PO SU
> mail: desolato...@163.com
> Majored in Statistics from SJTU
>
>
>
> At 2014-08-26 02:51:54, "MacQueen, Don" wrote:
>>Put simply,
>> .GlobalEnvstores objects you create
>> package:base contains functions and objects provided by R itself
>>
>>You don¹t need to use .GlobalEnv$a to use the variable named a. Just
>>is ³a² by itself.
>>
>> a <- 4
>> b <- 2*a
>>print(a)
>>print(b)
>>
>>Not necessary to use
>> print(.GlobalEnv$a)
>>
>>Similarly, to find an object in the base package, just type its name.
>>
>>I don¹t know what you are trying to do, or why you think you have to use
>>.GlobalEnv$a
>>But in more than 20 years of using R for many different tasks, I have
>>never had to do that.
>>
>>Furthermore, if you are new to R (which I would guess is the case), it
>>seems unlikely to me that you need to work with environments or use
>>attach() or assign(). In the vast majority of cases there are simpler ways
>>that are easier to understand.
>>
>>You are aware, I hope, that
>> > ls('.GlobalEnv')
>> > ls(1)
>> > ls()
>>all return the same result?
>>
>>
>>--
>>Don MacQueen
>>
>>Lawrence Livermore National Laboratory
>>7000 East Ave., L-627
>>Livermore, CA 94550
>>925-423-1062
>>
>>
>>
>>
>>
>>On 8/24/14, 11:07 PM, "PO SU" wrote:
>>
>>>
>>>
>>>Dear rusers,
>>>
>>>As we know, there are a lot of environments in the search() path,
>>>such as .Golbalenv and package:base .
>>>And i can just use .Golbalenv$a ,.Golbalenv$b to use the virable, but
>>>i must use as.envrionment("package:base") to find virable, i feel it not
>>>very convenient.
>>>
>>>
>>>For example, when i use the following codes to add a new env into the
>>>search() path.
>>>
>>>
>>>
tmp<-attach(NULL,name="new_name")
assign("a",2,envir=as.environment("new_name"))
a
>>>[1] 2
as.environment("new_name")$a
>>>[1] 2
>>> I must always convert the name to the environment, How can i just use
>>>the following form:
>>>
>>>
>>>
tmp<-attach(NULL,name="new_name")
assign("a",2,envir=new_name) #like using .GlobalEnv
a
>>>[1] 2
new_name$a
>>>
>>>[1] 2
>>>
>>>
>>>
>>>
>>>
>>>
>>>
>>>--
>>>
>>>PO SU
>>>mail: desolato...@163.com
>>>Majored in Statistics from SJTU
>>>__
>>>R-help@r-project.org mailing list
>>>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
> 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
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/po
Re: [R] StackRaster problem
Did you attach the raster package with library(raster)? It seems the
newest version of dismo does not depend on raster, so you will not be
able to use raster-functions if you only attach dismo.
This error message typically comes when R tries to use
utils:::stack.default instead of the stack-function defined in the
raster-package.
If this is not the case, please give the output from sessionInfo().
Cheers,
Jon
On 8/26/2014 3:21 PM, Guilherme Leite wrote:
Hi,
This is the process I want to do:
files <- list.files(path=paste(system.file(package="dismo"), '/ex',
sep=''), pattern='grd', full.names=TRUE )
