[R] different results between cor and ccf
Dear listers, I am working on a time series but find that for a given non-zero time lag correlations obtained by ccf and cor are different. x <- c(0.85472102802704641, 1.6008990694641689, 2.5019632258894835, 2.514654801253164, 3.3359198688206368, 3.5401357138398208, 2.6304117871193538, 3.6694074965420009, 3.9125153101706776, 4.4006592535478566, 3.0208991912866829, 2.959090589344433, 3.8434635568566056, 2.1683644330520457, 2.3060571563512973, 1.4680350663043942, 2.0346918622459054, 2.3674524446877538) y <- c(2.3085729270534765, 2.0809088217491416, 1.6249456563631131, 1.513338933177, 0.66754156827555422, 0.3080839731181978, 0.5265304299394, 0.89070463020837132, 0.71600791432232669, 0.82152341002975027, 0.22200290782700527, 0.6608410635137173, 0.90715232876618945, 0.45624062770725898, 0.35074487486980244, 1.1681750562971052, 1.6976462236079737, 0.88950230250556417) cc<-ccf(x,y) > cc Autocorrelations of series ‘X’, by lag -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 0.098 0.139 0.127 -0.043 -0.049 0.069 -0.237 -0.471 -0.668 -0.595 -0.269 -0.076 3 4 5 6 7 8 9 -0.004 0.123 0.272 0.283 0.401 0.435 0.454 cor(x,y) [1] -0.5948694 So far so good, but when I lag one of the series, I cannot find the same correlation as with ccf > cor(x[1:(length(x)-1)],y[2:length(y)]) [1] -0.7903428 ... where I expect -0.668 based on ccf Can anyone explain why ? Best, Patrick [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] glm.nb and Error in x[good, , drop = FALSE] * w : non-conformable arrays
Many thanks Ivan ! This is fairly clear to me, now... When I dumped the data.frame, I found strange to have a "table" declaration for deg, but was not able to judge if it was a problem or not (I would have expected something as "numeric") Your workaround is fine to me (I do not need to carry on with table attributes). Best, Patrick Le 21/04/2023 à 10:08, Ivan Krylov a écrit : > On Fri, 21 Apr 2023 09:02:37 +0200 > Patrick Giraudoux wrote: > >> I meet an error with glm.nb that I cannot explain the origin (and >> find a fix). The model I want to fit is the following: >> >> library(MASS) >> >> glm.nb(deg~offset(log(durobs))+zone,data=db) >> >> and the data.frame is dumped below. > Thank you for providing both the code and a small piece of data that > reproduces the error! > > (It almost worked. Your mailer automatically generated a plain text > version of the e-mail and put Unicode non-breaking spaces in there. R > considers it a syntax error to encounter any of the various Unicode > space-like characters outside string literals.) > >> deg = structure(c(0, 1, 0, 3, 0, 1, 0, 2, 1, 0, 3, 0, 0, 0, 4, 1, 0, >> 0, 0, 0, 4, 0, 0, 0, 4, 3, 2, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, >> 0, 3, 2, 3, 0, 1, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 2, >> 0, 0, 0, 2, 1, 0, 2, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 2, >> 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 2, 1, 0, 0, 1, 3, 2, >> 1, 2, 0, 0, 0, 0, 0, 1, 0, 0, 2, 0, 0, 0, 0, 0, 1, 2, 0, 0, 0, 0, 2, >> 1, 1, 0), dim = 135L, class = "table") > The problem is that `deg` is a table, which ends up making the > effective weights a table too. tables are arrays, and element-wise > product rules are stricter for them than for plain matrices. The code > makes use of the ability to take an element-wise product between a > matrix and a vector of the same length as the number of rows in the > matrix: > > matrix(1:12, 4) * 1:4 # works > matrix(1:12, 4) * as.array(1:4) # results in the same error > > # the right way to take products with an array is to make sure that the > # shapes match exactly > matrix(1:12, 4) * as.array(cbind(1:4, 1:4, 1:4)) > > One possible solution is to to remove all attributes from db$deg: > > db$deg <- as.vector(db$deg) > > This way the values of the expressions involved in glm.fit end up being > of the expected type, and the function completes successfully. > [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] glm.nb and Error in x[good, , drop = FALSE] * w : non-conformable arrays
Dear Listers,
I meet an error with glm.nb that I cannot explain the origin (and find a
fix). The model I want to fit is the following:
library(MASS)
glm.nb(deg~offset(log(durobs))+zone,data=db)
and the data.frame is dumped below.
Has anyone an idea about what the trouble comes from ? (except computing
leads to a non-conformable array somewhere... the question is why;
fitting goes through without any problem eg with a Poisson link)
Best,
Patrick
db <-
structure(list(deg = structure(c(0, 1, 0, 3, 0, 1, 0, 2, 1, 0,
3, 0, 0, 0, 4, 1, 0, 0, 0, 0, 4, 0, 0, 0, 4, 3, 2, 0, 0, 0, 0,
0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 3, 2, 3, 0, 1, 1, 0, 0, 0, 0, 1,
0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 2, 0, 0, 0, 2, 1, 0, 2, 0, 1, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 2, 1, 0, 1, 0, 0, 0, 0, 0,
0, 0, 0, 0, 1, 0, 0, 0, 2, 1, 0, 0, 1, 3, 2, 1, 2, 0, 0, 0, 0,
0, 1, 0, 0, 2, 0, 0, 0, 0, 0, 1, 2, 0, 0, 0, 0, 2, 1, 1, 0), dim = 135L,
class = "table"),
durobs = c(371, 371, 371, 371, 371, 371, 239, 266, 234, 71,
436, 407, 407, 414, 415, 418, 415, 329, 414, 414, 415, 330,
435, 436, 210, 436, 214, 436, 436, 210, 434, 438, 438, 402,
402, 289, 264, 264, 434, 435, 434, 434, 434, 434, 434, 427,
427, 427, 328, 422, 291, 412, 221, 417, 416, 416, 79, 322,
213, 440, 434, 462, 397, 457, 419, 406, 316, 392, 392, 392,
392, 392, 452, 386, 399, 305, 240, 404, 226, 226, 381, 385,
392, 388, 388, 391, 396, 392, 385, 385, 385, 237, 378, 378,
378, 381, 126, 315, 379, 314, 185, 313, 312, 301, 312, 312,
310, 310, 307, 306, 304, 455, 472, 472, 466, 467, 334, 565,
429, 429, 425, 422, 421, 419, 417, 417, 410, 405, 195, 422,
419, 419, 426, 426, 442), zone = c("MO1", "MO1", "MO1", "MO1",
"MO1", "MO1", "MO1", "MO1", "MO1", "MO1", "MO1", "MO1", "MO1",
"MO1", "MO1", "MO1", "MO1", "MO1", "MO1", "MO1", "MO1", "MO1",
"MO1", "MO1", "MO1", "MO1", "MO1", "MO1", "MO1", "MO1", "MO1",
"MO1", "MO1", "MO1", "MO1", "MO1", "MO1", "MO1", "MO1", "MO1",
"MO1", "MO1", "MO1", "MO1", "MO1", "MO1", "MO1", "MO1", "MO1",
"MO1", "MO1", "MO1", "MO1", "MO1", "MO1", "MO1", "MO1", "MO1",
"MO1", "MO1", "MO1", "MO1", "MO1", "MO1", "MO1", "MO1", "MO1",
"MO2", "MO2", "MO2", "MO2", "MO2", "MO2", "MO2", "MO2", "MO2",
"MO2", "MO2", "MO2", "MO2", "MO2", "MO2", "MO2", "MO2", "MO2",
"MO2", "MO2", "MO2", "MO2", "MO2", "MO2", "MO2", "MO2", "MO2",
"MO2", "MO2", "MO2", "MO2", "MO2", "MO2", "MO2", "MO2", "MO2",
"MO2", "MO2", "MO2", "MO2", "MO2", "MO2", "MO2", "MO2", "MO2",
"MO2", "MO2", "MO2", "MO2", "MO2", "MO2", "MO2", "MO2", "MO2",
"MO2", "MO2", "MO2", "MO2", "MO2", "MO2", "MO2", "MO2", "MO2",
"MO2", "MO2", "MO2", "MO2", "MO2")), row.names = c(NA, 135L
), class = "data.frame")
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Re: [R] readxl, read_excel: how colon (:) is read ?
This can be made using the TEXT (TEXTE in the French version) function
of Excel, hence:
TEXT(M2;"HH:MM")
Changes the time into text, and it can be imported from R as wanted.
Le 01/04/2022 à 08:34, Patrick Giraudoux a écrit :
> Absolutely correct ! I checked in Excel and when I change the format
> to "text", then I get in Excel the same fractional numbers as those
> obtained importing text from R... Hence the issue comes from Excel
> itself. Will find a way to change this format to text in Excel without
> avoiding such conversion...
> Thanks Andrew !
>
> Le 01/04/2022 à 08:26, Andrew Simmons a écrit :
>> Probably (but not entirely sure), Excel is storing your text as a
>> number of days, so 13:38 is a little more than half a day. Open your
>> spreadsheet in excel and save those columns as text instead of times,
>> that (should) fix your issue.
>>
>> On Fri, Apr 1, 2022, 02:12 Patrick Giraudoux
>> wrote:
>>
>> I have a unexpected behaviour reading times with colon from an Excel
>> file, using the package readxl.
>>
>> In an Excel sheet, I have a column with times in hours:minutes, e.g:
>>
>> Arrival_time
>> 13:39
>> 13:51
>>
>> When read from R with readxl::read_excel, this gives a tibble column
>> with full date by defaut being the last day of 1899. OK. Why not,
>> I know
>> that POSIX variables are starting in 1900 after R doc (however I
>> wonder
>> why here the defaut is one day before January 1, 1900
>>
>> > tmp$Arrival_time [1] "1899-12-31 13:39:00 UTC" "1899-12-31
>> 13:51:00 UTC"
>>
>> Well, this is not exactly what I want to. I do not care about the
>> year
>> and the day... Therefore I decided to import this column as "text"
>> explicitely (in order to manage it within R then). And this is
>> what I
>> get now:
>>
>> >
>>
>> read_excel("saisie_data_durban_rapaces_LPO.xlsx",sheet=2,col_types="text")
>> > tmp$Arrival_time [1] "0.568750009" "0.577083328"
>>
>> Can someone tell me what happens ?
>>
>> I would really appreciate to understand the trick...
>>
>>
>> [[alternative HTML version deleted]]
>>
>> __
>> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
>> https://stat.ethz.ch/mailman/listinfo/r-help
>> PLEASE do read the posting guide
>> http://www.R-project.org/posting-guide.html
>> <http://www.R-project.org/posting-guide.html>
>> and provide commented, minimal, self-contained, reproducible code.
>>
>
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Re: [R] readxl, read_excel: how colon (:) is read ?
Le 01/04/2022 à 08:40, Jeff Newmiller a écrit : > Both R and Excel assume a date is associated with every time object. In > Excel, when you show a date it is an integer number of days since 1899-12-31 > (due to a mistake made early in programming it). Whenever you show a time, it > it merely displaying the time portion (fraction of a day) of a date/time. The > date part of that value may or may not be 1899-12-31. > > With this in mind, you are tilting at windmills hoping to import a "pure > time" because no such thing exists in either program. You can choose to > render a `POSIXct` as showing only the time portion when you convert it to > character if you so choose. Thanks for the infos. Yes, this is exactly what I did yesterday with POSIXct > POSIXlt to go ahead. However I wanted to understand fully what happened, hence the call to the list. Jeff and Andrew, now eveything is clear to me thanks to you... [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] strptime, date and conversion of week number into POSIX
Thanks Uwe and Bert,
I got the essential now, and can manage. Date handling stays quite a
challenge with a variable number of weeks in a year, but I can
understand why. Means eye-control (or NA detection) of strptime
conversion stays necessary...
Best,
Patrick
Le 22/02/2021 à 17:09, Uwe Ligges a écrit :
> That monday does not exist. FOr the week before:
>
> strptime(paste0("2020-52","-1"),format="%Y-%W-%u")
> [1] "2020-12-28"
>
> One week later is no longer in 2020, so there is no 53th week.
