[R] Sorting data.frame and again sorting within data.frame
Dear R forum, I have a data.frame as defied below - df = data.frame(names = c(C, A, A, B, C, B, A, B, C), dates = c(4/15/2013, 4/13/2013, 4/15/2013, 4/13/2013, 4/13/2013, 4/15/2013, 4/14/2013, 4/14/2013,4/14/2013 ),values = c(10, 31, 31, 17, 11, 34, 102, 47, 29)) df names dates values 1 C 4/15/2013 10 2 A 4/13/2013 31 3 A 4/15/2013 31 4 B 4/13/2013 17 5 C 4/13/2013 11 6 B 4/15/2013 34 7 A 4/14/2013 102 8 B 4/14/2013 47 9 C 4/14/2013 29 I need to sort df first on names in increasing order and then further on dates in a decreasing order i.e. I need names dates values A 4/15/2013 31 A 4/14/2013 102 A 4/13/2013 31 B 4/15/2013 34 B 4/14/2013 47 B 4/13/2013 17 C 4/15/2013 10 C 4/14/2013 29 C 4/13/2013 11 I tried df_sorted = df[order(df$names, (as.Date(df$dates, %m/%d/%Y)), decreasing = TRUE),] df_sorted names dates values 1 C 4/15/2013 10 9 C 4/14/2013 29 5 C 4/13/2013 11 6 B 4/15/2013 34 8 B 4/14/2013 47 4 B 4/13/2013 17 3 A 4/15/2013 31 7 A 4/14/2013 102 2 A 4/13/2013 31 I need A to appear first with all three corresponding dates in decreasing order, then B and so on. Please guide. With regards Katherine [[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] how to transform string to Camel Case?
Dear all,
Given the following vector:
(z - c('R project', 'hello world', 'something Else'))
[1] R project hello worldsomething Else
I know how to obtain all capitals or all lower case letters:
tolower(z)
[1] r project hello worldsomething else
toupper(z)
[1] R PROJECT HELLO WORLDSOMETHING ELSE
I saw the tocamel() function in 'rapport', but it doesn't do what I
want to achieve as it actually proceeds to camelCase/CamelCase the
strings:
tocamel(z)
[1] RProject helloWorldsomethingElse
But how should I proceed to obtain Camel Case? Here's what I'd like to get:
c('R Project', 'Hello World', 'Something Else')
Regards,
Liviu
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Re: [R] how to transform string to Camel Case?
Hi,
There is an example of how to do do what you are looking for in ?toupper
Regards,
Pascal
On 04/15/2013 03:50 PM, Liviu Andronic wrote:
Dear all,
Given the following vector:
(z - c('R project', 'hello world', 'something Else'))
[1] R project hello worldsomething Else
I know how to obtain all capitals or all lower case letters:
tolower(z)
[1] r project hello worldsomething else
toupper(z)
[1] R PROJECT HELLO WORLDSOMETHING ELSE
I saw the tocamel() function in 'rapport', but it doesn't do what I
want to achieve as it actually proceeds to camelCase/CamelCase the
strings:
tocamel(z)
[1] RProject helloWorldsomethingElse
But how should I proceed to obtain Camel Case? Here's what I'd like to get:
c('R Project', 'Hello World', 'Something Else')
Regards,
Liviu
__
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Re: [R] how to transform string to Camel Case?
See for instance capitalize() in the R.utils package.
Henrik
On Apr 14, 2013 11:51 PM, Liviu Andronic landronim...@gmail.com wrote:
Dear all,
Given the following vector:
(z - c('R project', 'hello world', 'something Else'))
[1] R project hello worldsomething Else
I know how to obtain all capitals or all lower case letters:
tolower(z)
[1] r project hello worldsomething else
toupper(z)
[1] R PROJECT HELLO WORLDSOMETHING ELSE
I saw the tocamel() function in 'rapport', but it doesn't do what I
want to achieve as it actually proceeds to camelCase/CamelCase the
strings:
tocamel(z)
[1] RProject helloWorldsomethingElse
But how should I proceed to obtain Camel Case? Here's what I'd like to get:
c('R Project', 'Hello World', 'Something Else')
Regards,
Liviu
--
Do you know how to read?
http://www.alienetworks.com/srtest.cfm
http://goodies.xfce.org/projects/applications/xfce4-dict#speed-reader
Do you know how to write?
http://garbl.home.comcast.net/~garbl/stylemanual/e.htm#e-mail
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and provide commented, minimal, self-contained, reproducible code.
[[alternative HTML version deleted]]
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] how to transform string to Camel Case?
On Apr 15, 2013, at 08:50 , Liviu Andronic wrote:
Dear all,
Given the following vector:
(z - c('R project', 'hello world', 'something Else'))
[1] R project hello worldsomething Else
I know how to obtain all capitals or all lower case letters:
tolower(z)
[1] r project hello worldsomething else
toupper(z)
[1] R PROJECT HELLO WORLDSOMETHING ELSE
I saw the tocamel() function in 'rapport', but it doesn't do what I
want to achieve as it actually proceeds to camelCase/CamelCase the
strings:
tocamel(z)
[1] RProject helloWorldsomethingElse
But how should I proceed to obtain Camel Case? Here's what I'd like to get:
c('R Project', 'Hello World', 'Something Else')
That'll be capitalization, not camel case, camel-casing is when wordsHaveHumps
inTheMiddle.
As for actually solving your problem, I'd try googling for regular expression
to capitalize word and convert to use with gsub(). (I did do the googling, but
regexps being what they are - unreadable gibberish at first sight - I thought
I'd leave the hard work for others...). Or, as Pascal points out:
example(toupper).
-pd
Regards,
Liviu
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--
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Center for Statistics, Copenhagen Business School
Solbjerg Plads 3, 2000 Frederiksberg, Denmark
Phone: (+45)38153501
Email: pd@cbs.dk Priv: pda...@gmail.com
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Re: [R] Sorting data.frame and again sorting within data.frame
The examples in ?order show a method that could be applied if you avoid the decreasing argument and instead convert the Date to numeric for purposes of sorting. --- Jeff NewmillerThe . . Go Live... DCN:jdnew...@dcn.davis.ca.usBasics: ##.#. ##.#. Live Go... Live: OO#.. Dead: OO#.. Playing Research Engineer (Solar/BatteriesO.O#. #.O#. with /Software/Embedded Controllers) .OO#. .OO#. rocks...1k --- Sent from my phone. Please excuse my brevity. Katherine Gobin katherine_go...@yahoo.com wrote: Dear R forum, I have a data.frame as defied below - df = data.frame(names = c(C, A, A, B, C, B, A, B, C), dates = c(4/15/2013, 4/13/2013, 4/15/2013, 4/13/2013, 4/13/2013, 4/15/2013, 4/14/2013, 4/14/2013,4/14/2013 ),values = c(10, 31, 31, 17, 11, 34, 102, 47, 29)) df � names dates values 1 C 4/15/2013 10 2 A 4/13/2013 31 3 A 4/15/2013 31 4 B 4/13/2013 17 5 C 4/13/2013 11 6 B 4/15/2013 34 7 A 4/14/2013��� 102 8 B 4/14/2013 47 9 C 4/14/2013 29 I need to sort df first on names in increasing order and then further on dates in a decreasing order i.e. I need names��� dates��� values A��� 4/15/2013� 31 A��� 4/14/2013 102 A��� 4/13/2013� 31 B��� 4/15/2013� 34 B��� 4/14/2013� 47 B��� 4/13/2013� 17 C��� 4/15/2013� 10 C��� 4/14/2013� 29 C��� 4/13/2013� 11 I tried df_sorted = df[order(df$names, (as.Date(df$dates, %m/%d/%Y)), decreasing = TRUE),] df_sorted � names dates values 1 C 4/15/2013 10 9 C 4/14/2013 29 5 C 4/13/2013 11 6 B 4/15/2013 34 8 B 4/14/2013 47 4 B 4/13/2013 17 3 A 4/15/2013 31 7 A 4/14/2013��� 102 2 A 4/13/2013 31 I need A to appear first with all three corresponding dates in decreasing order, then B and so on. Please guide. With regards Katherine [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Imputation with SOM using kohonen package
I have a data set with 10 variables, and about 8000 instances (or
objects/rows/samples). In addition I have one more ('class') variable that
I have about 10 instances for, but for which I wish to impute values for.
I am a little confused how to go about doing this, mostly as I'm not
well-versed in it. Do I train the SOM with a data object that contains just
the first 10 variables (exclude the 'class' variable), then predict using
an object that has all of the variables (including the class variable)?
(I am using the kohonen package, and in general I am using the SOM
technique as a comparison to some other methods).
I don't know if providing some or all data is useful, please let me know if
you think it is.
# get the data
bw - read.csv(bw.csv)
# some missing values in data
bwm - data.frame(na.approx(bw, na.rm=FALSE, rule=2))
bw10 - bwm[, 1:10]
bw10.sc - scale(bw10)
bw.som - som(data=bw10.sc, grid=somgrid(25,20,'hexagonal')) # playing
with diff grid sizes
# the different plots of the som at this point show some interesting
features to me, but are quite difficult to interpret.
# there's much work needed here to understand it, but for now I want to see
if it's possible to impute values for another variable...
# here's where I lose it, missing values, trainY, don't get it.
bw.predict - predict(bw.som, newdata=scale(bw), trainX=???, trainY=???)
Ben.
[[alternative HTML version deleted]]
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Re: [R] how to transform string to Camel Case?
Dear Liviu,
I have just updated tocamel to have a new argument, so the development
version of the package would produce:
tocamel(z, upper = TRUE, sep = ' ')
[1] R Project Hello WorldSomething Else
Best,
Gergely
PS #1: to install the dev branch you might give a try to the devtools
package:
library(devtools)
install_github('rapport', 'rapporter', 'development')
PS #2: Alex (cc) pls verify.
On 15 April 2013 08:50, Liviu Andronic landronim...@gmail.com wrote:
Dear all,
Given the following vector:
(z - c('R project', 'hello world', 'something Else'))
[1] R project hello worldsomething Else
I know how to obtain all capitals or all lower case letters:
tolower(z)
[1] r project hello worldsomething else
toupper(z)
[1] R PROJECT HELLO WORLDSOMETHING ELSE
I saw the tocamel() function in 'rapport', but it doesn't do what I
want to achieve as it actually proceeds to camelCase/CamelCase the
strings:
tocamel(z)
[1] RProject helloWorldsomethingElse
But how should I proceed to obtain Camel Case? Here's what I'd like to get:
c('R Project', 'Hello World', 'Something Else')
Regards,
Liviu
--
Do you know how to read?
http://www.alienetworks.com/srtest.cfm
http://goodies.xfce.org/projects/applications/xfce4-dict#speed-reader
Do you know how to write?
http://garbl.home.comcast.net/~garbl/stylemanual/e.htm#e-mail
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[[alternative HTML version deleted]]
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[R] create an access file
Hi there, I have seen this post. https://stat.ethz.ch/pipermail/r-help/2007-June/133606.html have odbc installed in my machine. Now I have the following message: channel2 - odbcDriverConnect(test.mdb)Warning messages:1: In odbcDriverConnect(test.mdb) : [RODBC] ERROR: state IM002, code 0, message [Microsoft][ODBC Driver Manager] Data source name not found and no default driver specified2: In odbcDriverConnect(test.mdb) : [RODBC] ERROR: state 01S00, code 0, message [Microsoft][ODBC Driver Manager] Invalid connection string attribute3: In odbcDriverConnect(test.mdb) : ODBC connection failed Did I miss anything? -- Aileen L. View my Linkedin profile: http://au.linkedin.com/in/aileen2 Being happy doesn't mean you're perfect. It just means you've decided to look beyond the imperfections- K.B Indiana (age 14)http://www.boardofwisdom.com/default.asp?topic=1010search=K%2EB+Indiana+%28age+14%29 [[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] model frame and formula mismatch in model.matrix()
Hi Eva, you're right, it works with 50 variables. Then, how could I change this variable limit in the lm function? Thank you very much for your help. Julien. -- View this message in context: http://r.789695.n4.nabble.com/model-frame-and-formula-mismatch-in-model-matrix-tp4664093p4664226.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] create an access file
On 15/04/2013 07:12, Aileen Lin wrote: Hi there, I have seen this post. https://stat.ethz.ch/pipermail/r-help/2007-June/133606.html have odbc installed in my machine. Now I have the following message: channel2 - odbcDriverConnect(test.mdb)Warning messages:1: In odbcDriverConnect(test.mdb) : [RODBC] ERROR: state IM002, code 0, message [Microsoft][ODBC Driver Manager] Data source name not found and no default driver specified2: In odbcDriverConnect(test.mdb) : [RODBC] ERROR: state 01S00, code 0, message [Microsoft][ODBC Driver Manager] Invalid connection string attribute3: In odbcDriverConnect(test.mdb) : ODBC connection failed Did I miss anything? That is not doing what the URL you gave suggested. Please do follow it. -- Brian D. Ripley, rip...@stats.ox.ac.uk Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ 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] nearest stations in distance matrix
Dear R-user, Is there a way in R to locate the nearest 5 indices to a station, based on distances in a distance matrix. In other words i want to have nearest stations based on the distances in the matrix. The distance matrix, i have, has dimension 44*44. Thankyou very much in advance Elisa [[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] nearest stations in distance matrix
Dear Eliza, If you have the coordinates of the stations you can use the nnwhich() function from the spatstat package. Best regards, ir. Thierry Onkelinx Instituut voor natuur- en bosonderzoek / Research Institute for Nature and Forest team Biometrie Kwaliteitszorg / team Biometrics Quality Assurance Kliniekstraat 25 1070 Anderlecht Belgium + 32 2 525 02 51 + 32 54 43 61 85 thierry.onkel...@inbo.be www.inbo.be To call in the statistician after the experiment is done may be no more than asking him to perform a post-mortem examination: he may be able to say what the experiment died of. ~ Sir Ronald Aylmer Fisher The plural of anecdote is not data. ~ Roger Brinner The combination of some data and an aching desire for an answer does not ensure that a reasonable answer can be extracted from a given body of data. ~ John Tukey -Oorspronkelijk bericht- Van: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] Namens eliza botto Verzonden: maandag 15 april 2013 12:36 Aan: r-help@r-project.org Onderwerp: [R] nearest stations in distance matrix Dear R-user, Is there a way in R to locate the nearest 5 indices to a station, based on distances in a distance matrix. In other words i want to have nearest stations based on the distances in the matrix. The distance matrix, i have, has dimension 44*44. Thankyou very much in advance Elisa [[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. * * * * * * * * * * * * * D I S C L A I M E R * * * * * * * * * * * * * Dit bericht en eventuele bijlagen geven enkel de visie van de schrijver weer en binden het INBO onder geen enkel beding, zolang dit bericht niet bevestigd is door een geldig ondertekend document. The views expressed in this message and any annex are purely those of the writer and may not be regarded as stating an official position of INBO, as long as the message is not confirmed by a duly signed document. __ 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] Fwd: Dotchart per groups
