[R] Decision tree model using rpart ( classification
Hi Experts, I am new to R, using decision tree model for getting segmentation rules. A) Using behavioural data (attributes defining customer behaviour, ( example balances, number of accounts etc.) 1. Clustering: Cluster behavioural data to suitable number of clusters 2. Decision Tree: Using rpart classification tree for generating rules for segmentation using cluster number(cluster id) as target variable and variables from behavioural data as input variables. B) Using profile data (customers demographic data ) 1. Clustering: Cluster profile data to suitable number of clusters 2. Decision Tree: Using rpart classification tree for generating rules for segmentation using cluster number(cluster id) as target variable and variables from profile data as input variables. C) Using profile data (customers demographic data ) and deciles created based on behaviour 1. Deciles: Deciles customers to 10 groups based on some behavioural data 2. Decision Tree: Using rpart classification for generating rules for segmentation using Deciles as target variable and variables from profile data as input variables. In first two cases A and B decision tree model using rpart finish the execution in a minute or two, But in third case (C) it continues to run for infinite amount of time( monitored and running even after 14 hours). fit - rpart(decile ~., method=class,data=dtm_ip) Is there anything wrong with my approach? Thanks for the help in advance. -Ajit -- View this message in context: http://r.789695.n4.nabble.com/Decision-tree-model-using-rpart-classification-tp3989162p3989162.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] Reading parameters from dataframe and loading as objects
Hi List, I want to read several parameters from data frame and load them as object into R session, Is there any package or function in R for this?? Here is example param -c(clust_num, minsamp_size, maxsamp_size, min_pct, max_pct) value -c(15, 2, 20, 0.001, .999) data - data.frame ( cbind(param , value)) data param value 1clust_num 15 2 minsamp_size2 3 maxsamp_size 2e+05 4 min_pct 0.001 5 max_pct 0.999 My data contains many such parameters, I need to read each parameter and its value from the data and load it as objects in R session as below: clust_num - 15 minsamp_size -2 maxsamp_size -2e+05 min_pct -0.001 max_pct -0.999 The way right now I am doing it is as creating as many variables as parameters in the data frame and one observation for value of each parameter. example: clust_num minsamp_sizemaxsamp_sizemin_pct max_pct 15 2 20 0.001 0.999 data$ clust_num , data$minsamp_size, . Is there any better way for doing this? -- View this message in context: http://r.789695.n4.nabble.com/Reading-parameters-from-dataframe-and-loading-as-objects-tp3989150p3989150.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] Decision tree model using rpart ( classification
Hi Ajit, Please send the code you are running in each case. Contact Details:--- Contact me: tal.gal...@gmail.com | 972-52-7275845 Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) | www.r-statistics.com (English) -- On Fri, Nov 4, 2011 at 8:36 AM, aajit75 aaji...@yahoo.co.in wrote: Hi Experts, I am new to R, using decision tree model for getting segmentation rules. A) Using behavioural data (attributes defining customer behaviour, ( example balances, number of accounts etc.) 1. Clustering: Cluster behavioural data to suitable number of clusters 2. Decision Tree: Using rpart classification tree for generating rules for segmentation using cluster number(cluster id) as target variable and variables from behavioural data as input variables. B) Using profile data (customers demographic data ) 1. Clustering: Cluster profile data to suitable number of clusters 2. Decision Tree: Using rpart classification tree for generating rules for segmentation using cluster number(cluster id) as target variable and variables from profile data as input variables. C) Using profile data (customers demographic data ) and deciles created based on behaviour 1. Deciles: Deciles customers to 10 groups based on some behavioural data 2. Decision Tree: Using rpart classification for generating rules for segmentation using Deciles as target variable and variables from profile data as input variables. In first two cases A and B decision tree model using rpart finish the execution in a minute or two, But in third case (C) it continues to run for infinite amount of time( monitored and running even after 14 hours). fit - rpart(decile ~., method=class,data=dtm_ip) Is there anything wrong with my approach? Thanks for the help in advance. -Ajit -- View this message in context: http://r.789695.n4.nabble.com/Decision-tree-model-using-rpart-classification-tp3989162p3989162.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. [[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] Reading parameters from dataframe and loading as objects
Hi, Did you program in SAS previously? :) You don't really want data.frame but a named list (note that data frames are lists but for columns). mypars - as.list(value) names(mypars) - param mypars$max_pct If you really want to use data.frame you can assign rownames and access parameters through: datapars - data.frame(value=value) rownames(datapars) - param datapars[max_pct,value] But this really seems more complicated way ins't it? HTH Eric On 4 November 2011 07:24, Aher ajit.a...@cedar-consulting.com wrote: Hi List, I want to read several parameters from data frame and load them as object into R session, Is there any package or function in R for this?? Here is example param -c(clust_num, minsamp_size, maxsamp_size, min_pct, max_pct) value -c(15, 2, 20, 0.001, .999) data - data.frame ( cbind(param , value)) data param value 1clust_num 15 2 minsamp_size2 3 maxsamp_size 2e+05 4 min_pct 0.001 5 max_pct 0.999 My data contains many such parameters, I need to read each parameter and its value from the data and load it as objects in R session as below: clust_num - 15 minsamp_size -2 maxsamp_size -2e+05 min_pct -0.001 max_pct -0.999 The way right now I am doing it is as creating as many variables as parameters in the data frame and one observation for value of each parameter. example: clust_num minsamp_sizemaxsamp_sizemin_pct max_pct 15 2 20 0.001 0.999 data$ clust_num , data$minsamp_size, . Is there any better way for doing this? -- View this message in context: http://r.789695.n4.nabble.com/Reading-parameters-from-dataframe-and-loading-as-objects-tp3989150p3989150.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.htmlhttp://www.r-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Eric Lecoutre Consultant - Business Decision Business Intelligence Customer Intelligence [[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 count number of occurrences
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of uka
Sent: Thursday, November 03, 2011 4:10 PM
To: r-help@r-project.org
Subject: Re: [R] how to count number of occurrences
This was very helpful. Thank you very much. Just one question, I notice
that
it does not count the number of X's before the first Y. I want the
result be
1 4 0 0 0 5 0 0 0 0. I tried combining this output with the first value
of
rle output, but realized that rle doesn't give me the 0s. So, if my
first
observation was Y, then I want it to show that there are 0 Xs before
that.
Thank you again.