# The above finds all the files with extension "grd" in the
# examples ("ex") directory of the dismo package. You do not
# need such a complex statement to get your own files.
files
[1] "d:/temp/Rtmp8Mttik/Rinst1fe4202d2ea5/dismo/ex/bio1.grd"
[2] "d:/temp/Rtmp8Mttik/Rinst1fe4202d2ea5/dismo/ex/bio12.grd"
[3] "d:/temp/Rtmp8Mttik/Rinst1fe4202d2ea5/dismo/ex/bio16.grd"
[4] "d:/temp/Rtmp8Mttik/Rinst1fe4202d2ea5/dismo/ex/bio17.grd"
[5] "d:/temp/Rtmp8Mttik/Rinst1fe4202d2ea5/dismo/ex/bio5.grd"
[6] "d:/temp/Rtmp8Mttik/Rinst1fe4202d2ea5/dismo/ex/bio6.grd"
[7] "d:/temp/Rtmp8Mttik/Rinst1fe4202d2ea5/dismo/ex/bio7.grd"
[8] "d:/temp/Rtmp8Mttik/Rinst1fe4202d2ea5/dismo/ex/bio8.grd"
[9] "d:/temp/Rtmp8Mttik/Rinst1fe4202d2ea5/dismo/ex/biome.grd"
predictors <- stack(files)
predictors
class : RasterStack
dimensions : 192, 186, 35712, 9 (nrow, ncol, ncell, nlayers)
resolution : 0.5, 0.5 (x, y)
extent : -125, -32, -56, 40 (xmin, xmax, ymin, ymax)
coord. ref. : +proj=longlat +datum=WGS84 +ellps=WGS84 +towgs84=0,0,0
names : bio1, bio12, bio16, bio17, bio5, bio6, bio7, bio8, biome
min values : -23, 0, 0, 0, 61, -212, 60, -66, 1
max values : 289, 7682, 2458, 1496, 422, 242, 461, 323, 14
But what happen is:
files <- list.files(path=paste(system.file(package="dismo"), '/ex',
sep=''), pattern='grd', full.names=TRUE )
files
[1] "d:/temp/Rtmp8Mttik/Rinst1fe4202d2ea5/dismo/ex/bio1.grd"
[2] "d:/temp/Rtmp8Mttik/Rinst1fe4202d2ea5/dismo/ex/bio12.grd"
[3] "d:/temp/Rtmp8Mttik/Rinst1fe4202d2ea5/dismo/ex/bio16.grd"
[4] "d:/temp/Rtmp8Mttik/Rinst1fe4202d2ea5/dismo/ex/bio17.grd"
[5] "d:/temp/Rtmp8Mttik/Rinst1fe4202d2ea5/dismo/ex/bio5.grd"
[6] "d:/temp/Rtmp8Mttik/Rinst1fe4202d2ea5/dismo/ex/bio6.grd"
[7] "d:/temp/Rtmp8Mttik/Rinst1fe4202d2ea5/dismo/ex/bio7.grd"
[8] "d:/temp/Rtmp8Mttik/Rinst1fe4202d2ea5/dismo/ex/bio8.grd"
[9] "d:/temp/Rtmp8Mttik/Rinst1fe4202d2ea5/dismo/ex/biome.grd"
predictors <- stack(files)
Error in rep.int(names(x), lapply(x, length)) : invalid 'times' value
Do you know how to fix it?
Thank you,
Guilherme
[[alternative HTML version deleted]]
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--
Jon Olav Skøien
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Institute for Environment and Sustainability (IES)
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Via Fermi 2749, TP 100-01, I-21027 Ispra (VA), ITALY
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[R] Plot survreg and intcox
Dear R users, I'm trying to plot survival probability against time(in years) using "survreg" and "intcox". Please can you help me with this problem? My code shows below:>mod.reg1=survreg(s_new~type+sex+eye+preopiop+preopva,dist="weibull")>summary(mod.reg1) Call:survreg(formula = s_new ~ type + sex + eye + preopiop + preopva, dist = "weibull") Value Std. Error zp(Intercept) 40.539 20.582 1.970 4.89e-02typeTrab-6.606 4.279 -1.544 1.23e-01sexM -1.055 3.765 -0.280 7.79e-01eyeR-2.112 3.587 -0.589 5.56e-01preopiop-0.308 0.269 -1.147 2.52e-01preopva -0.461 1.771 -0.260 7.95e-01Log(scale) 2.058 0.285 7.222 5.12e-13 Scale= 7.83 Weibull distributionLoglik(model)= -78.7 Loglik(intercept only)= -81.4 Chisq= 5.37 on 5 degrees of freedom, p= 0.37 Number of Newton-Raphson Iterations: 10 n= 339 ->cox.fit=intcox(s_new~type+eye+sex+age+preopiop+preopva,data=glaucoma_new)>summary(cox.fit)Call:intcox(formula = s_new ~ type + eye + sex + age + preopiop + preopva, data = glaucoma_new) n= 339 coef exp(coef) se(coef) z Pr(>|z|)typeTrab 0.59391 1.81106 NA NA NAeyeR 0.28419 1.32868 NA NA NAsexM -0.11597 0.89050 NA NA NAage -0.06556 0.93655 NA NA NApreopiop 0.03903 1.03980 NA NA NApreopva -0.05517 0.94632 NA NA NA exp(coef) exp(-coef) lower .95 upper .95typeTrab1.8111 0.5522 NANAeyeR1.3287 0.7526NANAsexM 0.8905 1.1230NANAage 0.9365 1.0678NA NApreopiop1.0398 0.9617NANApreopva 0.9463 1.0567NANA Rsquare= NA (max possible= 0.327 )Likelihood ratio test= NA on 6 df, p=NAWald test= NA on 6 df, p=NAScore (logrank) test = NA on 6 df, p=NA Kind regards,Sid [[alternative HTML version deleted]] __ R-help@r-project.org mailing list 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] Weighted Mann-Whitney-Wilcoxon-Test
On Tuesday, August 19, 2014 9:46 PM Thomas Lumley wrote:
>> Is anyone aware of an(other) implementation in R?
> survey::svyranktest
Oh, that easy. Thanks a lot.
Actually, in my case the weights do not derive from some selection
probabilities, but your test works anyway.