Hi all, I would like to ask you for advice. I did a dotplot - using dotchart function. There are two localites (loc) with values 75 or 56 in my data ZZ. The f column has 4 levels: P1, S1, S8, R6. The dataframe is ordered by N value, pchloc value is assign to use pch in plot. head(ZZ) loc fN color ordered pchloc 98 75 S1 6.39 green 1 16 99 75 S8 6.44 blue 2 16 71 75 S8 6.60 blue 3 16 100 75 R6 6.64 black 4 16 70 75 S1 6.81 green 5 16 51 75 S8 6.83 blue 6 16 dput(ZZ) structure(list(loc = c(75L, 75L, 75L, 75L, 75L, 75L, 75L, 75L, 56L, 56L, 75L, 75L, 75L, 75L, 75L, 75L, 75L, 56L, 75L, 75L, 56L, 56L, 75L, 75L, 75L, 56L, 75L, 56L, 56L, 75L, 75L, 75L, 75L, 75L, 75L, 56L, 56L, 75L, 75L, 75L, 75L, 56L, 75L, 75L, 75L, 75L, 56L, 56L, 56L, 56L, 75L, 75L, 75L, 75L, 75L, 75L, 75L, 56L, 75L, 56L, 75L, 75L, 75L, 75L, 75L, 56L, 75L, 75L, 75L, 75L, 56L, 75L, 75L, 56L, 75L, 56L, 75L, 75L, 56L, 56L, 75L, 75L, 75L, 75L, 75L, 75L, 56L, 75L, 56L, 75L, 75L, 75L, 75L, 75L, 56L, 75L, 56L, 75L, 56L, 75L, 75L, 75L, 75L, 75L, 75L, 75L, 56L, 75L, 75L, 75L, 75L, 75L, 56L, 75L, 56L, 56L, 56L, 75L, 75L, 56L, 75L, 56L, 75L, 56L, 56L, 56L, 56L, 56L), f = structure(c(2L, 3L, 3L, 4L, 2L, 3L, 4L, 3L, 2L, 3L, 2L, 2L, 2L, 2L, 3L, 2L, 3L, 2L, 4L, 3L, 1L, 3L, 1L, 3L, 3L, 4L, 3L, 2L, 4L, 4L, 3L, 2L, 2L, 4L, 2L, 3L, 4L, 2L, 3L, 3L, 3L, 4L, 4L, 2L, 3L, 3L, 3L, 2L, 3L, 2L, 3L, 1L, 3L, 4L, 2L, 2L, 2L, 3L, 2L, 2L, 3L, 4L, 2L, 2L, 4L, 1L, 4L, 1L, 2L, 2L, 3L, 1L, 4L, 2L, 3L, 2L, 3L, 1L, 3L, 4L, 1L, 3L, 1L, 4L, 1L, 1L, 1L, 2L, 1L, 1L, 2L, 1L, 1L, 1L, 2L, 1L, 3L, 1L, 1L, 1L, 1L, 1L, 1L, 4L, 4L, 4L, 1L, 4L, 4L, 4L, 4L, 1L, 4L, 4L, 1L, 4L, 2L, 1L, 4L, 3L, 4L, 4L, 1L, 4L, 1L, 4L, 1L, 1L), .Label = c(P1, S1, S8, R6), class = factor), N = c(6.39, 6.44, 6.6, 6.64, 6.81, 6.83, 6.84, 6.91, 6.93, 6.93, 6.94, 6.95, 7.01, 7.04, 7.12, 7.16, 7.24, 7.26, 7.26, 7.27, 7.29, 7.34, 7.34, 7.34, 7.41, 7.42, 7.42, 7.43, 7.43, 7.43, 7.44, 7.44, 7.45, 7.47, 7.49, 7.51, 7.52, 7.53, 7.53, 7.54, 7.54, 7.55, 7.55, 7.57, 7.57, 7.58, 7.59, 7.61, 7.61, 7.65, 7.67, 7.68, 7.69, 7.69, 7.7, 7.72, 7.72, 7.73, 7.75, 7.76, 7.78, 7.78, 7.78, 7.8, 7.81, 7.86, 7.86, 7.88, 7.92, 7.94, 7.95, 7.95, 7.98, 8, 8, 8.01, 8.02, 8.04, 8.11, 8.16, 8.17, 8.22, 8.23, 8.23, 8.26, 8.27, 8.31, 8.35, 8.37, 8.37, 8.41, 8.44, 8.45, 8.47, 8.49, 8.52, 8.55, 8.59, 8.62, 8.71, 8.71, 8.72, 8.78, 8.85, 8.91, 8.95, 9.02, 9.07, 9.07, 9.1, 9.12, 9.17, 9.19, 9.34, 9.36, 9.46, 9.46, 9.52, 9.63, 9.82, 10.01, 10.33, 10.33, 10.49, 10.88, 11.01, 11.24, 12.6), color = c(green, blue, blue, black, green, blue, black, blue, green, blue, green, green, green, green, blue, green, blue, green, black, blue, red, blue, red, blue, blue, black, blue, green, black, black, blue, green, green, black, green, blue, black, green, blue, blue, blue, black, black, green, blue, blue, blue, green, blue, green, blue, red, blue, black, green, green, green, blue, green, green, blue, black, green, green, black, red, black, red, green, green, blue, red, black, green, blue, green, blue, red, blue, black, red, blue, red, black, red, red, red, green, red, red, green, red, red, red, green, red, blue, red, red, red, red, red, red, black, black, black, red, black, black, black, black, red, black, black, red, black, green, red, black, blue, black, black, red, black, red, black, red, red), ordered = c(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99, 100, 101, 102, 103, 104, 105, 106, 107, 108, 109, 110, 111, 112, 113, 114, 115, 116, 117, 118, 119, 120, 121, 122, 123, 124, 125, 126, 127, 128), pchloc = c(16, 16, 16, 16, 16, 16, 16, 16, 1, 1, 16, 16, 16, 16, 16, 16, 16, 1, 16, 16, 1, 1, 16, 16, 16, 1, 16, 1, 1, 16, 16, 16, 16, 16, 16, 1, 1, 16, 16, 16, 16, 1, 16, 16, 16, 16, 1, 1, 1, 1, 16, 16, 16, 16, 16, 16, 16, 1, 16, 1, 16, 16, 16, 16, 16, 1, 16, 16, 16, 16, 1, 16, 16, 1, 16, 1, 16, 16, 1, 1, 16, 16, 16, 16, 16, 16, 1, 16, 1, 16, 16, 16, 16, 16, 1, 16, 1, 16, 1, 16, 16, 16, 16, 16, 16, 16, 1, 16, 16, 16, 16, 16, 1, 16, 1, 1, 1, 16, 16, 1, 16, 1, 16, 1, 1, 1, 1, 1)), .Names = c(loc, f, N, color, ordered, pchloc), row.names = c(98L, 99L, 71L, 100L, 70L, 51L, 68L, 83L, 34L, 35L, 82L, 50L, 42L, 46L, 63L, 62L, 43L, 6L, 44L, 87L, 5L, 31L, 61L, 79L, 55L, 36L, 67L, 2L, 32L, 108L, 47L, 66L, 86L, 88L, 122L, 11L, 16L, 94L, 95L, 115L, 123L, 24L, 80L, 78L, 111L, 103L, 3L, 22L, 23L, 10L, 107L, 97L, 75L, 96L, 106L, 102L, 118L, 7L, 110L, 30L, 59L, 76L, 114L, 74L, 104L, 1L, 64L, 73L, 54L, 58L, 19L, 69L, 92L, 14L, 119L, 18L, 127L, 41L, 15L, 8L, 49L, 91L, 65L, 120L, 113L, 93L, 17L, 126L, 29L, 105L, 90L, 77L, 81L, 45L, 38L, 85L, 39L, 53L, 9L, 121L, 125L, 109L,
Re: [R] how to transform string to Camel Case?
On Mon, Apr 15, 2013 at 9:25 AM, Henrik Bengtsson h...@biostat.ucsf.edu wrote:
See for instance capitalize() in the R.utils package.
Unfortunately this also fails when NA values are present:
z - c(R project, hello world, something Else, NA)
R.utils::capitalize(z)
[1] R project Hello worldSomething Else NANA
Any pointers how to work around that? Thanks,
Liviu
Henrik
On Apr 14, 2013 11:51 PM, Liviu Andronic landronim...@gmail.com wrote:
Dear all,
Given the following vector:
(z - c('R project', 'hello world', 'something Else'))
[1] R project hello worldsomething Else
I know how to obtain all capitals or all lower case letters:
tolower(z)
[1] r project hello worldsomething else
toupper(z)
[1] R PROJECT HELLO WORLDSOMETHING ELSE
I saw the tocamel() function in 'rapport', but it doesn't do what I
want to achieve as it actually proceeds to camelCase/CamelCase the
strings:
tocamel(z)
[1] RProject helloWorldsomethingElse
But how should I proceed to obtain Camel Case? Here's what I'd like to
get:
c('R Project', 'Hello World', 'Something Else')
Regards,
Liviu
--
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Re: [R] how to transform string to Camel Case?
On Mon, Apr 15, 2013 at 9:20 AM, Pascal Oettli kri...@ymail.com wrote:
There is an example of how to do do what you are looking for in ?toupper
Unfortunately this fails when NA values are present:
z - c(R project, hello world, something Else, NA)
tocapwords(z)
[1] R Project Hello WorldSomething Else NANA
Liviu
Regards,
Pascal
On 04/15/2013 03:50 PM, Liviu Andronic wrote:
Dear all,
Given the following vector:
(z - c('R project', 'hello world', 'something Else'))
[1] R project hello worldsomething Else
I know how to obtain all capitals or all lower case letters:
tolower(z)
[1] r project hello worldsomething else
toupper(z)
[1] R PROJECT HELLO WORLDSOMETHING ELSE
I saw the tocamel() function in 'rapport', but it doesn't do what I
want to achieve as it actually proceeds to camelCase/CamelCase the
strings:
tocamel(z)
[1] RProject helloWorldsomethingElse
But how should I proceed to obtain Camel Case? Here's what I'd like to
get:
c('R Project', 'Hello World', 'Something Else')
Regards,
Liviu
--
Do you know how to read?
http://www.alienetworks.com/srtest.cfm
http://goodies.xfce.org/projects/applications/xfce4-dict#speed-reader
Do you know how to write?
http://garbl.home.comcast.net/~garbl/stylemanual/e.htm#e-mail
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] how to transform string to Camel Case?
On Mon, Apr 15, 2013 at 10:10 AM, Gergely Daróczi gerg...@snowl.net wrote: Dear Liviu, I have just updated tocamel to have a new argument, so the development version of the package would produce: tocamel(z, upper = TRUE, sep = ' ') [1] R Project Hello WorldSomething Else Thanks for putting this in. Unfortunately as with other approaches it fails on NA values: z - c(R project, hello world, something Else, NA) tocamel(z, sep=' ', upper=T) [1] R Project Hello WorldSomething Else NANA Can this be fixed? Thanks, Liviu __ 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 transform string to Camel Case?
Liviu Andronic landronim...@gmail.com writes:
On Mon, Apr 15, 2013 at 9:25 AM, Henrik Bengtsson h...@biostat.ucsf.edu
wrote:
See for instance capitalize() in the R.utils package.
Unfortunately this also fails when NA values are present:
z - c(R project, hello world, something Else, NA)
R.utils::capitalize(z)
[1] R project Hello worldSomething Else NANA
Any pointers how to work around that? Thanks,
would
R.utils::capitalize(z[is.character(z)])
work in your case?
Rainer
Liviu
Henrik
On Apr 14, 2013 11:51 PM, Liviu Andronic landronim...@gmail.com wrote:
Dear all,
Given the following vector:
(z - c('R project', 'hello world', 'something Else'))
[1] R project hello worldsomething Else
I know how to obtain all capitals or all lower case letters:
tolower(z)
[1] r project hello worldsomething else
toupper(z)
[1] R PROJECT HELLO WORLDSOMETHING ELSE
I saw the tocamel() function in 'rapport', but it doesn't do what I
want to achieve as it actually proceeds to camelCase/CamelCase the
strings:
tocamel(z)
[1] RProject helloWorldsomethingElse
But how should I proceed to obtain Camel Case? Here's what I'd like to
get:
c('R Project', 'Hello World', 'Something Else')
Regards,
Liviu
--
Do you know how to read?
http://www.alienetworks.com/srtest.cfm
http://goodies.xfce.org/projects/applications/xfce4-dict#speed-reader
Do you know how to write?
http://garbl.home.comcast.net/~garbl/stylemanual/e.htm#e-mail
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#secure method=pgpmime mode=sign
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Tel : +33 - (0)9 53 10 27 44
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Re: [R] how to transform string to Camel Case?
On 15 April 2013 14:10, Liviu Andronic landronim...@gmail.com wrote: On Mon, Apr 15, 2013 at 10:10 AM, Gergely Daróczi gerg...@snowl.net wrote: Dear Liviu, I have just updated tocamel to have a new argument, so the development version of the package would produce: tocamel(z, upper = TRUE, sep = ' ') [1] R Project Hello WorldSomething Else Thanks for putting this in. Unfortunately as with other approaches it fails on NA values: z - c(R project, hello world, something Else, NA) tocamel(z, sep=' ', upper=T) [1] R Project Hello WorldSomething Else NANA Can this be fixed? Thanks, Liviu I have added an extra check in the function for NA values before applying `paste` at https://github.com/Rapporter/rapport/compare/34ca6a35fb...a04abc8b21 Alex might not like it :) Example: tocamel(z, upper = TRUE, sep = ' ') [1] R Project Hello WorldSomething Else NA Best, Gergely [[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] Indices of lowest values in matrix
Dear R users,Sorry for such a basic question. I really need to know that how can i pick the indices of 5 lowest values from each row of a matrix with dimensions 12*12??Thank you very much in advance Elisa [[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] importing and merging many time series
On Sun, Apr 7, 2013 at 7:40 AM, Anton Lebedevich mab...@gmail.com wrote:
Hello.
I've got many (5-20k) files with time series in a text format like this:
1359635460 2.006747
1359635520 1.886745
1359635580 3.066988
1359635640 3.633578
1359635700 2.140082
1359635760 2.033564
1359635820 1.980123
1359635880 2.060131
1359635940 2.113416
1359636000 2.440172
First field is a unix timestamp, second is a float number. Its a text
export of http://graphite.readthedocs.org/en/latest/whisper.html
databases. Time series could have different resolutions, start/end
times, and possibly gaps inside.
Current way of importing them:
read.file - function(file.name) {
read.zoo(
file.name,
na.strings=None,
colClasses=c(integer, numeric),
col.names=c(time, basename(file.name)),
FUN=function(t) {as.POSIXct(t, origin=1970-01-01 00:00.00, tz=UTC)},
drop=FALSE)
}
load.metrics - function(path=.) {
do.call(merge.zoo, lapply(list.files(path, full.names=TRUE), read.file))
}
It works for 6k time series with 2k points in each, but fails with out
of memory error on 16Gb box when I try to import 10k time series with
10k points.
You're trying to merge 10,000 objects in a single call. I'm not
surprised you run out of RAM.
I've tried to make merging incremental by using Reduce but import speed
became unacceptable:
This is similar to growing an object in a for loop, which is also slow.
load.metrics - function(path=.) {
Reduce(
function(a, b) {
if (class(a) == character) {
a - read.file(a)
}
merge.zoo(a, read.file(b))
},
list.files(path, full.names=TRUE))
}
Is there faster and less memory consuming way to import and merge a lot
of time series?
Try something in between the two extremes (merging all objects at
once, versus merging every new object with the accumulated object).
For example, try merging 100-1000 objects at a time.
You might also benefit from converting your objects to xts, so you can
use xts' optimized merge. You can always convert the final object
back to zoo.
Regards,
Anton Lebedevich.
Best,
--
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FOSS Trading | www.fosstrading.com
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Re: [R] Indices of lowest values in matrix
On Apr 15, 2013, at 14:27 , eliza botto wrote: Dear R users,Sorry for such a basic question. I really need to know that how can i pick the indices of 5 lowest values from each row of a matrix with dimensions 12*12??Thank you very much in advance Something like this? m [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [1,] 1.2 -2.5 0.4 -0.2 1.4 -1.8 -1.6 -0.1 -0.1 -0.3 -0.2 -0.5 [2,] -0.2 -1.9 0.9 0.0 -1.0 0.2 -0.8 0.7 -1.5 -1.4 1.8 0.4 [3,] -0.9 0.7 0.4 -0.4 -1.6 0.6 1.1 -0.9 2.3 1.0 0.6 0.8 [4,] 1.4 -0.7 0.0 -2.3 0.4 1.0 1.5 -0.8 1.0 -0.1 0.6 0.3 [5,] 0.8 -0.1 1.3 -0.1 1.8 -0.7 -1.4 -0.9 0.6 0.7 0.3 1.4 [6,] 0.4 0.0 0.6 -0.4 0.7 -0.3 1.5 1.6 -1.9 -1.1 0.3 0.6 [7,] 0.5 -0.4 -1.9 -2.4 0.3 0.3 0.1 0.6 -0.9 0.2 0.0 1.4 [8,] -0.2 0.4 -1.4 -0.1 -0.1 1.2 0.2 0.0 -0.9 0.4 -0.3 -2.0 [9,] -0.3 1.0 -0.7 0.9 -1.7 -0.8 1.8 -0.4 0.1 0.2 0.4 1.2 [10,] -0.6 0.7 -0.1 0.3 1.4 0.7 -0.1 1.0 -0.8 0.9 0.0 -0.6 [11,] -0.1 0.2 0.4 -0.3 -1.5 -0.2 1.0 1.0 -0.5 -0.4 -1.2 1.1 [12,] -1.1 0.3 0.7 -0.9 0.2 -1.0 0.2 -1.1 -1.2 -0.3 0.8 -0.9 apply(m,1,order)[1:5,] [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [1,]2254794 125 9 5 9 [2,]69188 10336 111 1 [3,]7 10826499312 9 8 [4,] 1254 10262 118 310 6 [5,] 1073342 1111 7 4 4 -- Peter Dalgaard, Professor Center for Statistics, Copenhagen Business School Solbjerg Plads 3, 2000 Frederiksberg, Denmark Phone: (+45)38153501 Email: pd@cbs.dk Priv: pda...@gmail.com __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] how to transform string to Camel Case?