You should really provide the relevant context from previous posts so that
potential helpers don't need to go looking for it. That being said, you could
try something like
samp - c(X, Y, X, X, X, X, Y, Y, Y, Y, X, X, X, X,
X, Y, Y, Y, Y, Y)
diff(which(c('Y', samp)=='Y'))-1
Hope this is helpful,
Dan
Daniel J. Nordlund
Washington State Department of Social and Health Services
Planning, Performance, and Accountability
Research and Data Analysis Division
Olympia, WA 98504-5204
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[R] array manipulation
This worked example, hoping to be helpful, has been requested after a (my) further enquiry about array manipulation. I was looking for a command that is equivalent to repmat() in matlab and that could also be applied to array. (for Matlab users) The Matlal code was the following: temp_u=zeros(d,c,T); -creation of an array of dimensions d x c x T full of zeroes temp_u(1,:,:)=m_u;-filling the first row of each 'stratum' with the rows of the matrix 'm_u' temp_u=repmat(temp_u(1,:,:),d,1); -filling the remaining rows (full of zeroes) of 'temp_u' with copies of the corrensponding 1st row -- --- Simone Salvadei Faculty of Economics Department of Financial and Economic Studies and Quantitative Methods University of Rome Tor Vergata e-mail: simone.salva...@uniroma2.it federico.belo...@uniroma2.it url: http://www.economia.uniroma2.it/phd/econometricsempiricaleconomics/ http://www.econometrics.it/ --- [[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] replace double backslash with singel backslash
I want to replace \\ with \ in: str - C:\\DOKUME~1\\u0327336\\LOKALE~1\\Temp\\RtmpQ5NJ8X\\TIRIS_PICS\\1_Img.jpg and tried: gsub(, \\, str) but this removes the \\ without replacing them by \ Any help much appreciated, Kay - Kay Cichini Postgraduate student Institute of Botany Univ. of Innsbruck -- View this message in context: http://r.789695.n4.nabble.com/replace-double-backslash-with-singel-backslash-tp3989434p3989434.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] Decision tree model using rpart ( classification
Hi, Thanks for the responce, code for each case is as: c_c_factor - 0.001 min_obs_split - 80 A) fit - rpart(segment ~., method=class, control=rpart.control(minsplit=min_obs_split, cp=c_c_factor), data=Beh_cluster_out) B) fit - rpart(segment ~., method=class, control=rpart.control(minsplit=min_obs_split, cp=c_c_factor), data=profile_cluster_out) C) fit - rpart(decile ~., method=class, control=rpart.control(minsplit=min_obs_split, cp=c_c_factor), data=dtm_ip) In A and B target variable 'segment' is from the clustering data using same set of input variables , while in C target variable 'decile' is derived from behavioural variables and input variables are from profile data. Number of rows in the input table in all three cases are same. Regards, -Ajit -- View this message in context: http://r.789695.n4.nabble.com/Decision-tree-model-using-rpart-classification-tp3989162p3989320.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] array manipulation
This worked example, hoping to be helpful, has been requested after a (my) further enquiry about array manipulation. I was looking for a command that is equivalent to repmat() in matlab and that could also be applied to array. (for Matlab users) The Matlal code was the following: 1)temp_u=zeros(d,c,T); -creation of an array of dimensions d x c x T full of zeroes 2)temp_u(1,:,:)=m_u;-filling the first row of each 'stratum' with the rows of the matrix 'm_u' 3)temp_u=repmat(temp_u(1,:,:),d,1); -filling the remaining rows (full of zeroes) of 'temp_u' with copies of the corrensponding 1st row (what's happening if you are not a Matlab users) (numerical example d=2,c=4,T=4) 1)temp_u=zeros(d,c,T) temp_u(:,:,1) = 0 0 0 0 0 0 0 0 temp_u(:,:,2) = 0 0 0 0 0 0 0 0 temp_u(:,:,3) = 0 0 0 0 0 0 0 0 temp_u(:,:,4) = 0 0 0 0 0 0 0 0 2)temp_u(1,:,:)=m_u temp_u(:,:,1) = 0.96040.01560.02300.0009 0 0 0 0 temp_u(:,:,2) = 0.39060.29480.09810.2165 0 0 0 0 temp_u(:,:,3) = 0.53900.24820.11400.0988 0 0 0 0 temp_u(:,:,4) = 0.45460.26410.07940.2019 0 0 0 0 3)temp_u=repmat(temp_u(1,:,:),d,1) temp_u(:,:,1) = 0.96040.01560.02300.0009 0.96040.01560.02300.0009 temp_u(:,:,2) = 0.39060.29480.09810.2165 0.39060.29480.09810.2165 temp_u(:,:,3) = 0.53900.24820.11400.0988 0.53900.24820.11400.0988 temp_u(:,:,4) = 0.45460.26410.07940.2019 0.45460.26410.07940.2019 Now, in order to reply this exercise in R I used the following code: temp_u=array(0,dim=c(1,c,T)) temp_u[1,,]=m_u temp_u=kronecker(temp_u,matrix(rep(1,d),nr=d)) A special thank to David Winsemius,William Dunlap and Patrick Burns. I hope I have been helpful. --- Simone Salvadei Faculty of Economics Department of Financial and Economic Studies and Quantitative Methods University of Rome Tor Vergata e-mail: simone.salva...@uniroma2.it federico.belo...@uniroma2.it url: http://www.economia.uniroma2.it/phd/econometricsempiricaleconomics/ http://www.econometrics.it/ --- [[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] replace double backslash with singel backslash
what is the problem that you are trying to solve? you need the double \\ since they have a special meaning in quoted strings. in this case they represent a since backslash. if you really had a single one, then something like this '\n' would be a carriage return. Sent from my iPad On Nov 4, 2011, at 5:35, Kay Cichini kay.cich...@uibk.ac.at wrote: I want to replace \\ with \ in: str - C:\\DOKUME~1\\u0327336\\LOKALE~1\\Temp\\RtmpQ5NJ8X\\TIRIS_PICS\\1_Img.jpg and tried: gsub(, \\, str) but this removes the \\ without replacing them by \ Any help much appreciated, Kay - Kay Cichini Postgraduate student Institute of Botany Univ. of Innsbruck -- View this message in context: http://r.789695.n4.nabble.com/replace-double-backslash-with-singel-backslash-tp3989434p3989434.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.
Re: [R] Extract Data from Yahoo Finance
Deb, Sorry, you can't do that with getQuote because Yahoo does not make those data available historically. Generally, you will need to pay for historical bid/ask (tick) data. Best, -- Joshua Ulrich | FOSS Trading: www.fosstrading.com On Fri, Nov 4, 2011 at 6:10 AM, Deb Midya debmi...@yahoo.com wrote: Joshua, Thank you very much for your response. How can I use getQuote to download data for a stock for a certain period of time, say, 1 October 2011 to 31 October 2011. Once again, thank very much for the time you have given. Regards, Deb From: Joshua Ulrich josh.m.ulr...@gmail.com To: Deb Midya debmi...@yahoo.com Cc: R. Michael Weylandt michael.weyla...@gmail.com; r-help@r-project.org r-help@r-project.org Sent: Friday, 4 November 2011 4:12 PM Subject: Re: [R] Extract Data from Yahoo Finance Deb, See getQuote in the quantmod package. For example: getQuote(SPY) Be sure to read ?getQuote. Best, -- Joshua Ulrich | FOSS Trading: www.fosstrading.com On Thu, Nov 3, 2011 at 5:05 PM, Deb Midya debmi...@yahoo.com wrote: Michael, Thanks for your response. The link to the page is: http://www.gummy-stuff.org/Yahoo-data.htm I like to download the fields (mentioned under special tags) for a period of time and for a particular stock (or for a list of stocks). Once again, thank you very much for the time you have given. Regards, Deb From: R. Michael Weylandt michael.weyla...@gmail.com To: Deb Midya debmi...@yahoo.com Cc: r-help@r-project.org r-help@r-project.org Sent: Friday, 4 November 2011 12:13 AM Subject: Re: [R] Extract Data from Yahoo Finance The quantmod package can probably do what you are asking, but it's a little hard to be certain since you provide neither a list of all the fields you are actually talking about nor a link to the page with the fields in question. Michael On Thu, Nov 3, 2011 at 12:02 AM, Deb Midya debmi...@yahoo.com wrote: Hi R –users, I am using R-2.14.0 on Windows XP. May I request you to assist me for the following please. I like to extract all the fields (example: a : Ask, b : Bid, ……, w : 52-week Range, x: Stock Exchange) for certain period of time, say, 1 October 2011 to 31 October 2011. Is there any R-Package(s) any R- script please? Once again, thank you very much for the time you have given. Regards, Deb __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-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] Counting number of common elements between the rows of two different matrices
Hello
I'm trying to solve this problem without using a for loop but I have so
far failed to find a solution.
I have two matrices of K columns each, e.g. (K=5), and with numbers of
row N_A and N_B respectively
A = (1 5 3 8 15;
2 7 20 11 13;
12 19 20 21 43)
B = (2 6 30 8 16;
3 8 19 11 13)
(the actual matrices have hundreds of thousands of entry, that's why I'm
keen to avoid for loops)
And what I need to do is to apply a function which counts the number of
common elements between ANY row of A and ANY row of B, giving a result
like this:
A1 vs B1: 1 # (8 is a common element)
A1 vs B2: 1 # (8 is a common element)
A2 vs B1: 1 # (2 is a common element)
A2 vs B2: 1 # 11, 13 are common elements
Etc.
I've built a function that counts the number of common elements between
two vectors, based on the intersect function in the R manual
common_elements - function(x,y) length(y[match(x,y,nomatch=0)])
And a double loop who solves my problem would be something like
(pseudo-code)
For(i in 1:N_A){
for(j in 1:N_B){
ce(i,j)=common_elements(a(i),b(j))
}
}
Is there an efficient, clean way to do the same job and give as an
output a matrix N_A x N_B such as that above?
Thanks a lot for your help
Regards
Pietro
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[R] Select some, but not all, variables stepwise
Hi,
I would like to fit a linear model where some but not all explanators are
chosen stepwise - ie I definitely want to include some terms, but others only
if they are deemed significant (by AIC or whatever other approach is
available). For example if I wanted to definitely include x1 and x2, but only
include z1 and z2 if they are significant, something like this:
df - data.frame(y=c(4,2,6,7,3,9,5,7,6,2), x1=c(2,3,4,0,5,8,8,1,1,2),
x2=c(0,0,0,0,1,1,0,0,0,1), z1=c(0,1,0,0,0,1,1,0,1,1), z2=c(1,1,1,0,0,1,1,1,1,0))
model - lm(y ~ x1 + x2 + stepwise(z1 + z2))
Any help would be appreciated.
Cheers,
Andre
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Re: [R] Select some, but not all, variables stepwise
Resist the temptation. Stepwise analysis without shrinkage will ruin model
inferences without helping with predictive accuracy.
Frank
AndreE wrote:
Hi,
I would like to fit a linear model where some but not all explanators are
chosen stepwise - ie I definitely want to include some terms, but others
only if they are deemed significant (by AIC or whatever other approach is
available). For example if I wanted to definitely include x1 and x2, but
only include z1 and z2 if they are significant, something like this:
df - data.frame(y=c(4,2,6,7,3,9,5,7,6,2), x1=c(2,3,4,0,5,8,8,1,1,2),
x2=c(0,0,0,0,1,1,0,0,0,1), z1=c(0,1,0,0,0,1,1,0,1,1),
z2=c(1,1,1,0,0,1,1,1,1,0))
model - lm(y ~ x1 + x2 + stepwise(z1 + z2))
Any help would be appreciated.
