Re: [R] how to overwrite a Unary operator ?
On Oct 16, 2014, at 10:36 PM, PO SU wrote:
Tks for your advice, let the ++ problem alone, how to write an
Unary operator ? Is it permitted in R?
suchasa-2 , a%+2% will let a be 4 .
OK, that's just wrong. Oh, OK, just for fun, as it were:
inc - function(x)
{
eval.parent(substitute(x - x + 1))
}
inc(10)
Error in 10 - 10 + 1 : invalid (do_set) left-hand side to assignment
y=10
inc(y)
y
[1] 11
I just want to know it , i won't pollute r with it , because i know
what is r . : )
It's certainly permitted. Just look at all the overloadings of the +
operator in graphics packages. Look up the documentation on methods in
R.
Why not just use a well-behaved function, though?
.inc - function(x) x+1
.inc(10)
[1] 11
Then you won't be tempted to try 10 - .inc(10) because it just
wouldn't make sense.
--
David.
--
PO SU
mail: desolato...@163.com
Majored in Statistics from SJTU
At 2014-10-17 13:09:47, Rolf Turner r.tur...@auckland.ac.nz wrote:
On 17/10/14 17:29, PO SU wrote:
Dear expeRts,
Now i want to know how to implement an Unary operator like i++
in cpp's synax form.
e.g. 2++ will let 2 be 3 , a-2 ,a++ ,will let a be 3
I tried this :
'%++%'-function(x){
x-x+1
}
but it have problem, the biggest one is it seems the function need
twoparams like a%++%b , how to write a function needing just one
param?
TKS !
Just ***DON'T***. The ++ operator is useful only for those wish to
write code which is obscure to the point of incomprehensibility. It
makes C and its offspring write only languages.
If you are going to use R, use R and don't pollute it with such
abominations.
cheers,
Rolf Turner
--
Rolf Turner
Technical Editor ANZJS
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Re: [R] how to overwrite a Unary operator ?
On 17 Oct 2014, at 06:29, PO SU rhelpmaill...@163.com wrote: e.g. 2++ will let 2 be 3 That would not even work in C ... While I use this in C, I second Rolf on the general issue. Benno__ 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] Difference betweeen cor.test() and formula everyone says to use
This is pretty much standard. I'm quite sure that other stats packages do likewise and I wouldn't know who everyone is. It is not unheard of that textbook authors give suboptimal formulas in order not to confuse students, though. The basic point is that the t transformation gives the exact distribution under the null. Fisher's Z is only approximately normally distributed. The t transformation works because if beta is the regression coefficient of y on x, beta==0 iff rho==0, and we have exact theory for testing beta==0 by a t-test. Off-null, the t-approach does not readily transfer, so confidence intervals tend to be based on the Z-transformation. -Peter D. On 17 Oct 2014, at 02:20 , Joshua Wiley jwiley.ps...@gmail.com wrote: Hi Jeremy, I don't know about references, but this around. See for example: http://afni.nimh.nih.gov/sscc/gangc/tr.html the relevant line in cor.test is: STATISTIC - c(t = sqrt(df) * r/sqrt(1 - r^2)) You can convert *t*s to *r*s and vice versa. Best, Josh On Fri, Oct 17, 2014 at 10:32 AM, Jeremy Miles jeremy.mi...@gmail.com wrote: I'm trying to understand how cor.test() is calculating the p-value of a correlation. It gives a p-value based on t, but every text I've ever seen gives the calculation based on z. For example: data(cars) with(cars[1:10, ], cor.test(speed, dist)) Pearson's product-moment correlation data: speed and dist t = 2.3893, df = 8, p-value = 0.04391 alternative hypothesis: true correlation is not equal to 0 95 percent confidence interval: 0.02641348 0.90658582 sample estimates: cor 0.6453079 But when I use the regular formula: r - cor(cars[1:10, ])[1, 2] r.z - fisherz(r) se - se - 1/sqrt(10 - 3) z - r.z / se (1 - pnorm(z))*2 [1] 0.04237039 My p-value is different. The help file for cor.test doesn't (seem to) have any reference to this, and I can see in the source code that it is doing something different. I'm just not sure what. Thanks, Jeremy __ 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. -- Joshua F. Wiley Ph.D. Student, UCLA Department of Psychology http://joshuawiley.com/ Senior Analyst, Elkhart Group Ltd. http://elkhartgroup.com Office: 260.673.5518 [[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. -- Peter Dalgaard, Professor, Center for Statistics, Copenhagen Business School Solbjerg Plads 3, 2000 Frederiksberg, Denmark Phone: (+45)38153501 Email: pd@cbs.dk Priv: pda...@gmail.com __ 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] {car} outlierTest looses p/q values
Hi guys, I came across a strange phenomena and can't figure out why it happens by myself so here we go. I got a dataframe which consists of double numbers which I want to check, row-wise if there are outliers in the rows. So I iterate over the rows and create a glm using the numbers of that particular row. Which might look like this: case1) x1x2x3x4x5x6 x7 x8x9x10x11 0.00 3.91 0.00 0.00 0.00 68.03 40.39 0.00 0.00 0.00 4.11 or like this: case2) x1x2x3x4x5x6 x7 x8x9x10x11 1.00 1.001.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 5.34 or any other combination of double numbers... however, using a glm like this: glModel - glm(vector ~ some_other_meta_data_which_is_double_numbers) and testing it with: test.Res - outlierTest(glModel,digits=4,cutoff=Inf,n.max=Inf) I always get a result consisting of the desired p and q values but not if the vector I use looks like case2. There is no error message and the computation does not stop either. However, all p and q values are produced except for the last value x11. Any idea why this particular value gets dropped from the output of the outlierTest Method in the car package. Here is the sessioninfo: sessionInfo() R version 3.1.1 (2014-07-10) Platform: x86_64-redhat-linux-gnu (64-bit) locale: [1] LC_CTYPE=en_US.utf8 LC_NUMERIC=C [3] LC_TIME=en_US.utf8LC_COLLATE=en_US.utf8 [5] LC_MONETARY=en_US.utf8LC_MESSAGES=en_US.utf8 [7] LC_PAPER=en_US.utf8 LC_NAME=C [9] LC_ADDRESS=C LC_TELEPHONE=C [11] LC_MEASUREMENT=en_US.utf8 LC_IDENTIFICATION=C attached base packages: [1] stats graphics grDevices utils datasets methods base other attached packages: [1] ggplot2_1.0.0 car_2.0-21 RColorBrewer_1.0-5 iNEXT_1.0 [5] vegan_2.0-10 lattice_0.20-29permute_0.8-3 loaded via a namespace (and not attached): [1] colorspace_1.2-4 compiler_3.1.1 digest_0.6.4 grid_3.1.1 [5] gtable_0.1.2 labeling_0.3 MASS_7.3-33 munsell_0.4.2 [9] nnet_7.3-8 plyr_1.8.1 proto_0.3-10 Rcpp_0.11.2 [13] reshape2_1.4 scales_0.2.4 stringr_0.6.2tools_3.1.1 Any help is highly appreciated. Thanks Phil __ 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] need help for error incompatible dimensions
Hello, I m R user trying to map and find correlation between two entities i.e. time and usability of cup from log file, but its showing error while compilation. Is there any specific package(s) by means of which I can effectively run the code meant for different dimensions computation? I have attached .R file and log file (.txt) Please find attachment.. Regards, Swapnil Khobragade Larsen Toubro Infotech Ltd. Bldg. No.56, 1st Floor, Airoli, Navi Mumbai - 400708. e-mail ID: swapnil.khobrag...@lntinfotech.commailto:swapnil.khobrag...@lntinfotech.com The contents of this e-mail and any attachment(s) may contain confidential or privileged information for the intended recipient(s). Unintended recipients are prohibited from taking action on the basis of information in this e-mail and using or disseminating the information, and must notify the sender and delete it from their system. LT Infotech will not accept responsibility or liability for the accuracy or completeness of, or the presence of any virus or disabling code in this e-mail 20:47:15all 5.05 0.00 2.02 0.51 0.00 92.42 20:47:16all 4.02 0.00 2.01 0.00 0.00 93.97 20:47:17all 5.61 0.00 2.55 0.00 0.00 91.84 20:47:18all 3.55 0.00 1.52 0.00 0.00 94.92 20:47:19all 5.58 0.00 2.03 0.00 0.00 92.39 20:47:20all 2.05 0.00 3.08 0.00 0.00 94.87 20:47:21all 3.55 0.00 1.52 0.00 0.00 94.92 20:47:22all 5.64 0.00 2.56 0.00 0.00 91.79 20:47:23all 3.02 0.00 2.51 0.00 0.00 94.47 20:47:24all 3.08 0.00 2.05 0.00 0.00 94.87 20:47:25all 4.06 0.00 2.54 0.00 0.00 93.40 __ 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 overwrite a Unary operator ?