>
> Best,
> Uwe Ligges
>
>
>
>
>
> On 22.02.2021 16:15, Patrick Giraudoux wrote:
>> Sorry to answer to myself, but the format was clearly incorrect in the
>> previous post. It should read, refering to the 1th day of the week:
>>
>> strptime(paste0(mydate,"-1"),format="%Y-%W-%u")
>>
>> It converts better, but with a NA on week 53
>>
>>> strptime(paste0(pays$year_week,"-1"),format="%Y-%W-%u")
>> [1] "2020-01-06 CET" "2020-01-13 CET" "2020-01-20 CET"
>> "2020-01-27 CET"
>> [5] "2020-02-03 CET" "2020-02-10 CET" "2020-02-17 CET"
>> "2020-02-24 CET"
>> [9] "2020-03-02 CET" "2020-03-09 CET" "2020-03-16 CET"
>> "2020-03-23 CET"
>> [13] "2020-03-30 CEST" "2020-04-06 CEST" "2020-04-13 CEST"
>> "2020-04-20 CEST"
>> [17] "2020-04-27 CEST" "2020-05-04 CEST" "2020-05-11 CEST"
>> "2020-05-18 CEST"
>> [21] "2020-05-25 CEST" "2020-06-01 CEST" "2020-06-08 CEST"
>> "2020-06-15 CEST"
>> [25] "2020-06-22 CEST" "2020-06-29 CEST" "2020-07-06 CEST"
>> "2020-07-13 CEST"
>> [29] "2020-07-20 CEST" "2020-07-27 CEST" "2020-08-03 CEST"
>> "2020-08-10 CEST"
>> [33] "2020-08-17 CEST" "2020-08-24 CEST" "2020-08-31 CEST"
>> "2020-09-07 CEST"
>> [37] "2020-09-14 CEST" "2020-09-21 CEST" "2020-09-28 CEST"
>> "2020-10-05 CEST"
>> [41] "2020-10-12 CEST" "2020-10-19 CEST" "2020-10-26 CET" "2020-11-02
>> CET"
>> [45] "2020-11-09 CET" "2020-11-16 CET" "2020-11-23 CET" "2020-11-30
>> CET"
>> [49] "2020-12-07 CET" "2020-12-14 CET" "2020-12-21 CET" "2020-12-28
>> CET"
>> [53] NA "2021-01-04 CET" "2021-01-11 CET" "2021-01-18
>> CET"
>> [57] "2021-01-25 CET" "2021-02-01 CET" "2021-02-08 CET"
>> Warning message:
>> In strptime(paste0(pays$year_week, "-1"), format = "%Y-%W-%u") :
>> (0-based) yday 369 in year 2020 is invalid
>>
>>
>> Any idea on how to handle this ?
>>
>>
>>
>>
>> Le 22/02/2021 à 15:26, Patrick Giraudoux a écrit :
>>>
>>> Dear all,
>>>
>>> I have a trouble trying to convert dates given in character to POSIX.
>>> The date is expressed as a year then the week number e.g. "2020-01"
>>> (first week of 2020). I thought is can be converted as following:
>>>
>>> strptime(mydate,format="%Y-%W")
>>>
>>> %W refering to the week of the year as decimal number (00–53) using
>>> Monday as the first day of week (and typically with the first Monday
>>> of the year as day 1 of week 1), as indicated in the doc.
>>>
>>> However, I got this result, with the month fixed to 02 (february) and
>>> day 22 (only the year is converted correctly):
>>>
>>> strptime(mydate,format="%Y-%W") [1] "2020-02-22 CET" "2020-02-22 CET"
>>> "2020-02-22 CET" "2020-02-22 CET" [5] "2020-02-22 CET" "2020-02-22
>>> CET" "2020-02-22 CET" "2020-02-22 CET" [9] "2020-02-22 CET"
>>> "2020-02-22 CET" "2020-02-22 CET" "2020-02-22 CET" [13] "2020-02-22
>>> CET" "2020-02-22 CET" "2020-02-22 CET" "2020-02-22 CET" [17]
>>> "2020-02-22 CET" "2020-02-22 CET" "2020-02-22 CET" "2020-02-22 CET"
>>> [21] "2020-02-22 CET" "2020-02-22 CET" "2020-02-22 CET" "2020-02-22
>>> CET&qu
Re: [R] strptime, date and conversion of week number into POSIX
Sorry to answer to myself, but the format was clearly incorrect in the
previous post. It should read, refering to the 1th day of the week:
strptime(paste0(mydate,"-1"),format="%Y-%W-%u")
It converts better, but with a NA on week 53
> strptime(paste0(pays$year_week,"-1"),format="%Y-%W-%u")
[1] "2020-01-06 CET" "2020-01-13 CET" "2020-01-20 CET" "2020-01-27 CET"
[5] "2020-02-03 CET" "2020-02-10 CET" "2020-02-17 CET" "2020-02-24 CET"
[9] "2020-03-02 CET" "2020-03-09 CET" "2020-03-16 CET" "2020-03-23 CET"
[13] "2020-03-30 CEST" "2020-04-06 CEST" "2020-04-13 CEST" "2020-04-20 CEST"
[17] "2020-04-27 CEST" "2020-05-04 CEST" "2020-05-11 CEST" "2020-05-18 CEST"
[21] "2020-05-25 CEST" "2020-06-01 CEST" "2020-06-08 CEST" "2020-06-15 CEST"
[25] "2020-06-22 CEST" "2020-06-29 CEST" "2020-07-06 CEST" "2020-07-13 CEST"
[29] "2020-07-20 CEST" "2020-07-27 CEST" "2020-08-03 CEST" "2020-08-10 CEST"
[33] "2020-08-17 CEST" "2020-08-24 CEST" "2020-08-31 CEST" "2020-09-07 CEST"
[37] "2020-09-14 CEST" "2020-09-21 CEST" "2020-09-28 CEST" "2020-10-05 CEST"
[41] "2020-10-12 CEST" "2020-10-19 CEST" "2020-10-26 CET" "2020-11-02 CET"
[45] "2020-11-09 CET" "2020-11-16 CET" "2020-11-23 CET" "2020-11-30 CET"
[49] "2020-12-07 CET" "2020-12-14 CET" "2020-12-21 CET" "2020-12-28 CET"
[53] NA"2021-01-04 CET" "2021-01-11 CET" "2021-01-18 CET"
[57] "2021-01-25 CET" "2021-02-01 CET" "2021-02-08 CET"
Warning message:
In strptime(paste0(pays$year_week, "-1"), format = "%Y-%W-%u") :
(0-based) yday 369 in year 2020 is invalid
Any idea on how to handle this ?
Le 22/02/2021 à 15:26, Patrick Giraudoux a écrit :
>
> Dear all,
>
> I have a trouble trying to convert dates given in character to POSIX.
> The date is expressed as a year then the week number e.g. "2020-01"
> (first week of 2020). I thought is can be converted as following:
>
> strptime(mydate,format="%Y-%W")
>
> %W refering to the week of the year as decimal number (00–53) using
> Monday as the first day of week (and typically with the first Monday
> of the year as day 1 of week 1), as indicated in the doc.
>
> However, I got this result, with the month fixed to 02 (february) and
> day 22 (only the year is converted correctly):
>
> strptime(mydate,format="%Y-%W") [1] "2020-02-22 CET" "2020-02-22 CET"
> "2020-02-22 CET" "2020-02-22 CET" [5] "2020-02-22 CET" "2020-02-22
> CET" "2020-02-22 CET" "2020-02-22 CET" [9] "2020-02-22 CET"
> "2020-02-22 CET" "2020-02-22 CET" "2020-02-22 CET" [13] "2020-02-22
> CET" "2020-02-22 CET" "2020-02-22 CET" "2020-02-22 CET" [17]
> "2020-02-22 CET" "2020-02-22 CET" "2020-02-22 CET" "2020-02-22 CET"
> [21] "2020-02-22 CET" "2020-02-22 CET" "2020-02-22 CET" "2020-02-22
> CET" [25] "2020-02-22 CET" "2020-02-22 CET" "2020-02-22 CET"
> "2020-02-22 CET" [29] "2020-02-22 CET" "2020-02-22 CET" "2020-02-22
> CET" "2020-02-22 CET" [33] "2020-02-22 CET" "2020-02-22 CET"
> "2020-02-22 CET" "2020-02-22 CET" [37] "2020-02-22 CET" "2020-02-22
> CET" "2020-02-22 CET" "2020-02-22 CET" [41] "2020-02-22 CET"
> "2020-02-22 CET" "2020-02-22 CET" "2020-02-22 CET" [45] "2020-02-22
> CET" "2020-02-22 CET" "2020-02-22 CET" "2020-02-22 CET" [49]
> "2020-02-22 CET" "2020-02-22 CET" "2020-02-22 CET" "2020-02-22 CET"
> [53] "2020-02-22 CET" "2021-02-22 CET" "2021-02-22 CET" "2021-02-22
> CET" [57] "2021-02-22 CET" "2021-02-22 CET" "2021-02-22 CET"
>
> You'll find below a dump of "mydate" you can copy and paster if you
> need a try
>
> Any hint welcome...
>
> Best,
>
> Patrick
>
> mydate <-
> c("2020-01", "2020-02", "2020-03", "2020-04", "2020-05", "2020-06",
> "2020-07", "2020-08", "2020-09", "2020-10", "2020-11", "2020-12",
> "2020-13", "2020-14", "2020-15", "2020-16", "2020-17", "2020-18",
> "2020-19", "2020-20", "2020-21", "2020-22", "2020-23", "2020-24",
> "2020-25", "2020-26", "2020-27", "2020-28", "2020-29", "2020-30",
> "2020-31", "2020-32", "2020-33", "2020-34", "2020-35", "2020-36",
> "2020-37", "2020-38", "2020-39", "2020-40", "2020-41", "2020-42",
> "2020-43", "2020-44", "2020-45", "2020-46", "2020-47", "2020-48",
> "2020-49", "2020-50", "2020-51", "2020-52", "2020-53", "2021-01",
> "2021-02", "2021-03", "2021-04", "2021-05", "2021-06")
>
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[R] strptime, date and conversion of week number into POSIX
Dear all,
I have a trouble trying to convert dates given in character to POSIX.
The date is expressed as a year then the week number e.g. "2020-01"
(first week of 2020). I thought is can be converted as following:
strptime(mydate,format="%Y-%W")
%W refering to the week of the year as decimal number (00–53) using
Monday as the first day of week (and typically with the first Monday of
the year as day 1 of week 1), as indicated in the doc.
However, I got this result, with the month fixed to 02 (february) and
day 22 (only the year is converted correctly):
strptime(mydate,format="%Y-%W") [1] "2020-02-22 CET" "2020-02-22 CET"
"2020-02-22 CET" "2020-02-22 CET" [5] "2020-02-22 CET" "2020-02-22 CET"
"2020-02-22 CET" "2020-02-22 CET" [9] "2020-02-22 CET" "2020-02-22 CET"
"2020-02-22 CET" "2020-02-22 CET" [13] "2020-02-22 CET" "2020-02-22 CET"
"2020-02-22 CET" "2020-02-22 CET" [17] "2020-02-22 CET" "2020-02-22 CET"
"2020-02-22 CET" "2020-02-22 CET" [21] "2020-02-22 CET" "2020-02-22 CET"
"2020-02-22 CET" "2020-02-22 CET" [25] "2020-02-22 CET" "2020-02-22 CET"
"2020-02-22 CET" "2020-02-22 CET" [29] "2020-02-22 CET" "2020-02-22 CET"
"2020-02-22 CET" "2020-02-22 CET" [33] "2020-02-22 CET" "2020-02-22 CET"
"2020-02-22 CET" "2020-02-22 CET" [37] "2020-02-22 CET" "2020-02-22 CET"
"2020-02-22 CET" "2020-02-22 CET" [41] "2020-02-22 CET" "2020-02-22 CET"
"2020-02-22 CET" "2020-02-22 CET" [45] "2020-02-22 CET" "2020-02-22 CET"
"2020-02-22 CET" "2020-02-22 CET" [49] "2020-02-22 CET" "2020-02-22 CET"
"2020-02-22 CET" "2020-02-22 CET" [53] "2020-02-22 CET" "2021-02-22 CET"
"2021-02-22 CET" "2021-02-22 CET" [57] "2021-02-22 CET" "2021-02-22 CET"
"2021-02-22 CET"
You'll find below a dump of "mydate" you can copy and paster if you need
a try
Any hint welcome...