For completeness (and this time with a working example):
set.seed(seed = 123)
count.x <- NULL
count.y <- NULL
j <- 1
for (i in sample(x = 10:20, size = 20, replace = TRUE)){
count.x[j] <- sample(x = 0:i, size = 1)
count.y[j] <- i - count.x[j]
j <- j + 1
}
data <- data.frame(x.portion = (count.x/(count.x + count.y)),
y.portion = (count.y/(count.x + count.y)),
group = c(rep("A", 12), rep("B", 8)),
weight= (count.x + count.y)
)
I first considered the unweighted case with
library(package = survey)
design <- svydesign(ids = ~0, data = data)
svyranktest(formula = x.portion ~ group, design = design)
and compared it to the default Wilcoxon test by
wilcox.test(formula = x.portion ~ group, data = data, exact = FALSE, correct =
FALSE)
Or, if you prefer
library(package = coin)
wilcox_test(formula = x.portion ~ group, data = data)
The resulting p-values differ, as I understood due to an approximation in
package /survey/.
Now, finally, the weighted case:
design <- svydesign(ids = ~0, data = data, weights = ~weight)
svyranktest(formula = x.portion ~ group, design)
And, by the way, package /survey/ seems also to be the preferable way, if you
want to go for a parametric test. Once again, the unweighted case first:
design <- svydesign(ids = ~0, data = data)
svyttest(formula = x.portion ~ group, design)
And, yet again, the results differ from the default t test
t.test(formula = x.portion ~ group, data = data)
This time, I guess, it is due to the way standard errors are computed.
Finally (this time for real), the weighted case:
design <- svydesign(ids = ~0, data = data, weights = ~weight)
svyttest(x.portion ~ group, design)
Note that function /wtd.t.test/ from package /weights/ depends on the scale of
the weights, /svyttest/ not.
Thomas, one more time: thank you for your help.
Cheers,
Alex
--
Alexander Sommer
wissenschaftlicher Mitarbeiter
Technische Universität Dortmund
Fakultät Erziehungswissenschaft, Psychologie und Soziologie
Forschungsverbund Deutsches Jugendinstitut/Technische Universität Dortmund
Vogelpothsweg 78
44227 Dortmund
Telefon: +49 231 755-8189
Fax: +49 231 755-6553
E-Mail: alexander.som...@tu-dortmund.de
WWW: http://www.forschungsverbund.tu-dortmund.de/
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[R] StackRaster problem
Hi,
This is the process I want to do:
> files <- list.files(path=paste(system.file(package="dismo"), '/ex',
sep=''), pattern='grd', full.names=TRUE )
> # The above finds all the files with extension "grd" in the
> # examples ("ex") directory of the dismo package. You do not
> # need such a complex statement to get your own files.
> files
[1] "d:/temp/Rtmp8Mttik/Rinst1fe4202d2ea5/dismo/ex/bio1.grd"
[2] "d:/temp/Rtmp8Mttik/Rinst1fe4202d2ea5/dismo/ex/bio12.grd"
[3] "d:/temp/Rtmp8Mttik/Rinst1fe4202d2ea5/dismo/ex/bio16.grd"
[4] "d:/temp/Rtmp8Mttik/Rinst1fe4202d2ea5/dismo/ex/bio17.grd"
[5] "d:/temp/Rtmp8Mttik/Rinst1fe4202d2ea5/dismo/ex/bio5.grd"
[6] "d:/temp/Rtmp8Mttik/Rinst1fe4202d2ea5/dismo/ex/bio6.grd"
[7] "d:/temp/Rtmp8Mttik/Rinst1fe4202d2ea5/dismo/ex/bio7.grd"
[8] "d:/temp/Rtmp8Mttik/Rinst1fe4202d2ea5/dismo/ex/bio8.grd"
[9] "d:/temp/Rtmp8Mttik/Rinst1fe4202d2ea5/dismo/ex/biome.grd"
> predictors <- stack(files)
> predictors
class : RasterStack
dimensions : 192, 186, 35712, 9 (nrow, ncol, ncell, nlayers)
resolution : 0.5, 0.5 (x, y)
extent : -125, -32, -56, 40 (xmin, xmax, ymin, ymax)
coord. ref. : +proj=longlat +datum=WGS84 +ellps=WGS84 +towgs84=0,0,0
names : bio1, bio12, bio16, bio17, bio5, bio6, bio7, bio8, biome
min values : -23, 0, 0, 0, 61, -212, 60, -66, 1
max values : 289, 7682, 2458, 1496, 422, 242, 461, 323, 14
But what happen is:
> files <- list.files(path=paste(system.file(package="dismo"), '/ex',
sep=''), pattern='grd', full.names=TRUE )
> files
[1] "d:/temp/Rtmp8Mttik/Rinst1fe4202d2ea5/dismo/ex/bio1.grd"
[2] "d:/temp/Rtmp8Mttik/Rinst1fe4202d2ea5/dismo/ex/bio12.grd"
[3] "d:/temp/Rtmp8Mttik/Rinst1fe4202d2ea5/dismo/ex/bio16.grd"
[4] "d:/temp/Rtmp8Mttik/Rinst1fe4202d2ea5/dismo/ex/bio17.grd"
[5] "d:/temp/Rtmp8Mttik/Rinst1fe4202d2ea5/dismo/ex/bio5.grd"
[6] "d:/temp/Rtmp8Mttik/Rinst1fe4202d2ea5/dismo/ex/bio6.grd"
[7] "d:/temp/Rtmp8Mttik/Rinst1fe4202d2ea5/dismo/ex/bio7.grd"
[8] "d:/temp/Rtmp8Mttik/Rinst1fe4202d2ea5/dismo/ex/bio8.grd"
[9] "d:/temp/Rtmp8Mttik/Rinst1fe4202d2ea5/dismo/ex/biome.grd"
> predictors <- stack(files)
Error in rep.int(names(x), lapply(x, length)) : invalid 'times' value
Do you know how to fix it?