Tks for your alternative way's details. but like you mentioned in graphics
package, i still wonder how to overload an operator which can pass one param
like +2 .
There seems exists some examples for my needing. But i try to find them but
without any results.
can you show me some examples from it?
--
PO SU
mail: desolato...@163.com
Majored in Statistics from SJTU
At 2014-10-17 15:16:47, David Winsemius dwinsem...@comcast.net wrote:
On Oct 16, 2014, at 10:36 PM, PO SU wrote:
Tks for your advice, let the ++ problem alone, how to write an
Unary operator ? Is it permitted in R?
suchasa-2 , a%+2% will let a be 4 .
OK, that's just wrong. Oh, OK, just for fun, as it were:
inc - function(x)
{
eval.parent(substitute(x - x + 1))
}
inc(10)
Error in 10 - 10 + 1 : invalid (do_set) left-hand side to assignment
y=10
inc(y)
y
[1] 11
I just want to know it , i won't pollute r with it , because i know
what is r . : )
It's certainly permitted. Just look at all the overloadings of the +
operator in graphics packages. Look up the documentation on methods in
R.
Why not just use a well-behaved function, though?
.inc - function(x) x+1
.inc(10)
[1] 11
Then you won't be tempted to try 10 - .inc(10) because it just
wouldn't make sense.
--
David.
--
PO SU
mail: desolato...@163.com
Majored in Statistics from SJTU
At 2014-10-17 13:09:47, Rolf Turner r.tur...@auckland.ac.nz wrote:
On 17/10/14 17:29, PO SU wrote:
Dear expeRts,
Now i want to know how to implement an Unary operator like i++
in cpp's synax form.
e.g. 2++ will let 2 be 3 , a-2 ,a++ ,will let a be 3
I tried this :
'%++%'-function(x){
x-x+1
}
but it have problem, the biggest one is it seems the function need
twoparams like a%++%b , how to write a function needing just one
param?
TKS !
Just ***DON'T***. The ++ operator is useful only for those wish to
write code which is obscure to the point of incomprehensibility. It
makes C and its offspring write only languages.
If you are going to use R, use R and don't pollute it with such
abominations.
cheers,
Rolf Turner
--
Rolf Turner
Technical Editor ANZJS
__
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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, MD
Alameda, CA, USA
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[R] ggplot: Stacked bar/pie chart - Objects above the bar/pie
Hello,
I would like to draw a circle on top of a pie chart (The plot does not
need to fullfill scientific standards). The circle represents the
relation of a reference-value in comparison to the summed values of the
pie-pieces. To be able to do this I partly followed:
http://rpubs.com/RobinLovelace/11641 . But i have the problem of some
bar/pie pieces being placed above the stacked bar/outside of the pie
chart, see graph PA / PA + coord_polar.
System: $platform [1] i686-pc-linux-gnu; $version.string [1] R
version 3.1.1 (2014-07-10); ggplot2_0.9.3.1
require (ggplot2)
mdf - structure(list(
VG = structure(c(1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L),
.Label = c(A, B), class = factor),
variable = structure(c(4L, 3L, 5L, 1L, 2L, 1L, 4L, 3L, 2L, 5L),
.Label = c(a, b, c, d, e), class =
factor),
value = c(0, 8407346.56, 0, 124901773, 0, 184987520, 14612608,
3165030, 0, 0),
reference = c(0.75, 0.75, 0.75, 0.75, 0.75, 1.25, 1.25, 1.25, 1.25,
1.25)),
.Names = c(VG, variable, value, reference),
row.names = c(16L, 17L, 18L, 19L, 20L, 46L, 47L, 48L, 49L, 50L),
class = data.frame)
# Calculate position
pos - function (x) 0.5 * (cumsum(x) + cumsum(c(0, x[-length(x)])))
lmdf - dlply (mdf, VG)
lmdf - lapply (lmdf, function (X) { X$pos - pos (X$value)
return (X) })
mdf - rbind.data.frame (lmdf[[1]], lmdf[[2]])
# height (radius) of all pieces (variable) in one bar (pie) will be the
same:
mdf$rad - 1
# abline from the reference column in the dataframe
INTCA - unique (mdf$reference[mdf$VG==A])
PA - ggplot (mdf[mdf$VG==A,], aes (x = pos, y = rad)) +
geom_bar (stat = identity, aes (fill = variable, width = value)) +
geom_abline(intercept = INTCA, slope = 0)
PA
PA + coord_polar()
# Object c is placed above its position on the y axis
# abline from the reference column in the dataframe
INTCB - unique (mdf$reference[mdf$VG==B])
PB - ggplot (mdf[mdf$VG==B,], aes (x = pos, y = rad)) +
geom_bar (stat = identity, aes (fill = variable, width = value)) +
geom_abline(intercept = INTCB, slope = 0)
PB
PB + coord_polar()
# all objects are within rad = 1
###
Your help would be very much appreciated,
Gunnar
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Re: [R] need help for error incompatible dimensions
Hi Your .R attachment did not come through. Included txt file did not have headers. Maybe you just want cor(whatever is the name of your numeric object) Cheers Petr -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-bounces@r- project.org] On Behalf Of Swapnil Khobragade Sent: Friday, October 17, 2014 11:04 AM To: r-help@r-project.org Cc: skhobragad...@gmail.com Subject: [R] need help for error incompatible dimensions Hello, I m R user trying to map and find correlation between two entities i.e. time and usability of cup from log file, but its showing error while compilation. Is there any specific package(s) by means of which I can effectively run the code meant for different dimensions computation? I have attached .R file and log file (.txt) Please find attachment.. Regards, Swapnil Khobragade Larsen Toubro Infotech Ltd. Bldg. No.56, 1st Floor, Airoli, Navi Mumbai - 400708. e-mail ID: swapnil.khobrag...@lntinfotech.commailto:swapnil.khobragade@lntinfotec h.com The contents of this e-mail and any attachment(s) may contain confidential or privileged information for the intended recipient(s). Unintended recipients are prohibited from taking action on the basis of information in this e-mail and using or disseminating the information, and must notify the sender and delete it from their system. LT Infotech will not accept responsibility or liability for the accuracy or completeness of, or the presence of any virus or disabling code in this e-mail 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 intended only for its intended recipients. If you received this e-mail by mistake, please immediately inform its sender. Delete the contents of this e-mail with all attachments and its copies from your system. If you are not the intended recipient of this e-mail, you are not authorized to use, disseminate, copy or disclose this e-mail in any manner. The sender of this e-mail shall not be liable for any possible damage caused by modifications of the e-mail or by delay with transfer of the email. In case that this e-mail forms part of business dealings: - the sender reserves the right to end negotiations about entering into a contract in any time, for any reason, and without stating any reasoning. - if the e-mail contains an offer, the recipient is entitled to immediately accept such offer; The sender of this e-mail (offer) excludes any acceptance of the offer on the part of the recipient containing any amendment or variation. - the sender insists on that the respective contract is concluded only upon an express mutual agreement on all its aspects. - the sender of this e-mail informs that he/she is not authorized to enter into any contracts on behalf of the company except for cases in which he/she is expressly authorized to do so in writing, and such authorization or power of attorney is submitted to the recipient or the person represented by the recipient, or the existence of such authorization is known to the recipient of the person represented by the recipient. __ 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] assigning letter to a column
Hi,
I'm having trouble with assigning a letter to a column based on the value
of another column.
Since I have separate data files I've saved then into one folder and I'm
reading them in separately into the function.
The code is below.
#F= fast; S= slow; I1= Intermediate score 1; I2=Intermediate score 2
filename-list.files(pattern=*.txt)
filename
corloc- function(x){
x-read.table(filename[x], sep=\t, header=TRUE) #will extract the
relevant data file from folder 1998. ex. corloc(1) will return 1998
breeding year data
x[,ForS]-0 #new column
for (i in length(x$CORLOC)){ #this is the bit that I'm having a problem
with since it's not assigning the appropriate letter into the ForS column
ifelse(x$COR_LOC=3 | x$COR_LOC15.230, ForS-S,
ifelse(x$COR_LOC=15.230 | x$COR_LOC19.810, ForS-I1,
ifelse(x$COR_LOC=19.810 | x$COR_LOC25.540,
FS-I2,ForS-F)))}
print(x)
}
I've tried some of the solutions on stackoverflow but still was
unsuccessful.