Best,
Patrick
mydate <-
c("2020-01", "2020-02", "2020-03", "2020-04", "2020-05", "2020-06",
"2020-07", "2020-08", "2020-09", "2020-10", "2020-11", "2020-12",
"2020-13", "2020-14", "2020-15", "2020-16", "2020-17", "2020-18",
"2020-19", "2020-20", "2020-21", "2020-22", "2020-23", "2020-24",
"2020-25", "2020-26", "2020-27", "2020-28", "2020-29", "2020-30",
"2020-31", "2020-32", "2020-33", "2020-34", "2020-35", "2020-36",
"2020-37", "2020-38", "2020-39", "2020-40", "2020-41", "2020-42",
"2020-43", "2020-44", "2020-45", "2020-46", "2020-47", "2020-48",
"2020-49", "2020-50", "2020-51", "2020-52", "2020-53", "2021-01",
"2021-02", "2021-03", "2021-04", "2021-05", "2021-06")
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Re: [R] dir and pattern = ".r"
Ups... Thank you both. Indeed I must repeat lessons on regular expression... obviously forgotten... Le 28/02/2020 à 18:32, Jeremie Juste a écrit : Hello, you need dir(pattern="\\.r$",ignore.case=TRUE) remember that the pattern is a regular expression. so ".r" is [any single character]r. So basically it will give you any file that contains r (but not that starts with r) so you got what you expected with dir(pattern=".txt") just by chance. as or would have been printed as well HTH, Jeremie I have this directory contain listed at the page bottom Can somebody tell me why with dir(pattern = ".txt") dir(pattern = ".dbf") etc. I get exactly what I want (a vector with the file names correctly suffixed), but with dir(pattern = ".r") I get this: dir(pattern=".r") [1] "Article Predation" "BD Carto" [3] "broma1.txt" "broma3.txt" [5] "BromaBusesMilanMicha.xlsx" "Bromadiolone" [7] "BufferRenard.dbf" "BufferRenard.prj" [9] "BufferRenard.shp" "BufferRenard.shx" [11] "clpboard" "DatesDiurnes.txt" [13] "DatesDiurnes_plus.txt" "DatesDiurnes_plus.xlsx" [15] "DatesDiurnesFREDON" "DatesDiurnesFREDON_plus.txt" [17] "DatesDiurnesFREDON_plus.xlsx" "DatesDiurnesRegis.txt" [19] "DatesDiurnesRegis_plus.txt" "DatesDiurnesRegis_plus.xlsx" [21] "DatesNocturnes.txt" "DatesNocturnes_plus.txt" [23] "DatesNocturnes_plus.xlsx" "DatesNocturnesFREDON" [25] "DatesNocturnesFREDON_plus.txt" "DatesNocturnesFREDON_plus.xlsx" [27] "DatesNocturnesRegis.txt" "DatesNocturnesRegis_plus.txt" [29] "DatesNocturnesRegis_plus.xlsx" "Figures" [31] "ParcBuf300n.dbf" "ParcBuf300n.prj" [33] "ParcBuf300n.shp" "ParcBuf300n.shx" [35] "ParcBuf350n.dbf" "ParcBuf350n.prj" [37] "ParcBuf350n.shp" "ParcBuf350n.shx" [39] "Script_190517_preparation.r" "Script_190518_scores_AT.r" [41] "Script_190519_1430_cinetiques_IKA.r" "Script_190519_1622_transects_camp.r" [43] "Script_190530_1700_preparation2.r" "Script_190530_1903_cinetiques_IKA2.r" [45] "Script_190531_0922_scores_AT2.r" "Script_190531_1729_prey_resource.r" [47] "Script_190601_1509_graphIKAd.r" "Script_190601_1509_graphIKAd_old.r" [49] "Script_190601_1509_graphIKAn.r" "Script_190601_1509_graphIKAn_old.r" [51] "Script_190601_1955_spatial.r" "Script_190708_0930_distance.r" [53] "Script_190709_0930_impact renard.r" "Script_200117_cinetiques_article.r" [55] "Script_200119_source_stats_explore_diurne.r" "Script_200119_stats.r" [57] "Script_200122_spatial_distribution.r" "Script_200124_distance.r" [59] "Script_200124_distance_source_d.r" "Script_200124_distance_source_n.r" [61] "Script_200201_impacts_on_prey.R" "ScriptCompteLignes.r" [63] "Scripts_avant_200112.zip" "shinyPred" [65] "StudyArea.dbf" "StudyArea.prj" [67] "StudyArea.shp" "StudyArea.shx" [69] "SurfaceZoneEtude.dbf" "SurfaceZoneEtude.prj" [71] "SurfaceZoneEtude.shp" "SurfaceZoneEtude.shx" [73] "transects_camp.Rdata" How can I get all the files, only these files, suffixed with ".r" ? Thanks in advance, dir() [1] "Analyse_190523_baseline_190523_1506.docx" "Analyse_190523_baseline_190531_2110.docx" [3] "Analyse_190523_baseline_190531_2110.Rmd" "Analyse_190531_baseline.docx" [5] "Analyse_190531_baseline.Rmd" "Analyse_190531_baseline_cache" [7] "Analyse_190531_baseline_files" "Analyse_190603_spatial.docx" [9] "Analyse_190603_spatial.Rmd" "Analyse_190603_spatial_cache" [11] "Analyse_190603_spatial_files" "Article Predation" [13] "BD Carto" "Biblio" [15] "broma1.txt" "broma3.txt" [17] "BromaBusesMilanMicha.xlsx" "Bromadiolone" [19] "BufferRenard.dbf" "BufferRenard.prj" [21] "BufferRenard.shp" "BufferRenard.shx" [23] "clpboard" "DatesDiurnes.txt" [25] "DatesDiurnes_plus.txt" "DatesDiurnes_plus.xlsx" [27] "DatesDiurnesFREDON" "DatesDiurnesFREDON_plus.txt" [29] "DatesDiurnesFREDON_plus.xlsx" "DatesDiurnesRegis.txt" [31] "DatesDiurnesRegis_plus.txt" "DatesDiurnesRegis_plus.xlsx" [33] "DatesNocturnes.txt" "DatesNocturnes_plus.txt" [35] "DatesNocturnes_plus.xlsx" "DatesNocturnesFREDON" [37] "DatesNocturnesFREDON_plus.txt" "DatesNocturnesFREDON_plus.xlsx" [39] "DatesNocturnesRegis.txt" "DatesNocturnesRegis_plus.txt" [41] "DatesNocturnesRegis_plus.xlsx" "Figures" [43] "IKAZ_old.zip" "IKAZ1999.txt" [45] "IKAZ2000.txt" "IKAZ2007.txt" [47] "IKAZ2008.txt" "IKAZ2009.txt" [49] "IKAZ2010.txt" "IKAZ2011.txt" [51] "IKAZ2012.txt" "IKAZ2013.txt" [53] "IKAZ2014.txt" "IKAZ2015.txt" [55] "IKAZ2016.txt" "IKAZ2017.txt" [57] "IKAZ2018.txt" "ParcBuf300n.dbf" [59] "ParcBuf300n.prj" "ParcBuf300n.shp" [61] "ParcBuf300n.shx" "ParcBuf350n.dbf" [63] "ParcBuf350n.prj" "ParcBuf350n.shp" [65] "ParcBuf350n.shx" "Photos ZELAC" [67] "plot.ds.R" "plot.dsmodel.R" [69] "RData" "Script_190517_preparation.r" [71] "Script_190518_scores_AT.r" "Script_190519_1430_cinetiques_IKA.r" [73] "Script_190519_1622_transects_camp.r" "Script_190530_1700_preparation2.r" [75] "Script_190530_1903_cinetiques_IKA2.r" "Script_190531_0922_scores_AT2.r" [77] "Script_190531_1729_prey_resource.r"
[R] dir and pattern = ".r"
I have this directory contain listed at the page bottom Can somebody tell me why with dir(pattern = ".txt") dir(pattern = ".dbf") etc. I get exactly what I want (a vector with the file names correctly suffixed), but with dir(pattern = ".r") I get this: > dir(pattern=".r") [1] "Article Predation" "BD Carto" [3] "broma1.txt" "broma3.txt" [5] "BromaBusesMilanMicha.xlsx" "Bromadiolone" [7] "BufferRenard.dbf" "BufferRenard.prj" [9] "BufferRenard.shp" "BufferRenard.shx" [11] "clpboard" "DatesDiurnes.txt" [13] "DatesDiurnes_plus.txt" "DatesDiurnes_plus.xlsx" [15] "DatesDiurnesFREDON" "DatesDiurnesFREDON_plus.txt" [17] "DatesDiurnesFREDON_plus.xlsx" "DatesDiurnesRegis.txt" [19] "DatesDiurnesRegis_plus.txt" "DatesDiurnesRegis_plus.xlsx" [21] "DatesNocturnes.txt" "DatesNocturnes_plus.txt" [23] "DatesNocturnes_plus.xlsx" "DatesNocturnesFREDON" [25] "DatesNocturnesFREDON_plus.txt" "DatesNocturnesFREDON_plus.xlsx" [27] "DatesNocturnesRegis.txt" "DatesNocturnesRegis_plus.txt" [29] "DatesNocturnesRegis_plus.xlsx" "Figures" [31] "ParcBuf300n.dbf" "ParcBuf300n.prj" [33] "ParcBuf300n.shp" "ParcBuf300n.shx" [35] "ParcBuf350n.dbf" "ParcBuf350n.prj" [37] "ParcBuf350n.shp" "ParcBuf350n.shx" [39] "Script_190517_preparation.r" "Script_190518_scores_AT.r" [41] "Script_190519_1430_cinetiques_IKA.r" "Script_190519_1622_transects_camp.r" [43] "Script_190530_1700_preparation2.r" "Script_190530_1903_cinetiques_IKA2.r" [45] "Script_190531_0922_scores_AT2.r" "Script_190531_1729_prey_resource.r" [47] "Script_190601_1509_graphIKAd.r" "Script_190601_1509_graphIKAd_old.r" [49] "Script_190601_1509_graphIKAn.r" "Script_190601_1509_graphIKAn_old.r" [51] "Script_190601_1955_spatial.r" "Script_190708_0930_distance.r" [53] "Script_190709_0930_impact renard.r" "Script_200117_cinetiques_article.r" [55] "Script_200119_source_stats_explore_diurne.r" "Script_200119_stats.r" [57] "Script_200122_spatial_distribution.r" "Script_200124_distance.r" [59] "Script_200124_distance_source_d.r" "Script_200124_distance_source_n.r" [61] "Script_200201_impacts_on_prey.R" "ScriptCompteLignes.r" [63] "Scripts_avant_200112.zip" "shinyPred" [65] "StudyArea.dbf" "StudyArea.prj" [67] "StudyArea.shp" "StudyArea.shx" [69] "SurfaceZoneEtude.dbf" "SurfaceZoneEtude.prj" [71] "SurfaceZoneEtude.shp" "SurfaceZoneEtude.shx" [73] "transects_camp.Rdata" How can I get all the files, only these files, suffixed with ".r" ? Thanks in advance, > dir() [1] "Analyse_190523_baseline_190523_1506.docx" "Analyse_190523_baseline_190531_2110.docx" [3] "Analyse_190523_baseline_190531_2110.Rmd" "Analyse_190531_baseline.docx" [5] "Analyse_190531_baseline.Rmd" "Analyse_190531_baseline_cache" [7] "Analyse_190531_baseline_files" "Analyse_190603_spatial.docx" [9] "Analyse_190603_spatial.Rmd" "Analyse_190603_spatial_cache" [11] "Analyse_190603_spatial_files" "Article Predation" [13] "BD Carto" "Biblio" [15] "broma1.txt" "broma3.txt" [17] "BromaBusesMilanMicha.xlsx" "Bromadiolone" [19] "BufferRenard.dbf" "BufferRenard.prj" [21] "BufferRenard.shp" "BufferRenard.shx" [23] "clpboard" "DatesDiurnes.txt" [25] "DatesDiurnes_plus.txt" "DatesDiurnes_plus.xlsx" [27] "DatesDiurnesFREDON" "DatesDiurnesFREDON_plus.txt" [29] "DatesDiurnesFREDON_plus.xlsx" "DatesDiurnesRegis.txt" [31] "DatesDiurnesRegis_plus.txt" "DatesDiurnesRegis_plus.xlsx" [33] "DatesNocturnes.txt" "DatesNocturnes_plus.txt" [35] "DatesNocturnes_plus.xlsx" "DatesNocturnesFREDON" [37] "DatesNocturnesFREDON_plus.txt" "DatesNocturnesFREDON_plus.xlsx" [39] "DatesNocturnesRegis.txt" "DatesNocturnesRegis_plus.txt" [41] "DatesNocturnesRegis_plus.xlsx" "Figures" [43] "IKAZ_old.zip" "IKAZ1999.txt" [45] "IKAZ2000.txt" "IKAZ2007.txt" [47] "IKAZ2008.txt" "IKAZ2009.txt" [49] "IKAZ2010.txt" "IKAZ2011.txt" [51] "IKAZ2012.txt" "IKAZ2013.txt" [53] "IKAZ2014.txt" "IKAZ2015.txt" [55] "IKAZ2016.txt" "IKAZ2017.txt" [57] "IKAZ2018.txt" "ParcBuf300n.dbf" [59] "ParcBuf300n.prj" "ParcBuf300n.shp" [61] "ParcBuf300n.shx" "ParcBuf350n.dbf" [63] "ParcBuf350n.prj" "ParcBuf350n.shp" [65] "ParcBuf350n.shx" "Photos ZELAC" [67] "plot.ds.R" "plot.dsmodel.R" [69] "RData" "Script_190517_preparation.r" [71] "Script_190518_scores_AT.r" "Script_190519_1430_cinetiques_IKA.r" [73] "Script_190519_1622_transects_camp.r" "Script_190530_1700_preparation2.r" [75] "Script_190530_1903_cinetiques_IKA2.r" "Script_190531_0922_scores_AT2.r" [77] "Script_190531_1729_prey_resource.r" "Script_190601_1509_graphIKAd.r" [79] "Script_190601_1509_graphIKAd_old.r" "Script_190601_1509_graphIKAn.r" [81] "Script_190601_1509_graphIKAn_old.r" "Script_190601_1955_spatial.r" [83] "Script_190708_0930_distance.r" "Script_190709_0930_impact renard.r" [85] "Script_200117_cinetiques_article.r" "Script_200119_source_stats_explore_diurne.r" [87] "Script_200119_stats.r" "Script_200122_spatial_distribution.r" [89] "Script_200124_distance.r"
Re: [R] using a variable and a superscript in a legend
Would be nice to put those two way examples in the documentation of the
function 'expression' and 'bquote' in the next R version (we are in the
base) for other users ;-) I am sure many would enjoy.