Cheers,
Andre
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-
Frank Harrell
Department of Biostatistics, Vanderbilt University
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Re: [R] Plotting skewed normal distribution with a bar plot
Hi Michael, Thanks for pointing me fGarch. I actually started there, but it is not yet available for 2.13.2 so I went directly to the (sn-package). I've briefly explored your suggestion and think it will work. Thanks Steve On Nov 3, 2011 10:41 PM, R. Michael Weylandt michael.weyla...@gmail.com wrote: It seems like you'll need to apply some sort of MLE to estimate the parameters directly from the data before using dsn() to get the density. This might help with some of it: http://help.rmetrics.org/fGarch/html/snorm.html Michael On Thu, Nov 3, 2011 at 2:54 PM, steve_fried...@nps.gov wrote: Hi, I need to create a plot (type = h) and then overlay a skewed-normal curve on this distribution, but I'm not finding a procedure to accomplish this. I want to use the plot function here in order to control the bin distributions. I have explored the sn library and found the dsn function. dsn uses known location, scaling and shape parameters associated with a given input vector of probabilities. However, how can I calculate the skewed-normal curve if I don't know these parameters in advance? Is there another function to calculate the skew-normal, perhaps in a different package? I'm working with R 2.13.2 on a windows based machine. Steve Friedman Ph. D. Ecologist / Spatial Statistical Analyst Everglades and Dry Tortugas National Park 950 N Krome Ave (3rd Floor) Homestead, Florida 33034 steve_fried...@nps.gov Office (305) 224 - 4282 Fax (305) 224 - 4147 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] replace double backslash with singel backslash
On 11/4/2011 5:35 AM, Kay Cichini wrote: I want to replace \\ with \ in: str- C:\\DOKUME~1\\u0327336\\LOKALE~1\\Temp\\RtmpQ5NJ8X\\TIRIS_PICS\\1_Img.jpg and tried: gsub(, \\, str) but this removes the \\ without replacing them by \ You may be able to avoid this simply by using forward slashes in path names. Otherwise, ?normalizePath may help. -- Michael Friendly Email: friendly AT yorku DOT ca Professor, Psychology Dept. York University Voice: 416 736-5115 x66249 Fax: 416 736-5814 4700 Keele StreetWeb: http://www.datavis.ca Toronto, ONT M3J 1P3 CANADA __ 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] Counting number of common elements between the rows of two different matrices
Try this:
# create dummy data
a - matrix(sample(20, 50, TRUE), ncol = 5)
b - matrix(sample(20, 50, TRUE), ncol = 5)
# create combinations to test
x - expand.grid(seq(nrow(a)), seq(nrow(b)))
# test
result - mapply(function(m1, m2) any(a[m1, ] %in% b[m2, ])
, x[, 1]
, x[, 2]
)
# create the output matrix
result.m - matrix(result, nrow = nrow(a), ncol = nrow(b))
On Fri, Nov 4, 2011 at 8:51 AM, Parodi, Pietro pietro.par...@willis.com wrote:
Hello
I'm trying to solve this problem without using a for loop but I have so
far failed to find a solution.
I have two matrices of K columns each, e.g. (K=5), and with numbers of
row N_A and N_B respectively
A = (1 5 3 8 15;
2 7 20 11 13;
12 19 20 21 43)
B = (2 6 30 8 16;
3 8 19 11 13)
(the actual matrices have hundreds of thousands of entry, that's why I'm
keen to avoid for loops)
And what I need to do is to apply a function which counts the number of
common elements between ANY row of A and ANY row of B, giving a result
like this:
A1 vs B1: 1 # (8 is a common element)
A1 vs B2: 1 # (8 is a common element)
A2 vs B1: 1 # (2 is a common element)
A2 vs B2: 1 # 11, 13 are common elements
Etc.
I've built a function that counts the number of common elements between
two vectors, based on the intersect function in the R manual
common_elements - function(x,y) length(y[match(x,y,nomatch=0)])
And a double loop who solves my problem would be something like
(pseudo-code)
For(i in 1:N_A){
for(j in 1:N_B){
ce(i,j)=common_elements(a(i),b(j))
}
}
Is there an efficient, clean way to do the same job and give as an
output a matrix N_A x N_B such as that above?
Thanks a lot for your help
Regards
Pietro
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Re: [R] replace double backslash with singel backslash
Your str does not have any double backslashes to replace. You need to revisit the concept of escape characters in the documentation. In brief, every \\ in a quoted string is actually a single character as stored in memory. --- Jeff Newmiller The . . Go Live... DCN:jdnew...@dcn.davis.ca.us Basics: ##.#. ##.#. Live Go... Live: OO#.. Dead: OO#.. Playing Research Engineer (Solar/Batteries O.O#. #.O#. with /Software/Embedded Controllers) .OO#. .OO#. rocks...1k --- Sent from my phone. Please excuse my brevity. Kay Cichini kay.cich...@uibk.ac.at wrote: I want to replace \\ with \ in: str - C:\\DOKUME~1\\u0327336\\LOKALE~1\\Temp\\RtmpQ5NJ8X\\TIRIS_PICS\\1_Img.jpg and tried: gsub(, \\, str) but this removes the \\ without replacing them by \ Any help much appreciated, Kay - Kay Cichini Postgraduate student Institute of Botany Univ. of Innsbruck -- View this message in context: http://r.789695.n4.nabble.com/replace-double-backslash-with-singel-backslash-tp3989434p3989434.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. [[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] replace double backslash with singel backslash
I think that people are afraid to say You can't do that in R... But I think the real answer is: you can't do that in R. Although, it is helpful to understand Jeff's reply. I hadn't fully realized why this particular problem occurs before reading that. It's odd to me that // and / are both stored as /, but that makes sense given my experience in R. Also, the other replies are good advice, working with R's path functions or sticking with forward slashes is the way to go (don't fight the assimilation, the borg needs you). Personally, I think some of these seemingly small problems actually encumber R's mainstream adoption quite a bit, but then I'm not the one writing R. Plus, it's is still pretty dang awesome even with its minor annoyances. On Fri, Nov 4, 2011 at 8:45 AM, Jeff Newmiller jdnew...@dcn.davis.ca.uswrote: Your str does not have any double backslashes to replace. You need to revisit the concept of escape characters in the documentation. In brief, every \\ in a quoted string is actually a single character as stored in memory. --- Jeff Newmiller The . . Go Live... DCN:jdnew...@dcn.davis.ca.us Basics: ##.#. ##.#. Live Go... Live: OO#.. Dead: OO#.. Playing Research Engineer (Solar/Batteries O.O#. #.O#. with /Software/Embedded Controllers) .OO#. .OO#. rocks...1k --- Sent from my phone. Please excuse my brevity. Kay Cichini kay.cich...@uibk.ac.at wrote: I want to replace \\ with \ in: str - C:\\DOKUME~1\\u0327336\\LOKALE~1\\Temp\\RtmpQ5NJ8X\\TIRIS_PICS\\1_Img.jpg and tried: gsub(, \\, str) but this removes the \\ without replacing them by \ Any help much appreciated, Kay - Kay Cichini Postgraduate student Institute of Botany Univ. of Innsbruck -- View this message in context: http://r.789695.n4.nabble.com/replace-double-backslash-with-singel-backslash-tp3989434p3989434.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. [[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. [[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 write a shapefile with projection
Hi, I am trying to write a shapefile with projection. I have my data in a data.frame called try and consists in xy coordinates and a numerical attribute value z1. Libraries loaded are: sp, rgdal, raster, maptools head(try) x y z1 1 610237.1 3375751 8.221 2 610236.1 3375750 8.153 3 610236.1 3375749 8.275 4 610236.1 3375748 8.251 5 610236.1 3375747 8.217 6 610236.1 3375746 8.196 #Get the projection from a raster named llev I have loaded before: crs - projection(llev) # get a spatial point data frame from my data crest.sp - SpatialPointsDataFrame(try[,1:2], try, proj4string=CRS(crs)) summary(crest.sp) Object of class SpatialPointsDataFrame Coordinates: min max x 610235.1 610354.1 y 3374862.4 3375751.4 Is projected: TRUE proj4string : [+proj=utm +zone=15 +ellps=GRS80 +datum=NAD83 +units=m +no_defs +towgs84=0,0,0] Number of points: 890 Data attributes: x y z1 Min. : 610235 Min. : 3374862 Min.: 6.966 1st Qu.:610269 1st Qu.:3375085 1st Qu.:7.570 Median :610298 Median :3375307 Median :7.901 Mean :610300 Mean :3375307 Mean :7.882 3rd Qu.:610334 3rd Qu.:3375529 3rd Qu.:8.180 Max. :610354 Max. :3375751 Max. :8.756 #write the shapefile writePointsShape(crest.sp, I:/LA_levee/Shape/llev_crest_pts6) If I load this shapefile in ArcGIS is has no projection. I also looked at the write.shapefile command from shapefiles library and again I get a file without projection. Is there any way to write the projection for the shapefile in R? Probably I can do something in ArcGIS, but I would like to have my shapefile from R complete. Thanks, Monica __ 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 use 'prcomp' with CLUSPLOT?