On Mon, Apr 15, 2013 at 2:22 PM, Gergely Daróczi gerg...@snowl.net wrote: I have added an extra check in the function for NA values before applying `paste` at https://github.com/Rapporter/rapport/compare/34ca6a35fb...a04abc8b21 Alex might not like it :) Example: tocamel(z, upper = TRUE, sep = ' ') [1] R Project Hello WorldSomething Else NA Thanks! Now I'm getting somewhere: z - c(R project, hello world, something Else, something-else, NA) tocamel(z, sep=' ', upper=T, delim= ) [1] R Project Hello WorldSomething Else Something-else [5] NA I hope Alex likes your fixes. Thanks, Liviu __ 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] Sorting data.frame and again sorting within data.frame
library(plyr) arrange(df,names,desc(dates)) # names dates values #1 A 4/15/2013 31 #2 A 4/14/2013 102 #3 A 4/13/2013 31 #4 B 4/15/2013 34 #5 B 4/14/2013 47 #6 B 4/13/2013 17 #7 C 4/15/2013 10 #8 C 4/14/2013 29 #9 C 4/13/2013 11 #or df[with(df,order(names,desc(dates))),] # names dates values #3 A 4/15/2013 31 #7 A 4/14/2013 102 #2 A 4/13/2013 31 #6 B 4/15/2013 34 #8 B 4/14/2013 47 #4 B 4/13/2013 17 #1 C 4/15/2013 10 #9 C 4/14/2013 29 #5 C 4/13/2013 11 A.K. - Original Message - From: Katherine Gobin katherine_go...@yahoo.com To: r-help@r-project.org Cc: Sent: Monday, April 15, 2013 2:01 AM Subject: [R] Sorting data.frame and again sorting within data.frame Dear R forum, I have a data.frame as defied below - df = data.frame(names = c(C, A, A, B, C, B, A, B, C), dates = c(4/15/2013, 4/13/2013, 4/15/2013, 4/13/2013, 4/13/2013, 4/15/2013, 4/14/2013, 4/14/2013,4/14/2013 ),values = c(10, 31, 31, 17, 11, 34, 102, 47, 29)) df names dates values 1 C 4/15/2013 10 2 A 4/13/2013 31 3 A 4/15/2013 31 4 B 4/13/2013 17 5 C 4/13/2013 11 6 B 4/15/2013 34 7 A 4/14/2013 102 8 B 4/14/2013 47 9 C 4/14/2013 29 I need to sort df first on names in increasing order and then further on dates in a decreasing order i.e. I need names dates values A 4/15/2013 31 A 4/14/2013 102 A 4/13/2013 31 B 4/15/2013 34 B 4/14/2013 47 B 4/13/2013 17 C 4/15/2013 10 C 4/14/2013 29 C 4/13/2013 11 I tried df_sorted = df[order(df$names, (as.Date(df$dates, %m/%d/%Y)), decreasing = TRUE),] df_sorted names dates values 1 C 4/15/2013 10 9 C 4/14/2013 29 5 C 4/13/2013 11 6 B 4/15/2013 34 8 B 4/14/2013 47 4 B 4/13/2013 17 3 A 4/15/2013 31 7 A 4/14/2013 102 2 A 4/13/2013 31 I need A to appear first with all three corresponding dates in decreasing order, then B and so on. Please guide. With regards Katherine [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Sorting data.frame and again sorting within data.frame
You can also use: df[order(df$names,-xtfrm(df$dates),decreasing=FALSE),] # names dates values #3 A 4/15/2013 31 #7 A 4/14/2013 102 #2 A 4/13/2013 31 #6 B 4/15/2013 34 #8 B 4/14/2013 47 #4 B 4/13/2013 17 #1 C 4/15/2013 10 #9 C 4/14/2013 29 #5 C 4/13/2013 11 A.K. - Original Message - From: arun smartpink...@yahoo.com To: Katherine Gobin katherine_go...@yahoo.com Cc: R help r-help@r-project.org Sent: Monday, April 15, 2013 8:57 AM Subject: Re: [R] Sorting data.frame and again sorting within data.frame library(plyr) arrange(df,names,desc(dates)) # names dates values #1 A 4/15/2013 31 #2 A 4/14/2013 102 #3 A 4/13/2013 31 #4 B 4/15/2013 34 #5 B 4/14/2013 47 #6 B 4/13/2013 17 #7 C 4/15/2013 10 #8 C 4/14/2013 29 #9 C 4/13/2013 11 #or df[with(df,order(names,desc(dates))),] # names dates values #3 A 4/15/2013 31 #7 A 4/14/2013 102 #2 A 4/13/2013 31 #6 B 4/15/2013 34 #8 B 4/14/2013 47 #4 B 4/13/2013 17 #1 C 4/15/2013 10 #9 C 4/14/2013 29 #5 C 4/13/2013 11 A.K. - Original Message - From: Katherine Gobin katherine_go...@yahoo.com To: r-help@r-project.org Cc: Sent: Monday, April 15, 2013 2:01 AM Subject: [R] Sorting data.frame and again sorting within data.frame Dear R forum, I have a data.frame as defied below - df = data.frame(names = c(C, A, A, B, C, B, A, B, C), dates = c(4/15/2013, 4/13/2013, 4/15/2013, 4/13/2013, 4/13/2013, 4/15/2013, 4/14/2013, 4/14/2013,4/14/2013 ),values = c(10, 31, 31, 17, 11, 34, 102, 47, 29)) df names dates values 1 C 4/15/2013 10 2 A 4/13/2013 31 3 A 4/15/2013 31 4 B 4/13/2013 17 5 C 4/13/2013 11 6 B 4/15/2013 34 7 A 4/14/2013 102 8 B 4/14/2013 47 9 C 4/14/2013 29 I need to sort df first on names in increasing order and then further on dates in a decreasing order i.e. I need names dates values A 4/15/2013 31 A 4/14/2013 102 A 4/13/2013 31 B 4/15/2013 34 B 4/14/2013 47 B 4/13/2013 17 C 4/15/2013 10 C 4/14/2013 29 C 4/13/2013 11 I tried df_sorted = df[order(df$names, (as.Date(df$dates, %m/%d/%Y)), decreasing = TRUE),] df_sorted names dates values 1 C 4/15/2013 10 9 C 4/14/2013 29 5 C 4/13/2013 11 6 B 4/15/2013 34 8 B 4/14/2013 47 4 B 4/13/2013 17 3 A 4/15/2013 31 7 A 4/14/2013 102 2 A 4/13/2013 31 I need A to appear first with all three corresponding dates in decreasing order, then B and so on. Please guide. With regards Katherine [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Kruskal-Wallis
Hi, set.seed(25) myFile1-as.data.frame(matrix(sample(1:40,50,replace=TRUE),nrow=10)) row.names(myFile1)- LETTERS[1:10] groups - rep (0:1, c(3,2)) kruskal-apply(myFile1,1,kruskal.test,groups) p_kruskal - sapply(kruskal, function(x) x$p.value) p_kruskal # A B C D E F G #0.08326452 0.08326452 0.56370286 0.56370286 0.24821308 1. 0.08326452 # H I J #1. 0.37425932 0.56370286 #or sapply(seq_len(nrow(myFile1)),function(i) kruskal.test(unlist(myFile1[i,]),groups)$p.value) [1] 0.08326452 0.08326452 0.56370286 0.56370286 0.24821308 1. [7] 0.08326452 1. 0.37425932 0.56370286 A.K. - Original Message - From: Chintanu chint...@gmail.com To: R help r-help@r-project.org Cc: Sent: Monday, April 15, 2013 1:18 AM Subject: [R] Kruskal-Wallis Hi, I have got two groups of samples; and for every row, I wish to calculate Kruskal-Wallis' p-value. In the example below, and the stars () show where I am struggling to design and put things together. Any help would be appreciated. myFile - data.frame(Sample_1a = 1:10, Sample_1b = 2:11, Sample_1c = 3:12, Sample_2a=4:13, Sample_2b=7:16, row.names=LETTERS[1:10]) groups - rep (0:1, c(3,2)) kruskal - apply(myFile [1:nrow(myFile),], 1, kruskal.test, **) p_kruskal - sapply(kruskal, function(x) x$p.value) Thanks, Chintanu [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Kruskal-Wallis
One statistical point beyond A.K.'s well done response. As you should well know, Kruskal-Wallis is a non-parametric equivalent of ANOVA. However, you only have two groups and do not require an ANOVA approach. You could simply use a Mann-Whitney U (aka.. independent Wilcoxon) test using wilcox.test(). Charles On Mon, Apr 15, 2013 at 8:23 AM, arun smartpink...@yahoo.com wrote: Hi, set.seed(25) myFile1-as.data.frame(matrix(sample(1:40,50,replace=TRUE),nrow=10)) row.names(myFile1)- LETTERS[1:10] groups - rep (0:1, c(3,2)) kruskal-apply(myFile1,1,kruskal.test,groups) p_kruskal - sapply(kruskal, function(x) x$p.value) p_kruskal # A B C D E F G #0.08326452 0.08326452 0.56370286 0.56370286 0.24821308 1. 0.08326452 #H I J #1. 0.37425932 0.56370286 #or sapply(seq_len(nrow(myFile1)),function(i) kruskal.test(unlist(myFile1[i,]),groups)$p.value) [1] 0.08326452 0.08326452 0.56370286 0.56370286 0.24821308 1. [7] 0.08326452 1. 0.37425932 0.56370286 A.K. - Original Message - From: Chintanu chint...@gmail.com To: R help r-help@r-project.org Cc: Sent: Monday, April 15, 2013 1:18 AM Subject: [R] Kruskal-Wallis Hi, I have got two groups of samples; and for every row, I wish to calculate Kruskal-Wallis' p-value. In the example below, and the stars () show where I am struggling to design and put things together. Any help would be appreciated. myFile - data.frame(Sample_1a = 1:10, Sample_1b = 2:11, Sample_1c = 3:12, Sample_2a=4:13, Sample_2b=7:16, row.names=LETTERS[1:10]) groups - rep (0:1, c(3,2)) kruskal - apply(myFile [1:nrow(myFile),], 1, kruskal.test, **) p_kruskal - sapply(kruskal, function(x) x$p.value) Thanks, Chintanu [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Charles Determan Integrated Biosciences PhD Student University of Minnesota [[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] Overlay two stat_ecdf() plots
I want to plot two scdf-plots in the same graph. I have two input tables with one column each: Targets - read.table(/media/, sep=, header=T) NonTargets - read.table(/media/..., sep=, header=T) head(Targets) V1 1 3.160514 2 6.701948 3 4.093844 4 1.992014 5 1.604751 6 2.076802 head(NonTargets) V1 1 3.895934 2 1.990506 3 -1.746919 4 -3.451477 5 5.156554 6 1.195109 Targets.m - melt(Targets) head(Targets.m) variablevalue 1 V1 3.160514 2 V1 6.701948 3 V1 4.093844 4 V1 1.992014 5 V1 1.604751 6 V1 2.076802 NonTargets.m - melt(NonTargets) head(NonTargets.m) variable value 1 V1 3.895934 2 V1 1.990506 3 V1 -1.746919 4 V1 -3.451477 5 V1 5.156554 6 V1 1.195109 How do I proceed to plot them in one Graph using ecdf_stat() [[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] need help with R
hey there can i email questions to this address to get help with using R thanks sam [[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] (no subject)
Hi, I'm trying to decide between doing a FA or PCA and would appreciate some pointers. I've got a questionnaire with latent items which the participants answered on a Likert scale, and all I want to do at this point is to explore the data and extract a number of factors/components. Would FA or PCA be most appropriate in this case? Cheers, Hannah -- View this message in context: http://r.789695.n4.nabble.com/no-subject-tp4664248.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] Optimisation and NaN Errors using clm() and clmm()
Dear List, I am using both the clm() and clmm() functions from the R package 'ordinal'. I am fitting an ordinal dependent variable with 5 categories to 9 continuous predictors, all of which have been normalised (mean subtracted then divided by standard deviation), using a probit link function. From this global model I am generating a confidence set of 200 models using clm() and the 'glmulti' R package. This produces these errors: / model.2.10 - glmulti(as.factor(dependent) ~ predictor_1*predictor_2*predictor_3*predictor_4*predictor_5*predictor_6*predictor_7*predictor_8*predictor_9, data = database, fitfunc = clm, link = probit, method = g, crit = aicc, confsetsize = 200, marginality = TRUE) ... After 670 generations: Best model: as.factor(dependent)~1+predictor_1+predictor_2+predictor_3+predictor_4+predictor_5+predictor_6+predictor_8+predictor_9+predictor_4:predictor_3+predictor_6:predictor_2+predictor_8:predictor_5+predictor_9:predictor_1+predictor_9:predictor_4+predictor_9:predictor_5+predictor_9:predictor_6 Crit= 183.716706496392 Mean crit= 202.022138576506 Improvements in best and average IC have bebingo en below the specified goals. Algorithm is declared to have converged. Completed. There were 24 warnings (use warnings() to see them) warnings() Warning messages: 1: optimization failed: step factor reduced below minimum 2: optimization failed: step factor reduced below minimum 3: optimization failed: step factor reduced below minimum/ etc. I am then re-fitting each of the 200 models with the clmm() function, with 2 random factors (family nested within order). I get this error in a few of the re-fitted models: / model.2.glmm.2 - clmm(as.factor(dependent) ~ 1 + predictor_1 + predictor_2 + predictor_3 + predictor_6 + predictor_7 + predictor_8 + predictor_9 + predictor_6:predictor_2 + predictor_7:predictor_2 + predictor_7:predictor_3 + predictor_8:predictor_2 + predictor_9:predictor_1 + predictor_9:predictor_2 + predictor_9:predictor_3 + predictor_9:predictor_6 + predictor_9:predictor_7 + predictor_9:predictor_8+ (1|order/family), link = probit, data = database) summary(model.2.glmm.2) Cumulative Link Mixed Model fitted with the Laplace approximation formula: as.factor(dependent) ~ 1 + predictor_1 + predictor_2 + predictor_3 + predictor_6 + predictor_7 + predictor_8 + predictor_9 + predictor_6:predictor_2 + predictor_7:predictor_2 + predictor_7:predictor_3 + predictor_8:predictor_2 + predictor_9:predictor_1 + predictor_9:predictor_2 + predictor_9:predictor_3 + predictor_9:predictor_6 + predictor_9:predictor_7 + predictor_9:predictor_8 + (1 | order/family) data: database link threshold nobs logLik AIC niter max.grad cond.H probit flexible 103 -65.56 173.13 58(3225) 8.13e-06 4.3e+03 Random effects: Var Std.Dev family:order 7.493e-11 8.656e-06 order 1.917e-12 1.385e-06 Number of groups: family:order 12, order 4 Coefficients: Estimate Std. Error z value Pr(|z|) predictor_1 0.40802 0.78685 0.519 0.6041 predictor_2 0.02431 0.26570 0.092 0.9271 predictor_3 -0.84486 0.32056 -2.636 0.0084 ** predictor_6 0.65392 0.34348 1.904 0.0569 . predictor_7 0.71730 0.29596 2.424 0.0154 * predictor_8 -1.37692 0.75660 -1.820 0.0688 . predictor_9 0.15642 0.28969 0.540 0.5892 predictor_2:predictor_6 -0.46880 0.18829 -2.490 0.0128 * predictor_2:predictor_7 4.97365 0.82692 6.015 1.80e-09 *** predictor_3:predictor_7 -1.13192 0.46639 -2.427 0.0152 * predictor_2:predictor_8 -5.52913 0.88476 -6.249 4.12e-10 *** predictor_1:predictor_9 4.28519 NA NA NA predictor_2:predictor_9 -0.26558 0.10541 -2.520 0.0117 * predictor_3:predictor_9 -1.49790 NA NA NA predictor_6:predictor_9 -1.31538 NA NA NA predictor_7:predictor_9 -4.41998 NA NA NA predictor_8:predictor_9 3.99709 NA NA NA --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 Threshold coefficients: Estimate Std. Error z value 0|1 -0.2236 0.3072 -0.728 1|2 1.4229 0.3634 3.915 (211 observations deleted due to missingness) Warning message: In sqrt(diag(vc)[1:npar]) : NaNs produced/ I have tried a number of different approaches, each has its own problems. I have fixed these using various suggestions from online forums (eg https://stat.ethz.ch/pipermail/r-sig-mixed-models/2011q1/015328.html, https://stat.ethz.ch/pipermail/r-sig-mixed-models/2011q2/016165.html) and this is as good as I can get it. After the first stage (generating the model set with glmulti) I tested every model in the confidence set individually - there were no errors - but there was clearly a problem during the model selection process. Should I be worried? No errors appear in the top 5% of re-fitted models (which are the only ones I will be using) however I am concerned that errors may be indicative of a problem with my approach. A further worry is that the errors might be removing models that could otherwise be included. Any help would be much appreciated. Tom __ R-help@r-project.org mailing list
Re: [R] how to transform string to Camel Case?
HI,
You could use:
gsub((^|\\s+)([a-z]),\\1\\U\\2,z,perl=TRUE)
#[1] R Project Hello World Something Else
#or
gsub(\\b([a-z]),\\U\\1,z,perl=TRUE)
#[1] R Project Hello World Something Else
A.K.
- Original Message -
From: Liviu Andronic landronim...@gmail.com
To: r-help r-h...@stat.math.ethz.ch
Cc:
Sent: Monday, April 15, 2013 2:50 AM
Subject: [R] how to transform string to Camel Case?