Thank you,
Guilherme
[[alternative HTML version deleted]]
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[R] package 'gradientForest' and 'extendedForest'
Dear experts, I have 5 environmental predictors and abundance data (300 samples, 60 species, transformation: log(x + min(x,x > 0) and use the function 'gradientForest' to estimate (R�-weighted) predictor importance (regression trees). The resulting predictor importance in decreasing order is as follows: pred1, pred2, pred3, pred4, pred5. The two species with the highest R� (goodness-of-fit; output value 'result' of function 'gradientForest') are species 1 (R�=0.76), species 2 (R�=0.74), and species 3 (R�=0.72). To my understanding this means that the model (i.e. the predictor importance ranking) fits best to species 1, 2, and 3 in decreasing order. In a further step I want to know which predictors are the most important for selected species. Thus, I ran separate forests using the 'extendedForest' function with the same parameter settings (and the same set.seed()) as in the function call of 'gradientForest' for species 1, 2, and 3 (and others). Now the resulting predictor importance is (in decreasing order): species1: pred1, pred2, pred4, pred3, pred5; species2: pred1, pred4, pred2, pred5, pred3; species3: pred2, pred4, pred5, pred1, pred3. This seems strange to me, because I believed that the 'extendedForest' function should give similar predictor importance rankings as the 'gradientForest' predictor importance ranking for the species with the highest R� values obtained by 'gradientForest' . I'd be grateful for any help. Thanks a lot in anticipation. Best regards Thomas [[alternative HTML version deleted]] __ R-help@r-project.org mailing list 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] VennDiagram
Assuming your sample data is called dta: > table(dta$Results, dta$Analysis) A B C 1-5 1 1 0 20-50 1 0 0 4-7 0 0 1 8-9 0 1 1 David L. Carlson Department of Anthropology Texas A&M University -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Jurgens de Bruin Sent: Tuesday, August 26, 2014 4:08 AM To: João Azevedo Patrício Cc: r-help@r-project.org Subject: Re: [R] VennDiagram Hi,, Thanks for the link, I have tried that but it seems my data is in the wrong format for that to work. On 26 August 2014 10:35, João Azevedo Patrício wrote: > Em 26-08-2014 09:30, Jurgens de Bruin escreveu: > > Hi, >> >> I am new to R and dont use it very often so I would appreciate some help. >> >> I would like to create a VennDiagram based on the following, I have >> analyzed experimental results using different methods and captured >> the results of each analysis. >> I would like the VennDiagram to show the overlap in results for the >> different analysis. >> >> Sample data: >> Analysis Results >> A 1-5 >> B 8-9 >> C 4-7 >> B 1-5 >> A 20-50 >> C 8-9 >> >> So in this simple example analysis A and B have a single sample >> overlap and Analysis B and C also have a single overlap but C and A >> have no overlap. >> >> >> Hi, > > I don't know how to do it, but made a search and found this, see if it > helps you > > http://www.ats.ucla.edu/stat/r/faq/venn.htm > > -- > João Azevedo Patrício > Tel.: +31 91 400 53 63 > Portugal > @ http://tripaforra.bl.ee > > "Take 2 seconds to think before you act" > > __ > R-help@r-project.org mailing list > 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. > -- Regards/Groete/Mit freundlichen Grüßen/recuerdos/meilleures salutations/ distinti saluti/siong/duì yú/привет Jurgens de Bruin [[alternative HTML version deleted]] __ R-help@r-project.org mailing list 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 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] How to do t.test to rows of a dataframe using apply family function?