Le 20/10/2019 à 19:15, Patrick Giraudoux a écrit :
> Great ! You have helped to solve a problem on which I was sweating
> (sporadically, however) since months...
>
> Thanks,
>
> Best,
>
>
> Le 20/10/2019 à 18:29, Bert Gunter a écrit :
>> The legend must be "an expression vector."
>> c("Sans renard",bquote(.(densren) (ind./km)^2)) is not because the
>> first element is a character string.
>>
>> This works:
>>
>> plot(1:100,1:100,type="n")
>> legend(list(x=0,y=100),legend=c(expression("Sans
>> renard"),bquote(.(densren)
>> (ind./km)^2)),lty=c(1,2),col=c("black","red"),bty="n")
>>
>> Cheers,
>> Bert
>>
>>
>> Bert Gunter
>>
>> "The trouble with having an open mind is that people keep coming
>> along and sticking things into it."
>> -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
>>
>>
>> On Sun, Oct 20, 2019 at 9:02 AM Patrick Giraudoux
>> > <mailto:patrick.giraud...@univ-fcomte.fr>> wrote:
>>
>> Thanks Bert and Peter,
>>
>> Yes Bert, I was aware of the legend() function syntax, and just
>> quoting the legend argument within the function.
>>
>> However, Bert and Peter, I do not understand why it works with
>> your absolutely reproducible examples and not in the slightly
>> (not so slightly apparently) different context where I used it...
>>
>> densren=1.25
>> plot(1:100,1:100,type="n")
>> legend(list(x=0,y=100),legend=c("Sans renard",bquote(.(densren)
>> (ind./km)^2)),lty=c(1,2),col=c("black","red"),bty="n")
>>
>> densren=1.25
>> plot(1:100,1:100,type="n")
>> legend(list(x=0,y=100),legend=c("Sans renard",bquote(.(densren) *
>> " ind."/"km"^2)),lty=c(1,2),col=c("black","red"),bty="n"
>>
>> Probably because the result of bquote() is concatenated in a
>> character vector, but how to deal with this ?
>>
>> Best,
>>
>> Patrick
>>
>>
>>
>> Le 20/10/2019 à 16:42, Bert Gunter a écrit :
>>> Assuming you are using base graphics, your syntax for adding the
>>> legend appears to be wrong.
>>> legend() is a separate function, not a parameter of plot.default
>>> afaics.
>>>
>>> The following works for me:
>>>
>>> > densren <- 1.25
>>> > plot(1:10)
>>> > legend (x="center", legend =bquote(.(densren) (ind./km)^2))
>>>
>>> See ?legend
>>>
>>> Bert Gunter
>>>
>>> "The trouble with having an open mind is that people keep coming
>>> along and sticking things into it."
>>> -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
>>>
>>>
>>> On Sun, Oct 20, 2019 at 5:30 AM Patrick Giraudoux
>>> >> <mailto:patrick.giraud...@univ-fcomte.fr>> wrote:
>>>
>>> Dear listers,
>>>
>>> I am trying to pass an expression inlcuding a variable and a
>>> superpscript to a legend. What I want to obtain is e.g. with
>>> densren = 1.25
>>>
>>> 1.25 ind./km^2
>>>
>>> I have tried many variants of the following:
>>>
>>> legend=bquote(.(densren) (ind./km)^2)
>>>
>>> but if not errors, do obtain
>>>
>>> 1.25 (ind./km^2)
>>>
>>> hence not what I want (no parenthesis, 2 in superscript...)
>>>
>>> Any idea about a correct syntax to get what I need ?
>>>
>>> Best,
>>>
>>> Patrick
>>>
>>>
>>> [[alternative HTML version deleted]]
>>>
>>> __
>>> R-help@r-project.org <mailto:R-help@r-project.org> mailing
>>> list -- To UNSUBSCRIBE and more, see
>>> https://stat.ethz.ch/mailman/listinfo/r-help
>>> PLEASE do read the posting guide
>>> http://www.R-project.org/posting-guide.html
>>> and provide commented, minimal, self-contained, reproducible
>>> code.
>>>
>>
>
[[alternative HTML version deleted]]
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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] using a variable and a superscript in a legend
Now, we have two solutions working. This is great since I did not find
any example of the kind searching r-help archives and google...
Thanks !
Le 20/10/2019 à 19:31, Peter Dalgaard a écrit :
It's tricky, but I think what you want is
legend(list(x=0,y=100),
legend=as.expression(list(
"Sans renard",
bquote(.(densren) * " ind."/"km"^2)
)),
lty=c(1,2),col=c("black","red"),bty="n")
Generally, if you want a vector of unevaluated expressions, you need an object of mode
"expression", but you cannot create it directly with expression() because then
the bquote() is left unevaluated:
expression("Sans renard",bquote(.(densren) * " ind."/"km"^2))
expression("Sans renard", bquote(.(densren) * " ind."/"km"^2))
Putting the bquote on the outside _looks_ like it might work:
bquote(expression("Sans renard",.(densren) * " ind."/"km"^2))
expression("Sans renard", 1.25 * " ind."/"km"^2)
but that is not an "expression" object, but a call to expression() (!). Try it
and see.
Evaluating the call does actually work (notice that the printed value is
exactly the same, but the object is not):
eval(bquote(expression("Sans renard",.(densren) * " ind."/"km"^2)))
expression("Sans renard", 1.25 * " ind."/"km"^2)
but I think I prefer the as.expression(list()) construction.
An alternative tack is this:
e <- expression(0,0)
e[[1]] <- "sans renard"
e[[2]] <- bquote(.(densren) * " ind."/"km"^2)
plot(1:100,1:100,type="n")
legend(list(x=0,y=100),legend=e, lty=c(1,2),col=c("black","red"),bty="n")
On 20 Oct 2019, at 18:02 , Patrick Giraudoux
wrote:
Thanks Bert and Peter,
Yes Bert, I was aware of the legend() function syntax, and just quoting the
legend argument within the function.
However, Bert and Peter, I do not understand why it works with your absolutely
reproducible examples and not in the slightly (not so slightly apparently)
different context where I used it...
densren=1.25
plot(1:100,1:100,type="n")
legend(list(x=0,y=100),legend=c("Sans renard",bquote(.(densren)
(ind./km)^2)),lty=c(1,2),col=c("black","red"),bty="n")
densren=1.25
plot(1:100,1:100,type="n")
legend(list(x=0,y=100),legend=c("Sans renard",bquote(.(densren) * "
ind."/"km"^2)),lty=c(1,2),col=c("black","red"),bty="n"
Probably because the result of bquote() is concatenated in a character vector,
but how to deal with this ?
Best,
Patrick
Le 20/10/2019 à 16:42, Bert Gunter a écrit :
Assuming you are using base graphics, your syntax for adding the legend appears
to be wrong.
legend() is a separate function, not a parameter of plot.default afaics.
The following works for me:
densren <- 1.25
plot(1:10)
legend (x="center", legend =bquote(.(densren) (ind./km)^2))
See ?legend
Bert Gunter
"The trouble with having an open mind is that people keep coming along and sticking
things into it."
-- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
On Sun, Oct 20, 2019 at 5:30 AM Patrick Giraudoux
wrote:
Dear listers,
I am trying to pass an expression inlcuding a variable and a
superpscript to a legend. What I want to obtain is e.g. with densren = 1.25
1.25 ind./km^2
I have tried many variants of the following:
legend=bquote(.(densren) (ind./km)^2)
but if not errors, do obtain
1.25 (ind./km^2)
hence not what I want (no parenthesis, 2 in superscript...)
Any idea about a correct syntax to get what I need ?
Best,
Patrick
[[alternative HTML version deleted]]
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Re: [R] using a variable and a superscript in a legend
Great ! You have helped to solve a problem on which I was sweating
(sporadically, however) since months...
Thanks,
Best,
Le 20/10/2019 à 18:29, Bert Gunter a écrit :
> The legend must be "an expression vector."
> c("Sans renard",bquote(.(densren) (ind./km)^2)) is not because the
> first element is a character string.
>
> This works:
>
> plot(1:100,1:100,type="n")
> legend(list(x=0,y=100),legend=c(expression("Sans
> renard"),bquote(.(densren)
> (ind./km)^2)),lty=c(1,2),col=c("black","red"),bty="n")
>
> Cheers,
> Bert
>
>
> Bert Gunter
>
> "The trouble with having an open mind is that people keep coming along
> and sticking things into it."
> -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
>
>
> On Sun, Oct 20, 2019 at 9:02 AM Patrick Giraudoux
> <mailto:patrick.giraud...@univ-fcomte.fr>> wrote:
>
> Thanks Bert and Peter,
>
> Yes Bert, I was aware of the legend() function syntax, and just
> quoting the legend argument within the function.