Best,
Monaly.
[[alternative HTML version deleted]]
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Re: [R] Making a very specific heatmap in R (ggplot2?)
As the Posting Guide indicates, this is a text-only mailing list, and small, reproducible (self-contained) examples of R code are expected. You are not getting much response because what you sent is not what we see, and we cannot tweak code you do not share. --- Jeff NewmillerThe . . Go Live... DCN:jdnew...@dcn.davis.ca.usBasics: ##.#. ##.#. 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 October 16, 2014 7:02:37 PM PDT, Alexander Predeus pred...@gmail.com wrote: Hello All, I'm trying to figure out the (automated) way to generate heatmaps from simple data tables with annotated rows and columns. In the end, I need these files to be easily viewed in a browser. The initial data tables are simple; numbers are row-normalized (values are real numbers varying from -1.0 to +1.0), with rows numbered from a zero to a certain integer, and columns are named with (sometimes lengthy) strings. What I came up with so far is building the heatmap using standard facility with a set resolution of the final png file: Which sort of works sometimes: And fails pretty miserably other times: What I really would like to do, is to make a heatmap with set number of pixels per square, with a set font size (in pixels as well), and then save it as a .png with variable resolution (appropriate for each concrete heatmap). That way, a cell would be always say 10 by 10 pixels, and the text on the right is always 8 pixels tall. Alternatively, of course, it would be very neat to save it in some vector format easily interpreted by the browser, but I have an impression it is not an easy feat to accomplish. Thank you very much in advance. [[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] assigning letter to a column
I think it is doing exactly what you have told it to do, but that is probably
not what you want it to do.
First, you do not need a loop since the ifelse() function is vectorized. Read
the manual page and the examples carefully. Also you are coding ifelse() as if
it were the same as if() {} else() {}. Again you need to refer to the
documentation.
Second, this seems like a job for cut() not ifelse().
Third, look at your code. The first statement is x$COR_LOC=3 |
x$COR_LOC15.230 so everything greater than 3 will be coded as S. That is
probably all of your data. You probably want to use (and) instead of | (or).
It is not clear what you want to happen for values less than 3 but they will be
NA (missing).
Your entire ifelse() boils down to
set.seed(42)
x - data.frame(COR_LOC=runif(100, 0, 30))
x$ForS - cut(x$COR_LOC, breaks=c(3, 15.23, 19.81, 25.40, Inf),
labels=c(S, I1, I2, F), right=FALSE)
No loops, no ifelse's. Anything below 3 will
-
David L Carlson
Department of Anthropology
Texas AM University
College Station, TX 77840-4352
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Monaly Mistry
Sent: Friday, October 17, 2014 8:27 AM
To: r-help@r-project.org
Subject: [R] assigning letter to a column
Hi,
I'm having trouble with assigning a letter to a column based on the value
of another column.
Since I have separate data files I've saved then into one folder and I'm
reading them in separately into the function.
The code is below.
#F= fast; S= slow; I1= Intermediate score 1; I2=Intermediate score 2
filename-list.files(pattern=*.txt)
filename
corloc- function(x){
x-read.table(filename[x], sep=\t, header=TRUE) #will extract the
relevant data file from folder 1998. ex. corloc(1) will return 1998
breeding year data
x[,ForS]-0 #new column
for (i in length(x$CORLOC)){ #this is the bit that I'm having a problem
with since it's not assigning the appropriate letter into the ForS column
ifelse(x$COR_LOC=3 | x$COR_LOC15.230, ForS-S,
ifelse(x$COR_LOC=15.230 | x$COR_LOC19.810, ForS-I1,
ifelse(x$COR_LOC=19.810 | x$COR_LOC25.540,
FS-I2,ForS-F)))}
print(x)
}
I've tried some of the solutions on stackoverflow but still was
unsuccessful.
Best,
Monaly.
[[alternative HTML version deleted]]
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and provide commented, minimal, self-contained, reproducible code.
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Re: [R] assigning letter to a column
Minor correction, given your code, values less than 3 will be coded as S
since they are less than 15.23. In the code I suggested, values less than 3
will be coded as missing (NA).
David C
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of David L Carlson
Sent: Friday, October 17, 2014 9:15 AM
To: Monaly Mistry; r-help@r-project.org
Subject: Re: [R] assigning letter to a column
I think it is doing exactly what you have told it to do, but that is probably
not what you want it to do.
First, you do not need a loop since the ifelse() function is vectorized. Read
the manual page and the examples carefully. Also you are coding ifelse() as if
it were the same as if() {} else() {}. Again you need to refer to the
documentation.
Second, this seems like a job for cut() not ifelse().
Third, look at your code. The first statement is x$COR_LOC=3 |
x$COR_LOC15.230 so everything greater than 3 will be coded as S. That is
probably all of your data. You probably want to use (and) instead of | (or).
It is not clear what you want to happen for values less than 3 but they will be
NA (missing).
Your entire ifelse() boils down to
set.seed(42)
x - data.frame(COR_LOC=runif(100, 0, 30))
x$ForS - cut(x$COR_LOC, breaks=c(3, 15.23, 19.81, 25.40, Inf),
labels=c(S, I1, I2, F), right=FALSE)
No loops, no ifelse's. Anything below 3 will
-
David L Carlson
Department of Anthropology
Texas AM University
College Station, TX 77840-4352
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Monaly Mistry
Sent: Friday, October 17, 2014 8:27 AM
To: r-help@r-project.org
Subject: [R] assigning letter to a column
Hi,
I'm having trouble with assigning a letter to a column based on the value
of another column.
Since I have separate data files I've saved then into one folder and I'm
reading them in separately into the function.
The code is below.
#F= fast; S= slow; I1= Intermediate score 1; I2=Intermediate score 2
filename-list.files(pattern=*.txt)
filename
corloc- function(x){
x-read.table(filename[x], sep=\t, header=TRUE) #will extract the
relevant data file from folder 1998. ex. corloc(1) will return 1998
breeding year data
x[,ForS]-0 #new column
for (i in length(x$CORLOC)){ #this is the bit that I'm having a problem
with since it's not assigning the appropriate letter into the ForS column
ifelse(x$COR_LOC=3 | x$COR_LOC15.230, ForS-S,
ifelse(x$COR_LOC=15.230 | x$COR_LOC19.810, ForS-I1,
ifelse(x$COR_LOC=19.810 | x$COR_LOC25.540,
FS-I2,ForS-F)))}
print(x)
}
I've tried some of the solutions on stackoverflow but still was
unsuccessful.
Best,
Monaly.
[[alternative HTML version deleted]]
__
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and provide commented, minimal, self-contained, reproducible code.