Hello Jo,
Full disclosure: I don't know much about clustering/partition cluster
analysis/etc so I've only attacked this as an R problem. However, this
might get you going in the right direction:
df -
read.table(textConnection(PRVID,VAR1,VAR2,VAR3,VAR4,VAR5,VAR6,VAR7,VAR8,VAR9,VAR10,VAR11
PRV1,0,54463,53049,62847,75060,184925,0,0,0,0,0
PRV2,0,2100,76,131274,0,0,0,0,0,0,18
PRV3,967,0,0,0,0,0,0,0,0,3634,0
PRV4,817,18344,3274,9264,1862,0,0,141,0,0,0
PRV5,0,0,0,0,0,0,29044,0,0,0,0
PRV6,59,6924,825,3008,377,926,0,0,10156,0,
PRV7,11,24902,36040,47223,20086,0,0,749,415,0,0), header = T, sep =
,, stringsAsFactors = T)
closeAllConnections()
library(cluster)
mat - as.matrix(df[,-1])
newtble - prop.table(mat, 1) * 100
num.clust - 3
clusplotMW - cluster:::clusplot.default # Create a copy of the two
necessary functions for clusplot that route to princomp
mkCheckMW - cluster:::mkCheckX
body(mkCheckMW) - parse(text=gsub(princomp,
prcomp,deparse(body(mkCheckMW # replace princomp with prcomp in
our copy
body(clusplotMW) - parse(text=gsub(mkCheckX,
mkCheckMW,deparse(body(clusplotMW # route our clusplot to our
mkCheckX
clusplotMW(newtble, fitnw$cluster, color = T, shade = T, lines = 0)
Since you didn't provide a working example, I can't verify this, but
let me know if it works for you.
Michael
On Thu, Nov 3, 2011 at 8:10 PM, Jo Frabetti jfrabe...@sdsc.edu wrote:
Hello,
I have a large data set that has more columns than rows (sample data below).
I am trying to perform a partitioning cluster analysis and then plot that
using pca. I have tried using CLUSPLOT(), but that only allows for
'princomp' where I need 'prcomp' as I do not want to reduce my columns. Is
there a way to edit the CLUSPLOT() code to use 'prcomp', please?
# sample of my data
PRVID,VAR1,VAR2,VAR3,VAR4,VAR5,VAR6,VAR7,VAR8,VAR9,VAR10,VAR11
PRV1,0,54463,53049,62847,75060,184925,0,0,0,0,0
PRV2,0,2100,76,131274,0,0,0,0,0,0,18
PRV3,967,0,0,0,0,0,0,0,0,3634,0
PRV4,817,18344,3274,9264,1862,0,0,141,0,0,0
PRV5,0,0,0,0,0,0,29044,0,0,0,0
PRV6,59,6924,825,3008,377,926,0,0,10156,0,
PRV7,11,24902,36040,47223,20086,0,0,749,415,0,0
library(cluster)
fn = big.csv;
tbl = read.table(fn, header=TRUE, sep=,, row.names=1);
mat - as.matrix(tbl);
newtbl - prop.table(mat,1)*100;
num.clust - 3;
fitnw - kmeans(newtbl, num.clust);
clusplot(newtbl, fitnw$cluster, color=TRUE, shade=TRUE, lines=0, main=
paste('Principal Components plot - Kmeans ', clust.level, ' Clusters') )
Error in princomp.default(x, scores = TRUE, cor = ncol(x) != 2) :
'princomp' can only be used with more units than variables
Thank you for R and any assistance you may offer!
Jo
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[R] formatting a dataframe
http://r.789695.n4.nabble.com/file/n3989638/tab3.gif i want to create a dataframe like the one given in the pictures having names of sub categories of columns. the table may be presented either in the console or in an excel sheet. anyone please suggest me a technique. thanks in advance. -atanu -- View this message in context: http://r.789695.n4.nabble.com/formatting-a-dataframe-tp3989638p3989638.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] 12th Root of a Square (Transition) Matrix
I have tried this method, but the result is not working, at least not as I expect: I used the CreditMetrics package transition matrix rc - c(AAA, AA, A, BBB, BB, B, CCC, D) M - matrix(c(90.81, 8.33, 0.68, 0.06, 0.08, 0.02, 0.01, 0.01, 0.70, 90.65, 7.79, 0.64, 0.06, 0.13, 0.02, 0.01, 0.09, 2.27, 91.05, 5.52, 0.74, 0.26, 0.01, 0.06, 0.02, 0.33, 5.95, 85.93, 5.30, 1.17, 1.12, 0.18, 0.03, 0.14, 0.67, 7.73, 80.53, 8.84, 1.00, 1.06, 0.01, 0.11, 0.24, 0.43, 6.48, 83.46, 4.07, 5.20, 0.21, 0, 0.22, 1.30, 2.38, 11.24, 64.86, 19.79, 0, 0, 0, 0, 0, 0, 0, 100 )/100, 8, 8, dimnames = list(rc, rc), byrow = TRUE) then followed through with the steps: nth_root - X %*% L_star %*% X_inv But the check (going back 12 to the power again) doesn't yield the original matrix. Now some rounding errors can be expected, but I didn't expect a perfectly diagonal matrix, when the initial matrix isn't diagonal at all. round(nth_root^12,4) [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [1,] 0.9078 0. 0. 0. 0. 0. 0.0 [2,] 0. 0.9053 0. 0. 0. 0. 0.0 [3,] 0. 0. 0.9079 0. 0. 0. 0.0 [4,] 0. 0. 0. 0.8553 0. 0. 0.0 [5,] 0. 0. 0. 0. 0.7998 0. 0.0 [6,] 0. 0. 0. 0. 0. 0.8285 0.0 [7,] 0. 0. 0. 0. 0. 0. 0.64570 [8,] 0. 0. 0. 0. 0. 0. 0.1 Any takers - Christian Langkamp christian.langkamp-at-gmxpro.de -- View this message in context: http://r.789695.n4.nabble.com/12th-Root-of-a-Square-Transition-Matrix-tp2259736p3989618.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] nproc parameter in efpFunctional
The 2006 CSDA paper is really very informative, perhaps, I'm trying to understand the things lying beyond. If we have e.g. k=3, then taking nproc=3 for the functional maxBB we get a critical value (boundary) maxBB$computeCritval(0.05,nproc=3) [1] 1.544421, and this for nproc=NULL (Bonferroni approximation) will be maxBB$computeCritval(0.05) [1] 1.358099. Aggregating 3 Brownian bridges first over components, we obtain time series process. Now, we wonder if maximum value of the process (aggregation over time) lies over boundary. Which boundary - 1.544421 or 1.358099 - should one take? They look too different and, for instance, lead to unfair computing of empirical size (as rejection rate of null hypothesis) or empirical power (as acception rate of alternative). -- View this message in context: http://r.789695.n4.nabble.com/nproc-parameter-in-efpFunctional-tp3972419p3989598.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] Plotting skewed normal distribution with a bar plot
Steve, I don't profess to be an expert in R (by ANY means), but I've used the sn.mle function in package sn to do something similar. So, for example, if you have data (y), you can do the following: y = c(5,4,6,4,5,6,7,7,55,64,23,13,1,4,1,14,1,15,11,12) sn.mle(y=y) I just made up this stream of data, so please forgive if it doesn't make sense. The function will produce a number of things, including the mles for the mean, sd, and skew, as well as a histogram with the density overlayed onto it. Maybe this helps? Hope so. Greg -Original Message- From: R. Michael Weylandt michael.weyla...@gmail.com Sent: Nov 3, 2011 10:39 PM To: steve_fried...@nps.gov Cc: r-help@r-project.org Subject: Re: [R] Plotting skewed normal distribution with a bar plot It seems like you'll need to apply some sort of MLE to estimate the parameters directly from the data before using dsn() to get the density. This might help with some of it: http://help.rmetrics.org/fGarch/html/snorm.html Michael On Thu, Nov 3, 2011 at 2:54 PM, steve_fried...@nps.gov wrote: Hi, I need to create a plot (type = h) and then overlay a skewed-normal curve on this distribution, but I'm not finding a procedure to accomplish this. I want to use the plot function here in order to control the bin distributions. I have explored the sn library and found the dsn function. dsn uses known location, scaling and shape parameters associated with a given input vector of probabilities. However, how can I calculate the skewed-normal curve if I don't know these parameters in advance? Is there another function to calculate the skew-normal, perhaps in a different package? I'm working with R 2.13.2 on a windows based machine. Steve Friedman Ph. D. Ecologist / Spatial Statistical Analyst Everglades and Dry Tortugas National Park 950 N Krome Ave (3rd Floor) Homestead, Florida 33034 steve_fried...@nps.gov Office (305) 224 - 4282 Fax (305) 224 - 4147 __ 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] How to write a shapefile with projection - problem solved
Hi, Sorry i have put such a detailed question to the list about writing a shapefile with projection. I realized that if i use writeOGR from rgdal and not the other write shapefile functions i can get a shapefile with projection recognized by ArcGIS. The command is (in case anybody wonders): writeOGR(crest.sp, I:\\LA_levee\\Shape, llev_crest_pts6, driver = ESRI Shapefile) where crest.sp is a spatial point data frame with projection. Thanks, Monica __ 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] Select some, but not all, variables stepwise
Hi Andre
I don´t know if it will work, but I´ve tried the MuMIn package, were you can
evaluate
all possible models (usin for example AIC) at one time. Maybe you can focus on
comparing
those models which retain the explanators you want.