>
> However, Bert and Peter, I do not understand why it works with
> your absolutely reproducible examples and not in the slightly (not
> so slightly apparently) different context where I used it...
>
> densren=1.25
> plot(1:100,1:100,type="n")
> legend(list(x=0,y=100),legend=c("Sans renard",bquote(.(densren)
> (ind./km)^2)),lty=c(1,2),col=c("black","red"),bty="n")
>
> densren=1.25
> plot(1:100,1:100,type="n")
> legend(list(x=0,y=100),legend=c("Sans renard",bquote(.(densren) *
> " ind."/"km"^2)),lty=c(1,2),col=c("black","red"),bty="n"
>
> Probably because the result of bquote() is concatenated in a
> character vector, but how to deal with this ?
>
> Best,
>
> Patrick
>
>
>
> Le 20/10/2019 à 16:42, Bert Gunter a écrit :
>> Assuming you are using base graphics, your syntax for adding the
>> legend appears to be wrong.
>> legend() is a separate function, not a parameter of plot.default
>> afaics.
>>
>> The following works for me:
>>
>> > densren <- 1.25
>> > plot(1:10)
>> > legend (x="center", legend =bquote(.(densren) (ind./km)^2))
>>
>> See ?legend
>>
>> Bert Gunter
>>
>> "The trouble with having an open mind is that people keep coming
>> along and sticking things into it."
>> -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
>>
>>
>> On Sun, Oct 20, 2019 at 5:30 AM Patrick Giraudoux
>> > <mailto:patrick.giraud...@univ-fcomte.fr>> wrote:
>>
>> Dear listers,
>>
>> I am trying to pass an expression inlcuding a variable and a
>> superpscript to a legend. What I want to obtain is e.g. with
>> densren = 1.25
>>
>> 1.25 ind./km^2
>>
>> I have tried many variants of the following:
>>
>> legend=bquote(.(densren) (ind./km)^2)
>>
>> but if not errors, do obtain
>>
>> 1.25 (ind./km^2)
>>
>> hence not what I want (no parenthesis, 2 in superscript...)
>>
>> Any idea about a correct syntax to get what I need ?
>>
>> Best,
>>
>> Patrick
>>
>>
>> [[alternative HTML version deleted]]
>>
>> __
>> R-help@r-project.org <mailto:R-help@r-project.org> mailing
>> list -- To UNSUBSCRIBE and more, see
>> https://stat.ethz.ch/mailman/listinfo/r-help
>> PLEASE do read the posting guide
>> http://www.R-project.org/posting-guide.html
>> and provide commented, minimal, self-contained, reproducible
>> code.
>>
>
[[alternative HTML version deleted]]
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Re: [R] using a variable and a superscript in a legend
Thanks Bert and Peter,
Yes Bert, I was aware of the legend() function syntax, and just quoting
the legend argument within the function.
However, Bert and Peter, I do not understand why it works with your
absolutely reproducible examples and not in the slightly (not so
slightly apparently) different context where I used it...
densren=1.25
plot(1:100,1:100,type="n")
legend(list(x=0,y=100),legend=c("Sans renard",bquote(.(densren)
(ind./km)^2)),lty=c(1,2),col=c("black","red"),bty="n")
densren=1.25
plot(1:100,1:100,type="n")
legend(list(x=0,y=100),legend=c("Sans renard",bquote(.(densren) * "
ind."/"km"^2)),lty=c(1,2),col=c("black","red"),bty="n"
Probably because the result of bquote() is concatenated in a character
vector, but how to deal with this ?
Best,
Patrick
Le 20/10/2019 à 16:42, Bert Gunter a écrit :
> Assuming you are using base graphics, your syntax for adding the
> legend appears to be wrong.
> legend() is a separate function, not a parameter of plot.default afaics.
>
> The following works for me:
>
> > densren <- 1.25
> > plot(1:10)
> > legend (x="center", legend =bquote(.(densren) (ind./km)^2))
>
> See ?legend
>
> Bert Gunter
>
> "The trouble with having an open mind is that people keep coming along
> and sticking things into it."
> -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
>
>
> On Sun, Oct 20, 2019 at 5:30 AM Patrick Giraudoux
> <mailto:patrick.giraud...@univ-fcomte.fr>> wrote:
>
> Dear listers,
>
> I am trying to pass an expression inlcuding a variable and a
> superpscript to a legend. What I want to obtain is e.g. with
> densren = 1.25
>
> 1.25 ind./km^2
>
> I have tried many variants of the following:
>
> legend=bquote(.(densren) (ind./km)^2)
>
> but if not errors, do obtain
>
> 1.25 (ind./km^2)
>
> hence not what I want (no parenthesis, 2 in superscript...)
>
> Any idea about a correct syntax to get what I need ?
>
> Best,
>
> Patrick
>
>
> [[alternative HTML version deleted]]
>
> __
> R-help@r-project.org <mailto:R-help@r-project.org> mailing list --
> To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
[[alternative HTML version deleted]]
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Re: [R] using a variable and a superscript in a legend
Thanks Eric. I got it too already (and already tried some variations based on it), but to my understanding it does not include a variable whose contents is used in the expression as in the case submitted... Le 20/10/2019 à 14:56, Eric Berger a écrit : > I did a Google search on > > R plot superscript in legend > > and the first search result was > https://stackoverflow.com/questions/20453408/superscript-r-squared-for-legend > > which looks like it might address your question. > > On Sun, Oct 20, 2019 at 3:30 PM Patrick Giraudoux > <mailto:patrick.giraud...@univ-fcomte.fr>> wrote: > > Dear listers, > > I am trying to pass an expression inlcuding a variable and a > superpscript to a legend. What I want to obtain is e.g. with > densren = 1.25 > > 1.25 ind./km^2 > > I have tried many variants of the following: > > legend=bquote(.(densren) (ind./km)^2) > > but if not errors, do obtain > > 1.25 (ind./km^2) > > hence not what I want (no parenthesis, 2 in superscript...) > > Any idea about a correct syntax to get what I need ? > > Best, > > Patrick > > > [[alternative HTML version deleted]] > > __ > R-help@r-project.org <mailto:R-help@r-project.org> mailing list -- > To UNSUBSCRIBE and more, see > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide > http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. > [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] using a variable and a superscript in a legend
Dear listers, I am trying to pass an expression inlcuding a variable and a superpscript to a legend. What I want to obtain is e.g. with densren = 1.25 1.25 ind./km^2 I have tried many variants of the following: legend=bquote(.(densren) (ind./km)^2) but if not errors, do obtain 1.25 (ind./km^2) hence not what I want (no parenthesis, 2 in superscript...) Any idea about a correct syntax to get what I need ? Best, Patrick [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] troubles with foreign:read.dbf
Ashes of my head and all those sorts of things...
If I was a totally a newbie in R, I could claim for some sort of excuse,
but it is definitely not the case, even.
Thanks !
Patrick
Le 20/04/2019 à 19:13, Eric Berger a écrit :
> You seem to have a typo.
> In the case that works your filename is "Mailles_2011a.dbf"
> but in the case that fails your filename is "Mailles_2011a.shp"
> (different extensions)
>
> HTH,
> Eric
>
>
> On Sat, Apr 20, 2019 at 8:00 PM Patrick Giraudoux
> <mailto:patrick.giraud...@univ-fcomte.fr>> wrote:
>
> Dear listers,
>
> I am using the package foreign function read.dbf and meet the
> following
> issue:
>
> i<-"Mailles_2011a.dbf"
>
> dbf<-read.dbf(i)
>
> works well BUT
>
> if I have a vector such as
>
> files <- c("Mailles_2011a.shp", "Mailles_2011p.shp",
> "Mailles_2012a.shp", "Mailles_2012p.shp", "Mailles_2013a.shp",
> "Mailles_2013p.shp", "Mailles_2014p.shp", "Mailles_2015a.shp",
> "Mailles_2015p.shp", "Mailles_2016p.shp")
>
> for(i in files) {
> dbf<-read.dbf(i)
> names(dbf)
> }
>
> gives the following error message:
>
> Error in read.dbf(i) : unable to open DBF file
>
> Same error with e.g.
>
> dbf<-read.dbf(files[1])
>
> Any idea about what's happening?
>
> Patrick
>
>
> [[alternative HTML version deleted]]
>
> __
> R-help@r-project.org <mailto:R-help@r-project.org> mailing list --
> To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
[[alternative HTML version deleted]]
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[R] troubles with foreign:read.dbf
Dear listers,
I am using the package foreign function read.dbf and meet the following
issue:
i<-"Mailles_2011a.dbf"
dbf<-read.dbf(i)
works well BUT
if I have a vector such as
files <- c("Mailles_2011a.shp", "Mailles_2011p.shp",
"Mailles_2012a.shp", "Mailles_2012p.shp", "Mailles_2013a.shp",
"Mailles_2013p.shp", "Mailles_2014p.shp", "Mailles_2015a.shp",
"Mailles_2015p.shp", "Mailles_2016p.shp")
for(i in files) {
dbf<-read.dbf(i)
names(dbf)
}
gives the following error message:
Error in read.dbf(i) : unable to open DBF file
Same error with e.g.
dbf<-read.dbf(files[1])
Any idea about what's happening?