__
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Re: [R] {car} outlierTest looses p/q values
Dear Phil, After reading your posting several times, I still don't understand what you did. As usual, having a reproducible example illustrating the error would be a great help. I do have a guess about the source of the error: glm() failed in some way for the problematic case. Best, John John Fox, Professor McMaster University Hamilton, Ontario, Canada http://socserv.mcmaster.ca/jfox/ On Fri, 17 Oct 2014 12:53:30 +0200 Phil chobop...@gmail.com wrote: Hi guys, I came across a strange phenomena and can't figure out why it happens by myself so here we go. I got a dataframe which consists of double numbers which I want to check, row-wise if there are outliers in the rows. So I iterate over the rows and create a glm using the numbers of that particular row. Which might look like this: case1) x1x2x3x4x5x6 x7 x8x9x10x11 0.00 3.91 0.00 0.00 0.00 68.03 40.39 0.00 0.00 0.00 4.11 or like this: case2) x1x2x3x4x5x6 x7 x8x9x10x11 1.00 1.001.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 5.34 or any other combination of double numbers... however, using a glm like this: glModel - glm(vector ~ some_other_meta_data_which_is_double_numbers) and testing it with: test.Res - outlierTest(glModel,digits=4,cutoff=Inf,n.max=Inf) I always get a result consisting of the desired p and q values but not if the vector I use looks like case2. There is no error message and the computation does not stop either. However, all p and q values are produced except for the last value x11. Any idea why this particular value gets dropped from the output of the outlierTest Method in the car package. Here is the sessioninfo: sessionInfo() R version 3.1.1 (2014-07-10) Platform: x86_64-redhat-linux-gnu (64-bit) locale: [1] LC_CTYPE=en_US.utf8 LC_NUMERIC=C [3] LC_TIME=en_US.utf8LC_COLLATE=en_US.utf8 [5] LC_MONETARY=en_US.utf8LC_MESSAGES=en_US.utf8 [7] LC_PAPER=en_US.utf8 LC_NAME=C [9] LC_ADDRESS=C LC_TELEPHONE=C [11] LC_MEASUREMENT=en_US.utf8 LC_IDENTIFICATION=C attached base packages: [1] stats graphics grDevices utils datasets methods base other attached packages: [1] ggplot2_1.0.0 car_2.0-21 RColorBrewer_1.0-5 iNEXT_1.0 [5] vegan_2.0-10 lattice_0.20-29permute_0.8-3 loaded via a namespace (and not attached): [1] colorspace_1.2-4 compiler_3.1.1 digest_0.6.4 grid_3.1.1 [5] gtable_0.1.2 labeling_0.3 MASS_7.3-33 munsell_0.4.2 [9] nnet_7.3-8 plyr_1.8.1 proto_0.3-10 Rcpp_0.11.2 [13] reshape2_1.4 scales_0.2.4 stringr_0.6.2tools_3.1.1 Any help is highly appreciated. Thanks Phil __ 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 overwrite a Unary operator ?
You may be interested in looking at Reference Classes/objects (see
?setRefClass). This is a form of OO programming that is more similar
to C++ and Java. You could create a counter object that you could
then increment with syntax like:
x$inc()
x$inc(5)
The first would increment by the default (1), the second would then
increment by 5.
On Fri, Oct 17, 2014 at 2:06 AM, PO SU rhelpmaill...@163.com wrote:
Tks for your alternative way's details. but like you mentioned in graphics
package, i still wonder how to overload an operator which can pass one param
like +2 .
There seems exists some examples for my needing. But i try to find them but
without any results.
can you show me some examples from it?
--
PO SU
mail: desolato...@163.com
Majored in Statistics from SJTU
At 2014-10-17 15:16:47, David Winsemius dwinsem...@comcast.net wrote:
On Oct 16, 2014, at 10:36 PM, PO SU wrote:
Tks for your advice, let the ++ problem alone, how to write an
Unary operator ? Is it permitted in R?
suchasa-2 , a%+2% will let a be 4 .
OK, that's just wrong. Oh, OK, just for fun, as it were:
inc - function(x)
{
eval.parent(substitute(x - x + 1))
}
inc(10)
Error in 10 - 10 + 1 : invalid (do_set) left-hand side to assignment
y=10
inc(y)
y
[1] 11
I just want to know it , i won't pollute r with it , because i know
what is r . : )
It's certainly permitted. Just look at all the overloadings of the +
operator in graphics packages. Look up the documentation on methods in
R.
Why not just use a well-behaved function, though?
.inc - function(x) x+1
.inc(10)
[1] 11
Then you won't be tempted to try 10 - .inc(10) because it just
wouldn't make sense.
--
David.
--
PO SU
mail: desolato...@163.com
Majored in Statistics from SJTU
At 2014-10-17 13:09:47, Rolf Turner r.tur...@auckland.ac.nz wrote:
On 17/10/14 17:29, PO SU wrote:
Dear expeRts,
Now i want to know how to implement an Unary operator like i++
in cpp's synax form.
e.g. 2++ will let 2 be 3 , a-2 ,a++ ,will let a be 3
I tried this :
'%++%'-function(x){
x-x+1
}
but it have problem, the biggest one is it seems the function need
twoparams like a%++%b , how to write a function needing just one
param?
TKS !
Just ***DON'T***. The ++ operator is useful only for those wish to
write code which is obscure to the point of incomprehensibility. It
makes C and its offspring write only languages.
If you are going to use R, use R and don't pollute it with such
abominations.
cheers,
Rolf Turner
--
Rolf Turner
Technical Editor ANZJS
__
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, MD
Alameda, CA, USA
__
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.
--
Gregory (Greg) L. Snow Ph.D.
538...@gmail.com
__
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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] assigning letter to a column
Thank you for your help David, will make sure to check the documentation.
Best,
Monaly
On Fri, Oct 17, 2014 at 3:27 PM, David L Carlson dcarl...@tamu.edu wrote:
Minor correction, given your code, values less than 3 will be coded as S
since they are less than 15.23. In the code I suggested, values less than 3
will be coded as missing (NA).
David C
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf Of David L Carlson
Sent: Friday, October 17, 2014 9:15 AM
To: Monaly Mistry; r-help@r-project.org
Subject: Re: [R] assigning letter to a column
I think it is doing exactly what you have told it to do, but that is
probably not what you want it to do.
First, you do not need a loop since the ifelse() function is vectorized.
Read the manual page and the examples carefully. Also you are coding
ifelse() as if it were the same as if() {} else() {}. Again you need to
refer to the documentation.
Second, this seems like a job for cut() not ifelse().
Third, look at your code. The first statement is x$COR_LOC=3 |
x$COR_LOC15.230 so everything greater than 3 will be coded as S. That is
probably all of your data. You probably want to use (and) instead of |
(or). It is not clear what you want to happen for values less than 3 but
they will be NA (missing).
Your entire ifelse() boils down to
set.seed(42)
x - data.frame(COR_LOC=runif(100, 0, 30))
x$ForS - cut(x$COR_LOC, breaks=c(3, 15.23, 19.81, 25.40, Inf),
labels=c(S, I1, I2, F), right=FALSE)
No loops, no ifelse's. Anything below 3 will
-
David L Carlson
Department of Anthropology
Texas AM University
College Station, TX 77840-4352
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf Of Monaly Mistry
Sent: Friday, October 17, 2014 8:27 AM
To: r-help@r-project.org
Subject: [R] assigning letter to a column
Hi,
I'm having trouble with assigning a letter to a column based on the value
of another column.
Since I have separate data files I've saved then into one folder and I'm
reading them in separately into the function.
The code is below.
#F= fast; S= slow; I1= Intermediate score 1; I2=Intermediate score 2
filename-list.files(pattern=*.txt)
filename
corloc- function(x){
x-read.table(filename[x], sep=\t, header=TRUE) #will extract the
relevant data file from folder 1998. ex. corloc(1) will return 1998
breeding year data
x[,ForS]-0 #new column
for (i in length(x$CORLOC)){ #this is the bit that I'm having a problem
with since it's not assigning the appropriate letter into the ForS column
ifelse(x$COR_LOC=3 | x$COR_LOC15.230, ForS-S,
ifelse(x$COR_LOC=15.230 | x$COR_LOC19.810, ForS-I1,
ifelse(x$COR_LOC=19.810 | x$COR_LOC25.540,
FS-I2,ForS-F)))}
print(x)
}
I've tried some of the solutions on stackoverflow but still was
unsuccessful.
Best,
Monaly.
[[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.
[[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] how to overwrite a Unary operator ?
On Oct 17, 2014, at 1:06 AM, PO SU wrote:
Tks for your alternative way's details. but like you mentioned in graphics
package, i still wonder how to overload an operator which can pass one param
like +2 .
There seems exists some examples for my needing. But i try to find them but
without any results.
can you show me some examples from it?
I think this might be a case of if you don't know enough to do it, then you
don't know why you shouldn't do it. (Or vice versa?) I did search for a
relevant fortune to support my impression, but the various entries for
fortune(parse) and fortune(eval) didn't seem to hit the mark.
library(ggplot2)
help(pack=ggplot2) # scroll to bottom of page
ggplot2:::`+.gg` # this shows the S3 method of adding an operator based on
the S3 method dispatch.
I was able to emulate that example to create a C-like, unary `+` operator for a
new class, but I'm not willing to put it in print for fear that my karmic
account might be depleted.
--
David.
--
PO SU
mail: desolato...@163.com
Majored in Statistics from SJTU
At 2014-10-17 15:16:47, David Winsemius dwinsem...@comcast.net wrote:
On Oct 16, 2014, at 10:36 PM, PO SU wrote:
Tks for your advice, let the ++ problem alone, how to write an
Unary operator ? Is it permitted in R?
suchasa-2 , a%+2% will let a be 4 .