Best wishes
Francisco
On Fri, 4 Nov 2011 13:06:09 +, Andre Easom wrote
Hi,
I would like to fit a linear model where some but not all explanators are
chosen stepwise - ie I definitely want to include some terms, but others only
if they are deemed significant (by AIC or whatever other approach is
available)
. For example if I wanted to definitely include x1 and x2, but only include
z1 and z2 if they are significant, something like this:
df - data.frame(y=c(4,2,6,7,3,9,5,7,6,2), x1=c(2,3,4,0,5,8,8,1,1,2),
x2=c(0,0,
0,0,1,1,0,0,0,1), z1=c(0,1,0,0,0,1,1,0,1,1), z2=c(1,1,1,0,0,1,1,1,1,0)) model
- lm(y ~ x1 + x2 + stepwise(z1 + z2))
Any help would be appreciated.
Cheers,
Andre
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Centro de Investigaciones en Ecosistemas
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[R] How to delete only those rows in a dataframe in which all records are missing
Hi, Imagine I have the following data frame: a - c(1,NA,3) b - c(2,NA,NA) c - data.frame(cbind(a,b)) c a b 1 1 2 2 NA NA 3 3 NA I want to delete the second row. If I use na.omit, that would also affect the third row. I tried to use a loop and an ifelse clause with is.na to get R identify that row in which all records are missing, as opposed to the first row in which no records are missing or the third one, in which only one record is missing. How can I get R identify the row in which all records are missing? Or, how can I get R delete/omit only this row? Thanks in advance, José José Iparraguirre Chief Economist Age UK T 020 303 31482 E jose.iparragui...@ageuk.org.ukmailto:jose.iparragui...@ageuk.org.uk Tavis House, 1- 6 Tavistock Square London, WC1H 9NB www.ageuk.org.ukhttp://www.ageuk.org.uk | ageukblog.org.ukhttp://ageukblog.org.uk/ | @AgeUKPAhttp://twitter.com/ageukpa Age UK Improving later life www.ageuk.org.uk --- Age UK is a registered charity and company limited by guarantee, (registered charity number 1128267, registered company number 6825798). Registered office: Tavis House, 1-6 Tavistock Square, London WC1H 9NA. For the purposes of promoting Age UK Insurance, Age UK is an Appointed Representative of Age UK Enterprises Limited, Age UK is an Introducer Appointed Representative of JLT Benefit Solutions Limited and Simplyhealth Access for the purposes of introducing potential annuity and health cash plans customers respectively. Age UK Enterprises Limited, JLT Benefit Solutions Limited and Simplyhealth Access are all authorised and regulated by the Financial Services Authority. -- This email and any files transmitted with it are confidential and intended solely for the use of the individual or entity to whom they are addressed. If you receive a message in error, please advise the sender and delete immediately. Except where this email is sent in the usual course of our business, any opinions expressed in this email are those of the author and do not necessarily reflect the opinions of Age UK or its subsidiaries and associated companies. Age UK monitors all e-mail transmissions passing through its network and may block or modify mails which are deemed to be unsuitable. Age Concern England (charity number 261794) and Help the Aged (charity number 272786) and their trading and other associated companies merged on 1st April 2009. Together they have formed the Age UK Group, dedicated to improving the lives of people in later life. The three national Age Concerns in Scotland, Northern Ireland and Wales have also merged with Help the Aged in these nations to form three registered charities: Age Scotland, Age NI, Age Cymru. [[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] survfit function?
Hi, I am working on fitting a proportional hazard model to predict the probability of default for mortgage loans. I have a question regarding survfit function. My historical data set is a pool of loans with monthly observed default status for the next 24 months. The data is left truncated (delayed entry to observation window after the loan is opened) and right censored. I would like to fit the model with time varying covariate such as unemployment rates and time constant variables at loan application, and then use the model to predict the probability of default in the next 24 month for the pool of loans we have right now, by using function survfit. When loans are outside of the observed time window, is it reliable to use survfit function to do the prediction? If its not reliable, how to deal with this problem? Is there another way to set the model? Any thoughts are appreciated. Thanks so much in advance. Ying(Cindy) [[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] Counting number of common elements between the rows of two different matrices
Jim
I tried that and it works. Thank you very much for your help!
Regards
Pietro
-Original Message-
From: jim holtman [mailto:jholt...@gmail.com]
Sent: 04 November 2011 13:38
To: Parodi, Pietro
Cc: r-help@r-project.org
Subject: Re: [R] Counting number of common elements between the rows of two
different matrices
Try this:
# create dummy data
a - matrix(sample(20, 50, TRUE), ncol = 5)
b - matrix(sample(20, 50, TRUE), ncol = 5)
# create combinations to test
x - expand.grid(seq(nrow(a)), seq(nrow(b)))
# test
result - mapply(function(m1, m2) any(a[m1, ] %in% b[m2, ])
, x[, 1]
, x[, 2]
)
# create the output matrix
result.m - matrix(result, nrow = nrow(a), ncol = nrow(b))
On Fri, Nov 4, 2011 at 8:51 AM, Parodi, Pietro pietro.par...@willis.com wrote:
Hello
I'm trying to solve this problem without using a for loop but I have so
far failed to find a solution.
I have two matrices of K columns each, e.g. (K=5), and with numbers of
row N_A and N_B respectively
A = (1 5 3 8 15;
2 7 20 11 13;
12 19 20 21 43)
B = (2 6 30 8 16;
3 8 19 11 13)
(the actual matrices have hundreds of thousands of entry, that's why I'm
keen to avoid for loops)
And what I need to do is to apply a function which counts the number of
common elements between ANY row of A and ANY row of B, giving a result
like this:
A1 vs B1: 1 # (8 is a common element)
A1 vs B2: 1 # (8 is a common element)
A2 vs B1: 1 # (2 is a common element)
A2 vs B2: 1 # 11, 13 are common elements
Etc.
I've built a function that counts the number of common elements between
two vectors, based on the intersect function in the R manual
common_elements - function(x,y) length(y[match(x,y,nomatch=0)])
And a double loop who solves my problem would be something like
(pseudo-code)
For(i in 1:N_A){
for(j in 1:N_B){
ce(i,j)=common_elements(a(i),b(j))
}
}
Is there an efficient, clean way to do the same job and give as an
output a matrix N_A x N_B such as that above?