Patrick
[[alternative HTML version deleted]]
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[R] Read text files with Chinese characters
Dear listers, There is number of requests about reading Chinese characters from Excel or text files. I had to cope with the issue and wrote a small manual about it. It might not be an optimal solution, but at least it works :-) One can download the pdf at: https://chrono-environnement.univ-fcomte.fr/personnes/annuaire/article/giraudoux-patrick?lang=en#chinese Cheers, Patrick [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Getting a 404 error reading CRAN mirror repository
Dear all, I try to get a CRAN mirror repository working on my Ubuntu trusty plateform. So, including e.g.: deb https://mirror.ibcp.fr/pub/CRAN/bin/linux/ubuntu trusty universe in /etc/apt/source.lst However, on every mirror tried I get: Err https://mirror.ibcp.fr trusty/universe amd64 Packages HttpError404 Err https://mirror.ibcp.fr trusty/universe i386 Packages HttpError404 Can someone figure out what is going wrong ? Best, Patrick [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] AIC models are not all fitted to the same number of observation
Hi, Using lme from the package nlme 3.1-103, I meet a strange warning. I am trying to compare to models with: library(nlme) lmez6=lme(lepus~vulpes,random=~1|troncon/an,data=ika_z6_test) lmez60=lme(lepus~1,random=~1|troncon/an,data=ika_z6_test) Both have the same number of observations and groups: lmez6 Linear mixed-effects model fit by REML Data: ika_z6_test Log-restricted-likelihood: -2267.756 Fixed: lepus ~ vulpes (Intercept) vulpes 1.35017117 0.04722338 Random effects: Formula: ~1 | troncon (Intercept) StdDev: 0.8080261 Formula: ~1 | an %in% troncon (Intercept) Residual StdDev:1.086611 0.4440076 Number of Observations: 1350 Number of Groups: troncon an %in% troncon 1691350 lmez60 Linear mixed-effects model fit by REML Data: ika_z6_test Log-restricted-likelihood: -2266.869 Fixed: lepus ~ 1 (Intercept) 1.435569 Random effects: Formula: ~1 | troncon (Intercept) StdDev: 0.8139646 Formula: ~1 | an %in% troncon (Intercept) Residual StdDev:1.086843 0.4445815 Number of Observations: 1350 Number of Groups: troncon an %in% troncon 1691350 ...but when I want to compare their AIC, I get: AIC(lmez6,lmez60) df AIC lmez6 5 4545.511 lmez60 4 4541.737 Warning message: In AIC.default(lmez6, lmez60) : models are not all fitted to the same number of observations Has anybody an explanation about this strange warning ? To what extent this warning may limit the conclusions that could be drawn from AIC comparison ? Thanks in advance, Patrick __ 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] AIC models are not all fitted to the same number of observation
Le 21/03/2012 10:56, Patrick Giraudoux a écrit : Hi, Using lme from the package nlme 3.1-103, I meet a strange warning. I am trying to compare to models with: library(nlme) lmez6=lme(lepus~vulpes,random=~1|troncon/an,data=ika_z6_test) lmez60=lme(lepus~1,random=~1|troncon/an,data=ika_z6_test) Both have the same number of observations and groups: lmez6 Linear mixed-effects model fit by REML Data: ika_z6_test Log-restricted-likelihood: -2267.756 Fixed: lepus ~ vulpes (Intercept) vulpes 1.35017117 0.04722338 Random effects: Formula: ~1 | troncon (Intercept) StdDev: 0.8080261 Formula: ~1 | an %in% troncon (Intercept) Residual StdDev:1.086611 0.4440076 Number of Observations: 1350 Number of Groups: troncon an %in% troncon 1691350 lmez60 Linear mixed-effects model fit by REML Data: ika_z6_test Log-restricted-likelihood: -2266.869 Fixed: lepus ~ 1 (Intercept) 1.435569 Random effects: Formula: ~1 | troncon (Intercept) StdDev: 0.8139646 Formula: ~1 | an %in% troncon (Intercept) Residual StdDev:1.086843 0.4445815 Number of Observations: 1350 Number of Groups: troncon an %in% troncon 1691350 ...but when I want to compare their AIC, I get: AIC(lmez6,lmez60) df AIC lmez6 5 4545.511 lmez60 4 4541.737 Warning message: In AIC.default(lmez6, lmez60) : models are not all fitted to the same number of observations Has anybody an explanation about this strange warning ? To what extent this warning may limit the conclusions that could be drawn from AIC comparison ? Thanks in advance, Patrick Sorry to go on on the thread, I have created, but the trouble I meet is above my level in stats... Actually, not using AIC but an anova approach, I get a more informative message: anova(lmez6, lmez60) Model df AIC BIClogLik Test L.Ratio p-value lmez6 1 5 4545.511 4571.543 -2267.756 lmez60 2 4 4541.737 4562.566 -2266.869 1 vs 2 1.774036 0.1829 Warning message: In anova.lme(lmez6, lmez60) : Fitted objects with different fixed effects. REML comparisons are not meaningful. And fubbling a bit more, I disclosed that this was an effect of fitting the model using REML. If fitted using ML, things are going (apparently) smoothly: lmez6=lme(lepus~vulpes,random=~1|troncon/an,data=ika_z6_test,method=ML) lmez60=lme(lepus~1,random=~1|troncon/an,data=ika_z6_test,method=ML) anova(lmez6, lmez60) Model df AIC BIClogLik Test L.Ratio p-value lmez6 1 5 4536.406 4562.445 -2263.203 lmez60 2 4 4538.262 4559.093 -2265.131 1 vs 2 3.856102 0.0496 AIC(lmez6,lmez60) df AIC lmez6 5 4536.406 lmez60 4 4538.262 Now I have the following problem. What I understood from Pinheiro and Bates's book and some forums, is that ML estimations are biased to some extent tending to underestimate variance parameters. So probably not to recommend however results looks consistent here. Thus, I am lost. The two models looks to me clearly embedded (one is just a null model with the only intercept to estimate and the other with intercept + one independent variable (numeric), both have the same random effects, the same response variable and the same number of observations). Warnings, from this point of view sounds inconsistent. They are probably not, but beyond my understanding... Any idea ? Patrick __ 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] some CRAN mirrors not accessible
Dear all, Since some weeks, look like the following CRAN mirrors are no longer accessible for package update: http://cran.univ-lyon1.fr http://mirror.ibcp.fr/pub/CRAN/ update.packages(ask='graphics',checkBuilt=TRUE) Warning: unable to access index for repository http://cran.univ-lyon1.fr/bin/windows/contrib/2.14 update.packages(ask='graphics',checkBuilt=TRUE) Warning: unable to access index for repository http://mirror.ibcp.fr/pub/CRAN/bin/windows/contrib/2.14 Furthermore, attempting to connect via a web browser gives However, I don't know how to get in touch with the webmasters in charge. Any idea about hos to signal the trouble ? PG __ 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] some CRAN mirrors not accessible
Le 11/12/2011 18:57, Prof Brian Ripley a écrit : First, see the status links in the first para of http://cran.r-project.org/mirrors.html . So it seems that the issue is not that the mirror is not accessible, but the part you are looking for is not current/available. yes indeed. For one R for Windows contrib deny access, and for the other has no 2.14/ folder Second, the mirror list is http://cran.r-project.org/CRAN_mirrors.csv and that lists the maintainers. It is also in your R distribution, in directory 'doc' (but that is as old as your distribution, and mirrors do change). Got it ! Thanks. Also fubbling in the chooseCRANmirror() doc I also discovered that getCRANmirrors() returns a data.frame with the maintainer address; looking into the function it either straight reads http://cran.r-project.org/CRAN_mirrors.csv, and if no connection reads de csv file stored in the local directory 'doc'. Now the info about accessibility/availability has been conveyed to the maintainers. Thanks again, On 11/12/2011 17:43, Patrick Giraudoux wrote: Dear all, Since some weeks, look like the following CRAN mirrors are no longer accessible for package update: http://cran.univ-lyon1.fr http://mirror.ibcp.fr/pub/CRAN/ update.packages(ask='graphics',checkBuilt=TRUE) Warning: unable to access index for repository http://cran.univ-lyon1.fr/bin/windows/contrib/2.14 update.packages(ask='graphics',checkBuilt=TRUE) Warning: unable to access index for repository http://mirror.ibcp.fr/pub/CRAN/bin/windows/contrib/2.14 Furthermore, attempting to connect via a web browser gives However, I don't know how to get in touch with the webmasters in charge. Any idea about hos to signal the trouble ? PG __ 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] symbols and legend: how to harmonize point size ?
Hi, I was wondering if it is possible to harmonize the ouput of symbols() and legend() both from the graphics package. Let us take this example: x-runif(10) y-runif(10) z-runif(10) leg-round(seq(min(z),max(z),l=4),2) # 4 values rounded up to 2 decimals for the legend symbols(x,y,circles=z,inches=0.2) legend(topright,legend=leg,pch=1,pt.cex=leg/max(leg)*2) # multiplied by 2 arbitrarily just to make it visible Actually, what I want to do is to pass to pt.cex a value which would make the biggest circle in the legend (leg/max(leg) = 1) exactly the same size as the one specified in symbols (here 0.2 inches). I suppose this is possible using par(cin) but I cannot figure out how to do it properly. Any hint appreciated, Best, Patrick __ 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 write a shapefile with projection
Hi, Sorry i have put such a detailed question to the list about writing a shapefile with projection. I realized that if i use writeOGR from rgdal and not the other write shapefile functions i can get a shapefile with projection recognized by ArcGIS. The command is (in case anybody wonders): ?writeOGR(crest.sp, I:\\LA_levee\\Shape, llev_crest_pts6, driver = ESRI Shapefile) where crest.sp is a spatial point data frame with projection. Thanks, Monica Indeed. writePointsShape() does not write the projection file, but using the function showWKT from rgdal, you can also create one like that: writePointsShape(crest.sp,crest) cat(showWKT(proj4string(crest.sp)),file=crest.prj) Patrick __ 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] logistic regression where the independant variable is a ratio
Dear Lister, I have collected data in 6 geographical areas on prevalence of a parasite in humans and in foxes. The results are expressed as a number of positive or negative cases in human and foxes in the following data.frame: Pvtab - structure(list(posHum = c(3, 5, 3, 17, 0, 4), negHum = c(32631, 16293, 27988, 231282, 53215, 51046), posFox = c(18, 23, 18, 191, 12, 55), negFox = c(14, 24, 62, 105, 55, 43)), .Names = c(posHum, negHum, posFox, negFox), row.names = c(zone 1, zone 2, zone 3, zone 4, zone 5, zone 6), class = data.frame) I want to check a possible link between prevalences in humans (the reponse variable) and prevalences in foxes (the independant variable). I though about a logistic regression of the form: pvFox-Pvtab$posFox/(Pvtab$posFox+Pvtab$negFox) # computes the prevalence in foxes for each area mod0-mod0-glm(cbind(Pvtab$posHum,Pvtab$negHum)~pvFox,family=binomial) But in this cas the number of foxes that have been used to compute the prevalence estimate in foxes (pvFox) is deliberatly not taken into account in the model. I can hardly figure out how to do it (weighing the model with the square root of the number of fox in each area ?). Any advise appreciated about how to model a prevalence as a response of another prevalence at best. Patrick __ 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] how to use # in a rd doc in url address
I am writing a rd doc, and need to use # in a url adress. This would make:
\url{http://www..org/myfolder/#myanchor}
Of course, I suppose this will not work because # is a special character
starting a comment line in the rd dialect. I did not found a similar
example in Writing R exentions. I am not sure bout using \dQuote{a
quotation}), and use \sQuote and \dQuote correctly. Does anyone know how
to get the thing right ?
Patrick
[[alternative HTML version deleted]]
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Re: [R] how to use # in a rd doc in url address
Daniel Malter a écrit :
x=\url{http://www..org/myfolder/#myanchor};
print(x,quote=F)
Does this work for you?
Daniel
I am not working on consol mode (which would make your suggestion
straight applicable), but writing a rd documentationn (the documentation
that comes out with the command ?myfunction). The rd file has a Latex
style syntax and I just want to insert the url within this
documentation. Eg.
\details{
You may want to connect to \url{http://www..org/myfolder/#myanchor}
}
I am not sure one can define a variable and print it in such context...
Best
Patrick
-
cuncta stricte discussurus
-
-Ursprüngliche Nachricht-
Von: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] Im
Auftrag von Patrick Giraudoux
Gesendet: Wednesday, November 11, 2009 12:15 PM
An: r-help@r-project.org
Betreff: [R] how to use # in a rd doc in url address
I am writing a rd doc, and need to use # in a url adress. This would make:
\url{http://www..org/myfolder/#myanchor}
Of course, I suppose this will not work because # is a special character
starting a comment line in the rd dialect. I did not found a similar
example in Writing R exentions. I am not sure bout using \dQuote{a
quotation}), and use \sQuote and \dQuote correctly. Does anyone know how to
get the thing right ?
Patrick
[[alternative HTML version deleted]]
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and provide commented, minimal, self-contained, reproducible code.
__
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] how to use # in a rd doc in url address
Duncan Murdoch a écrit :
On 11/11/2009 12:15 PM, Patrick Giraudoux wrote:
I am writing a rd doc, and need to use # in a url adress. This
would make:
\url{http://www..org/myfolder/#myanchor}
That should work.
Of course, I suppose this will not work because # is a special
character starting a comment line in the rd dialect.
That's not correct. # is only special in R code, and with \url{} the
text is considered as verbatim text, i.e. only \, %, { and } are special.
I did not found a similar
example in Writing R exentions. I am not sure bout using \dQuote{a
quotation}), and use \sQuote and \dQuote correctly. Does anyone know
how to get the thing right ?
I don't understand this question.
You answered it above... There is no reason for using special
quotation considering your reminder: with \url{} the text is
considered as verbatim text
Thanks for the focus,
Best,
Patrick
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Re: [R] how to use # in a rd doc in url address
Patrick Giraudoux a écrit :
Duncan Murdoch a écrit :
On 11/11/2009 12:15 PM, Patrick Giraudoux wrote:
I am writing a rd doc, and need to use # in a url adress. This
would make:
\url{http://www..org/myfolder/#myanchor}
That should work.
Of course, I suppose this will not work because # is a special
character starting a comment line in the rd dialect.
That's not correct. # is only special in R code, and with \url{} the
text is considered as verbatim text, i.e. only \, %, { and } are
special.
I did not found a similar
example in Writing R exentions. I am not sure bout using \dQuote{a
quotation}), and use \sQuote and \dQuote correctly. Does anyone know
how to get the thing right ?
I don't understand this question.