OK, that's just wrong. Oh, OK, just for fun, as it were:
inc - function(x)
{
eval.parent(substitute(x - x + 1))
}
inc(10)
Error in 10 - 10 + 1 : invalid (do_set) left-hand side to assignment
y=10
inc(y)
y
[1] 11
I just want to know it , i won't pollute r with it , because i know
what is r . : )
It's certainly permitted. Just look at all the overloadings of the +
operator in graphics packages. Look up the documentation on methods in
R.
Why not just use a well-behaved function, though?
.inc - function(x) x+1
.inc(10)
[1] 11
Then you won't be tempted to try 10 - .inc(10) because it just
wouldn't make sense.
--
David.
--
PO SU
mail: desolato...@163.com
Majored in Statistics from SJTU
At 2014-10-17 13:09:47, Rolf Turner r.tur...@auckland.ac.nz wrote:
On 17/10/14 17:29, PO SU wrote:
Dear expeRts,
Now i want to know how to implement an Unary operator like i++
in cpp's synax form.
e.g. 2++ will let 2 be 3 , a-2 ,a++ ,will let a be 3
I tried this :
'%++%'-function(x){
x-x+1
}
but it have problem, the biggest one is it seems the function need
twoparams like a%++%b , how to write a function needing just one
param?
TKS !
Just ***DON'T***. The ++ operator is useful only for those wish to
write code which is obscure to the point of incomprehensibility. It
makes C and its offspring write only languages.
If you are going to use R, use R and don't pollute it with such
abominations.
cheers,
Rolf Turner
--
Rolf Turner
Technical Editor ANZJS
__
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, MD
Alameda, CA, USA
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] {car} outlierTest looses p/q values
Dear Phil, Yes, that's a bit clearer. One can invent data configurations where certain studentized residuals are undefined. For example, try the following: y - c(0, 0, 0, 0, 0, 1) x - 1:6 xx - (1:6 - 3.5)^2 rstudent(lm(y ~ x)) rstudent(lm(y ~ xx)) plot(x, y) plot(xx, y) The plots should clarify what's going on. I'm copying to r-help since the discussion began there. I hope this helps, John On Fri, 17 Oct 2014 18:35:59 +0200 Philip Stevens chobop...@gmail.com wrote: Dear John, Thank you for your fast reply. Unfortunately I am out of office right now but I will get back to you on Monday asap with a toy example and some code. Meanwhile let me try to explain further. Basically not the glm but the outlierTest afterwards fails. And it only fails if all values, used to set up the glm, are exactly 1 except for one value which can be arbitrary large. I construct the glm from a vector of doubles (the target values) and a vector of other numeric values(metadata). (So this should be fit - glm(targetVector ~ metadata) ) And want to check if one of those doubles(in the target vector) is an outlier using outlierTest(fit). The residuals are calculated by the glm but the outlierTest does not report a p nor a q value for the one value in the target vector which is not 1. And I can't figure out why... I hope this makes it a bit clearer. Anyways I will come back to you on Monday. Best, Phil Am 17.10.2014 17:43 schrieb John Fox j...@mcmaster.ca: Dear Phil, After reading your posting several times, I still don't understand what you did. As usual, having a reproducible example illustrating the error would be a great help. I do have a guess about the source of the error: glm() failed in some way for the problematic case. Best, John John Fox, Professor McMaster University Hamilton, Ontario, Canada http://socserv.mcmaster.ca/jfox/ On Fri, 17 Oct 2014 12:53:30 +0200 Phil chobop...@gmail.com wrote: Hi guys, I came across a strange phenomena and can't figure out why it happens by myself so here we go. I got a dataframe which consists of double numbers which I want to check, row-wise if there are outliers in the rows. So I iterate over the rows and create a glm using the numbers of that particular row. Which might look like this: case1) x1x2x3x4x5x6 x7 x8x9x10x11 0.00 3.91 0.00 0.00 0.00 68.03 40.39 0.00 0.00 0.00 4.11 or like this: case2) x1x2x3x4x5x6 x7 x8x9x10x11 1.00 1.001.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 5.34 or any other combination of double numbers... however, using a glm like this: glModel - glm(vector ~ some_other_meta_data_which_is_double_numbers) and testing it with: test.Res - outlierTest(glModel,digits=4,cutoff=Inf,n.max=Inf) I always get a result consisting of the desired p and q values but not if the vector I use looks like case2. There is no error message and the computation does not stop either. However, all p and q values are produced except for the last value x11. Any idea why this particular value gets dropped from the output of the outlierTest Method in the car package. Here is the sessioninfo: sessionInfo() R version 3.1.1 (2014-07-10) Platform: x86_64-redhat-linux-gnu (64-bit) locale: [1] LC_CTYPE=en_US.utf8 LC_NUMERIC=C [3] LC_TIME=en_US.utf8LC_COLLATE=en_US.utf8 [5] LC_MONETARY=en_US.utf8LC_MESSAGES=en_US.utf8 [7] LC_PAPER=en_US.utf8 LC_NAME=C [9] LC_ADDRESS=C LC_TELEPHONE=C [11] LC_MEASUREMENT=en_US.utf8 LC_IDENTIFICATION=C attached base packages: [1] stats graphics grDevices utils datasets methods base other attached packages: [1] ggplot2_1.0.0 car_2.0-21 RColorBrewer_1.0-5 iNEXT_1.0 [5] vegan_2.0-10 lattice_0.20-29permute_0.8-3 loaded via a namespace (and not attached): [1] colorspace_1.2-4 compiler_3.1.1 digest_0.6.4 grid_3.1.1 [5] gtable_0.1.2 labeling_0.3 MASS_7.3-33 munsell_0.4.2 [9] nnet_7.3-8 plyr_1.8.1 proto_0.3-10 Rcpp_0.11.2 [13] reshape2_1.4 scales_0.2.4 stringr_0.6.2tools_3.1.1 Any help is highly appreciated. Thanks Phil __ 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. John Fox, Professor McMaster University Hamilton,
Re: [R] Difference betweeen cor.test() and formula everyone says to use
The distribution of the statistic $ndf * r^2 / (1-r^2)$ with the true value $\rho = zero$ follows an $F(1,ndf)$ distribution. So the t-test is the correct test for $\rho=0$. Fisher's z is an asymptotically normal transformation for any value of $\rho$. Thus Fisher's z is better for testing $\rho= \rho_0 $ or $\rho_1 = \rho_2$. The two statistics will not be equivalent at $\rho=0$ because the statistics are based on different assumptions. Jeremy Miles jeremy.mi...@gmail.com Sent by: r-help-boun...@r-project.org 10/16/2014 07:32 PM To r-help r-help@r-project.org, cc Subject [R] Difference betweeen cor.test() and formula everyone says to use I'm trying to understand how cor.test() is calculating the p-value of a correlation. It gives a p-value based on t, but every text I've ever seen gives the calculation based on z. For example: data(cars) with(cars[1:10, ], cor.test(speed, dist)) Pearson's product-moment correlation data: speed and dist t = 2.3893, df = 8, p-value = 0.04391 alternative hypothesis: true correlation is not equal to 0 95 percent confidence interval: 0.02641348 0.90658582 sample estimates: cor 0.6453079 But when I use the regular formula: r - cor(cars[1:10, ])[1, 2] r.z - fisherz(r) se - se - 1/sqrt(10 - 3) z - r.z / se (1 - pnorm(z))*2 [1] 0.04237039 My p-value is different. The help file for cor.test doesn't (seem to) have any reference to this, and I can see in the source code that it is doing something different. I'm just not sure what. Thanks, Jeremy __ 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. [[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] Microsoft fighting against cxxfunction() and winning