Thanks a lot for your help
Regards
Pietro
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Re: [R] How to delete only those rows in a dataframe in which all records are missing
Perhaps something like this will work. df[!(rowSums(is.na(df))==NCOL(df)),] Michael On Fri, Nov 4, 2011 at 9:27 AM, Jose Iparraguirre jose.iparragui...@ageuk.org.uk wrote: Hi, Imagine I have the following data frame: a - c(1,NA,3) b - c(2,NA,NA) c - data.frame(cbind(a,b)) c a b 1 1 2 2 NA NA 3 3 NA I want to delete the second row. If I use na.omit, that would also affect the third row. I tried to use a loop and an ifelse clause with is.na to get R identify that row in which all records are missing, as opposed to the first row in which no records are missing or the third one, in which only one record is missing. How can I get R identify the row in which all records are missing? Or, how can I get R delete/omit only this row? Thanks in advance, José José Iparraguirre Chief Economist Age UK T 020 303 31482 E jose.iparragui...@ageuk.org.ukmailto:jose.iparragui...@ageuk.org.uk Tavis House, 1- 6 Tavistock Square London, WC1H 9NB www.ageuk.org.ukhttp://www.ageuk.org.uk | ageukblog.org.ukhttp://ageukblog.org.uk/ | @AgeUKPAhttp://twitter.com/ageukpa Age UK Improving later life www.ageuk.org.uk --- Age UK is a registered charity and company limited by guarantee, (registered charity number 1128267, registered company number 6825798). Registered office: Tavis House, 1-6 Tavistock Square, London WC1H 9NA. For the purposes of promoting Age UK Insurance, Age UK is an Appointed Representative of Age UK Enterprises Limited, Age UK is an Introducer Appointed Representative of JLT Benefit Solutions Limited and Simplyhealth Access for the purposes of introducing potential annuity and health cash plans customers respectively. Age UK Enterprises Limited, JLT Benefit Solutions Limited and Simplyhealth Access are all authorised and regulated by the Financial Services Authority. -- This email and any files transmitted with it are confidential and intended solely for the use of the individual or entity to whom they are addressed. If you receive a message in error, please advise the sender and delete immediately. Except where this email is sent in the usual course of our business, any opinions expressed in this email are those of the author and do not necessarily reflect the opinions of Age UK or its subsidiaries and associated companies. Age UK monitors all e-mail transmissions passing through its network and may block or modify mails which are deemed to be unsuitable. Age Concern England (charity number 261794) and Help the Aged (charity number 272786) and their trading and other associated companies merged on 1st April 2009. Together they have formed the Age UK Group, dedicated to improving the lives of people in later life. The three national Age Concerns in Scotland, Northern Ireland and Wales have also merged with Help the Aged in these nations to form three registered charities: Age Scotland, Age NI, Age Cymru. [[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] Select some, but not all, variables stepwise
Thanks Francisco - I've actually realized that ?step can do pretty much exactly what I want. Andre -- View this message in context: http://r.789695.n4.nabble.com/Select-some-but-not-all-variables-stepwise-tp3990002p3990516.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] replace double backslash with singel backslash
A) you wrote // when you meant \\. This behavior does not apply to forward slash. B) if you want the ability to represent any arbitrary character in a string, some version of this feature is required. Visual Basic doesn't allow arbitrary characters unless you concatenate chr() function calls, though it does allow you to embed quotes () by repeating them (), sort of like \\, but much less useful. Learning to understand escaping is a significant step, because it occurs in both the parsing of string literals and in regular expressions. Since specifying a regex often involves using a string literal to specify a regex, you have to escape the escape characters. Once you realize how useful this is you never want to return to the simplicity of VB, and simplifying the R language doesn't seem so attractive anymore. So, just remember that \ is special in strings, and go reread the documentation whenever it crops up to figure out what should go after it to get a particular character into the string, and remember that a regex gets processed by two different sets of rules one after the other. --- Jeff Newmiller The . . Go Live... DCN:jdnew...@dcn.davis.ca.us Basics: ##.#. ##.#. Live Go... Live: OO#.. Dead: OO#.. Playing Research Engineer (Solar/Batteries O.O#. #.O#. with /Software/Embedded Controllers) .OO#. .OO#. rocks...1k --- Sent from my phone. Please excuse my brevity. Gene Leynes gleyne...@gmail.com wrote: I think that people are afraid to say You can't do that in R... But I think the real answer is: you can't do that in R. Although, it is helpful to understand Jeff's reply. I hadn't fully realized why this particular problem occurs before reading that. It's odd to me that // and / are both stored as /, but that makes sense given my experience in R. Also, the other replies are good advice, working with R's path functions or sticking with forward slashes is the way to go (don't fight the assimilation, the borg needs you). Personally, I think some of these seemingly small problems actually encumber R's mainstream adoption quite a bit, but then I'm not the one writing R. Plus, it's is still pretty dang awesome even with its minor annoyances. On Fri, Nov 4, 2011 at 8:45 AM, Jeff Newmiller jdnew...@dcn.davis.ca.us wrote: Your str does not have any double backslashes to replace. You need to revisit the concept of escape characters in the documentation. In brief, every \\ in a quoted string is actually a single character as stored in memory. --- Jeff Newmiller The . . Go Live... DCN:jdnew...@dcn.davis.ca.us Basics: ##.#. ##.#. Live Go... Live: OO#.. Dead: OO#.. Playing Research Engineer (Solar/Batteries O.O#. #.O#. with /Software/Embedded Controllers) .OO#. .OO#. rocks...1k --- Sent from my phone. Please excuse my brevity. Kay Cichini kay.cich...@uibk.ac.at wrote: I want to replace \\ with \ in: str - C:\\DOKUME~1\\u0327336\\LOKALE~1\\Temp\\RtmpQ5NJ8X\\TIRIS_PICS\\1_Img.jpg and tried: gsub(, \\, str) but this removes the \\ without replacing them by \ Any help much appreciated, Kay - Kay Cichini Postgraduate student Institute of Botany Univ. of Innsbruck -- View this message in context: http://r.789695.n4.nabble.com/replace-double-backslash-with-singel-backslash-tp3989434p3989434.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. [[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. [[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] Select some, but not all, variables stepwise
Why do any of that? Those procedures are statistically invalid. Frank AndreE wrote: Thanks Francisco - I've actually realized that ?step can do pretty much exactly what I want. Andre - Frank Harrell Department of Biostatistics, Vanderbilt University -- View this message in context: http://r.789695.n4.nabble.com/Select-some-but-not-all-variables-stepwise-tp3990002p3990598.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] zoo performance regression noticed (1.6-5 is faster...)
Good morning, I have discovered what I believe to be a performance regression between Zoo 1.6x and Zoo 1.7-6 in the application of rollapply. On zoo 1.6x, rollapply of my function over my data takes about 20 minutes. Using 1.7-6, the same code takes about 6 hours. R --version R version 2.13.1 (2011-07-08) Copyright (C) 2011 The R Foundation for Statistical Computing ISBN 3-900051-07-0 Platform: x86_64-pc-linux-gnu (64-bit) Two versions of zoo 1.6 run *fast* On one machine I am running less /usr/lib64/R/library/zoo/DESCRIPTION Package: zoo Version: 1.6-3 Date: 2010-04-23 Title: Z's ordered observations ... Packaged: 2010-04-23 07:28:47 UTC; zeileis Repository: CRAN Date/Publication: 2010-04-23 07:43:54 Built: R 2.10.1; ; 2010-04-25 06:41:34 UTC; unix (Thankfully I forgot to upgrade.packages() on this machine!) On the other Package: zoo Version: 1.6-5 Date: 2011-04-08 ... Packaged: 2011-04-08 17:13:47 UTC; zeileis Repository: CRAN Date/Publication: 2011-04-08 17:27:47 Built: R 2.13.1; ; 2011-11-04 15:49:54 UTC; unix I have stripped out zoo 1.7-6 from all my machines. I tried to ensure all libraries were identical on the two machines (using lsof), and after finally downgrading zoo I got the second machine to be as fast as the first, so I am quite certain the difference in speed is down to the Zoo version used. My code runs a fairly simple function over a time series using the following call to process a year of 30s data (9 columns, about a million rows): vals - rollapply(data=ts.data[,c(n.3.cols, o.3.cols,volocc.cols)] ,width=40 ,FUN=rolling.function.fn(n.cols=n.3.cols,o.cols=o.3.cols,vo.cols=volocc.cols) ,by.column=FALSE ,align='right') (The rolling.function.fn call returns a function that is initialized with the initial call above (a trick I learned from Javascript)) If this is a known situation with the new 1.7 generation Zoo, my apologies and I'll go away. If my code could be turned into a useful test, I'd be happy to help out as much as I'm able. Given the extreme runtime difference though, I thought I should offer my help in this case, since zoo is such a useful package in my work. Regards, James Marca pgp9V1vTe92wd.pgp Description: PGP signature __ 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] zoo performance regression noticed (1.6-5 is faster...)