Hi
That is because do.call is one function and rbind another. For both functions
you can find its help page and read:
rbind
vectors or matrices. These can be given as named arguments. Other R objects
will be coerced as appropriate: see sections ‘Details’ and ‘Value’.
The type of a matrix result determined from the highest type of any of the
inputs in the hierarchy raw < logical < integer < double < complex < character
< list .
do.call
what
either a function or a non-empty character string naming the function to be
called.
args
a list of arguments to the function call. The names attribute of args gives the
argument names.
Actually inner lapply was not necessary and simple
do.call("rbind", lll)
gives the same result.
From what I understand from help page, when argument in rbind is list of lists
it returns matrix of lists which are printed as
[,1] [,2]
lll List,9 List,9
do.call takes rbind and uses it on each node of list and this node is coerced
to matrix, which is suitable for changing to data frame.
as.data.frame(do.call("rbind", lll))
It is some magic of R which I use although I do not know much about its
internals. If you are interested, the source code is available.
Regards
Petr
From: my1stbox [mailto:my1st...@163.com]
Sent: Tuesday, August 26, 2014 12:24 PM
To: PIKAL Petr
Cc: R Help
Subject: Re: RE: RE: [R] How to do t.test to rows of a dataframe using apply
family function?
Thank you so much!
Why do do.call(rbind,lapply(lll,rbind)) and rbind(lapply(lll,rbind)) act so
differently? What is the tricky part of that?
> do.call(rbind,lapply(lll,rbind))
statistic parameter p.value conf.int estimate null.value alternative
[1,] 2.775282 37.99977 0.008509272 Numeric,2 Numeric,2 0 "two.sided"
[2,] -1.498133 37.31294 0.1425116 Numeric,2 Numeric,2 0 "two.sided"
method data.name
[1,] "Welch Two Sample t-test" "rnorm(20) and rnorm(20)"
[2,] "Welch Two Sample t-test" "rnorm(20) and rnorm(20)"
> rbind(lapply(lll,rbind))
[,1] [,2]
[1,] List,9 List,9
2014-08-26
发件人：PIKAL Petr mailto:petr.pi...@precheza.cz>>
发送时间：2014-08-26 17:58
主题：RE: RE: [R] How to do t.test to rows of a dataframe using apply family
function?
收件人："my1stbox"mailto:my1st...@163.com>>
抄送："R Help"mailto:r-help@r-project.org>>
Hi
Please reply also to rhelp, maybe others can offer better answer.
Well, are you aware that by unlist you get all results as character values and
not numbers?
Personally I would use lapply and do.call on final result to get it as data
frame.
Here is example
Some model list data
lll<-vector("list",length=2)
lll
[[1]]
NULL
[[2]]
NULL
lll[[1]]<-t.test(rnorm(20), rnorm(20))
lll[[2]]<-t.test(rnorm(20), rnorm(20))
str(unlist(lll))
Named chr [1:22] "0.0091599930484716" "37.9972278626102" ...
- attr(*, "names")= chr [1:22] "statistic.t" "parameter.df" "p.value"
"conf.int1" ...
do.call(rbind, lapply(lll, rbind))
statistic parameter p.value conf.int estimate null.value alternative
[1,] 0.009159993 37.99723 0.9927394 Numeric,2 Numeric,2 0 "two.sided"
[2,] 0.1256267 32.93057 0.9007913 Numeric,2 Numeric,2 0 "two.sided"
methoddata.name
[1,] "Welch Two Sample t-test" "rnorm(20) and rnorm(20)"
[2,] "Welch Two Sample t-test" "rnorm(20) and rnorm(20)"
Regards
Petr
From: my1stbox [mailto:my1st...@163.com]
Sent: Tuesday, August 26, 2014 10:44 AM
To: PIKAL Petr
Subject: Re: RE: [R] How to do t.test to rows of a dataframe using apply family
function?
Hi Petr，
Thank you! I use unlist() and rbind() because I want the result to be in the
form of a matrix.
Regards
Allen
2014-08-26
发件人：PIKAL Petr mailto:petr.pi...@precheza.cz>>
发送时间：2014-08-26 14:33
主题：RE: [R] How to do t.test to rows of a dataframe using apply family function?
收件人："my1stbox"mailto:my1st...@163.com>>,"R
Help"mailto:r-help@r-project.org>>
抄送：
Hi
If your function works for you why to bother with apply? If it does not give
you required results, please post some data and show us what is expected.
I would remove unlist from your function and declare list for storing data,
which seems to me more natural for storing results.
t.test.list=function(df,paired){
lll<-vector("list", length=nrow(df))
for(i in 1:nrow(df)){
lll[[i]]=t.test(df[i,grp1],df[i,grp2],paired=paired)
}
names(lll)=rownames(df)
lll
> }
Regards
Petr
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-bounces@r-
> project.org] On Behalf Of my1stbox
> Sent: Tuesday, August 26, 2014 3:26 AM
> To: R Help
> Subject: [R] How to do t.test to rows of a dataframe using apply family
> function?