Running Windows 7, 64 bit (but also including, I think, 32 bit R files) RStudio 0.98.501 win-library 3.1 Getting error messages, probably related to add_path, for the following example script; I did look within RStudio and on the internet but can't understand the errors, let alone fix them, so any hints would be appreciated. library(inline) library(Rcpp) library(devtools) f2 - cxxfunction( signature(x = integer, y = numeric ) , ' return wrap( asint(x) * asdouble(y) ) ; ', plugin = Rcpp ) fx( 2L, 5 ) # add_path(Rtools\\bin,after=0) # add_path(Rtools\\gcc-4.6.3\\bin,after=1) # get_path() 1) If I run as is, the error message is: Error in compileCode(f, code, language = language, verbose = verbose) : Compilation ERROR, function(s)/method(s) not created! Warning message: running command 'make -f C:/PROGRA~1/R/R-31~1.1/etc/x64/Makeconf -f C:/PROGRA~1/R/R-31~1.1/share/make/winshlib.mk SHLIB_LDFLAGS='$(SHLIB_CXXLDFLAGS)' SHLIB_LD='$(SHLIB_CXXLD)' SHLIB=file1a2c862738.dll WIN=64 TCLBIN=64 OBJECTS=file1a2c862738.o' had status 127 In addition: Warning message: running command 'C:/PROGRA~1/R/R-31~1.1/bin/x64/R CMD SHLIB file1a2c862738.cpp 2 file1a2c862738.cpp.err.txt' had status 1 2) If I uncomment the commented lines I get a longer, more interesting, even less comprehensible error message: Error in compileCode(f, code, language = language, verbose = verbose) : Compilation ERROR, function(s)/method(s) not created! cygwin warning: MS-DOS style path detected: C:/PROGRA~1/R/R-31~1.1/etc/x64/Makeconf Preferred POSIX equivalent is: /cygdrive/c/PROGRA~1/R/R-31~1.1/etc/x64/Makeconf CYGWIN environment variable option nodosfilewarning turns off this warning. Consult the user's guide for more details about POSIX paths: http://cygwin.com/cygwin-ug-net/using.html#using-pathnames Syntax error: EOF in backquote substitution make: *** [file20681e531a0b.o] Error 2 Warning message: running command 'make -f C:/PROGRA~1/R/R-31~1.1/etc/x64/Makeconf -f C:/PROGRA~1/R/R-31~1.1/share/make/winshlib.mk SHLIB_LDFLAGS='$(SHLIB_CXXLDFLAGS)' SHLIB_LD='$(SHLIB_CXXLD)' SHLIB=file20681e531a0b.dll WIN=64 TCLBIN=64 OBJECTS=file20681e531a0b.o' had status 2 In addition: Warning message: running command 'C:/PROGRA~1/R/R-31~1.1/bin/x64/R CMD SHLIB file20681e531a0b.cpp 2 file20681e531a0b.cpp.err.txt' had status 1 3) If I try to read cygwin-ug-net/using.html they don't seem to address 64 bit Windows. Any suggestions? Thank you, Ray Sachs __ 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] [survMisc]: error message in examples of comp()
Hello, list members, I have tried to contact the maintainer of the survMisc package, but my message could not be delivered. I can't get survMisc package to work as expected or documented, respectively. Maybe one of you has an idea regarding the following problem: When, e.g., trying to executed the first few lines of code in the examples section of comp() (one of the package functions), i.e., data(kidney,package=KMsurv) s1 - survfit(Surv(time=time, event=delta) ~ type, data=kidney) comp(s1) I receive the following error message: Error in parse(text = t2) : text:1:8: unexpected input 1: paste( `` ^ Could this be an encoding-related issue inside the code? Does somebody have an idea where to look for a solution for this problem? Below you'll find my survMisc version and R session info. Thanks for any support! Kind regards -- Gerrit - Dr. Gerrit Eichner Mathematical Institute, Room 212 gerrit.eich...@math.uni-giessen.de Justus-Liebig-University Giessen Tel: +49-(0)641-99-32104 Arndtstr. 2, 35392 Giessen, Germany Fax: +49-(0)641-99-32109http://www.uni-giessen.de/cms/eichner - citation( survMisc) To cite package .survMisc. in publications use: Chris Dardis (2014). survMisc: Miscellaneous functions for survival data.. R package version 0.4.2. http://CRAN.R-project.org/package=survMisc sessionInfo() R version 3.1.1 (2014-07-10) Platform: x86_64-w64-mingw32/x64 (64-bit) locale: [1] LC_COLLATE=German_Germany.1252 LC_CTYPE=German_Germany.1252 [3] LC_MONETARY=German_Germany.1252 LC_NUMERIC=C [5] LC_TIME=German_Germany.1252 attached base packages: [1] splines stats graphics grDevices utils datasets methods [8] base other attached packages: [1] survMisc_0.4.2 rpart_4.1-8 data.table_1.9.2 ggplot2_1.0.0 [5] survival_2.37-7 fortunes_1.5-2 loaded via a namespace (and not attached): [1] colorspace_1.2-4 digest_0.6.4 gam_1.09.1 grid_3.1.1 [5] gtable_0.1.2 km.ci_0.5-2 KMsurv_0.1-5 MASS_7.3-33 [9] munsell_0.4.2plyr_1.8.1 proto_0.3-10 Rcpp_0.11.2 [13] reshape2_1.4 scales_0.2.4 stringr_0.6.2tools_3.1.1 __ 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] [survMisc]: error message in examples of comp()
On Oct 17, 2014, at 1:39 PM, Gerrit Eichner wrote:
Hello, list members,
I have tried to contact the maintainer of the survMisc package, but my
message could not be delivered. I can't get survMisc package to work as
expected or documented, respectively. Maybe one of you has an idea regarding
the following problem:
When, e.g., trying to executed the first few lines of code in the examples
section of comp() (one of the package functions), i.e.,
data(kidney,package=KMsurv)
s1 - survfit(Surv(time=time, event=delta) ~ type, data=kidney)
comp(s1)
I receive the following error message:
Error in parse(text = t2) : text:1:8: unexpected input
1: paste( ``
^
I get the same error on a Mac in an English locale. I'm afraid the code in the
offending function `.getTne` appears somewhat cobbled together. When I try to
debug, I get a single row, 1920 column matrix for n1 which is numeric with all
entries == 1. Object `t1` cannot be found.
{
s - Es - .SD - NULL
n1 - names(dt1)
f1 - function(x) paste(dQuote(paste(x, =, sep = )),
, get(, dQuote(x), ), sep = )
t1 - paste(sapply(n1, f1), sep = ,)
if (length(t1) 1) {
t1[2:length(t1)] - sub(\, \_, t1[2:length(t1)],
fixed = TRUE)
}
t1 - paste(t1, collapse = ,)
t2 - paste(paste(, t1, , sep=''))
# next line throw the error.
p1 - parse(text = t2)
q - quote(eval(p1))
dt1[, `:=`(s, as.factor(eval(q, envir = .SD)))]
stopifnot(attr(model.response(mf), type) == right)
y - data.table(unclass(model.response(mf, numeric)))
I'm afraid I have been o help.
--
David.
Could this be an encoding-related issue inside the code? Does somebody have
an idea where to look for a solution for this problem?
Below you'll find my survMisc version and R session info. Thanks for any
support!
Kind regards -- Gerrit
-
Dr. Gerrit Eichner Mathematical Institute, Room 212
gerrit.eich...@math.uni-giessen.de Justus-Liebig-University Giessen
Tel: +49-(0)641-99-32104 Arndtstr. 2, 35392 Giessen, Germany
Fax: +49-(0)641-99-32109http://www.uni-giessen.de/cms/eichner
-
citation( survMisc)
To cite package .survMisc. in publications use:
Chris Dardis (2014). survMisc: Miscellaneous functions for survival
data.. R package version 0.4.2.
http://CRAN.R-project.org/package=survMisc
sessionInfo()
R version 3.1.1 (2014-07-10)
Platform: x86_64-w64-mingw32/x64 (64-bit)
locale:
[1] LC_COLLATE=German_Germany.1252 LC_CTYPE=German_Germany.1252
[3] LC_MONETARY=German_Germany.1252 LC_NUMERIC=C
[5] LC_TIME=German_Germany.1252
attached base packages:
[1] splines stats graphics grDevices utils datasets methods
[8] base
other attached packages:
[1] survMisc_0.4.2 rpart_4.1-8 data.table_1.9.2 ggplot2_1.0.0
[5] survival_2.37-7 fortunes_1.5-2
loaded via a namespace (and not attached):
[1] colorspace_1.2-4 digest_0.6.4 gam_1.09.1 grid_3.1.1
[5] gtable_0.1.2 km.ci_0.5-2 KMsurv_0.1-5 MASS_7.3-33
[9] munsell_0.4.2plyr_1.8.1 proto_0.3-10 Rcpp_0.11.2
[13] reshape2_1.4 scales_0.2.4 stringr_0.6.2tools_3.1.1
__
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] [survMisc]: error message in examples of comp()
If you execute the command
options(useFancyQuotes=FALSE)
before calling comp(s1) that error should go away.
dQuote() was being used inappropriately - its value depends on the
option useFancyQuotes so should really only be used for messages or
text that machines don't have to parse.