Dear all,
Given the following vector:
(z - c('R project', 'hello world', 'something Else'))
[1] R project hello world something Else
I know how to obtain all capitals or all lower case letters:
tolower(z)
[1] r project hello world something else
toupper(z)
[1] R PROJECT HELLO WORLD SOMETHING ELSE
I saw the tocamel() function in 'rapport', but it doesn't do what I
want to achieve as it actually proceeds to camelCase/CamelCase the
strings:
tocamel(z)
[1] RProject helloWorld somethingElse
But how should I proceed to obtain Camel Case? Here's what I'd like to get:
c('R Project', 'Hello World', 'Something Else')
Regards,
Liviu
--
Do you know how to read?
http://www.alienetworks.com/srtest.cfm
http://goodies.xfce.org/projects/applications/xfce4-dict#speed-reader
Do you know how to write?
http://garbl.home.comcast.net/~garbl/stylemanual/e.htm#e-mail
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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] need help with R
Yes, but do read: http://www.R-project.org/posting-guide.html first. Sarah On Mon, Apr 15, 2013 at 2:09 AM, sam tingey sam.ting...@gmail.com wrote: hey there can i email questions to this address to get help with using R thanks sam __ 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. -- Sarah Goslee http://www.functionaldiversity.org __ 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] Cross validation for Naive Bayes and Bayes Networks
Hi Guilherme, On Sun, Apr 14, 2013 at 11:48 PM, Guilherme Ferraz de Arruda gu...@yahoo.com.br wrote: Hi, I need to classify, using Naive Bayes and Bayes Networks, and estimate their performance using cross validation. How can I do this? I tried the bnlearn package for Bayes Networks, althought I need to get more indexes, not only the error rate (precision, sensitivity, ...). You can do that using the object returned by bn.cv(), because it contains the predicted values and the indexes of the corresponding observations in the original data, for each fold. It's just a matter to reassemble observed and predicted class labels and compute your metrics. I also tried the *e1071* package, but I could not find a way to do cross-validation. You might be able to trick the tune() function to do it, but I am not sure. Marco -- Marco Scutari, Ph.D. Research Associate, Genetics Institute (UGI) University College London (UCL), United Kingdom __ 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] Indices of lowest values in matrix
Hi,
TRy this:
set.seed(30)
mat1- matrix(sample(1:50,12*12,replace=TRUE),ncol=12)
mat1
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12]
[1,] 5 28 13 4 47 3 30 16 3 39 42 9
[2,] 25 44 15 21 9 5 32 5 4 33 30 21
[3,] 19 12 24 2 7 37 36 3 13 9 33 33
[4,] 22 47 13 14 9 29 20 3 1 25 14 29
[5,] 16 13 41 22 15 39 50 44 10 50 32 32
[6,] 8 42 24 32 21 31 45 32 41 44 22 31
[7,] 45 31 31 39 4 25 1 19 28 16 20 48
[8,] 12 22 12 25 45 10 5 7 31 27 18 30
[9,] 49 8 25 10 6 30 8 29 14 10 48 44
[10,] 8 30 31 35 4 50 25 33 7 15 17 38
[11,] 4 2 15 40 30 49 44 14 3 33 41 12
[12,] 20 17 34 14 43 2 40 50 42 42 23 20
#If you need the lowest values:
sapply(seq_len(nrow(mat1)),function(i) {x1- mat1[i,]; x1[order(x1)][1:5]})
# [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12]
#[1,] 3 4 2 1 10 8 1 5 6 4 2 2
#[2,] 3 5 3 3 13 21 4 7 8 7 3 14
#[3,] 4 5 7 9 15 22 16 10 8 8 4 17
#[4,] 5 9 9 13 16 24 19 12 10 15 12 20
#[5,] 9 15 12 14 22 31 20 12 10 17 14 20
#indices
sapply(seq_len(nrow(mat1)),function(i) {order(mat1[i,])})[1:5,]
# [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12]
#[1,] 6 9 4 9 9 1 7 7 5 5 2 6
#[2,] 9 6 8 8 2 5 5 8 2 9 9 4
#[3,] 4 8 5 5 5 11 10 6 7 1 1 2
#[4,] 1 5 10 3 1 3 8 1 4 10 12 1
#[5,] 12 3 2 4 4 6 11 3 10 11 8 12
A.K.
- Original Message -
From: eliza botto eliza_bo...@hotmail.com
To: r-help@r-project.org r-help@r-project.org
Cc:
Sent: Monday, April 15, 2013 8:27 AM
Subject: [R] Indices of lowest values in matrix
Dear R users,Sorry for such a basic question. I really need to know that how
can i pick the indices of 5 lowest values from each row of a matrix with
dimensions 12*12??Thank you very much in advance
Elisa
[[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.
__
R-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] need help with R
Sam tingey wrote: can i email questions to this address to get help with using R Sarah Goslee replied; Yes, but do read: http://www.R-project.org/posting-guide.html Dieter Menne thought: Catch 42. When you are so good that you can create a reproducible example without using R, you only will post answers. -- View this message in context: http://r.789695.n4.nabble.com/need-help-with-R-tp4664259p4664264.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] (no subject)
On Apr 15, 2013, at 14:30 , ilovestats wrote: Hi, I'm trying to decide between doing a FA or PCA and would appreciate some pointers. I've got a questionnaire with latent items which the participants answered on a Likert scale, and all I want to do at this point is to explore the data and extract a number of factors/components. Would FA or PCA be most appropriate in this case? Cheers, Hannah Not really an R question, is it? Stats.StackExchange.com is - that way! In terms of theory, PCA is essentially FA with the same residual variance in all responses. With all-Likert scales, it is unlikely that there will be much of a difference. In practical terms: - factanal can diverge (Heywood cases) which is a bit of a bother on the other hand - factor rotation is based on factanal() output; may require a little extra diddling to work with prcomp(). I think I'd try factanal() first, and if it acts up, switch to prcomp(). -- Peter Dalgaard, Professor Center for Statistics, Copenhagen Business School Solbjerg Plads 3, 2000 Frederiksberg, Denmark Phone: (+45)38153501 Email: pd@cbs.dk Priv: pda...@gmail.com __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] as.date not reading Hour/Minute am/pm Data
Hi, Try: x- c(2/26/13 11:59 PM, 2/25/13 10:25 AM) strptime(x,%m/%d/%y %I:%M %p) #[1] 2013-02-26 23:59:00 2013-02-25 10:25:00 #or library(lubridate) parse_date_time(x,%m/%d/%y %I:%M %p) # 2 parsed with %m/%d/%y %I:%M %p #[1] 2013-02-26 23:59:00 UTC 2013-02-25 10:25:00 UTC A.K. I'm a newbie to R but not programming. I've researched online and through this forum and am unsure of why I'm not getting a as.date object to only store a limited amount of data. The date is in the format of: 2/26/13 11:59 PM in a .csv file with over 5000 data points. My as.date function reads: gld - read.csv(c:/data/gbpdata.csv, stringsAsFactors=F) gld_dates - as.Date(gld[,1], format=%m/%d/%y %H:%M %p) gld_dates: [36775] 2013-04-03 2013-04-03 2013-04-03 2013-04-03 2013-04-03 2013-04-03 2013-04-03 2013-04-03 2013-04-03 What am I doing wrong? Thanks in advance for your replies. __ 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] use of simulate.Arima (forecast package)
I would like to simulate some SARIMA models, e.g. a SARIMA (1,0,1)(1,0,1)[4] process. I installed the package 'forecast', where the function simulate.Arima should do what I am trying to do. I am not able to understand how it works Could somebody help me with an example? thank you Stefano Sofia AVVISO IMPORTANTE: Questo messaggio di posta elettronica può contenere informazioni confidenziali, pertanto è destinato solo a persone autorizzate alla ricezione. I messaggi di posta elettronica per i client di Regione Marche possono contenere informazioni confidenziali e con privilegi legali. Se non si è il destinatario specificato, non leggere, copiare, inoltrare o archiviare questo messaggio. Se si è ricevuto questo messaggio per errore, inoltrarlo al mittente ed eliminarlo completamente dal sistema del proprio computer. Ai sensi dell’art. 6 della DGR n. 1394/2008 si segnala che, in caso di necessità ed urgenza, la risposta al presente messaggio di posta elettronica può essere visionata da persone estranee al destinatario. IMPORTANT NOTICE: This e-mail message is intended to be received only by persons entitled to receive the confidential information it may contain. E-mail messages to clients of Regione Marche may contain information that is confidential and legally privileged. Please do not read, copy, forward, or store this message unless you are an intended recipient of it. If you have received this message in error, please forward it to the sender and delete it completely from your computer system. __ 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] Using R commander within R Studio (Mac)
Hi. I am a R studio user and would like to use RCommander within R studio. Each time I try to install and use Rcommander, I see the following message: Loading Tcl/Tk interface ... Error : .onLoad failed in loadNamespace() for 'tcltk', details: call: dyn.load(file, DLLpath = DLLpath, ...) error: unable to load shared object '/Library/Frameworks/R.framework/Versions/2.15/Resources/library/tcltk/libs/x86_64/tcltk.so': dlopen(/Library/Frameworks/R.framework/Versions/2.15/Resources/library/tcltk/libs/x86_64/tcltk.so, 10): Library not loaded: /usr/local/lib/libtcl8.5.dylib Referenced from: /Library/Frameworks/R.framework/Versions/2.15/Resources/library/tcltk/libs/x86_64/tcltk.so Reason: image not found In addition: Warning message: package Rcmdr was built under R version 2.15.3 Error: package/namespace load failed for Rcmdr I went to get X11 and installed it as I believed this was the problem, but I get the same message. Would anyone get me some clues on what is going on? [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] how to transform string to Camel Case?
On Mon, Apr 15, 2013 at 2:50 AM, Liviu Andronic landronim...@gmail.com wrote:
Dear all,
Given the following vector:
(z - c('R project', 'hello world', 'something Else'))
[1] R project hello worldsomething Else
I know how to obtain all capitals or all lower case letters:
tolower(z)
[1] r project hello worldsomething else
toupper(z)
[1] R PROJECT HELLO WORLDSOMETHING ELSE
I saw the tocamel() function in 'rapport', but it doesn't do what I
want to achieve as it actually proceeds to camelCase/CamelCase the
strings:
tocamel(z)
[1] RProject helloWorldsomethingElse
But how should I proceed to obtain Camel Case? Here's what I'd like to get:
c('R Project', 'Hello World', 'Something Else')
Here is a one liner using gsubfn from the gsubfn package
(http://gsubfn.googlecode.com). gsubfn is like gsubfn except the
second argument can be a function (or a list or a proto object). The
regular expression here matches a space followed by any character.
For each such match gsubfn will call the function denoted by the
second argument using the two parenthesized portions of the regular
expression as the two arguments. It supports an optional formula
notation to express the function for compactness so ... ~ toupper(..2)
denotes function(...) toupper(..2) which is equivalent to function(x,
y) toupper(y) . It replaces the input with the output of that
function. Finally we check if any of the components of the input are
NA and replace those with NA in the output:
replace(gsubfn(( )(.), ... ~ toupper(..2), z), is.na(z), NA)
[1] RProject helloWorldsomethingElse NA
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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 R commander within R Studio (Mac)
Dear Helene, I don't think that your problem has anything to do with RStudio. I suspect that you haven't installed Tcl/Tk for X-Windows on your Mac, as is required by the tcltk package for versions of R on the Mac prior to 3.0.0. Please read the Rcmdr installation instructions at http://socserv.socsci.mcmaster.ca/jfox/Misc/Rcmdr/installation-notes.html for detailed information (and see in particular the link to Mac OS X installation notes for the R Commander under older versions of R). That said, I suggest that upgrade to R 3.0.0. You should also know that while the Rcmdr runs under RStudio, there are some problems, particularly with the RStudio graphics device. I hope this helps, John John Fox Sen. William McMaster Prof. of Social Statistics Department of Sociology McMaster University Hamilton, Ontario, Canada http://socserv.mcmaster.ca/jfox/ On Mon, 15 Apr 2013 10:19:24 -0400 h moore helene.mo...@ryerson.ca wrote: Hi. I am a R studio user and would like to use RCommander within R studio. Each time I try to install and use Rcommander, I see the following message: Loading Tcl/Tk interface ... Error : .onLoad failed in loadNamespace() for 'tcltk', details: call: dyn.load(file, DLLpath = DLLpath, ...) error: unable to load shared object '/Library/Frameworks/R.framework/Versions/2.15/Resources/library/tcltk/libs/x86_64/tcltk.so': dlopen(/Library/Frameworks/R.framework/Versions/2.15/Resources/library/tcltk/libs/x86_64/tcltk.so, 10): Library not loaded: /usr/local/lib/libtcl8.5.dylib Referenced from: /Library/Frameworks/R.framework/Versions/2.15/Resources/library/tcltk/libs/x86_64/tcltk.so Reason: image not found In addition: Warning message: package Rcmdr was built under R version 2.15.3 Error: package/namespace load failed for Rcmdr I went to get X11 and installed it as I believed this was the problem, but I get the same message. Would anyone get me some clues on what is going on? [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] how to transform string to Camel Case?
On Mon, Apr 15, 2013 at 10:56 AM, Gabor Grothendieck
ggrothendi...@gmail.com wrote:
On Mon, Apr 15, 2013 at 2:50 AM, Liviu Andronic landronim...@gmail.com
wrote:
Dear all,
Given the following vector:
(z - c('R project', 'hello world', 'something Else'))
[1] R project hello worldsomething Else
I know how to obtain all capitals or all lower case letters:
tolower(z)
[1] r project hello worldsomething else
toupper(z)
[1] R PROJECT HELLO WORLDSOMETHING ELSE
I saw the tocamel() function in 'rapport', but it doesn't do what I
want to achieve as it actually proceeds to camelCase/CamelCase the
strings:
tocamel(z)
[1] RProject helloWorldsomethingElse
But how should I proceed to obtain Camel Case? Here's what I'd like to get:
c('R Project', 'Hello World', 'Something Else')
Here is a one liner using gsubfn from the gsubfn package
(http://gsubfn.googlecode.com). gsubfn is like gsubfn except the
second argument can be a function (or a list or a proto object). The
regular expression here matches a space followed by any character.
For each such match gsubfn will call the function denoted by the
second argument using the two parenthesized portions of the regular
expression as the two arguments. It supports an optional formula
notation to express the function for compactness so ... ~ toupper(..2)
denotes function(...) toupper(..2) which is equivalent to function(x,
y) toupper(y) . It replaces the input with the output of that
function. Finally we check if any of the components of the input are
NA and replace those with NA in the output:
replace(gsubfn(( )(.), ... ~ toupper(..2), z), is.na(z), NA)
[1] RProject helloWorldsomethingElse NA
Above gsubfn is like gsubfn should read gsubfn is like gsub
Also here is a further simplification:
replace(gsubfn( (.), toupper, z), is.na(z), NA)
[1] RProject helloWorldsomethingElse NA
--
Statistics Software Consulting
GKX Group, GKX Associates Inc.
tel: 1-877-GKX-GROUP
email: ggrothendieck at gmail.com
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Indices of lowest values in matrix
Dear Arun,
Thankyou very much. it worked out
:D
Elisa
Date: Mon, 15 Apr 2013 07:19:40 -0700
From: smartpink...@yahoo.com
Subject: Re: [R] Indices of lowest values in matrix
To: eliza_bo...@hotmail.com
Hi Elisa,
I am not sure how you got that values for the first rows of mat1. Is it from
other dataset
set.seed(30)
mat1- matrix(sample(1:50,12*12,replace=TRUE),ncol=12)
mat1[1,]
# [1] 5 28 13 4 47 3 30 16 3 39 42 9
#So, here the lowest 5 values are 3,3,4,5,9
sort(mat1[1,],index.return=TRUE)
#$x
#[1] 3 3 4 5 9 13 16 28 30 39 42 47
#$ix
#[1] 6 9 4 1 12 3 8 2 7 10 11 5
From my previous solution:
sapply(seq_len(nrow(mat1)),function(i) {order(mat1[i,])})[1:5,1]
#[1] 6 9 4 1 12 #which are the column numbers.
A.K.