You answered it above... There is no reason for using special
quotation considering your reminder: with \url{} the text is
considered as verbatim text
Thanks for the focus,
Best,
Patrick
Yes, can confirmed it works perfect without any complication... Good
lesson. Being used to prepare oneself to the worst, one over-anticipates
it, but occasionally it does not happen
Cheers,
Patrick
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[R] rd doc truncated with R 2.10.0
Hi,
I am routinely compiling a package and since I have moved to R 2.10.0,
it troncates some section texts in the doc:
With the following section in the rd file:
\details{
The function calls gpsbabel via the system. The gpsbabel program must
be present and on the user's PATH for the function to work see
http://www.gpsbabel.org/. The function has been tested on the
following Garmin GPS devices: Etrex Summit, Etrex Vista Cx and GPSmap 60CSx.
}
...compiling under R 2.9.2 (rcmd build --binary --auto-zip pgirmess) I
get this
Details:
The function calls gpsbabel via the system. The gpsbabel program
must be present and on the user's PATH for the function to work,
see http://www.gpsbabel.org/. The function has been tested on
the following Garmin GPS devices: Etrex Summit, Etrex Vista Cx and
GPSmap 60CSx.
and compiling now under R 2.10.0
Details:
The function has been tested on the following Garmin GPS devices:
Etrex Summit, Etrex Vista Cx and GPSmap 60CSx. The function calls
gpsbabel via the system. The gpsbabel program must be presen
Has anyone an explanation and a workaround ?
Best,
Patrick
__
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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] rd doc truncated with R 2.10.0
Duncan Murdoch a écrit :
On 08/11/2009 12:07 PM, Patrick Giraudoux wrote:
Hi,
I am routinely compiling a package and since I have moved to R
2.10.0, it troncates some section texts in the doc:
With the following section in the rd file:
\details{
The function calls gpsbabel via the system. The gpsbabel program
must be present and on the user's PATH for the function to work see
http://www.gpsbabel.org/. The function has been tested on the
following Garmin GPS devices: Etrex Summit, Etrex Vista Cx and GPSmap
60CSx.
}
...compiling under R 2.9.2 (rcmd build --binary --auto-zip pgirmess)
I get this
Details:
The function calls gpsbabel via the system. The gpsbabel program
must be present and on the user's PATH for the function to work,
see http://www.gpsbabel.org/. The function has been tested on
the following Garmin GPS devices: Etrex Summit, Etrex Vista Cx and
GPSmap 60CSx.
and compiling now under R 2.10.0
Details:
The function has been tested on the following Garmin GPS devices:
Etrex Summit, Etrex Vista Cx and GPSmap 60CSx. The function calls
gpsbabel via the system. The gpsbabel program must be presen
Has anyone an explanation and a workaround ?
You will need to make the complete file available to us to diagnose
this. Is it in pgirmess 1.4.0? Which topic?
Duncan Murdoch
OK. Will send it offlist.
Patrick
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Re: [R] What happen for Negative binomial link in Lmer
Seems the message below and the thread have reveived no attention/answer. The output presented is quite tricky. Looks like if lmer (lme4 0.9975-10) has accepted a negative binomial link with reasonable estimates, although it was not designed for... What can one think about result validity ? Best Patrick Message: 34 Date: Thu, 29 Oct 2009 06:51:24 -0700 (PDT) From: E. Robardet e.robar...@gmail.com Subject: Re: [R] What happen for Negative binomial link in Lmer fonction? To: r-help@r-project.org Message-ID: 26113408.p...@talk.nabble.com Content-Type: text/plain; charset=us-ascii Thank you for your answers, I have an exemple of that i was using: m1a-lmer(atpos~ninter+saison+milieu*zone+(1|code),family=neg.bin(0.429),method=Laplace,data=manu) summary(m1a) Generalized linear mixed model fit using Laplace Formula: atpos ~ ninter + saison + milieu * zone + (1 | code) Data: manu Family: Negative Binomial(log link) AIC BIC logLik deviance 125.1 147.6 -54.57109.1 I think It was the version lme4 0.9975-10. Unfortunately, I have this version no more available on my computer.. I wonder if this old results are still valid.. Ben Bolker wrote: ROBARDET Emmanuelle wrote: Dear R users, I'm performing some GLMMs analysis with a negative binomial link. I already performed such analysis some months ago with the lmer() function but when I tried it today I encountered this problem: Erreur dans famType(glmFit$family) : unknown GLM family: 'Negative Binomial' Does anyone know if the negative binomial family has been removed from this function? I really appreciate any response. Emmanuelle I would be extremely surprised if this worked in the past; to the best of my knowledge the negative binomial family has never been implemented in lmer. One could in principle do a glmmPQL fit with the negative binomial family (with a fixed value of the overdispersion parameter). glmmADMB is another option. Can you say which version etc. you were using??? Follow-ups should probably be sent to r-sig-mixed-mod...@r-project.org -- View this message in context: http://www.nabble.com/What-happen-for-Negative-binomial-link-in-Lmer-fonction--tp26013041p26113408.html Sent from the R help mailing list archive at Nabble.com. [[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] lmer and negative binomial family
Dear listers, One of my former students is trying to fit a model of the negative binomial family with lmer. In the past (two years ago), the following call was working well: m1a-lmer(mapos~ninter+saison+milieu*zone+(1|code),family=neg.bin(0.451),REML=TRUE,data=manu) But now (R version 2.9.2 and lme4 version 0.999375-32), that gives (even with the library MASS loaded): m1a-lmer(mapos~ninter+saison+milieu*zone+(1|code),family=neg.bin(0.451),REML=TRUE,data=manu) Error in famType(glmFit$family) : unknown GLM family: 'Negative Binomial' Any idea about what happens ? Patrick [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] lmer and negative binomial family
Dear listers, One of my former students is trying to fit a model of the negative binomial family with lmer. In the past (two years ago), the following call was working well: m1a-lmer(mapos~ninter+saison+milieu*zone+(1|code),family=neg.bin(0.451),REML=TRUE,data=manu) But now (R version 2.9.2 and lme4 version 0.999375-32), that gives (even with the library MASS loaded): m1a-lmer(mapos~ninter+saison+milieu*zone+(1|code),family=neg.bin(0.451),REML=TRUE,data=manu) Error in famType(glmFit$family) : unknown GLM family: 'Negative Binomial' Any idea about what happens ? Patrick [[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] lmer and negative binomial family
Patrick Giraudoux a écrit : Dear listers, One of my former students is trying to fit a model of the negative binomial family with lmer. In the past (two years ago), the following call was working well: m1a-lmer(mapos~ninter+saison+milieu*zone+(1|code),family=neg.bin(0.451),REML=TRUE,data=manu) But now (R version 2.9.2 and lme4 version 0.999375-32), that gives (even with the library MASS loaded): m1a-lmer(mapos~ninter+saison+milieu*zone+(1|code),family=neg.bin(0.451),REML=TRUE,data=manu) Error in famType(glmFit$family) : unknown GLM family: 'Negative Binomial' Any idea about what happens ? Patrick Oups. Sorry to reply to myself, but the answer was here: http://www.nabble.com/What-happen-for-Negative-binomial-link-in-Lmer-fonction--td26013041.html [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] decimal troubles ?
Dear all, I have some trouble with the number of decimals in R (currently R 2.9.0). For instance: options()$digits [1] 3 let me hope that I will get three digits where useful when a number is printed. BUT: 44.25+31.1+50 [1] 125 No way to get the right result 125.35 Can anybody tell me what's happens ? Patrick __ 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] decimal troubles ?
Shame on me... I confused digits and decimals Thanks anyway to make me come to the English basics... Patrick Peter Alspach a écrit : Tena koe Patrick If you want more than three digits, change the options: options(digits=7) 44.25+31.1+50 [1] 125.35 HTH Peter Alspach -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Patrick Giraudoux Sent: Tuesday, 12 May 2009 8:08 p.m. To: r-help@r-project.org Subject: [R] decimal troubles ? Dear all, I have some trouble with the number of decimals in R (currently R 2.9.0). For instance: options()$digits [1] 3 let me hope that I will get three digits where useful when a number is printed. BUT: 44.25+31.1+50 [1] 125 No way to get the right result 125.35 Can anybody tell me what's happens ? Patrick __ 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. The contents of this e-mail are confidential and may be subject to legal privilege. If you are not the intended recipient you must not use, disseminate, distribute or reproduce all or any part of this e-mail or attachments. If you have received this e-mail in error, please notify the sender and delete all material pertaining to this e-mail. Any opinion or views expressed in this e-mail are those of the individual sender and may not represent those of The New Zealand Institute for Plant and Food Research Limited. [[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] nls, convergence and starting values
Patrick Burns a écrit : Patrick Giraudoux wrote: Bert Gunter a écrit : Based on a simple scatterplot of pourcma vs transat, a 4 parameter logistic looks like wild overfitting, and that may be the source of your problems. Given the huge scatter, a straight line is about as much as would seem sensible. I think this falls into the Why ever would you want to do such a thing? category. -- Bert Right, well, the general idea was just to show that the straight line was the best model indeed (in the other data sets, with model comparison, the logistic one was clearly shown to be the best... ). Can the fact that convergence cannot be obtained be an acceptable and sufficient reason to select the null model (the straight line) ? It is my experience that convergence problems are often encountered when the model makes little sense. I'm not so sure that non-convergence on its own is a good reason to reject the model. That is, to answer your specific question, I think it is acceptable but not sufficient. Patrick Burns patr...@burns-stat.com +44 (0)20 8525 0696 http://www.burns-stat.com (home of The R Inferno and A Guide for the Unwilling S User) OK. Thanks for this opinion. Actually I was sharing it intuitively but facing such situation for the first time, was quite unconfortable to make a decision (and still I am). We are touching epistemology... and maybe a bit far from purely technical thus from the R list issues. Tanks again, anyway, Patrick __ 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] nls, convergence and starting values
in non linear modelling finding appropriate starting values is something like an art... (maybe from somewhere in Crawley , 2007) Here a colleague and I just want to compare different response models to a null model. This has worked OK for almost all the other data sets except that one (dumped below). Whatever our trials and algorithms, even subsetting data (to check if some singular point was the cause of the mess), we do not reach convergence... or screw up with singular gradients (?) etc... eg: nls(pourcma~SSlogis(transat, Asym, xmid, scal), start=c(Asym=30, xmid=0.07, scal=0.02),data=bdd, weights=sqrt(nbfeces),trace=T,alg=plinear) As anyone a hint about an alternate approach to fit a model ? Or an idea to get evidence that such model cannot be fitted to the data bdd - structure(list(transat = c(0.0697, 0.13079, 0.314265, 0.241613, 0.039319, 0, 0, 0, 0, 0, 0.0805, 0.41, 0.30585, 0.27465, 0.06085, 0.09114, 0.05766, 0.036983, 0.093186, 0.046624, 0, 0, 0, 0, 0.000616, 0, 0.0025, 0.0325, 0.03125, 0.04599, 0.38398, 0.524505, 0.450337, 0.061831, 0.133926, 0.091806, 0.00928, 0.25114, 0.3074, 0.431056, 0.026158), transma = c(0.04141, 0.01599, 0.101803, 0.002378, 0.039319, 0.00472459016393443, 0.0031016393442623, 0.000178524590163934, 0.00255704918032787, 0.000346229508196721, 0.0665, 0.012, 0.0553, 0.0045, 0.0056, 0.00155, 0.00124, 0.011966, 0.001736, 0.004712, 3.62903225806452e-05, 9.79838709677419e-05, 2.20161290322581e-05, 0.00462, 0.01006444, 0.00213, 0.046, 0.005, 0.01195, 0.07154, 0.08468, 0.141182, 0.086578, 0.027959, 0.003159, 0.003081, 0.13862, 0.00754, 0.078648, 0.068324, 0.025288), nbfeces = c(22L, 26L, 43L, 30L, 35L, 25L, 21L, 36L, 34L, 37L, 23L, 32L, 40L, 35L, 30L, 16L, 25L, 37L, 37L, 34L, 31L, 35L, 41L, 31L, 34L, 39L, 5L, 14L, 31L, 13L, 21L, 34L, 32L, 36L, 36L, 40L, 31L, 35L, 39L, 29L, 32L), pourcma = c(50, 34.6153846153846, 27.9069767441860, 43.3, 65.7142857142857, 32, 28.5714285714286, 22.2, 50, 10.8108108108108, 26.0869565217391, 40.625, 12.5, 22.8571428571429, 43.3, 6.25, 4, 10.8108108108108, 16.2162162162162, 23.5294117647059, 25.8064516129032, 45.7142857142857, 39.0243902439024, 25.8064516129032, 41.7, 27.5, 20, 14.2857142857143, 22.5806451612903, 15.3846153846154, 38.0952380952381, 17.6470588235294, 78.125, 61.1, 25, 37.5, 22.5806451612903, 40, 17.9487179487179, 41.3793103448276, 50), pourcat = c(22.7272727272727, 30.7692307692308, 41.8604651162791, 56.7, 5.71428571428571, 0, 0, 0, 0, 0, 30.4347826086957, 15.625, 45, 74.2857142857143, 13.3, 50, 12, 18.9189189189189, 27.0270270270270, 20.5882352941176, 0, 0, 0, 0, 0, 5, 40, 0, 0, 7.69230769230769, 9.52380952380952, 38.2352941176471, 59.375, 5.56, 41.7, 42.5, 9.67741935483871, 14.2857142857143, 51.2820512820513, 79.3103448275862, 6.25)), .Names = c(transat, transma, nbfeces, pourcma, pourcat), class = data.frame, row.names = c(NA, -41L)) __ 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] nls, convergence and starting values