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Fri, Oct 17, 2014 at 5:34 PM, David Winsemius dwinsem...@comcast.net wrote:
On Oct 17, 2014, at 1:39 PM, Gerrit Eichner wrote:
Hello, list members,
I have tried to contact the maintainer of the survMisc package, but my
message could not be delivered. I can't get survMisc package to work as
expected or documented, respectively. Maybe one of you has an idea regarding
the following problem:
When, e.g., trying to executed the first few lines of code in the examples
section of comp() (one of the package functions), i.e.,
data(kidney,package=KMsurv)
s1 - survfit(Surv(time=time, event=delta) ~ type, data=kidney)
comp(s1)
I receive the following error message:
Error in parse(text = t2) : text:1:8: unexpected input
1: paste( ``
^
I get the same error on a Mac in an English locale. I'm afraid the code in
the offending function `.getTne` appears somewhat cobbled together. When I
try to debug, I get a single row, 1920 column matrix for n1 which is numeric
with all entries == 1. Object `t1` cannot be found.
{
s - Es - .SD - NULL
n1 - names(dt1)
f1 - function(x) paste(dQuote(paste(x, =, sep = )),
, get(, dQuote(x), ), sep = )
t1 - paste(sapply(n1, f1), sep = ,)
if (length(t1) 1) {
t1[2:length(t1)] - sub(\, \_, t1[2:length(t1)],
fixed = TRUE)
}
t1 - paste(t1, collapse = ,)
t2 - paste(paste(, t1, , sep=''))
# next line throw the error.
p1 - parse(text = t2)
q - quote(eval(p1))
dt1[, `:=`(s, as.factor(eval(q, envir = .SD)))]
stopifnot(attr(model.response(mf), type) == right)
y - data.table(unclass(model.response(mf, numeric)))
I'm afraid I have been o help.
--
David.
Could this be an encoding-related issue inside the code? Does somebody have
an idea where to look for a solution for this problem?
Below you'll find my survMisc version and R session info. Thanks for any
support!
Kind regards -- Gerrit
-
Dr. Gerrit Eichner Mathematical Institute, Room 212
gerrit.eich...@math.uni-giessen.de Justus-Liebig-University Giessen
Tel: +49-(0)641-99-32104 Arndtstr. 2, 35392 Giessen, Germany
Fax: +49-(0)641-99-32109http://www.uni-giessen.de/cms/eichner
-
citation( survMisc)
To cite package .survMisc. in publications use:
Chris Dardis (2014). survMisc: Miscellaneous functions for survival
data.. R package version 0.4.2.
http://CRAN.R-project.org/package=survMisc
sessionInfo()
R version 3.1.1 (2014-07-10)
Platform: x86_64-w64-mingw32/x64 (64-bit)
locale:
[1] LC_COLLATE=German_Germany.1252 LC_CTYPE=German_Germany.1252
[3] LC_MONETARY=German_Germany.1252 LC_NUMERIC=C
[5] LC_TIME=German_Germany.1252
attached base packages:
[1] splines stats graphics grDevices utils datasets methods
[8] base
other attached packages:
[1] survMisc_0.4.2 rpart_4.1-8 data.table_1.9.2 ggplot2_1.0.0
[5] survival_2.37-7 fortunes_1.5-2
loaded via a namespace (and not attached):
[1] colorspace_1.2-4 digest_0.6.4 gam_1.09.1 grid_3.1.1
[5] gtable_0.1.2 km.ci_0.5-2 KMsurv_0.1-5 MASS_7.3-33
[9] munsell_0.4.2plyr_1.8.1 proto_0.3-10 Rcpp_0.11.2
[13] reshape2_1.4 scales_0.2.4 stringr_0.6.2tools_3.1.1
__
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-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] Comparing values of different columns: Error: level sets of factors are different
Hi, R listers,
I��m trying to compare a value of a row in a column to values of previous rows
in another column in a loop. ��i�� is just from first row to the last row. j is
an another looping controller indicating the rows that row[i] will be compared
to and j will be rows before row[i]. I want to compare phone_1[i] with
phone_2[j] and vise versa.
for (i in 1: nrow(df)) {
if (df$Incremental[i] == 1) {mark[i] - 1}
else {for (j in (i �C df$Incremental[i] + 1) : (i - 1)) {
if ((df$Phone_1[i] != df$Phone_1[i] == df$Phone_2[j]) |
(df$Phone_2[i] != df$Phone_2[i] == df$Phone_1[j])) {
mark[i] - mark[j]}
else {mark[i] - mark[i-1] + 1}
}
}
However I��m getting an error with phone_1[i] and phone_2[j], indicating that
��level sets of factors are different��. I��m not sure what I need to do to fix
it.
Thanks!
[[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] Comparing values of different columns: Error: level sets of factors are different
On Oct 17, 2014, at 1:21 PM, Chichi Shu wrote:
Hi, R listers,
I¡¯m trying to compare a value of a row in a column to values of previous
rows in another column in a loop. ¡°i¡± is just from first row to the last
row. j is an another looping controller indicating the rows that row[i] will
be compared to and j will be rows before row[i]. I want to compare phone_1[i]
with phone_2[j] and vise versa.
for (i in 1: nrow(df)) {
if (df$Incremental[i] == 1) {mark[i] - 1}
else {for (j in (i ¨C df$Incremental[i] + 1) : (i - 1)) {
if ((df$Phone_1[i] != df$Phone_1[i] == df$Phone_2[j]) |
(df$Phone_2[i] != df$Phone_2[i] == df$Phone_1[j])) {
mark[i] - mark[j]}
else {mark[i] - mark[i-1] + 1}
}
}
However I¡¯m getting an error with phone_1[i] and phone_2[j], indicating that
¡°level sets of factors are different¡±. I¡¯m not sure what I need to do to
fix it.
Factor fixing. An ancient, aRcane sport.
Perhaps:
lev2 - unique( c( levels(phone_1), levels(phone_2) ) )
phone_1 - factor(phone_1, levels=lev2)
phone_2 - factor(phone_2, levels=lev2)
Thanks!
[[alternative HTML version deleted]]
And HTML posting is not well supported. Learn to post in that most ancient of
computer tongues, plain text.