>
> Hi All，
> How to do t.test to rows (with two levels, or in other words, groups
> of samples) of a dataframe using apply family function? I have done
> that using function and loops. And my overall purpose is to calculate
> DE genes fr
Re: [R] How to do t.test to rows of a dataframe using apply family function?
Thank you so much!
Why do do.call(rbind,lapply(lll,rbind)) and rbind(lapply(lll,rbind)) act so
differently? What is the tricky part of that?
> do.call(rbind,lapply(lll,rbind))
statistic parameter p.value conf.int estimate null.value alternative
[1,] 2.775282 37.99977 0.008509272 Numeric,2 Numeric,2 0 "two.sided"
[2,] -1.498133 37.31294 0.1425116 Numeric,2 Numeric,2 0 "two.sided"
method data.name
[1,] "Welch Two Sample t-test" "rnorm(20) and rnorm(20)"
[2,] "Welch Two Sample t-test" "rnorm(20) and rnorm(20)"
> rbind(lapply(lll,rbind))
[,1] [,2]
[1,] List,9 List,9
2014-08-26
发件人：PIKAL Petr
发送时间：2014-08-26 17:58
主题：RE: RE: [R] How to do t.test to rows of a dataframe using apply family
function?
收件人："my1stbox"
抄送："R Help"
Hi
Please reply also to rhelp, maybe others can offer better answer.
Well, are you aware that by unlist you get all results as character values and
not numbers?
Personally I would use lapply and do.call on final result to get it as data
frame.
Here is example
Some model list data
lll<-vector("list",length=2)
lll
[[1]]
NULL
[[2]]
NULL
lll[[1]]<-t.test(rnorm(20), rnorm(20))
lll[[2]]<-t.test(rnorm(20), rnorm(20))
str(unlist(lll))
Named chr [1:22] "0.0091599930484716" "37.9972278626102" ...
- attr(*, "names")= chr [1:22] "statistic.t" "parameter.df" "p.value"
"conf.int1" ...
do.call(rbind, lapply(lll, rbind))
statistic parameter p.value conf.int estimate null.value alternative
[1,] 0.009159993 37.99723 0.9927394 Numeric,2 Numeric,2 0 "two.sided"
[2,] 0.1256267 32.93057 0.9007913 Numeric,2 Numeric,2 0 "two.sided"
methoddata.name
[1,] "Welch Two Sample t-test" "rnorm(20) and rnorm(20)"
[2,] "Welch Two Sample t-test" "rnorm(20) and rnorm(20)"
Regards
Petr
From: my1stbox [mailto:my1st...@163.com]
Sent: Tuesday, August 26, 2014 10:44 AM
To: PIKAL Petr
Subject: Re: RE: [R] How to do t.test to rows of a dataframe using apply family
function?
Hi Petr，
Thank you! I use unlist() and rbind() because I want the result to be in the
form of a matrix.
Regards
Allen
2014-08-26
发件人：PIKAL Petr
发送时间：2014-08-26 14:33
主题：RE: [R] How to do t.test to rows of a dataframe using apply family function?
收件人："my1stbox","R Help"
抄送：
Hi
If your function works for you why to bother with apply? If it does not give
you required results, please post some data and show us what is expected.
I would remove unlist from your function and declare list for storing data,
which seems to me more natural for storing results.
t.test.list=function(df,paired){
lll<-vector("list", length=nrow(df))
for(i in 1:nrow(df)){
lll[[i]]=t.test(df[i,grp1],df[i,grp2],paired=paired)
}
names(lll)=rownames(df)
lll
> }
Regards
Petr
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of my1stbox
> Sent: Tuesday, August 26, 2014 3:26 AM
> To: R Help
> Subject: [R] How to do t.test to rows of a dataframe using apply family
> function?
>
> Hi All，
> How to do t.test to rows (with two levels, or in other words, groups
> of samples) of a dataframe using apply family function? I have done
> that using function and loops. And my overall purpose is to calculate
> DE genes from two groups of miRNA microarray samples. And I know limma
> can do that, but it seems limma doesn't support paired t-test.
>
> t.test.df=function(df,paired){
> df.t=c()
> for(i in 1:nrow(df)){
>
> df.t=rbind(df.t,unlist(t.test(df[i,grp1],df[i,grp2],paired=paired)))
> }
> rownames(df.t)=rownames(df)
> df.t
> }
>
> Bests!
> Allen Chiu
>
> 2014-08-26
> [[alternative HTML version deleted]]
>
> __
> R-help@r-project.org mailing list
> 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.
Tento e-mail a jakékoliv k němu připojené dokumenty jsou důvěrné a jsou určeny
pouze jeho adresátům.