Bert Gunter a écrit : Based on a simple scatterplot of pourcma vs transat, a 4 parameter logistic looks like wild overfitting, and that may be the source of your problems. Given the huge scatter, a straight line is about as much as would seem sensible. I think this falls into the Why ever would you want to do such a thing? category. -- Bert Right, well, the general idea was just to show that the straight line was the best model indeed (in the other data sets, with model comparison, the logistic one was clearly shown to be the best... ). Can the fact that convergence cannot be obtained be an acceptable and sufficient reason to select the null model (the straight line) ? Patrick __ 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] Comparison of age categories using contrasts
Dear listers, I would like to compare the levels of a factor with 8 age categories (0,10] (10,20] (20,30] (30,40] (40,50] (50,60] (60,70] (70,90] (however, the factor has not been ordered yet). The default in glm is cont.treatment (for unordered factors) and that leads to compare each level to the first one. I would rather prefer to compare the 2nd to the 1st, the 3rd to the 2nd, the 4th to the 3rd, etc... My understanding is that cont.poly may make the trick, eg specified like this: mod3-glm(AE~agecat, family=binomial,data=qinghai2, contrasts=list(agecat=contr.poly)) but I am not sure to be right. Would be grateful if a true statistician can confirm or fire me... and before definitive fire tell me how to manage with this... __ 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] Comparison of age categories using contrasts
Greg Snow a écrit : One approach is to create your own contrasts matrix: mycmat - diag(8) mycmat[ row(mycmat) == col(mycmat) + 1 ] - -1 mycmati - solve(mycmat) contrasts(agefactor) - mycmati[,-1] Now when you use agefactor, the intercept will be the first age group and the slopes will be the differences between the pairs of groups (make sure that the order of the levels of agefactor is correct). The difference between this method and the contr.sdif function in MASS is how the intercept will end up being interpreted (and the dimnames). Hope this helps, Actually, exactly what I needed including the reference to contr.sdif in MASS I did not spot before (although I am a faithful reader of the yellow book... but so many things still escape to me). Again thanks a lot. Patrick [[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] Comparison of age categories using contrasts
Dear listers, I would like to compare the levels of a factor with 8 age categories (0,10] (10,20] (20,30] (30,40] (40,50] (50,60] (60,70] (70,90] (however, the factor has not been ordered yet). The default in glm is cont.treatment (for unordered factors) and that leads to compare each level to the first one. I would rather prefer to compare the 2nd to the 1st, the 3rd to the 2nd, the 4th to the 3rd, etc... My understanding is that cont.poly may make the trick, eg specified like this: mod3-glm(AE~agecat, family=binomial,data=qinghai2, contrasts=list(agecat=contr.poly)) but I am not sure to be right. Would be grateful if a true statistician can confirm or fire me... and before definitive fire tell me how to manage with this... __ 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] Coordinate systems for geostatistics in R (imicola)
If you use the spatial objects provided by the sp-package (http://cran.r-project.org/web/packages/sp/vignettes/sp.pdf) you transform your data to other projections using the spTransform package. Thus you will need the rgdal package in complement (it actually includes spTransform). This function is extremely convenient: you can manage coordinate transformations extremely easily for common systems (WGS84, UTM) within the R environment. __ 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] svIDE and Tinn-R
Probably an old moon since evoqued one year ago in this link:
http://tolstoy.newcastle.edu.au/R/e2/help/07/04/15738.html
but I have recently re-installed Tinn-R with R 2.7.0 and forgot to insert
options(warn=-1)
library(svIDE)
...
options(warn=0)
in Rprofile.site... and could see that we have still the same warning
launching R:
Warning messages:
1: '\A' is an unrecognized escape in a character string
2: unrecognized escape removed from ;for Options\AutoIndent: 0=Off,
1=follow language scoping and 2=copy from previous line\n
3: In grep(paste([{]TclEval , topic, [}], sep = ),
tclvalue(.Tcl(dde services TclEval {})), :
argument 'useBytes = TRUE' will be ignored
I wonder how far it may be problematical ?
Patrick
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and provide commented, minimal, self-contained, reproducible code.
[R] AIC and anova, lme
Dear listers, Here we have a strange result we can hardly cope with. We want to compare a null mixed model with a mixed model with one independent variable. lmmedt1-lme(mediane~1, random=~1|site, na.action=na.omit, data=bdd2) lmmedt9-lme(mediane~log(0.0001+transat), random=~1|site, na.action=na.omit, data=bdd2) Using the Akaike Criterion and selMod of the package pgirmess gives the following output: selMod(list(lmmedt1,lmmedt9)) model LL K N2K AIC deltAIC w_i AICc deltAICc w_ic 2 log(1e-04 + transat) 44.63758 4 7.5 -81.27516 0.00 0.65 -79.67516 0.00 0.57 11 43.02205 3 10.0 -80.04410 1.231069 0.35 -79.12102 0.554146 0.43 The usual conclusion would be that the two models are equivalent and to keep the null model for parsimony (!). However, an anova shows that the variable 'log(1e-04 + transat)' is significantly different from 0 in model 2 (lmmedt9) anova(lmmedt9) numDF denDF F-value p-value (Intercept) 120 289.43109 .0001 log(1e-04 + transat) 120 31.18446 .0001 Has anyone an opinion about what looks like a paradox here ? Patrick __ 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] AIC and anova, lme
ian white a écrit : Patrick, The likelihoods of two models fitted using REML cannot be compared unless the fixed effects are the same in the two models. Many thanks for this reminder. Shame on me: it recalls me that this subject may have been already largely discussed on this list. Now, I can search the archives specifically with the REML issue... All the best, Patrick On Tue, 2008-02-26 at 14:38 +0100, Patrick Giraudoux wrote: Dear listers, Here we have a strange result we can hardly cope with. We want to compare a null mixed model with a mixed model with one independent variable. lmmedt1-lme(mediane~1, random=~1|site, na.action=na.omit, data=bdd2) lmmedt9-lme(mediane~log(0.0001+transat), random=~1|site, na.action=na.omit, data=bdd2) Using the Akaike Criterion and selMod of the package pgirmess gives the following output: selMod(list(lmmedt1,lmmedt9)) model LL K N2K AIC deltAIC w_i AICc deltAICc w_ic 2 log(1e-04 + transat) 44.63758 4 7.5 -81.27516 0.00 0.65 -79.67516 0.00 0.57 11 43.02205 3 10.0 -80.04410 1.231069 0.35 -79.12102 0.554146 0.43 The usual conclusion would be that the two models are equivalent and to keep the null model for parsimony (!). However, an anova shows that the variable 'log(1e-04 + transat)' is significantly different from 0 in model 2 (lmmedt9) anova(lmmedt9) numDF denDF F-value p-value (Intercept) 120 289.43109 .0001 log(1e-04 + transat) 120 31.18446 .0001 Has anyone an opinion about what looks like a paradox here ? Patrick __ 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] random location in polygons sp spsample splancs csr
Dear all, I had to place points at random, one in each of larger number of polygons (actually in objects of class 'SpatialPolygonsDataFrame' , see sp library), and tried first to do it using spsample (from sp). Surprisingly, every 5-15 trials, the output was a NULL value. The doc says that ' this may occur when trying to hit a small and awkwardly shaped polygon in a large bounding box with a small number of points', but in my case, the shapes were not really awkward, and the bounding box just the smallest rectangle including the shape, just the number of points was 1 in each polygon. Thus I tried csr (from splancs) after having extracted the polygon coordinates of each shape from the Spatial object, and everything went smoothly, with hit success every trial. Has anybody (anybody will probably be Edzer or/and Roger...) an idea why here splancs looks like outperforming spsample ? Patrick __ 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] random location in polygons sp spsample splancs csr
Thanks for those detailed explanation and the time taken to write them. The spsample methods for polygons have an iter= argument that can be used to make then try harder, did you try it (with what values - the help page senctence you quote is from the iter= description)? Yes sure, I went up to 10, but no success. Could you provide an example with a set.seed() value that does what you say, or at least the code you used? The easiest way is to send the data and the script off list. I will do it. Did you try asking for multiple points and then choosing a single point at random? This would be equivalent to increasing iter while asking for a single point. I did not try this one Actually, I found my way out easy with csr() in splancs, and did not fight too much with spsample. My question on the list was just for general information PS. Perhaps R-sig-geo is a more appropriate list? I was wondering too... and chose r-help because I though the question was of 'general' interest enough. This is debatable indeed... Thank you anyway for your answer, and see you in a few minutes off list... Cheers, Patrick __ 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] change in I(x) ? in R 2.6.0
Dear listers, I am trying to use an old script which was working well in the previous R version. It looks like if it no longer works in R.6.0. I have a model of the form: glm(nath2$Positif ~ n + yearday + x + y + I(x^2) + I(y^2) + yearday:x + yearday:y, family = poisson, data = nath2) and want to get predicts from a data.frame whose column names are: names(data1) [1] x y yearday n x and y are geographical coordinates, yearday is equal to 120 and n equal 100; they are all numerics: sapply(data1,is.numeric) x y yearday n TRUE TRUE TRUE TRUE when I use the function predict: z1-predict(mod1b,newdata=data1,type=response) Error: variables ‘I(x^2)’, ‘I(y^2)’ were specified with different types from the fit Any idea about what goes wrong ? Patrick __ 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] R-2.6.0 and RWinEdt
Installed and tested right now. Works fine, no problem. Thanks, Patrick Uwe Ligges a écrit : Patrick Giraudoux wrote: Dear Listers, I have just installed R-2.6.0 and the RWinEdt package 1.7-6 under Windows XP. wait for 1.7-7 which should appear on CRAN real soon now. Uwe The R-WinEdt menu well appears at launching (the command library(RWinEdt) is in .Rprofile), but WinEdt is NOT started automatically (this was not the case in the earlier versions of R). When WinEdt is started by hand (eg double-click on a RWinEdt alias after R launching), syntax highlighting and connexion to R works well. Any idea about how to fix this and get WinEdt automatically started when library(RWinEdt) is called? Patrick __ 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] R-2.6.0 and RWinEdt
Dear Listers, I have just installed R-2.6.0 and the RWinEdt package 1.7-6 under Windows XP. The R-WinEdt menu well appears at launching (the command library(RWinEdt) is in .Rprofile), but WinEdt is NOT started automatically (this was not the case in the earlier versions of R). When WinEdt is started by hand (eg double-click on a RWinEdt alias after R launching), syntax highlighting and connexion to R works well. Any idea about how to fix this and get WinEdt automatically started when library(RWinEdt) is called? Patrick __ 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.