On Fri, Nov 4, 2011 at 12:34 PM, James Marca jma...@translab.its.uci.edu wrote: Good morning, I have discovered what I believe to be a performance regression between Zoo 1.6x and Zoo 1.7-6 in the application of rollapply. On zoo 1.6x, rollapply of my function over my data takes about 20 minutes. Using 1.7-6, the same code takes about 6 hours. R --version R version 2.13.1 (2011-07-08) Copyright (C) 2011 The R Foundation for Statistical Computing ISBN 3-900051-07-0 Platform: x86_64-pc-linux-gnu (64-bit) Two versions of zoo 1.6 run *fast* On one machine I am running less /usr/lib64/R/library/zoo/DESCRIPTION Package: zoo Version: 1.6-3 Date: 2010-04-23 Title: Z's ordered observations ... Packaged: 2010-04-23 07:28:47 UTC; zeileis Repository: CRAN Date/Publication: 2010-04-23 07:43:54 Built: R 2.10.1; ; 2010-04-25 06:41:34 UTC; unix (Thankfully I forgot to upgrade.packages() on this machine!) On the other Package: zoo Version: 1.6-5 Date: 2011-04-08 ... Packaged: 2011-04-08 17:13:47 UTC; zeileis Repository: CRAN Date/Publication: 2011-04-08 17:27:47 Built: R 2.13.1; ; 2011-11-04 15:49:54 UTC; unix I have stripped out zoo 1.7-6 from all my machines. I tried to ensure all libraries were identical on the two machines (using lsof), and after finally downgrading zoo I got the second machine to be as fast as the first, so I am quite certain the difference in speed is down to the Zoo version used. My code runs a fairly simple function over a time series using the following call to process a year of 30s data (9 columns, about a million rows): vals - rollapply(data=ts.data[,c(n.3.cols, o.3.cols,volocc.cols)] ,width=40 ,FUN=rolling.function.fn(n.cols=n.3.cols,o.cols=o.3.cols,vo.cols=volocc.cols) ,by.column=FALSE ,align='right') (The rolling.function.fn call returns a function that is initialized with the initial call above (a trick I learned from Javascript)) If this is a known situation with the new 1.7 generation Zoo, my apologies and I'll go away. If my code could be turned into a useful test, I'd be happy to help out as much as I'm able. Given the extreme runtime difference though, I thought I should offer my help in this case, since zoo is such a useful package in my work. This was a known problem and was fixed but if its still there then there must be some other condition under which it can occur as well. If you can provide a small self contained reproducible example it would help in tracking it down. -- Statistics Software Consulting GKX Group, GKX Associates Inc. tel: 1-877-GKX-GROUP email: ggrothendieck at 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] Decision tree model using rpart ( classification
Could you please repeat the error massage you get for C ? Contact Details:--- Contact me: tal.gal...@gmail.com | 972-52-7275845 Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) | www.r-statistics.com (English) -- On Fri, Nov 4, 2011 at 10:32 AM, aajit75 aaji...@yahoo.co.in wrote: Hi, Thanks for the responce, code for each case is as: c_c_factor - 0.001 min_obs_split - 80 A) fit - rpart(segment ~., method=class, control=rpart.control(minsplit=min_obs_split, cp=c_c_factor), data=Beh_cluster_out) B) fit - rpart(segment ~., method=class, control=rpart.control(minsplit=min_obs_split, cp=c_c_factor), data=profile_cluster_out) C) fit - rpart(decile ~., method=class, control=rpart.control(minsplit=min_obs_split, cp=c_c_factor), data=dtm_ip) In A and B target variable 'segment' is from the clustering data using same set of input variables , while in C target variable 'decile' is derived from behavioural variables and input variables are from profile data. Number of rows in the input table in all three cases are same. Regards, -Ajit -- View this message in context: http://r.789695.n4.nabble.com/Decision-tree-model-using-rpart-classification-tp3989162p3989320.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. [[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] zoo performance regression noticed (1.6-5 is faster...)
On Fri, Nov 4, 2011 at 12:56 PM, Gabor Grothendieck ggrothendi...@gmail.com wrote: On Fri, Nov 4, 2011 at 12:34 PM, James Marca jma...@translab.its.uci.edu wrote: Good morning, I have discovered what I believe to be a performance regression between Zoo 1.6x and Zoo 1.7-6 in the application of rollapply. On zoo 1.6x, rollapply of my function over my data takes about 20 minutes. Using 1.7-6, the same code takes about 6 hours. R --version R version 2.13.1 (2011-07-08) Copyright (C) 2011 The R Foundation for Statistical Computing ISBN 3-900051-07-0 Platform: x86_64-pc-linux-gnu (64-bit) Two versions of zoo 1.6 run *fast* On one machine I am running less /usr/lib64/R/library/zoo/DESCRIPTION Package: zoo Version: 1.6-3 Date: 2010-04-23 Title: Z's ordered observations ... Packaged: 2010-04-23 07:28:47 UTC; zeileis Repository: CRAN Date/Publication: 2010-04-23 07:43:54 Built: R 2.10.1; ; 2010-04-25 06:41:34 UTC; unix (Thankfully I forgot to upgrade.packages() on this machine!) On the other Package: zoo Version: 1.6-5 Date: 2011-04-08 ... Packaged: 2011-04-08 17:13:47 UTC; zeileis Repository: CRAN Date/Publication: 2011-04-08 17:27:47 Built: R 2.13.1; ; 2011-11-04 15:49:54 UTC; unix I have stripped out zoo 1.7-6 from all my machines. I tried to ensure all libraries were identical on the two machines (using lsof), and after finally downgrading zoo I got the second machine to be as fast as the first, so I am quite certain the difference in speed is down to the Zoo version used. My code runs a fairly simple function over a time series using the following call to process a year of 30s data (9 columns, about a million rows): vals - rollapply(data=ts.data[,c(n.3.cols, o.3.cols,volocc.cols)] ,width=40 ,FUN=rolling.function.fn(n.cols=n.3.cols,o.cols=o.3.cols,vo.cols=volocc.cols) ,by.column=FALSE ,align='right') (The rolling.function.fn call returns a function that is initialized with the initial call above (a trick I learned from Javascript)) If this is a known situation with the new 1.7 generation Zoo, my apologies and I'll go away. If my code could be turned into a useful test, I'd be happy to help out as much as I'm able. Given the extreme runtime difference though, I thought I should offer my help in this case, since zoo is such a useful package in my work. This was a known problem and was fixed but if its still there then there must be some other condition under which it can occur as well. If you can provide a small self contained reproducible example it would help in tracking it down. -- Statistics Software Consulting GKX Group, GKX Associates Inc. tel: 1-877-GKX-GROUP email: ggrothendieck at gmail.com Also, as a workaround you can try this to use an old rollapply in a new version of zoo: library(zoo) source(http://r-forge.r-project.org/scm/viewvc.php/*checkout*/pkg/zoo/R/rollapply.R?revision=817root=zoo;) rollapply(...whatever...) -- Statistics Software Consulting GKX Group, GKX Associates Inc. tel: 1-877-GKX-GROUP email: ggrothendieck at 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] query about counting rows of a dataframe
On Nov 3, 2011, at 12:28 PM, Stefano Sofia wrote: Dear R users, I have got the following data frame, called my_df: gender day_birth month_birth year_birth labour 1 F 22 10 2001 1 2 M29 10 2001 2 3 M 1 11 2001 1 4 F 3 11 2001 1 5 M 3 11 2001 2 6 F 4 11 2001 1 7 F 4 11 2001 2 8 F 5 12 2001 2 9 M 22 14 2001 2 10 F 29 13 2001 2 ... I need to count data in different ways: 1. count the births for each day (having 0 when necessary) independently from the value of the labour column xtabs sometimes give better results. If you want all 31 days then make day_birth a factor with levels=1:31) xtabs( ~ day_birth + month_birth + year_birth, data=dat) , , year_birth = 2001 month_birth day_birth 10 11 12 13 14 1 0 1 0 0 0 3 0 2 0 0 0 4 0 2 0 0 0 5 0 0 1 0 0 22 1 0 0 0 1 29 1 0 0 1 0 2. count the births for each day (having 0 when necessary), divided by the value of labour (which can have two valuers, 1 or 2) Cannot figure out what is being asked here. What to do with the two values? Just count them? This would give a partitioned count xtabs( labour==1 ~ day_birth + month_birth , data=dat) month_birth day_birth 10 11 12 13 14 1 0 1 0 0 0 3 0 1 0 0 0 4 0 1 0 0 0 5 0 0 0 0 0 22 1 0 0 0 0 29 0 0 0 0 0 xtabs( labour==2 ~ day_birth + month_birth , data=dat) month_birth day_birth 10 11 12 13 14 1 0 0 0 0 0 3 0 1 0 0 0 4 0 1 0 0 0 5 0 0 1 0 0 22 0 0 0 0 1 29 1 0 0 1 0 3. count the births for each day of all the years (i.e. the 22nd of October of all the years present in the data frame) independently from the value of labour If I understand correctly: xtabs( ~ day_birth + month_birth + year_birth, data=dat) , , year_birth = 2001 month_birth day_birth 10 11 12 13 14 1 0 1 0 0 0 3 0 2 0 0 0 4 0 2 0 0 0 5 0 0 1 0 0 22 1 0 0 0 1 29 1 0 0 1 0 4. count the births for each day of all the years (i.e. the 22nd of October of all the years present in the data frame), divided by the value of labour Again confusing. Do you mean to use separate tables for labour==1 and labour==2? Perhaps context to explain what these values represent. Some of us are concrete. The results of xtabs are tables and can be divided like matrices. I tried with the command table(my_df$year_birth, my_df$month_birth, my_df$day_birth) which satisfies (partially) question numer 1 (I am not able to have 0 in the not available days). Is there a smart way to do that without invoking too many loops? thank you for your help David Winsemius, MD Heritage Laboratories West Hartford, CT __ 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] Plotting skewed normal distribution with a bar plot