Jestliže jste obdržel(a) tento e-mail omylem, informujte laskavě neprodleně
jeho odesílatele. Obsah tohoto emailu i s přílohami a jeho kopie vymažte ze
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- vyhrazuje si odesílatel právo ukončit kdykoliv jednání o uzavření smlouvy, a
to z jakéhokoliv důvodu i bez uvedení důvodu.
- a obsahuje-li nabídku, je adresát oprávněn nabídku bezodkladně přijmout;
Odesílatel tohoto e-mailu (nabídky) vylučuje přijetí nabídky ze strany příjemce
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Re: [R] How to do t.test to rows of a dataframe using apply family function?
Hi
Please reply also to rhelp, maybe others can offer better answer.
Well, are you aware that by unlist you get all results as character values and
not numbers?
Personally I would use lapply and do.call on final result to get it as data
frame.
Here is example
Some model list data
lll<-vector("list",length=2)
lll
[[1]]
NULL
[[2]]
NULL
lll[[1]]<-t.test(rnorm(20), rnorm(20))
lll[[2]]<-t.test(rnorm(20), rnorm(20))
str(unlist(lll))
Named chr [1:22] "0.0091599930484716" "37.9972278626102" ...
- attr(*, "names")= chr [1:22] "statistic.t" "parameter.df" "p.value"
"conf.int1" ...
do.call(rbind, lapply(lll, rbind))
statistic parameter p.value conf.int estimate null.value alternative
[1,] 0.009159993 37.99723 0.9927394 Numeric,2 Numeric,2 0 "two.sided"
[2,] 0.1256267 32.93057 0.9007913 Numeric,2 Numeric,2 0 "two.sided"
methoddata.name
[1,] "Welch Two Sample t-test" "rnorm(20) and rnorm(20)"
[2,] "Welch Two Sample t-test" "rnorm(20) and rnorm(20)"
Regards
Petr
From: my1stbox [mailto:my1st...@163.com]
Sent: Tuesday, August 26, 2014 10:44 AM
To: PIKAL Petr
Subject: Re: RE: [R] How to do t.test to rows of a dataframe using apply family
function?
Hi Petr，
Thank you! I use unlist() and rbind() because I want the result to be in the
form of a matrix.
Regards
Allen
2014-08-26
发件人：PIKAL Petr mailto:petr.pi...@precheza.cz>>
发送时间：2014-08-26 14:33
主题：RE: [R] How to do t.test to rows of a dataframe using apply family function?
收件人："my1stbox"mailto:my1st...@163.com>>,"R
Help"mailto:r-help@r-project.org>>
抄送：
Hi
If your function works for you why to bother with apply? If it does not give
you required results, please post some data and show us what is expected.
I would remove unlist from your function and declare list for storing data,
which seems to me more natural for storing results.
t.test.list=function(df,paired){
lll<-vector("list", length=nrow(df))
for(i in 1:nrow(df)){
lll[[i]]=t.test(df[i,grp1],df[i,grp2],paired=paired)
}
names(lll)=rownames(df)
lll
> }
Regards
Petr
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-bounces@r-
> project.org] On Behalf Of my1stbox
> Sent: Tuesday, August 26, 2014 3:26 AM
> To: R Help
> Subject: [R] How to do t.test to rows of a dataframe using apply family
> function?
>
> Hi All，
> How to do t.test to rows (with two levels, or in other words, groups
> of samples) of a dataframe using apply family function? I have done
> that using function and loops. And my overall purpose is to calculate
> DE genes from two groups of miRNA microarray samples. And I know limma
> can do that, but it seems limma doesn't support paired t-test.
>
> t.test.df=function(df,paired){
> df.t=c()
> for(i in 1:nrow(df)){
>
> df.t=rbind(df.t,unlist(t.test(df[i,grp1],df[i,grp2],paired=paired)))
> }
> rownames(df.t)=rownames(df)
> df.t
> }
>
> Bests!
> Allen Chiu
>
> 2014-08-26
> [[alternative HTML version deleted]]
>
> __
> R-help@r-project.org mailing list
> 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.
Tento e-mail a jakékoliv k němu připojené dokumenty jsou důvěrné a jsou určeny
pouze jeho adresátům.
Jestliže jste obdržel(a) tento e-mail omylem, informujte laskavě neprodleně
jeho odesílatele. Obsah tohoto emailu i s přílohami a jeho kopie vymažte ze
svého systému.
Nejste-li zamýšleným adresátem tohoto emailu, nejste oprávněni tento email
jakkoliv užívat, rozšiřovat, kopírovat či zveřejňovat.
Odesílatel e-mailu neodpovídá za eventuální škodu způsobenou modifikacemi či
zpožděním přenosu e-mailu.
V případě, že je tento e-mail součástí obchodního jednání:
- vyhrazuje si odesílatel právo ukončit kdykoliv jednání o uzavření smlouvy, a
to z jakéhokoliv důvodu i bez uvedení důvodu.