From: eliza botto eliza_bo...@hotmail.com
To: smartpink...@yahoo.com smartpink...@yahoo.com
Sent: Monday, April 15, 2013 10:09 AM
Subject: RE: [R] Indices of lowest values in matrix
Thankyou very much.
one additional question is that what if i am interest to find that along the
row 1 which are 5 lowest values? more precisely, for row 1 of mat1 the five
lowest values are
3,5,8,19,21
therefore, i should have
1,2,8,5,4 (the column numbers which contain these values)
Thanks in advance
Elisa
Date: Mon, 15 Apr 2013 06:53:32 -0700
From: smartpink...@yahoo.com
Subject: Re: [R] Indices of lowest values in matrix
To: eliza_bo...@hotmail.com
CC: r-help@r-project.org
Hi,
TRy this:
set.seed(30)
Ámat1- matrix(sample(1:50,12*12,replace=TRUE),ncol=12)
mat1
Á [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12]
Á[1,]ÁÁÁ 5ÁÁ 28ÁÁ 13ÁÁÁ 4ÁÁ 47ÁÁÁ 3ÁÁ 30ÁÁ 16ÁÁÁ 3ÁÁÁ 39ÁÁÁ 42 9
Á[2,]ÁÁ 25ÁÁ 44ÁÁ 15ÁÁ 21ÁÁÁ 9ÁÁÁ 5ÁÁ 32ÁÁÁ 5ÁÁÁ 4ÁÁÁ 33ÁÁÁ 30ÁÁÁ 21
Á[3,]ÁÁ 19ÁÁ 12ÁÁ 24ÁÁÁ 2ÁÁÁ 7ÁÁ 37ÁÁ 36ÁÁÁ 3ÁÁ 13 9ÁÁÁ 33ÁÁÁ 33
Á[4,]ÁÁ 22ÁÁ 47ÁÁ 13ÁÁ 14ÁÁÁ 9ÁÁ 29ÁÁ 20ÁÁÁ 3ÁÁÁ 1ÁÁÁ 25ÁÁÁ 14ÁÁÁ 29
Á[5,]ÁÁ 16ÁÁ 13ÁÁ 41ÁÁ 22ÁÁ 15ÁÁ 39ÁÁ 50ÁÁ 44ÁÁ 10ÁÁÁ 50ÁÁÁ 32ÁÁÁ 32
Á[6,]ÁÁÁ 8ÁÁ 42ÁÁ 24ÁÁ 32ÁÁ 21ÁÁ 31ÁÁ 45ÁÁ 32ÁÁ 41ÁÁÁ 44ÁÁÁ 22ÁÁÁ 31
Á[7,]ÁÁ 45ÁÁ 31ÁÁ 31ÁÁ 39ÁÁÁ 4ÁÁ 25ÁÁÁ 1ÁÁ 19ÁÁ 28ÁÁÁ 16ÁÁÁ 20ÁÁÁ 48
Á[8,]ÁÁ 12ÁÁ 22ÁÁ 12ÁÁ 25ÁÁ 45ÁÁ 10ÁÁÁ 5ÁÁÁ 7ÁÁ 31ÁÁÁ 27ÁÁÁ 18ÁÁÁ 30
Á[9,]ÁÁ 49ÁÁÁ 8ÁÁ 25ÁÁ 10ÁÁÁ 6ÁÁ 30ÁÁÁ 8ÁÁ 29ÁÁ 14ÁÁÁ 10ÁÁÁ 48ÁÁÁ 44
[10,]ÁÁÁ 8ÁÁ 30ÁÁ 31ÁÁ 35ÁÁÁ 4ÁÁ 50ÁÁ 25ÁÁ 33ÁÁÁ 7ÁÁÁ 15ÁÁÁ 17ÁÁÁ 38
[11,]ÁÁÁ 4ÁÁÁ 2ÁÁ 15ÁÁ 40ÁÁ 30ÁÁ 49ÁÁ 44ÁÁ 14ÁÁÁ 3ÁÁÁ 33ÁÁÁ 41ÁÁÁ 12
[12,]ÁÁ 20ÁÁ 17ÁÁ 34ÁÁ 14ÁÁ 43ÁÁÁ 2ÁÁ 40ÁÁ 50ÁÁ 42ÁÁÁ 42ÁÁÁ 23ÁÁÁ 20
#If you need the lowest values:
Ásapply(seq_len(nrow(mat1)),function(i) {x1- mat1[i,]; x1[order(x1)][1:5]})
#ÁÁÁ [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12]
#[1,]ÁÁÁ 3ÁÁÁ 4ÁÁÁ 2ÁÁÁ 1ÁÁ 10ÁÁÁ 8ÁÁÁ 1ÁÁÁ 5ÁÁÁ 6 4 2 2
#[2,]ÁÁÁ 3ÁÁÁ 5ÁÁÁ 3ÁÁÁ 3ÁÁ 13ÁÁ 21ÁÁÁ 4ÁÁÁ 7ÁÁÁ 8 7 3ÁÁÁ 14
#[3,]ÁÁÁ 4ÁÁÁ 5ÁÁÁ 7ÁÁÁ 9ÁÁ 15ÁÁ 22ÁÁ 16ÁÁ 10ÁÁÁ 8 8 4ÁÁÁ 17
#[4,]ÁÁÁ 5ÁÁÁ 9ÁÁÁ 9ÁÁ 13ÁÁ 16ÁÁ 24ÁÁ 19ÁÁ 12ÁÁ 10ÁÁÁ 15ÁÁÁ 12ÁÁÁ 20
#[5,]ÁÁÁ 9ÁÁ 15ÁÁ 12ÁÁ 14ÁÁ 22ÁÁ 31ÁÁ 20ÁÁ 12ÁÁ 10ÁÁÁ 17ÁÁÁ 14ÁÁÁ 20
#indices
sapply(seq_len(nrow(mat1)),function(i) {order(mat1[i,])})[1:5,]
Á # Á [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12]
#[1,]ÁÁÁ 6ÁÁÁ 9ÁÁÁ 4ÁÁÁ 9ÁÁÁ 9ÁÁÁ 1ÁÁÁ 7ÁÁÁ 7ÁÁÁ 5 5 2 6
#[2,]ÁÁÁ 9ÁÁÁ 6ÁÁÁ 8ÁÁÁ 8ÁÁÁ 2ÁÁÁ 5ÁÁÁ 5ÁÁÁ 8ÁÁÁ 2 9 9 4
#[3,]ÁÁÁ 4ÁÁÁ 8ÁÁÁ 5ÁÁÁ 5ÁÁÁ 5ÁÁ 11ÁÁ 10ÁÁÁ 6ÁÁÁ 7 1 1 2
#[4,]ÁÁÁ 1ÁÁÁ 5ÁÁ 10ÁÁÁ 3ÁÁÁ 1ÁÁÁ 3ÁÁÁ 8ÁÁÁ 1ÁÁÁ 4ÁÁÁ 10ÁÁÁ 12 1
#[5,]ÁÁ 12ÁÁÁ 3ÁÁÁ 2ÁÁÁ 4ÁÁÁ 4ÁÁÁ 6ÁÁ 11ÁÁÁ 3ÁÁ 10ÁÁÁ 11 8ÁÁÁ 12
A.K.
- Original Message -
From: eliza botto eliza_bo...@hotmail.com
To: r-help@r-project.org r-help@r-project.org
Cc:
Sent: Monday, April 15, 2013 8:27 AM
Subject: [R] Indices of lowest values in matrix
Dear R users,Sorry for such a basic question. I really need to know that
how can i pick the indices of 5 lowest values from each row of a matrix
with dimensions 12*12??Thank you very much in advance
Elisa
ÁÁÁ ÁÁÁ ÁÁÁ Á ÁÁÁ ÁÁÁ Á
ÁÁÁ [[alternative HTML version deleted]]
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Re: [R] RJSONIO Installation
On 13.04.2013 08:22, Jeff Newmiller wrote: R CMD build will only work on Windows if you have Rtools installed. Visit your local CRAN mirror and read... this is not exactly trivial. Not exactly true: pure R code / data packages typically R CMD build and R CMD INSTALL without those tools installed. And, if you disable the rebuilt of vignettes, R CMD build will always work without the tools installed, ad far as I know. Best, Uwe Ligges --- Jeff NewmillerThe . . Go Live... DCN:jdnew...@dcn.davis.ca.usBasics: ##.#. ##.#. Live Go... Live: OO#.. Dead: OO#.. Playing Research Engineer (Solar/BatteriesO.O#. #.O#. with /Software/Embedded Controllers) .OO#. .OO#. rocks...1k --- Sent from my phone. Please excuse my brevity. John Henry johnmghe...@gmail.com wrote: Trying to install rjsonio and I've run into a couple of issues. 1. When installing from R, it's starts to download, then I get * installing *source* package 'RJSONIO' ... ERROR: configuration failed for package 'RJSONIO' * removing 'C:/Program Files/R/R-3.0.0/library/RJSONIO' 2. I've tried similar steps on the cmd(I'm win7). I receive the same message, starts to install then, configuration failed. 3. Looking into the 'read_me' doc that comes with rjsonio it mentions libjson C libraries. They give a special command if you don't have this installed, which I tried, but it didn't work. At this point I want to generate the libjson libraries to try the other 'read_me' suggestions or just get it to work from R. FYI, on the R CMD I've tried --binary and I've tried --build, neither work. Please help, running out of options. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ 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] Is it possible to create Dual y axes with ggplot2 package?
Hi, Is it possible to create Dual y axes with ggplot2 package? If so, is there any sample codes that reaches (or partially) this goal? The author of the package, Dr Hadley Wickham, seem to discourage the dual axes, but some remedies were provided by other users (If I am wrong, please correct me). I have done some homework: This is the discussion which I find most helpful. https://groups.google.com/forum/?fromgroups=#!topic/ggplot2/tImExPZg61o This is a nice graph with dual axes http://blog.revolutionanalytics.com/2010/01/r-package-growth.html Codes are provided here http://onertipaday.blogspot.tw/2010/01/scatter-plot-with-4-axes-labels-and.html Is it possible to modify the code so that it produces ggplot2 style dual-axis graph? Thanks, Miao [[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] group data
What if more generally that the group name doest have anything to do with the ID, eg. for ID=AL1 and AL2, I want to name the group as Key1, how can I approach that? Thanks, On Thu, Apr 11, 2013 at 11:54 AM, Rui Barradas ruipbarra...@sapo.pt wrote: Hello, Try the following. dat - read.table(text = ID Value AL1 1 AL2 2 CA1 3 CA4 4 , header = TRUE, stringsAsFactors = FALSE) dat$State - substr(dat$ID, 1, 2) Note that this dependes on having State being defined by the first two characters of ID. Hope this helps, Rui Barradas Em 11-04-2013 19:37, Ye Lin escreveu: Hey, I have a dataset and I want to identify the records by groups for further use in ggplot. Here is a sample data: ID Value AL1 1 AL2 2 CA1 3 CA4 4 I want to identify all the records that in the same state (AL1 AND A2), group them as AL, and do the same for CA1 and CA4. How can I have an output like: ID Value State AL1 1 AL AL2 2 AL CA1 3 CA CA4 4 CA Thanks for your help! [[alternative HTML version deleted]] __** R-help@r-project.org mailing list https://stat.ethz.ch/mailman/**listinfo/r-helphttps://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. [[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] Sorting data.frame and again sorting within data.frame
On Apr 14, 2013, at 11:01 PM, Katherine Gobin wrote: Dear R forum, I have a data.frame as defied below - df = data.frame(names = c(C, A, A, B, C, B, A, B, C), dates = c(4/15/2013, 4/13/2013, 4/15/2013, 4/13/2013, 4/13/2013, 4/15/2013, 4/14/2013, 4/14/2013,4/14/2013 ),values = c(10, 31, 31, 17, 11, 34, 102, 47, 29)) df names dates values 1 C 4/15/2013 10 2 A 4/13/2013 31 3 A 4/15/2013 31 4 B 4/13/2013 17 5 C 4/13/2013 11 6 B 4/15/2013 34 7 A 4/14/2013102 8 B 4/14/2013 47 9 C 4/14/2013 29 I need to sort df first on names in increasing order and then further on dates in a decreasing order i.e. I need So far no one has pointed out that these are not really Dates in the R sense and will not sort correctly if any of the proposed methods are applied to sequences that extend beyond6 months, i.e, until October forward. You would be advised to convert to real Date-classed variables. ?strptime ?as.Date -- David namesdatesvalues A4/15/2013 31 A4/14/2013 102 A4/13/2013 31 B4/15/2013 34 B4/14/2013 47 B4/13/2013 17 C4/15/2013 10 C4/14/2013 29 C4/13/2013 11 I tried df_sorted = df[order(df$names, (as.Date(df$dates, %m/%d/%Y)), decreasing = TRUE),] df_sorted names dates values 1 C 4/15/2013 10 9 C 4/14/2013 29 5 C 4/13/2013 11 6 B 4/15/2013 34 8 B 4/14/2013 47 4 B 4/13/2013 17 3 A 4/15/2013 31 7 A 4/14/2013102 2 A 4/13/2013 31 I need A to appear first with all three corresponding dates in decreasing order, then B and so on. Please guide. With regards Katherine [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. David Winsemius Alameda, CA, USA __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] regression with paired left-censored data
HI
I am trying to analyse data which is left-censored (i.e. has values below the
detection limit). I have been using the NADA package of R to derive summary
statistics and do some regression. I am now trying to carry out regression on
paired data where both my X and Y have left-censored data within them.
I have tried various commands in R:
rega = cenreg(Cen(conc, cens_ind) ~ Gp_ident))
with all X and Y data stacked and using a group identifier to look at the
differences
this doesn't take account of the paired data though.
I have also tried splitting the data and regessing one on the other
rega = cenreg(Cen(conc1, censind1) ~ Cen(conc2,censind2))
which doesn't work.
Does anyone know of a command that will work - or perhaps suggest another
package that I could use?
I have also looked at multiple imputation packages but they all seem to impute
data depending on other columns - whereas I would want to impute data between
zero and the censored value.
Any guidance/advice would be very much appreciated.
Laura
Dr Laura MacCalman Msci MSc PhD Gradstat
Senior Statistician
Institute of Occupational Medicine
Research Avenue North
Riccarton
Edinburgh
EH14 4AP
Tel: 0131 449 8078
Fax: 0131 449 8084
Mob: 07595 054 881
Email: laura.maccal...@iom-world.org
Web: http://www.iom-world.org
--
The Institute of Occupational Medicine (IOM) is a company limited by guarantee,
registered in Scotland (No.SC123972) and a Registered Scottish Charity
(No.SC000365). IOM Consulting Ltd is a wholly owned subsidiary of IOM and a
private limited company registered in Scotland (No. SC205670). Registered
Office: Research Avenue North, Riccarton, Edinburgh, EH14 4AP, Tel +44 (0)131
449 8000.