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-es] [Grupo Usuarios de R de Madrid]: Siguiente reunión 15-octubre....(*Material Disponible*)
Hola, Ya está disponible todo el material que presentamos durante la reunión del pasado miércoles. Los detalles los podéis encontrar aquí: http://madrid.r-es.org/miercoles-15-de-octubre-2014/ Saludos, Carlos Ortega www.qualityexcellence.es El 15 de octubre de 2014, 10:06, Carlos Ortega c...@qualityexcellence.es escribió: Hola, Un a breve nota para recordar que hoy tendremos la reunión del Grupo de Usuarios de R de Madrid. - La agenda es la misma que anunciamos. - La reunión será en la Facultad de Educación (UNED) en la sala Fernández Huertas, planta 1ª. Los detalles de la convocatoria los podéis encontrar en el portal del grupo: http://madrid.r-es.org/ Gracias, Carlos Ortega www.qualityexcellence.es El 8 de octubre de 2014, 10:47, Carlos Ortega c...@qualityexcellence.es escribió: Buenas a todos, Ya tenemos programada la agenda de la siguiente reunión del grupo. De forma escueta: - *Presentaciones:* - Pedro Valverde https://www.linkedin.com/profile/view?id=232087352authType=NAME_SEARCHauthToken=qJG6locale=es_ESsrchid=51802411410559294801srchindex=1srchtotal=139trk=vsrp_people_res_nametrkInfo=VSRPsearchId%3A51802411410559294801%2CVSRPtargetId%3A232087352%2CVSRPcmpt%3Aprimary : “Modelización de los salarios en España”. - Pedro Concejero: https://www.linkedin.com/profile/view?id=8293681authType=NAME_SEARCHauthToken=VdNxlocale=en_USsrchid=51802411412718027835srchindex=1srchtotal=3trk=vsrp_people_res_nametrkInfo=VSRPsearchId%3A51802411412718027835%2CVSRPtargetId%3A8293681%2CVSRPcmpt%3Aprimary“decidiR – teoría del prospecto – paquete pt http://cran.r-project.org/web/packages/pt/index.html “. - Carlos Ortega: https://www.linkedin.com/in/carlosortegafernandez “Otros diveRtimentos sobre Money-is-a-gas”. - *Píldoras:* - Jorge Ayuso https://www.linkedin.com/profile/view?id=229197095authType=NAME_SEARCHauthToken=GvWulocale=es_ESsrchid=51802411412718420247srchindex=1srchtotal=33trk=vsrp_people_res_nametrkInfo=VSRPsearchId%3A51802411412718420247%2CVSRPtargetId%3A229197095%2CVSRPcmpt%3Aprimary : “OpenCpu http://cran.r-project.org/web/packages/opencpu/index.html y Docker https://www.docker.com/“. Más detalles aquí: http://madrid.r-es.org/miercoles-15-de-octubre-2014/ -- Saludos, Carlos Ortega www.qualityexcellence.es -- Saludos, Carlos Ortega www.qualityexcellence.es -- Saludos, Carlos Ortega www.qualityexcellence.es [[alternative HTML version deleted]] ___ R-help-es mailing list R-help-es@r-project.org https://stat.ethz.ch/mailman/listinfo/r-help-es
[R-es] Fwd: Question about GEE and interaction plot
Hello all! I am trying to analyze a dataset using the GEE. This is what I have so far C=read.table(Cnarinofinal4.txt,header=T) attach(C) names(C) library(geepack) Form - formula(Conteo~Grupo+Tiempo+Comportamiento+Comportamiento*Grupo+Comportamiento*Hora) M4 - geeglm(Form,family = poisson, id =Individuo, corstr = exchangeable) anova(M4) 1) I want to know if someboy knows how to perform post hoc comparisons for significant terms, in this case I have a significance result for the interaction Comportamiento*Hora and the factor Comportamiento. 2) I want to perform and interaction plot, I tried with the following code: interaction.plot(C$Comportamiento,C$Tiempo,C$Conteo) Nevertheless, I would like to plot the graph in a decreasing order by the mean and also add the error bars. For the first approach I do not have available info, and for the second one I found that data must be sorted before using the function summarySE from the ggplot2 library, nevertheless it does not seem to work anymore. Any help you can provide will be welcome! HoraIndividuo Comportamiento Tiempo Grupo Conteo 0 1 Exploración NOCHE MACHO 1 0 1 interacciones NOCHE MACHO 0 0 1 Mantenimiento NOCHE MACHO 0 0 1 AlimentaciónNOCHE MACHO 5 0 1 ReproducciónNOCHE MACHO 0 0 1 Reposo NOCHE MACHO 2 0 1 Exacavación NOCHE MACHO 0 0 2 Exploración NOCHE MACHO 0 0 2 interacciones NOCHE MACHO 0 0 2 Mantenimiento NOCHE MACHO 0 0 2 AlimentaciónNOCHE MACHO 0 0 2 ReproducciónNOCHE MACHO 0 0 2 Reposo NOCHE MACHO 7 0 2 Exacavación NOCHE MACHO 0 0 3 Exploración NOCHE MACHO 1 0 3 interacciones NOCHE MACHO 0 0 3 Mantenimiento NOCHE MACHO 0 0 3 AlimentaciónNOCHE MACHO 0 0 3 ReproducciónNOCHE MACHO 0 0 3 Reposo NOCHE MACHO 7 0 3 Exacavación NOCHE MACHO 0 0 4 Exploración NOCHE MACHO 3 0 4 interacciones NOCHE MACHO 0 0 4 Mantenimiento NOCHE MACHO 0 0 4 AlimentaciónNOCHE MACHO 1 0 4 ReproducciónNOCHE MACHO 0 0 4 Reposo NOCHE MACHO 5 0 4 Exacavación NOCHE MACHO 0 0 5 Exploración NOCHE MACHO 1 0 5 interacciones NOCHE MACHO 0 0 5 Mantenimiento NOCHE MACHO 0 0 5 AlimentaciónNOCHE MACHO 1 0 5 ReproducciónNOCHE MACHO 0 0 5 Reposo NOCHE MACHO 8 0 5 Exacavación NOCHE MACHO 0 0 6 Exploración NOCHE MACHO 3 0 6 interacciones NOCHE MACHO 0 0 6 Mantenimiento NOCHE MACHO 0 0 6 AlimentaciónNOCHE MACHO 0 0 6 ReproducciónNOCHE MACHO 0 0 6 Reposo NOCHE MACHO 2 0 6 Exacavación NOCHE MACHO 0 0 7 Exploración NOCHE HEMBRA 3 0 7 interacciones NOCHE HEMBRA 0 0 7 Mantenimiento NOCHE HEMBRA 0 0 7 AlimentaciónNOCHE HEMBRA 0 0 7 ReproducciónNOCHE HEMBRA 0 0 7 Reposo NOCHE HEMBRA 5 0 7 Exacavación NOCHE HEMBRA 0 0 8 Exploración NOCHE HEMBRA 2 0 8 interacciones NOCHE HEMBRA 0 0 8 Mantenimiento NOCHE HEMBRA 0 0 8 AlimentaciónNOCHE HEMBRA 0 0 8 ReproducciónNOCHE HEMBRA 0 0 8 Reposo NOCHE HEMBRA 2 0 8 Exacavación NOCHE HEMBRA 0 0 9 Exploración NOCHE HEMBRA 2 0 9 interacciones NOCHE HEMBRA 0 0 9 Mantenimiento NOCHE HEMBRA 0 0 9 AlimentaciónNOCHE HEMBRA 0 0 9 ReproducciónNOCHE HEMBRA 0 0 9 Reposo NOCHE HEMBRA 5 0 9 Exacavación NOCHE HEMBRA 0 0 10 Exploración NOCHE HEMBRA 6 0 10 interacciones NOCHE HEMBRA 0 0 10 Mantenimiento NOCHE HEMBRA 0 0 10 AlimentaciónNOCHE HEMBRA 0 0 10 ReproducciónNOCHE HEMBRA 0 0 10 Reposo NOCHE HEMBRA 2 0 10 Exacavación NOCHE HEMBRA 0 0 11 Exploración NOCHE HEMBRA 1 0 11 interacciones NOCHE HEMBRA 0 0 11 Mantenimiento NOCHE HEMBRA 0 0 11 AlimentaciónNOCHE HEMBRA 2 0 11 ReproducciónNOCHE HEMBRA 0 0 11 Reposo NOCHE HEMBRA 4 0 11 Exacavación NOCHE HEMBRA 0 0 12 Exploración NOCHE HEMBRA 2 0 12 interacciones NOCHE HEMBRA 0 0 12
Re: [R-es] Ayuda prueba de significancia funcion nls()
Hola, ¿qué tal? Hu... no conozco ese modelo que tratas de ajustar demasiado bien, pero creo que el parámetro A significa demasiadas cosas (y depende de demasiadas cosas) a la vez. Por ejemplo, seguro, está muy correlacionado con alfa porque opera como una especie de término independiente de lo que hay dentro de exp y su valor depende mucho, por ejemplo, de si normalizas o no la edad. Yo buscaría una formulación o reparametrización alternativa. Pero en este caso en particular, parecía claro que el problema estaba en la falta de ajuste en esa cola poco relevante. Un saludo, Carlos J. Gil Bellosta http://www.datanalytics.com El día 16 de octubre de 2014, 22:12, Boris Polanco borisc...@gmail.com escribió: Hola, he vuelto a realizar el ajuste con la sugerencia que me diste y ha funcionado, pero mi pregunta es si para con otro modelo similar no me sale un parámetro no significativo, cómo interpreto el no rechazo de la hipótesis nula? saludos El 15 de octubre de 2014, 17:44, Carlos J. Gil Bellosta c...@datanalytics.com escribió: Hola, ¿qué tal? Yo eliminaría las observaciones por encima de los noventaitantos (¿o cien?) años. Son solo ruido, como puede observarse. Estarán basadas en un porcentaje nimio de la muestra. Igual entonces cambian las cosas. Un saludo, Carlos J. Gil Bellosta http://www.datanalytics.com El día 15 de octubre de 2014, 23:18, Boris Polanco borisc...@gmail.com escribió: Hola, estoy haciendo un estudio de la desaceleración de la mortalidad y tengo dudas con la función nls(), adjunto un archivo donde he redactado mi inquietud. saludos cordiales Boris Polanco ___ R-help-es mailing list R-help-es@r-project.org https://stat.ethz.ch/mailman/listinfo/r-help-es ___ R-help-es mailing list R-help-es@r-project.org https://stat.ethz.ch/mailman/listinfo/r-help-es