- a obsahuje-li nabídku, je adresát oprávněn nabídku bezodkladně přijmout;
Odesílatel tohoto e-mailu (nabídky) vylučuje přijetí nabídky ze strany příjemce
s dodatkem či odchylkou.
- trvá odesílatel na tom, že příslušná smlouva je uzavřena teprve výslovným
dosažením shody na všech jejích náležitostech.
- odesílatel tohoto emailu informuje, že není oprávněn uzavírat za společnost
žádné smlouvy s výjimkou případů, kdy k tomu byl písemně zmocněn nebo písemně
pověřen a takové pověření nebo plná moc byly adresátovi tohoto emailu případně
osobě, kterou adresát zastupuje, předloženy nebo jejich existence je adresátovi
či osobě jím zastoupené známá.
This e-mail and any documents attached to it may be confidential and are
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If you received this e-mail by mistake, please immediately inform its sender.
Delete t
[R] What would a typical miRNA microarray analysis workflow look like?
Hi all, What would a typical miRNA microarray analysis workflow look like? Say just a test group of 5 replicates from bladder cancer tumor and corresponding control group from normal tissue of the same patients. What could I do to make the analysis seems more sophisticated. I have done differential expression analysis, target gene prediction, and GO, pathway enrichments of target genes. What else could I do? It would be better if you can specify some packages or codes. Regards, Allen PS: Is paired (t)-test necessery? I found that DE gene number from paired t-test is about half of that from normal limma functions( which I don't know how to do paired test if it could). 2014-08-26 [[alternative HTML version deleted]] __ R-help@r-project.org mailing list 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] VennDiagram
Hi,, Thanks for the link, I have tried that but it seems my data is in the wrong format for that to work. On 26 August 2014 10:35, João Azevedo Patrício wrote: > Em 26-08-2014 09:30, Jurgens de Bruin escreveu: > > Hi, >> >> I am new to R and dont use it very often so I would appreciate some help. >> >> I would like to create a VennDiagram based on the following, I have >> analyzed experimental results using different methods and captured the >> results of each analysis. >> I would like the VennDiagram to show the overlap in results for the >> different analysis. >> >> Sample data: >> Analysis Results >> A 1-5 >> B 8-9 >> C 4-7 >> B 1-5 >> A 20-50 >> C 8-9 >> >> So in this simple example analysis A and B have a single sample overlap >> and >> Analysis B and C also have a single overlap but C and A have no overlap. >> >> >> Hi, > > I don't know how to do it, but made a search and found this, see if it > helps you > > http://www.ats.ucla.edu/stat/r/faq/venn.htm > > -- > João Azevedo Patrício > Tel.: +31 91 400 53 63 > Portugal > @ http://tripaforra.bl.ee > > "Take 2 seconds to think before you act" > > __ > R-help@r-project.org mailing list > 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. > -- Regards/Groete/Mit freundlichen Grüßen/recuerdos/meilleures salutations/ distinti saluti/siong/duì yú/привет Jurgens de Bruin [[alternative HTML version deleted]] __ R-help@r-project.org mailing list 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] VennDiagram
Em 26-08-2014 09:30, Jurgens de Bruin escreveu: Hi, I am new to R and dont use it very often so I would appreciate some help. I would like to create a VennDiagram based on the following, I have analyzed experimental results using different methods and captured the results of each analysis. I would like the VennDiagram to show the overlap in results for the different analysis. Sample data: Analysis Results A 1-5 B 8-9 C 4-7 B 1-5 A 20-50 C 8-9 So in this simple example analysis A and B have a single sample overlap and Analysis B and C also have a single overlap but C and A have no overlap. Hi, I don't know how to do it, but made a search and found this, see if it helps you http://www.ats.ucla.edu/stat/r/faq/venn.htm -- João Azevedo Patrício Tel.: +31 91 400 53 63 Portugal @ http://tripaforra.bl.ee "Take 2 seconds to think before you act" __ R-help@r-project.org mailing list 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] VennDiagram
Hi, I am new to R and dont use it very often so I would appreciate some help. I would like to create a VennDiagram based on the following, I have analyzed experimental results using different methods and captured the results of each analysis. I would like the VennDiagram to show the overlap in results for the different analysis. Sample data: Analysis Results A 1-5 B 8-9 C 4-7 B 1-5 A 20-50 C 8-9 So in this simple example analysis A and B have a single sample overlap and Analysis B and C also have a single overlap but C and A have no overlap. -- Regards/Groete/Mit freundlichen Grüßen/recuerdos/meilleures salutations/ distinti saluti/siong/duì yú/привет Jurgens de Bruin [[alternative HTML version deleted]] __ R-help@r-project.org mailing list 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.