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[R] Latent variable manual construction
Dear all, I had a question around how to manually construct latent variables. I'll provide a little background…. Basically I have a SEM with 17 items, 6 latent variables and 615 rows of data. I have all my model data (loadings, t values etc.) from lavaan, as well as correlation matrices. I wanted to get p-values for my correlation matrix (items and latent variables) and understand I can't get this from lavaan. So I've done this for the items manually in r, but not sure how to construct/calculate the latent variables so that I can then determine the p-values. Alternatively maybe there'a much easier way. As constructing the correlation matrices again when already have them from lavaan seems a little unnecessary, however, need to get these p-values somehow! Any help much appreciated! Best, Taman __ 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] Convert results from print(survfit(formula, ...)) into a matrix or data frame
Hello All, Below is some sample survival analysis code. I'd like to able to get the results from print(gehan.surv) into a matrix or data frame, so I can manipulate them and then create a table using odfWeave. Trouble is, I'm not quite sure how make such a conversion using the results from a print method. Is there some simple way of doing this? Thanks, Paul require(survival) require(MASS) attach(gehan) gehan.surv - survfit(Surv(time, cens) ~ treat, data= gehan, conf.type = log-log) print(gehan.surv) __ 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] Sorting data.frame and again sorting within data.frame
Yes, that would be because she converted to Date on the fly in her example, and so apparently did not need this reminder. --- Jeff NewmillerThe . . Go Live... DCN:jdnew...@dcn.davis.ca.usBasics: ##.#. ##.#. Live Go... Live: OO#.. Dead: OO#.. Playing Research Engineer (Solar/BatteriesO.O#. #.O#. with /Software/Embedded Controllers) .OO#. .OO#. rocks...1k --- Sent from my phone. Please excuse my brevity. David Winsemius dwinsem...@comcast.net wrote: On Apr 14, 2013, at 11:01 PM, Katherine Gobin wrote: Dear R forum, I have a data.frame as defied below - df = data.frame(names = c(C, A, A, B, C, B, A, B, C), dates = c(4/15/2013, 4/13/2013, 4/15/2013, 4/13/2013, 4/13/2013, 4/15/2013, 4/14/2013, 4/14/2013,4/14/2013 ),values = c(10, 31, 31, 17, 11, 34, 102, 47, 29)) df names dates values 1 C 4/15/2013 10 2 A 4/13/2013 31 3 A 4/15/2013 31 4 B 4/13/2013 17 5 C 4/13/2013 11 6 B 4/15/2013 34 7 A 4/14/2013102 8 B 4/14/2013 47 9 C 4/14/2013 29 I need to sort df first on names in increasing order and then further on dates in a decreasing order i.e. I need So far no one has pointed out that these are not really Dates in the R sense and will not sort correctly if any of the proposed methods are applied to sequences that extend beyond6 months, i.e, until October forward. You would be advised to convert to real Date-classed variables. ?strptime ?as.Date __ 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] Convert results from print(survfit(formula, ...)) into a matrix or data frame
On Apr 15, 2013, at 11:31 AM, Paul Miller pjmiller...@yahoo.com wrote: Hello All, Below is some sample survival analysis code. I'd like to able to get the results from print(gehan.surv) into a matrix or data frame, so I can manipulate them and then create a table using odfWeave. Trouble is, I'm not quite sure how make such a conversion using the results from a print method. Is there some simple way of doing this? Thanks, Paul require(survival) require(MASS) attach(gehan) gehan.surv - survfit(Surv(time, cens) ~ treat, data= gehan, conf.type = log-log) print(gehan.surv) Hi Paul, If I am correctly understanding the output that you want to save, which is the summary results table output: gehan.surv Call: survfit(formula = Surv(time, cens) ~ treat, data = gehan, conf.type = log-log) records n.max n.start events median 0.95LCL 0.95UCL treat=6-MP 2121 21 9 23 13 NA treat=control 2121 21 21 8 4 11 You can use: summary(gehan.surv)$table records n.max n.start events median 0.95LCL 0.95UCL treat=6-MP 2121 21 9 23 13 NA treat=control 2121 21 21 8 4 11 str(summary(gehan.surv)$table) num [1:2, 1:7] 21 21 21 21 21 21 9 21 23 8 ... - attr(*, dimnames)=List of 2 ..$ : chr [1:2] treat=6-MP treat=control ..$ : chr [1:7] records n.max n.start events ... See ?summary.survfit and note the 'table' part of the Value section. Regards, Marc Schwartz __ 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] Convert results from print(survfit(formula, ...)) into a matrix or data frame
Hi Marc, Oh, yes. That is quite simple. Silly me for not recognizing this. Thanks very much for your help. Paul __ 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] Remove Rows Based on Factor
Dear R Helpers, I did a search for deleting rows based on conditions but wasn't able to find an example that addressed the error that I am getting. I am hoping that this is a simple syntax phenomenon that somebody else knows off the top of their head. My apologies for not providing a reproducible example but I think that the information given will allow someone to give me a hint. I want to delete the rows of the data frame ZZ where Index is earlier that Jan 1 of 2007. That Index column is a factor. When I tired a couple of different methods, I got the error shown below. Can anybody tell me what I am doing wrong? I would really appreciate it. --John Sparks str(ZZ) 'data.frame': 1584 obs. of 7 variables: $ Index : Factor w/ 1583 levels 2006-04-07,2006-04-10,..: 1 2 3 4 5 6 7 8 9 10 ... $ Open: num 17.5 17.6 16.8 17.2 17 ... $ High: num 18.2 17.6 17.2 17.2 17.1 ... $ Low : num 17.3 16.8 16.8 16.8 16.6 ... $ Close : num 17.5 16.8 17.1 16.8 16.7 ... $ Volume : num 23834500 2916000 1453700 991400 967400 ... $ Adjusted: num 16.8 16.2 16.4 16.2 16 ... test-ZZ[ZZ$Index2007-01-01,] Warning message: In Ops.factor(ZZ$Index, 2007-01-01) : not meaningful for factors test-subset(ZZ,Index2007-01-01) Warning message: In Ops.factor(Index, 2007 - 1 - 1) : not meaningful for factors __ 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] Sorting data.frame and again sorting within data.frame
On Apr 15, 2013, at 9:33 AM, Jeff Newmiller wrote: Yes, that would be because she converted to Date on the fly in her example, and so apparently did not need this reminder. I apologize, Iobviously missed that. So the answer was simply to put a minus sign in front of the as.Date() expression. And that was what you were hoping she would see when she looked at the second and third examples help page that even had helpful comments. As you suggested: df_sorted = df[order(df$names, -as.numeric( as.Date(df$dates, %m/%d/%Y) ), ] -- David --- Jeff NewmillerThe . . Go Live... DCN:jdnew...@dcn.davis.ca.usBasics: ##.#. ##.#. Live Go... Live: OO#.. Dead: OO#.. Playing Research Engineer (Solar/BatteriesO.O#. #.O#. with /Software/Embedded Controllers) .OO#. .OO#. rocks...1k --- Sent from my phone. Please excuse my brevity. David Winsemius dwinsem...@comcast.net wrote: On Apr 14, 2013, at 11:01 PM, Katherine Gobin wrote: Dear R forum, I have a data.frame as defied below - df = data.frame(names = c(C, A, A, B, C, B, A, B, C), dates = c(4/15/2013, 4/13/2013, 4/15/2013, 4/13/2013, 4/13/2013, 4/15/2013, 4/14/2013, 4/14/2013,4/14/2013 ),values = c(10, 31, 31, 17, 11, 34, 102, 47, 29)) df names dates values 1 C 4/15/2013 10 2 A 4/13/2013 31 3 A 4/15/2013 31 4 B 4/13/2013 17 5 C 4/13/2013 11 6 B 4/15/2013 34 7 A 4/14/2013102 8 B 4/14/2013 47 9 C 4/14/2013 29 I need to sort df first on names in increasing order and then further on dates in a decreasing order i.e. I need So far no one has pointed out that these are not really Dates in the R sense and will not sort correctly if any of the proposed methods are applied to sequences that extend beyond6 months, i.e, until October forward. You would be advised to convert to real Date-classed variables. ?strptime ?as.Date David Winsemius Alameda, CA, USA __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Remove Rows Based on Factor
On Apr 15, 2013, at 9:58 AM, Sparks, John James wrote: Dear R Helpers, I did a search for deleting rows based on conditions but wasn't able to find an example that addressed the error that I am getting. I am hoping that this is a simple syntax phenomenon that somebody else knows off the top of their head. My apologies for not providing a reproducible example but I think that the information given will allow someone to give me a hint. I want to delete the rows of the data frame ZZ where Index is earlier that Jan 1 of 2007. That Index column is a factor. When I tired a couple of different methods, I got the error shown below. Can anybody tell me what I am doing wrong? I would really appreciate it. --John Sparks str(ZZ) 'data.frame': 1584 obs. of 7 variables: $ Index : Factor w/ 1583 levels 2006-04-07,2006-04-10,..: 1 2 3 4 5 6 7 8 9 10 ... $ Open: num 17.5 17.6 16.8 17.2 17 ... $ High: num 18.2 17.6 17.2 17.2 17.1 ... $ Low : num 17.3 16.8 16.8 16.8 16.6 ... $ Close : num 17.5 16.8 17.1 16.8 16.7 ... $ Volume : num 23834500 2916000 1453700 991400 967400 ... $ Adjusted: num 16.8 16.2 16.4 16.2 16 ... test-ZZ[ZZ$Index2007-01-01,] It should work using string comparisons: test-ZZ[as.character(ZZ$Index) 2007-01-01, ] -- David Warning message: In Ops.factor(ZZ$Index, 2007-01-01) : not meaningful for factors test-subset(ZZ,Index2007-01-01) Warning message: In Ops.factor(Index, 2007 - 1 - 1) : not meaningful for factors __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. David Winsemius Alameda, CA, USA __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Remove Rows Based on Factor
Hi, May be this helps: ZZ- data.frame(Index=c(2006-04-07,2006-04-08,2007-01-01,2007-01-01,2008-01-04,2008-01-04),Open=c(17.5,17.6,16.8,17.2,17.3,17.5),High=c(18.2,17.6,17.2,17.2,17.5,17.6),Low=c(17.3,16.8,16.8,16.8,16.2,15.8),Close=c(17.5,16.8,17.1,16.8,15.9,16.2),Volume=c(23834500,2916000,1453700,991400,443522,38990),Adjusted=c(16.8,16.2,16.4,16.2,16.5,17.6)) ZZ$Index- as.Date(ZZ$Index,format=%Y-%m-%d) library(xts) ZZ1-xts(ZZ[,-1],ZZ[,1]) ZZ1 # Open High Low Close Volume Adjusted #2006-04-07 17.5 18.2 17.3 17.5 23834500 16.8 #2006-04-08 17.6 17.6 16.8 16.8 2916000 16.2 #2007-01-01 16.8 17.2 16.8 17.1 1453700 16.4 #2007-01-01 17.2 17.2 16.8 16.8 991400 16.2 #2008-01-04 17.3 17.5 16.2 15.9 443522 16.5 #2008-01-04 17.5 17.6 15.8 16.2 38990 17.6 Suppose, I wanted to get only the rows starting from 2006-04-08 ZZ1['2006-04-08/'] # Open High Low Close Volume Adjusted #2006-04-08 17.6 17.6 16.8 16.8 2916000 16.2 #2007-01-01 16.8 17.2 16.8 17.1 1453700 16.4 #2007-01-01 17.2 17.2 16.8 16.8 991400 16.2 #2008-01-04 17.3 17.5 16.2 15.9 443522 16.5 #2008-01-04 17.5 17.6 15.8 16.2 38990 17.6 A.K. - Original Message - From: Sparks, John James jspa...@uic.edu To: r-help@r-project.org Cc: Sent: Monday, April 15, 2013 12:58 PM Subject: [R] Remove Rows Based on Factor Dear R Helpers, I did a search for deleting rows based on conditions but wasn't able to find an example that addressed the error that I am getting. I am hoping that this is a simple syntax phenomenon that somebody else knows off the top of their head. My apologies for not providing a reproducible example but I think that the information given will allow someone to give me a hint. I want to delete the rows of the data frame ZZ where Index is earlier that Jan 1 of 2007. That Index column is a factor. When I tired a couple of different methods, I got the error shown below. Can anybody tell me what I am doing wrong? I would really appreciate it. --John Sparks str(ZZ) 'data.frame': 1584 obs. of 7 variables: $ Index : Factor w/ 1583 levels 2006-04-07,2006-04-10,..: 1 2 3 4 5 6 7 8 9 10 ... $ Open : num 17.5 17.6 16.8 17.2 17 ... $ High : num 18.2 17.6 17.2 17.2 17.1 ... $ Low : num 17.3 16.8 16.8 16.8 16.6 ... $ Close : num 17.5 16.8 17.1 16.8 16.7 ... $ Volume : num 23834500 2916000 1453700 991400 967400 ... $ Adjusted: num 16.8 16.2 16.4 16.2 16 ... test-ZZ[ZZ$Index2007-01-01,] Warning message: In Ops.factor(ZZ$Index, 2007-01-01) : not meaningful for factors test-subset(ZZ,Index2007-01-01) Warning message: In Ops.factor(Index, 2007 - 1 - 1) : not meaningful for factors __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Remove Rows Based on Factor
Hi, In addition: You can subset based on time or date: ZZ1['2006'] # Open High Low Close Volume Adjusted #2006-04-07 17.5 18.2 17.3 17.5 23834500 16.8 #2006-04-08 17.6 17.6 16.8 16.8 2916000 16.2 ZZ1['2007-01/2008-01'] # Open High Low Close Volume Adjusted #2007-01-01 16.8 17.2 16.8 17.1 1453700 16.4 #2007-01-01 17.2 17.2 16.8 16.8 991400 16.2 #2008-01-04 17.3 17.5 16.2 15.9 443522 16.5 #2008-01-04 17.5 17.6 15.8 16.2 38990 17.6 ZZ1['2007/2008'] # Open High Low Close Volume Adjusted #2007-01-01 16.8 17.2 16.8 17.1 1453700 16.4 #2007-01-01 17.2 17.2 16.8 16.8 991400 16.2 #2008-01-04 17.3 17.5 16.2 15.9 443522 16.5 #2008-01-04 17.5 17.6 15.8 16.2 38990 17.6 ZZ2- data.frame(Index=c(2006-04-07 02:30:00,2006-04-08 02:00:00,2007-01-01 01:30:00,2007-01-01 02:30:00,2008-01-04 03:00:00,2008-01-04 03:30:00),Open=c(17.5,17.6,16.8,17.2,17.3,17.5),High=c(18.2,17.6,17.2,17.2,17.5,17.6),Low=c(17.3,16.8,16.8,16.8,16.2,15.8),Close=c(17.5,16.8,17.1,16.8,15.9,16.2),Volume=c(23834500,2916000,1453700,991400,443522,38990),Adjusted=c(16.8,16.2,16.4,16.2,16.5,17.6),stringsAsFactors=FALSE) ZZ2$Index- as.POSIXct(ZZ2$Index,format=%Y-%m-%d %H:%M:%S) library(xts) ZZ3-xts(ZZ2[,-1],ZZ2[,1]) ZZ3[T02:30/T03:00] # Open High Low Close Volume Adjusted #2006-04-07 02:30:00 17.5 18.2 17.3 17.5 23834500 16.8 #2007-01-01 02:30:00 17.2 17.2 16.8 16.8 991400 16.2 #2008-01-04 03:00:00 17.3 17.5 16.2 15.9 443522 16.5 A.K. - Original Message - From: arun smartpink...@yahoo.com To: Sparks, John James jspa...@uic.edu Cc: R help r-help@r-project.org Sent: Monday, April 15, 2013 1:32 PM Subject: Re: [R] Remove Rows Based on Factor Hi, May be this helps: ZZ- data.frame(Index=c(2006-04-07,2006-04-08,2007-01-01,2007-01-01,2008-01-04,2008-01-04),Open=c(17.5,17.6,16.8,17.2,17.3,17.5),High=c(18.2,17.6,17.2,17.2,17.5,17.6),Low=c(17.3,16.8,16.8,16.8,16.2,15.8),Close=c(17.5,16.8,17.1,16.8,15.9,16.2),Volume=c(23834500,2916000,1453700,991400,443522,38990),Adjusted=c(16.8,16.2,16.4,16.2,16.5,17.6)) ZZ$Index- as.Date(ZZ$Index,format=%Y-%m-%d) library(xts) ZZ1-xts(ZZ[,-1],ZZ[,1]) ZZ1 # Open High Low Close Volume Adjusted #2006-04-07 17.5 18.2 17.3 17.5 23834500 16.8 #2006-04-08 17.6 17.6 16.8 16.8 2916000 16.2 #2007-01-01 16.8 17.2 16.8 17.1 1453700 16.4 #2007-01-01 17.2 17.2 16.8 16.8 991400 16.2 #2008-01-04 17.3 17.5 16.2 15.9 443522 16.5 #2008-01-04 17.5 17.6 15.8 16.2 38990 17.6 Suppose, I wanted to get only the rows starting from 2006-04-08 ZZ1['2006-04-08/'] # Open High Low Close Volume Adjusted #2006-04-08 17.6 17.6 16.8 16.8 2916000 16.2 #2007-01-01 16.8 17.2 16.8 17.1 1453700 16.4 #2007-01-01 17.2 17.2 16.8 16.8 991400 16.2 #2008-01-04 17.3 17.5 16.2 15.9 443522 16.5 #2008-01-04 17.5 17.6 15.8 16.2 38990 17.6 A.K. - Original Message - From: Sparks, John James jspa...@uic.edu To: r-help@r-project.org Cc: Sent: Monday, April 15, 2013 12:58 PM Subject: [R] Remove Rows Based on Factor Dear R Helpers, I did a search for deleting rows based on conditions but wasn't able to find an example that addressed the error that I am getting. I am hoping that this is a simple syntax phenomenon that somebody else knows off the top of their head. My apologies for not providing a reproducible example but I think that the information given will allow someone to give me a hint. I want to delete the rows of the data frame ZZ where Index is earlier that Jan 1 of 2007. That Index column is a factor. When I tired a couple of different methods, I got the error shown below. Can anybody tell me what I am doing wrong? I would really appreciate it. --John Sparks str(ZZ) 'data.frame': 1584 obs. of 7 variables: $ Index : Factor w/ 1583 levels 2006-04-07,2006-04-10,..: 1 2 3 4 5 6 7 8 9 10 ... $ Open : num 17.5 17.6 16.8 17.2 17 ... $ High : num 18.2 17.6 17.2 17.2 17.1 ... $ Low : num 17.3 16.8 16.8 16.8 16.6 ... $ Close : num 17.5 16.8 17.1 16.8 16.7 ... $ Volume : num 23834500 2916000 1453700 991400 967400 ... $ Adjusted: num 16.8 16.2 16.4 16.2 16 ... test-ZZ[ZZ$Index2007-01-01,] Warning message: In Ops.factor(ZZ$Index, 2007-01-01) : not meaningful for factors test-subset(ZZ,Index2007-01-01) Warning message: In Ops.factor(Index, 2007 - 1 - 1) : not meaningful for factors __ 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,
Re: [R] group data
Hi,
You could do this:
dat1- read.table(text=
ID Value
AL1 1
AL2 2
CA1 3
CA4 4
,sep=,header=TRUE,stringsAsFactors=FALSE)
lst1-split(dat1,gsub(\\d+,,dat1$ID))
res-do.call(rbind,lapply(seq_along(lst1),function(i) {x1-lst1[[i]];
x1$group- paste0(Key,i);x1}))
res
# ID Value group
#1 AL1 1 Key1
#2 AL2 2 Key1
#3 CA1 3 Key2
#4 CA4 4 Key2
A.K.
- Original Message -
From: Ye Lin ye...@lbl.gov
To: Rui Barradas ruipbarra...@sapo.pt
Cc: R help r-help@r-project.org
Sent: Monday, April 15, 2013 11:50 AM
Subject: Re: [R] group data
What if more generally that the group name doest have anything to do with
the ID, eg. for ID=AL1 and AL2, I want to name the group as Key1, how can
I approach that?
Thanks,
On Thu, Apr 11, 2013 at 11:54 AM, Rui Barradas ruipbarra...@sapo.pt wrote:
Hello,
Try the following.
dat - read.table(text =
ID Value
AL1 1
AL2 2
CA1 3
CA4 4
, header = TRUE, stringsAsFactors = FALSE)
dat$State - substr(dat$ID, 1, 2)
Note that this dependes on having State being defined by the first two
characters of ID.
Hope this helps,
Rui Barradas
Em 11-04-2013 19:37, Ye Lin escreveu:
Hey,
I have a dataset and I want to identify the records by groups for further
use in ggplot.
Here is a sample data:
ID Value
AL1 1
AL2 2
CA1 3
CA4 4
I want to identify all the records that in the same state (AL1 AND A2),
group them as AL, and do the same for CA1 and CA4. How can I have an
output like:
ID Value State
AL1 1 AL
AL2 2 AL
CA1 3 CA
CA4 4 CA
Thanks for your help!
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Re: [R] regression with paired left-censored data
I would probably start with maximum likelihood estimation.
I suppose you could impute X and Y separately using ros() from the NADA
package, and then run you ordinary regression on the imputed values.
Obviously, this ignores any relationship between X and Y, since each is
imputed independently of the other. I have no idea whether ordinary
inferences on the parameter estimates would be valid. Probably not.
Probably, MLE would be better.
-Don
--
Don MacQueen
Lawrence Livermore National Laboratory
7000 East Ave., L-627
Livermore, CA 94550
925-423-1062
On 4/15/13 8:55 AM, Laura MacCalman laura.maccal...@iom-world.org
wrote:
HI
I am trying to analyse data which is left-censored (i.e. has values below
the detection limit). I have been using the NADA package of R to derive
summary statistics and do some regression. I am now trying to carry out
regression on paired data where both my X and Y have left-censored data
within them.
I have tried various commands in R:
rega = cenreg(Cen(conc, cens_ind) ~ Gp_ident))
with all X and Y data stacked and using a group identifier to look at the
differences
this doesn't take account of the paired data though.
I have also tried splitting the data and regessing one on the other
rega = cenreg(Cen(conc1, censind1) ~ Cen(conc2,censind2))
which doesn't work.
Does anyone know of a command that will work - or perhaps suggest another
package that I could use?
I have also looked at multiple imputation packages but they all seem to
impute data depending on other columns - whereas I would want to impute
data between zero and the censored value.
Any guidance/advice would be very much appreciated.
Laura
Dr Laura MacCalman Msci MSc PhD Gradstat
Senior Statistician
Institute of Occupational Medicine
Research Avenue North
Riccarton
Edinburgh
EH14 4AP
Tel: 0131 449 8078
Fax: 0131 449 8084
Mob: 07595 054 881
Email: laura.maccal...@iom-world.org
Web: http://www.iom-world.org
--
The Institute of Occupational Medicine (IOM) is a company limited by
guarantee, registered in Scotland (No.SC123972) and a Registered Scottish
Charity (No.SC000365). IOM Consulting Ltd is a wholly owned subsidiary of
IOM and a private limited company registered in Scotland (No. SC205670).
Registered Office: Research Avenue North, Riccarton, Edinburgh, EH14 4AP,
Tel +44 (0)131 449 8000.
This email and any files transmitted with it are confide...{{dropped:18}}
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Re: [R] regression with paired left-censored data
On Apr 15, 2013, at 8:55 AM, Laura MacCalman wrote:
HI
I am trying to analyse data which is left-censored (i.e. has values below the
detection limit). I have been using the NADA package of R to derive summary
statistics and do some regression. I am now trying to carry out regression on
paired data where both my X and Y have left-censored data within them.
I have tried various commands in R:
rega = cenreg(Cen(conc, cens_ind) ~ Gp_ident))
with all X and Y data stacked and using a group identifier to look at the
differences
this doesn't take account of the paired data though.
I have also tried splitting the data and regessing one on the other
rega = cenreg(Cen(conc1, censind1) ~ Cen(conc2,censind2))
which doesn't work.
Does anyone know of a command that will work - or perhaps suggest another
package that I could use?
I have also looked at multiple imputation packages but they all seem to
impute data depending on other columns - whereas I would want to impute data
between zero and the censored value.
Any guidance/advice would be very much appreciated.
The `survival::Surv` function allows left censoring and the `coxph` function
allows strata or clusters to be specified. My understanding is that for many
years analysts used the Cox regression machinery to crank out conditional
logistic regression by creating two-member (or 1+n) member strata/cluster. I'm
wondering if that could be made to work here, since this seems even closer to a
real survival analysis problem.
This response from Terry Therneau (looked up with Markmail) to a question about
a right-censored situation suggests that the use of cluster() rather than
strata() might be be more powerful:
http://markmail.org/message/c2oqqd34nujxvuvi?q=list:org%2Er-project%2Er-help+paired+survival+coxph+strata
--
David.
Laura
Dr Laura MacCalman Msci MSc PhD Gradstat
Senior Statistician
Institute of Occupational Medicine
Research Avenue North
Riccarton
Edinburgh
EH14 4AP
Tel: 0131 449 8078
Fax: 0131 449 8084
Mob: 07595 054 881
Email: laura.maccal...@iom-world.org
Web: http://www.iom-world.org
--
The Institute of Occupational Medicine (IOM) is a company limited by
guarantee, registered in Scotland (No.SC123972) and a Registered Scottish
Charity (No.SC000365). IOM Consulting Ltd is a wholly owned subsidiary of IOM
and a private limited company registered in Scotland (No. SC205670).
Registered Office: Research Avenue North, Riccarton, Edinburgh, EH14 4AP, Tel
+44 (0)131 449 8000.
This email and any files transmitted with it are confide...{{dropped:18}}
__
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and provide commented, minimal, self-contained, reproducible code.
David Winsemius
Alameda, CA, USA
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[R] Smallest Space Analysis help
Hello fellow R users, I need to perform Guttman's Smallest Space Analysis, which is a type of Non-metric Multidimensional Scaling. Not wanting to reinvent the wheel i decided to look for a package which may have an implementation of it, but no success at all. Do you guys know of any package that perform SSA? I found 'isoMDS' and 'sammon', both on the MASS package, but though they perform MDS, both don't perform SSA. Thanks a lot for any info. Diego Queiroz [[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] Removing multiple elements from a vector or list
Hi, vec1- letters vec1[!grepl(b|r|x,alp)] # [1] a c d e f g h i j k l m n o p q s t u #[20] v w y z vec1[!vec1%in% c(b,r,x) ] # [1] a c d e f g h i j k l m n o p q s t u #[20] v w y z alp-lapply(seq_along(vec1),function(i) vec1[i]) res-alp[!grepl(b|r|x,alp)] unlist(res) # [1] a c d e f g h i j k l m n o p q s t u #[20] v w y z unlist(alp[!alp%in%c(b,r,x)]) #[1] a c d e f g h i j k l m n o p q s t u #[20] v w y z A.K. If I for example have a list (or vector) that contains all the letters in the alphabet. alp - list(a,b,c,...z) this is of course not the exact code How can I remove multiple elements at one time without knowing their location in the list. Say I want to remove b,r,x? Same question if apl is a vector I have tried alp - alp[-c(b,r,x)] this does not work Thanks, Johnny __ 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] View saved workspace commands
R version : 2.15.2
Windows: 7 32-bit
Hello,
I was trying to load a saved workspace image from the working directory.
I issued the command load('image_name') in R console. It did not throw any
error, but then it returned the prompt. I want to view the commands (or
history) I typed there, how do I do that? Pressing up arrow does not bring
any of this workspace command, it rather brings older ones, same as the
output of history(). Any help will be appreciated.
N.B. To save the workspace I pressed the save button on RGui.
Thanks.
--
Sourabh Sinha, PhD
Department of Physics
Arizona State University
Tempe, AZ 85287
USA
http://phy.asu.edu/faculty.php?name=ssinha5
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[R] matching multiple fields from a matrix
I have been trying many ways to match 2 separate fields in a matrix. Here is a simplified version of the matrix: site1 depth1 year1 site2 depth2 year2 10 30 1860NA NA NA NA NA NA 50 30 1860 Basically I am trying to identify the sites which have a common year and depth from 2 datasets. What I would like to do is match all of the year1 field to year2 field and the depth1 field and to depth2 field. Then I would like to output site1, site2, depth, and year. I have been trying if loops, which(), isTRUE(), etc. but I have not come up with anything that works. Any help would be greatly appreciated. Jeremy -- View this message in context: http://r.789695.n4.nabble.com/matching-multiple-fields-from-a-matrix-tp4664309.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Can not load Rcmdr
Hello, I have the same problem on my mac and it doesn't work to load Rcmdr.. library(Rcmdr) Error : .onAttach failed in attachNamespace() for 'Rcmdr', details: call: structure(.External(.C_dotTclObjv, objv), class = tclObj) error: [tcl] invalid command name image. Error: package or namespace load failed for 'Rcmdr' sessionInfo() R version 3.0.0 (2013-04-03) Platform: x86_64-apple-darwin10.8.0 (64-bit) locale: [1] en_US.UTF-8/en_US.UTF-8/en_US.UTF-8/C/en_US.UTF-8/en_US.UTF-8 attached base packages: [1] tcltk stats graphics grDevices utils datasets methods base other attached packages: [1] car_2.0-16 nnet_7.3-6 MASS_7.3-26 loaded via a namespace (and not attached): [1] Rcmdr_1.9-6 tools_3.0.0 What can I do to load the package? Best regards, Ludwig -- Ludwig Asam Hauptstraße 3 86438 Kissing Tel.: 08233 210866 Handy: 0163 3133304 [[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] Removing multiple elements from a vector or list
On Apr 15, 2013, at 2:30 PM, arun wrote: Hi, vec1- letters alp - as.list(letters) would have constructed the vector that was described. vec1[!grepl(b|r|x,alp)] # [1] a c d e f g h i j k l m n o p q s t u #[20] v w y z vec1[!vec1%in% c(b,r,x) ] # [1] a c d e f g h i j k l m n o p q s t u #[20] v w y z alp-lapply(seq_along(vec1),function(i) vec1[i]) res-alp[!grepl(b|r|x,alp)] unlist(res) # [1] a c d e f g h i j k l m n o p q s t u #[20] v w y z unlist(alp[!alp%in%c(b,r,x)]) #[1] a c d e f g h i j k l m n o p q s t u #[20] v w y z A.K. All good. There would be an additional way to do this if the list were first assigned names. (At the moment the list only has positions for reference.) alp - setNames(alp, letters[1:26]) # Something like that initial code would succeed: alp - alp[ !names(alp) %in% c(b,r,x)] The result is still a list. If I for example have a list (or vector) that contains all the letters in the alphabet. alp - list(a,b,c,...z) this is of course not the exact code How can I remove multiple elements at one time without knowing their location in the list. Say I want to remove b,r,x? Same question if apl is a vector I have tried alp - alp[-c(b,r,x)] You _cannot_use_negative_indexing_with_names (or values). You could have used logical indexing: alp [ sapply(alp, function(x) !x %in% c(b,r,x) ) ] # OR perhaps the most compact solution offered so far; numeric indexing with the minus unary operator: alp[ -grep(b|r|x, alp) ] -- David Winsemius Alameda, CA, USA __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Can not load Rcmdr
On Apr 15, 2013, at 1:51 PM, Ludwig Asam wrote: Hello, I have the same problem on my mac and it doesn't work to load Rcmdr.. library(Rcmdr) Error : .onAttach failed in attachNamespace() for 'Rcmdr', details: call: structure(.External(.C_dotTclObjv, objv), class = tclObj) error: [tcl] invalid command name image. Error: package or namespace load failed for 'Rcmdr' sessionInfo() R version 3.0.0 (2013-04-03) Platform: x86_64-apple-darwin10.8.0 (64-bit) locale: [1] en_US.UTF-8/en_US.UTF-8/en_US.UTF-8/C/en_US.UTF-8/en_US.UTF-8 attached base packages: [1] tcltk stats graphics grDevices utils datasets methods base other attached packages: [1] car_2.0-16 nnet_7.3-6 MASS_7.3-26 loaded via a namespace (and not attached): [1] Rcmdr_1.9-6 tools_3.0.0 You already have Rcmdr loaded What can I do to load the package? Start a new session an load it again? Maybe you have an .Rprofile file that is loading it at Startup? Best regards, Ludwig -- David Winsemius Alameda, CA, USA __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] nearest stations in distance matrix
Presumably --- based on the question she actually *asked* --- she does
NOT have the coordinates of the stations; only the distance matrix.
I believe that the following will do the job: Let M be the distance matrix.
diag(M) - Inf
nn5 - apply(M,1,function(x){((1:length(x))[order(x)])[1:5]})
Check:
require(spatstat)
set.seed(42)
X - runifpoint(44)
M - crossdist(X,X)
diag(M) - Inf
nn5 - apply(M,1,function(x){((1:length(x))[order(x)])[1:5]})
plot(X)
plot(X[1],add=TRUE,chars=20,cols=red)
plot(X[nn5[,1]],add=TRUE,chars=20,cols=blue)
# Looks right to me.
cheers,
Rolf Turner
On 15/04/13 22:42, ONKELINX, Thierry wrote:
Dear Eliza,
If you have the coordinates of the stations you can use the nnwhich() function
from the spatstat package.
Best regards,
ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek / Research Institute for Nature and
Forest
team Biometrie Kwaliteitszorg / team Biometrics Quality Assurance
Kliniekstraat 25
1070 Anderlecht
Belgium
+ 32 2 525 02 51
+ 32 54 43 61 85
thierry.onkel...@inbo.be
www.inbo.be
To call in the statistician after the experiment is done may be no more than
asking him to perform a post-mortem examination: he may be able to say what the
experiment died of.
~ Sir Ronald Aylmer Fisher
The plural of anecdote is not data.
~ Roger Brinner
The combination of some data and an aching desire for an answer does not ensure
that a reasonable answer can be extracted from a given body of data.
~ John Tukey
-Oorspronkelijk bericht-
Van: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] Namens
eliza botto
Verzonden: maandag 15 april 2013 12:36
Aan: r-help@r-project.org
Onderwerp: [R] nearest stations in distance matrix
Dear R-user,
Is there a way in R to locate the nearest 5 indices to a station, based on
distances in a distance matrix. In other words i want to have nearest stations
based on the distances in the matrix. The distance matrix, i have, has
dimension 44*44.
Thankyou very much in advance
Elisa
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[R] Fwd: Imputation with SOM using kohonen package
Trying re-send as plain text.
I have a data set with 10 variables, and about 8000 instances (or
objects/rows/samples). In addition I have one more ('class') variable
that I have about 10 instances for, but for which I wish to impute
values for.
I am a little confused how to go about doing this, mostly as I'm not
well-versed in it. Do I train the SOM with a data object that contains
just the first 10 variables (exclude the 'class' variable), then
predict using an object that has all of the variables (including the
class variable)?
(I am using the kohonen package, and in general I am using the SOM
technique as a comparison to some other methods).
I don't know if providing some or all data is useful, please let me
know if you think it is.
# get the data
bw - read.csv(bw.csv)
# some missing values in data
bwm - data.frame(na.approx(bw, na.rm=FALSE, rule=2))
bw10 - bwm[, 1:10]
bw10.sc - scale(bw10)
bw.som - som(data=bw10.sc, grid=somgrid(25,20,'hexagonal')) #
playing with diff grid sizes
# the different plots of the som at this point show some interesting
features to me, but are quite difficult to interpret.
# there's much work needed here to understand it, but for now I want
to see if it's possible to impute values for another variable...
# here's where I lose it, missing values, trainY, don't get it.
bw.predict - predict(bw.som, newdata=scale(bw), trainX=???, trainY=???)
Ben.
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Re: [R] nearest stations in distance matrix
There was a bit of cranial wind-passing in my previous message.
(A bit of redundancy.)
My solution should have read:
diag(M) - Inf
nn5 - apply(M,1,function(x){order(x)[1:5]})
What I wrote gave the right answer; there was just a bunch of
unnecessary ring-around-the-rosy playing in it.
Sorry 'bout that.
cheers,
Rolf Turner
-Oorspronkelijk bericht-
Van: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] Namens
eliza botto
Verzonden: maandag 15 april 2013 12:36
Aan: r-help@r-project.org
Onderwerp: [R] nearest stations in distance matrix
Dear R-user,
Is there a way in R to locate the nearest 5 indices to a station, based on
distances in a distance matrix. In other words i want to have nearest stations
based on the distances in the matrix. The distance matrix, i have, has
dimension 44*44.
Thankyou very much in advance
Elisa
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Re: [R] View saved workspace commands
On 15.04.2013 21:53, Sourabh Sinha wrote:
R version : 2.15.2
Windows: 7 32-bit
Hello,
I was trying to load a saved workspace image from the working directory.
I issued the command load('image_name') in R console. It did not throw any
error, but then it returned the prompt. I want to view the commands (or
history) I typed there, how do I do that? Pressing up arrow does not bring
any of this workspace command, it rather brings older ones, same as the
output of history(). Any help will be appreciated.
To load the workspace, use load(), for the history, use loadhistory().
Uwe Ligges
N.B. To save the workspace I pressed the save button on RGui.
Thanks.
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Re: [R] create an access file
worked it out. Thank you. misunderstood the post earlier. Aileen On 15 April 2013 20:17, Prof Brian Ripley rip...@stats.ox.ac.uk wrote: On 15/04/2013 07:12, Aileen Lin wrote: Hi there, I have seen this post. https://stat.ethz.ch/**pipermail/r-help/2007-June/**133606.htmlhttps://stat.ethz.ch/pipermail/r-help/2007-June/133606.html have odbc installed in my machine. Now I have the following message: channel2 - odbcDriverConnect(test.mdb)**Warning messages:1: In odbcDriverConnect(test.mdb) : [RODBC] ERROR: state IM002, code 0, message [Microsoft][ODBC Driver Manager] Data source name not found and no default driver specified2: In odbcDriverConnect(test.mdb) : [RODBC] ERROR: state 01S00, code 0, message [Microsoft][ODBC Driver Manager] Invalid connection string attribute3: In odbcDriverConnect(test.mdb) : ODBC connection failed Did I miss anything? That is not doing what the URL you gave suggested. Please do follow it. -- Brian D. Ripley, rip...@stats.ox.ac.uk Professor of Applied Statistics, http://www.stats.ox.ac.uk/~**ripley/http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 -- Aileen L. View my Linkedin profile: http://au.linkedin.com/in/aileen2 Being happy doesn't mean you're perfect. It just means you've decided to look beyond the imperfections- K.B Indiana (age 14)http://www.boardofwisdom.com/default.asp?topic=1010search=K%2EB+Indiana+%28age+14%29 [[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] create an access file
Hi there,
I tried the command in https://stat.ethz.ch/**pipermail/r-help/2007-June/**
133606.html https://stat.ethz.ch/pipermail/r-help/2007-June/133606.html
In R, it works OK. In Rstudio, there seems to be a problem:
library(RODBC)ch - odbcDriverConnect(Driver={Microsoft Access Driver
(*.mdb)})Warning messages:1: In odbcDriverConnect(Driver={Microsoft
Access Driver (*.mdb)}) :
[RODBC] ERROR: Could not SQLDriverConnect2: In
odbcDriverConnect(Driver={Microsoft Access Driver (*.mdb)}) :
ODBC connection failed
I use R 2.15.3 and Rstudio 0.97.336. Did I miss anything?
Regards,
Aileen
On 15 April 2013 20:17, Prof Brian Ripley rip...@stats.ox.ac.uk wrote:
On 15/04/2013 07:12, Aileen Lin wrote:
Hi there,
I have seen this post.
https://stat.ethz.ch/**pipermail/r-help/2007-June/**133606.htmlhttps://stat.ethz.ch/pipermail/r-help/2007-June/133606.html
have odbc installed in my machine. Now I have the following message:
channel2 - odbcDriverConnect(test.mdb)**Warning messages:1: In
odbcDriverConnect(test.mdb) :
[RODBC] ERROR: state IM002, code 0, message [Microsoft][ODBC Driver
Manager] Data source name not found and no default driver specified2:
In odbcDriverConnect(test.mdb) :
[RODBC] ERROR: state 01S00, code 0, message [Microsoft][ODBC Driver
Manager] Invalid connection string attribute3: In
odbcDriverConnect(test.mdb) : ODBC connection failed
Did I miss anything?
That is not doing what the URL you gave suggested. Please do follow it.
--
Brian D. Ripley, rip...@stats.ox.ac.uk
Professor of Applied Statistics,
http://www.stats.ox.ac.uk/~**ripley/http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel: +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UKFax: +44 1865 272595
--
Aileen L.
View my Linkedin profile: http://au.linkedin.com/in/aileen2
Being happy doesn't mean you're perfect. It just means you've decided to
look beyond the imperfections- K.B Indiana (age
14)http://www.boardofwisdom.com/default.asp?topic=1010search=K%2EB+Indiana+%28age+14%29
[[alternative HTML version deleted]]
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Re: [R] Can not load Rcmdr
Dear Lugwig, Have you followed the steps in the Rcmdr installation notes at http://socserv.socsci.mcmaster.ca/jfox/Misc/Rcmdr/installation-notes.html? In particular, have you installed X-Windows? Mac OS X Mountain Lion does not come with X-Windows installed. Best, John John Fox Sen. William McMaster Prof. of Social Statistics Department of Sociology McMaster University Hamilton, Ontario, Canada http://socserv.mcmaster.ca/jfox/ On Mon, 15 Apr 2013 22:51:49 +0200 Ludwig Asam asamlu...@gmx.de wrote: Hello, I have the same problem on my mac and it doesn't work to load Rcmdr.. library(Rcmdr) Error : .onAttach failed in attachNamespace() for 'Rcmdr', details: call: structure(.External(.C_dotTclObjv, objv), class = tclObj) error: [tcl] invalid command name image. Error: package or namespace load failed for 'Rcmdr' sessionInfo() R version 3.0.0 (2013-04-03) Platform: x86_64-apple-darwin10.8.0 (64-bit) locale: [1] en_US.UTF-8/en_US.UTF-8/en_US.UTF-8/C/en_US.UTF-8/en_US.UTF-8 attached base packages: [1] tcltk stats graphics grDevices utils datasets methods base other attached packages: [1] car_2.0-16 nnet_7.3-6 MASS_7.3-26 loaded via a namespace (and not attached): [1] Rcmdr_1.9-6 tools_3.0.0 What can I do to load the package? Best regards, Ludwig -- Ludwig Asam Hauptstraße 3 86438 Kissing Tel.: 08233 210866 Handy: 0163 3133304 [[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] create an access file
You missed that this is not the RStudio support forum.
You probably also missed that R comes in both 32-bit and 64-bit flavors in the
Windows binary distribution, and you have to use the one that matches the ODBC
drivers you want to use. You can specify which version RStudio is to use.
---
Jeff NewmillerThe . . Go Live...
DCN:jdnew...@dcn.davis.ca.usBasics: ##.#. ##.#. Live Go...
Live: OO#.. Dead: OO#.. Playing
Research Engineer (Solar/BatteriesO.O#. #.O#. with
/Software/Embedded Controllers) .OO#. .OO#. rocks...1k
---
Sent from my phone. Please excuse my brevity.
Aileen Lin aileenshanhong@gmail.com wrote:
Hi there,
I tried the command in
https://stat.ethz.ch/**pipermail/r-help/2007-June/**
133606.html
https://stat.ethz.ch/pipermail/r-help/2007-June/133606.html
In R, it works OK. In Rstudio, there seems to be a problem:
library(RODBC)ch - odbcDriverConnect(Driver={Microsoft Access Driver
(*.mdb)})Warning messages:1: In odbcDriverConnect(Driver={Microsoft
Access Driver (*.mdb)}) :
[RODBC] ERROR: Could not SQLDriverConnect2: In
odbcDriverConnect(Driver={Microsoft Access Driver (*.mdb)}) :
ODBC connection failed
I use R 2.15.3 and Rstudio 0.97.336. Did I miss anything?
Regards,
Aileen
On 15 April 2013 20:17, Prof Brian Ripley rip...@stats.ox.ac.uk
wrote:
On 15/04/2013 07:12, Aileen Lin wrote:
Hi there,
I have seen this post.
https://stat.ethz.ch/**pipermail/r-help/2007-June/**133606.htmlhttps://stat.ethz.ch/pipermail/r-help/2007-June/133606.html
have odbc installed in my machine. Now I have the following message:
channel2 - odbcDriverConnect(test.mdb)**Warning messages:1: In
odbcDriverConnect(test.mdb) :
[RODBC] ERROR: state IM002, code 0, message [Microsoft][ODBC
Driver
Manager] Data source name not found and no default driver
specified2:
In odbcDriverConnect(test.mdb) :
[RODBC] ERROR: state 01S00, code 0, message [Microsoft][ODBC
Driver
Manager] Invalid connection string attribute3: In
odbcDriverConnect(test.mdb) : ODBC connection failed
Did I miss anything?
That is not doing what the URL you gave suggested. Please do follow
it.
--
Brian D. Ripley, rip...@stats.ox.ac.uk
Professor of Applied Statistics,
http://www.stats.ox.ac.uk/~**ripley/http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel: +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UKFax: +44 1865 272595
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] converting blank cells to NAs
You can use na.strings= in read.table() or read.csv() library(stringr) vec1-unlist(str_split(readLines(textConnection(3,7,11,,12,14,15,,17,18,19)),,)) vec1[vec1==]- NA vec1 # [1] 3 7 11 NA 12 14 15 NA 17 18 19 #or scan(text=3,7,11,,12,14,15,,17,18,19,sep=,) #Read 11 items #[1] 3 7 11 NA 12 14 15 NA 17 18 19 set.seed(25) arr1- array(sample(letters,40,replace=TRUE),dim=c(5,4,2)) arr1[5,3,2]- arr1[arr1==]- NA arr1[,,2] # [,1] [,2] [,3] [,4] #[1,] b h v k #[2,] l b j y #[3,] c p g j #[4,] e n c e #[5,] h g NA d arr2- array(1:40,dim=c(5,4,2)) arr2[5,3,2]- arr2[arr2==]- NA arr2- apply(arr2,c(2,3),as.numeric) arr2[,,2] # [,1] [,2] [,3] [,4] #[1,] 21 26 31 36 #[2,] 22 27 32 37 #[3,] 23 28 33 38 #[4,] 24 29 34 39 #[5,] 25 30 NA 40 A.K. Is there a way to convert blank cells in a vector or array into NAs? Alternatively, is there a test for blank cells the way one can use is.na to test for NAs? Thanks in advance. __ 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 add a row vector in a dataframe
Hi,
May be this helps you.
#Using
set.seed(12345)
S=10
simdata - replicate(S, generate(250))
lstpshat-lapply(seq_len(ncol(simdata)),function(i)
{glm.t-glm(t~x1+x2+x3+x4+x5+x6+x7+I(x2^2)+I(x4^2)+I(x7^2)+x1:x3+x2:x4+x3:x5+x4:x6+x5:x7+x1:x6+x2:x3+x3:x4+x4:x5+x5:x6,family=binomial,data=simdata[,i]);
pshat- predict(glm.t,type=response)})
simdata1-rbind(simdata,pshat=lstpshat)
pdf(hist1.pdf)
lapply(seq_len(ncol(simdata1)),function(i){ x1- simdata1[,i];
pshat0-x1$pshat[x1$t==0];pshat1- x1$pshat[x1$t==1];
hist(pshat1,xlim=c(0,1),col=rgb(0.7,0,0,0.5));
hist(pshat0,add=T,col=rgb(0,0,1,0.3))})
dev.off()
(You need to change the colors as per your requirements)
A.K.
thanks! it really helps!
anyway, how will i have a histogram of the lstpshat basing on
the value of t, that is, fot t=1, the color is red, for t=0, the color
is blue and for their overlap, the color is green? thanks a lot!
set.seed(12345)
S=1000
generate - function(size) {
x1 - rnorm(size, mean=0, sd=1)
x2 - rnorm(size, mean=0, sd=1)
x3 - rnorm(size, mean=0, sd=1)
x4 - rnorm(size, mean=0, sd=1)
x5 - rnorm(size, mean=0, sd=1)
x6 - rnorm(size, mean=0, sd=1)
x7 - rnorm(size, mean=0, sd=1)
x8 - rnorm(size, mean=0, sd=1)
x9 - rnorm(size, mean=0, sd=1)
x10 - rnorm(size, mean=0, sd=1)
e-rnorm(size, mean=0, sd=1)
t_trueps - (1 + exp( -(b0 + b1*x1 + b2*x2 + b3*x3 + b4*x4 + b5*x5 + b6*x6 +
b7*x7
+ b2*x2*x2 + b4*x4*x4 + b7*x7*x7 + b1*0.5*x1*x3 + b2*0.7*x2*x4 +b3*0.5*x3*x5
+ b4*0.7*x4*x6 + b5*0.5*x5*x7 + b1*0.5*x1*x6 + b2*0.7*x2*x3 + b3*0.5*x3*x4
+ b4*0.5*x4*x5 + b5*0.5*x5*x6) ) )^-1
prob.exposure - runif(size)
t - ifelse(t_trueps prob.exposure, 1, 0)
y - a0 + a1*x1 + a2*x2 + a3*x3 + a4*x4 +a5*x8 + a6*x9 + a7*x10 + g1*t + e
sim - as.data.frame(cbind(x1, x2, x3 ,x4, x5, x6, x7, x8, x9, x10, t, y))
return(sim)
}
b0 - 0.05
b1 - 0.95
b2 - -0.25
b3 - 0.6
b4 - -0.4
b5 - -0.8
b6 - -0.5
b7 - 0.7
a0 - -3.85
a1 - 0.3
a2 - -0.36
a3 - -0.73
a4 - -0.2
a5 - 0.71
a6 - -0.19
a7 - 0.26
g1 - -0.4
simdata - replicate(S, generate(3000))
lstpshat-lapply(seq_len(ncol(simdata)),function(i)
{glm.t-glm(t~x1+x2+x3+x4+x5+x6+x7+I(x2^2)+I(x4^2)+I(x7^2)+x1:x3+x2:x4+x3:x5+x4:x6+x5:x7+x1:x6+x2:x3+x3:x4+x4:x5+x5:x6,family=binomial,data=simdata[,i]);
pshat- predict(glm.t,type=response)})
simdata1-rbind(simdata,pshat=lstpshat)
simdata.ps1- simdata1
simdata.ps1[]-do.call(c,lapply(seq_len(ncol(simdata1)),function(i)
lapply(simdata1[,i],function(x) x[simdata1[,i]$t==1])))
lstm1- lapply(seq_len(ncol(simdata.ps1)),function(i)
{dat-do.call(data.frame,lapply(simdata.ps1[,i],function(x)
x));if(nrow(dat)!=0) {glm.1-glm(y~x1+x2+x3+x4+x8+x9+x10,data=dat)}
else NULL; glm.1; m1- predict(glm.1)})
simdata.ps0- simdata1
simdata.ps0[]-do.call(c,lapply(seq_len(ncol(simdata1)),function(i)
lapply(simdata1[,i],function(x) x[simdata1[,i]$t==0])))
lstm0-lapply(seq_len(ncol(simdata.ps0)),function(i)
{dat-do.call(data.frame,lapply(simdata.ps0[,i],function(x)
x));if(nrow(dat)!=0) {glm.0-glm(y~x1+x2+x3+x4+x8+x9+x10,data=dat)}
else NULL; glm.0; m0- predict(glm.0)})
simdata.psm1- rbind(simdata.ps1,m1=lstm1)
simdata.psm0- rbind(simdata.ps0,m0=lstm0)
hist1.pdf
Description: Adobe PDF document
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Re: [R] matching multiple fields from a matrix
HI, May be this helps: dat1- read.table(text= site1 depth1 year1 site2 depth2 year2 10 30 1860 NA NA NA NA NA NA 50 30 1860 10 20 1850 11 20 1850 11 25 1950 12 25 1960 10 NA 1870 12 30 1960 11 25 1880 15 22 1890 14 22 1890 14 25 1880 ,sep=,header=TRUE,stringsAsFactors=FALSE) res-merge(dat1[,1:3],dat1[,4:6],by.x=c(depth1,year1),by.y=c(depth2,year2)) names(res)[1:2]- gsub(\\d+,,names(res))[1:2] na.omit(res) # depth year site1 site2 #1 20 1850 10 11 #2 22 1890 14 15 #3 25 1880 11 14 #4 30 1860 10 50 A.K. - Original Message - From: jercrowley jcrowl...@mtech.edu To: r-help@r-project.org Cc: Sent: Monday, April 15, 2013 5:07 PM Subject: [R] matching multiple fields from a matrix I have been trying many ways to match 2 separate fields in a matrix. Here is a simplified version of the matrix: site1 depth1 year1 site2 depth2 year2 10 30 1860 NA NA NA NA NA NA 50 30 1860 Basically I am trying to identify the sites which have a common year and depth from 2 datasets. What I would like to do is match all of the year1 field to year2 field and the depth1 field and to depth2 field. Then I would like to output site1, site2, depth, and year. I have been trying if loops, which(), isTRUE(), etc. but I have not come up with anything that works. Any help would be greatly appreciated. Jeremy -- View this message in context: http://r.789695.n4.nabble.com/matching-multiple-fields-from-a-matrix-tp4664309.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-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] HMM Package parameter estimation
Hi, I am having difficulties estimating the parameters of a HMM using the HMM package. I have simulated a sequence of observations from a known HMM. When I estimate the parameters of a HMM using these simulated observations the parameters are not at all close to the known ones. I realise the estimated parameters are not going to be exactly the same as the known/true parameters, but these are nowhere close. Below is my code used. Any ideas or possible suggestions regarding this issue would be greatly appreciated? library(HMM) ## DECLARE PARAMETERS OF THE KNOWN MODEL states = c(1,2,3) symbols = c(1,2) startProb = c(0.5,0.25,0.25) transProb = matrix(c(0.8,0.05,0.15,0.2,0.6,0.2,0.2,0.3,0.5),3,3,TRUE) emissionProb = matrix(c(0.9,0.1,0.2,0.8,0.7,0.3), 3,2,TRUE) # CREATE THE KNOWN MODEL hmmTrue = initHMM(states, symbols, startProb, transProb , emissionProb) # SIMULATE 1000 OBSERVATIONS OF THE KNOWN MODEL observation = simHMM(hmmTrue, 1000) obs = observation$observation #ESTIMATE A MODEL USING THE OBSERVATIONS GENERATED FROM THE KNOWN MODEL hmmInit = initHMM(states, symbols, c(1/3,1/3,1/3)) hmmFit = baumWelch(hmmInit, obs) #The parameters of hmmTrue and hmmFit are not at all alike, why is this? Kind Regards, Richard [[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] rdagum(Library (VGAM))
Hello! I am working with a procedure where I need to draw one observation at a time from dagum distribution. I used the function defined in the VGAM library. when I did, rdagum(n=1, 0.2,1,4.5)) it says: Error in qdagum(runif(n), shape1.a = shape1.a, scale = scale, shape2.p = shape2.p) : object 'Scale' not found. In the description of dagum in R website I found that the sample size, n, that we need to draw using this function, should be strictly 1. Can you please let me know what should I do in this case? Thank you very much. Shant [[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.