You might want to check again: I'm running fGarch on 2.13.2, Mac OSX 10.5.8. Michael On Fri, Nov 4, 2011 at 9:36 AM, Steve Friedman skfgla...@gmail.com wrote: Hi Michael, Thanks for pointing me fGarch. I actually started there, but it is not yet available for 2.13.2 so I went directly to the (sn-package). I've briefly explored your suggestion and think it will work. Thanks Steve On Nov 3, 2011 10:41 PM, R. Michael Weylandt michael.weyla...@gmail.com wrote: It seems like you'll need to apply some sort of MLE to estimate the parameters directly from the data before using dsn() to get the density. This might help with some of it: http://help.rmetrics.org/fGarch/html/snorm.html Michael On Thu, Nov 3, 2011 at 2:54 PM, steve_fried...@nps.gov wrote: Hi, I need to create a plot (type = h) and then overlay a skewed-normal curve on this distribution, but I'm not finding a procedure to accomplish this. I want to use the plot function here in order to control the bin distributions. I have explored the sn library and found the dsn function. dsn uses known location, scaling and shape parameters associated with a given input vector of probabilities. However, how can I calculate the skewed-normal curve if I don't know these parameters in advance? Is there another function to calculate the skew-normal, perhaps in a different package? I'm working with R 2.13.2 on a windows based machine. Steve Friedman Ph. D. Ecologist / Spatial Statistical Analyst Everglades and Dry Tortugas National Park 950 N Krome Ave (3rd Floor) Homestead, Florida 33034 steve_fried...@nps.gov Office (305) 224 - 4282 Fax (305) 224 - 4147 __ 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] barplot as histogram
Hello: I'm dealing with an issue currently that I'm not sure the best way to approach. I've got a very large (10G+) dataset that I'm trying to create a histogram for. I don't seem to be able to use hist directly as I can not create an R vector of size greater than 2.2G. I considered condensing the data previous to loading it into R and just plotting the frequencies as a barplot; unfortunately, barplot does not support plotting the values according to a set of x-axis positions. What I have is something similar to: ys - c(12,3,7,22,10) xs - c(1,30,35,39,60) and I'd like the bars (ys) to appear at the positions described by xs. I can get this to work on smaller sets by filling zero values in for missing ys for the entire range of xs but in my case this would again create a vector too large for R. Is there another way to use the two vectors to create a simulated frequency histogram? Is there a way to create a histogram object (as returned by hist) from the condensed data so that plot would handle it correctly? Thanks in advance, Jesse __ 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] barplot as histogram
Perhaps plot(xs, ys, type = h, lwd = 3) will work? I'm not sure that a direct call to hist(, plot = F) will get around the data problems. If you type getAnywhere(hist.default) you can see the code that runs hist(): perhaps you can extract the working bits you need. Michael On Fri, Nov 4, 2011 at 2:04 PM, Jesse Brown jesse.r.br...@lmco.com wrote: Hello: I'm dealing with an issue currently that I'm not sure the best way to approach. I've got a very large (10G+) dataset that I'm trying to create a histogram for. I don't seem to be able to use hist directly as I can not create an R vector of size greater than 2.2G. I considered condensing the data previous to loading it into R and just plotting the frequencies as a barplot; unfortunately, barplot does not support plotting the values according to a set of x-axis positions. What I have is something similar to: ys - c(12,3,7,22,10) xs - c(1,30,35,39,60) and I'd like the bars (ys) to appear at the positions described by xs. I can get this to work on smaller sets by filling zero values in for missing ys for the entire range of xs but in my case this would again create a vector too large for R. Is there another way to use the two vectors to create a simulated frequency histogram? Is there a way to create a histogram object (as returned by hist) from the condensed data so that plot would handle it correctly? Thanks in advance, Jesse __ 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] barplot as histogram
On 04/11/2011 2:04 PM, Jesse Brown wrote: Hello: I'm dealing with an issue currently that I'm not sure the best way to approach. I've got a very large (10G+) dataset that I'm trying to create a histogram for. I don't seem to be able to use hist directly as I can not create an R vector of size greater than 2.2G. I considered condensing the data previous to loading it into R and just plotting the frequencies as a barplot; unfortunately, barplot does not support plotting the values according to a set of x-axis positions. What I have is something similar to: ys- c(12,3,7,22,10) xs- c(1,30,35,39,60) and I'd like the bars (ys) to appear at the positions described by xs. I can get this to work on smaller sets by filling zero values in for missing ys for the entire range of xs but in my case this would again create a vector too large for R. Is there another way to use the two vectors to create a simulated frequency histogram? Is there a way to create a histogram object (as returned by hist) from the condensed data so that plot would handle it correctly? Follow your own last suggestion. Take a small subset of your data, and calculate x - hist(data, plot=FALSE) str(x) will show you the structure of the object in x. Modify the entries to reflect your full dataset, and then plot(x) will show it. Duncan Murdoch __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Random multinomial variable
I need some help interpreting the following code which is part of a mutlilevel model simulation with 2 levels. I've put in comments with my understanding of the code, but I'm not sure how [i2id] is functioning. It's defined in another part of the program as l2id-rep(c(1:n2),each=n1) which looks like a number corresponding to the level 1 and 2 sample size (n1, n2). Can someone explain what [i2id] is telling x[,3] and x[,4]? macpred-rmultinom(n2,1,c(0.15,0.30,0.55)) ## generate one multinomial variable of length=1 with probabilities of .15, .30 and .55 and do this n2 times x[,3]-macpred[1,][l2id] ##assign the first mutinomial value to column 3 x[,4]-macpred[2,][l2id] ##assign the second multinomial value to column 4 __ 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] Fit continuous distribution to truncated empirical values
On Nov 3, 2011, at 7:54 AM, Michele Mazzucco wrote: Hi all, I am trying to fit a distribution to some data about survival times. I am interested only in a specific interval, e.g., while the data lies in the interval (0,, 600), I want the best for the interval (0,..., 24). I have tried both fitdistr (MASS package) and fitdist (from the fitdistrplus package), but I could not get them working, e.g. fitdistr(left, weibull, upper=24) Error in optim(x = c(529L, 528L, 527L, 526L, 525L, 524L, 523L, 522L, 521L, : L-BFGS-B needs finite values of 'fn' In addition: Warning message: In dweibull(x, shape, scale, log) : NaNs produced Am I doing something wrong? You didn't supply data to test, but shouldn't you supply a lower bound if you want to fit weibull? It is, after all, bounded at 0. left - c(529L, 528L, 527L, 526L, 525L, 524L, 523L, 522L, 521L, 50*runif(100)) fitdistr(left, weibull, upper=24) Error in optim(x = c(529, 528, 527, 526, 525, 524, 523, 522, 521, 18.3964251773432, : L-BFGS-B needs finite values of 'fn' In addition: Warning message: In dweibull(x, shape, scale, log) : NaNs produced fitdistr(left, weibull, upper=24, lower=0.5) shape scale 0.58195013 24. ( 0.04046087) ( 3.38621367) Thanks, Michele p.s. I have seen similar posts, e.g., http://tolstoy.newcastle.edu.au/R/help/05/02/11558.html , but I am not sure whether I can apply the same approach here. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. David Winsemius, MD Heritage Laboratories West Hartford, CT __ 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] barplot as histogram
I believe that plot(..., type='h') will do the trick. I had tried that earlier but forgot to play with the lwd parameter. Incidentally, I didn't know about getAnywhere(hist.default) - really handy. I was reading the code to find the details. Thanks! Jesse R. Michael Weylandt wrote: Perhaps plot(xs, ys, type = h, lwd = 3) will work? I'm not sure that a direct call to hist(, plot = F) will get around the data problems. If you type getAnywhere(hist.default) you can see the code that runs hist(): perhaps you can extract the working bits you need. Michael On Fri, Nov 4, 2011 at 2:04 PM, Jesse Brown jesse.r.br...@lmco.com wrote: Hello: I'm dealing with an issue currently that I'm not sure the best way to approach. I've got a very large (10G+) dataset that I'm trying to create a histogram for. I don't seem to be able to use hist directly as I can not create an R vector of size greater than 2.2G. I considered condensing the data previous to loading it into R and just plotting the frequencies as a barplot; unfortunately, barplot does not support plotting the values according to a set of x-axis positions. What I have is something similar to: ys - c(12,3,7,22,10) xs - c(1,30,35,39,60) and I'd like the bars (ys) to appear at the positions described by xs. I can get this to work on smaller sets by filling zero values in for missing ys for the entire range of xs but in my case this would again create a vector too large for R. Is there another way to use the two vectors to create a simulated frequency histogram? Is there a way to create a histogram object (as returned by hist) from the condensed data so that plot would handle it correctly? Thanks in advance, Jesse __ 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] 12th Root of a Square (Transition) Matrix
On Nov 4, 2011, at 6:34 AM, clangkamp wrote: I have tried this method, but the result is not working, at least not as I expect: I used the CreditMetrics package transition matrix rc - c(AAA, AA, A, BBB, BB, B, CCC, D) M - matrix(c(90.81, 8.33, 0.68, 0.06, 0.08, 0.02, 0.01, 0.01, 0.70, 90.65, 7.79, 0.64, 0.06, 0.13, 0.02, 0.01, 0.09, 2.27, 91.05, 5.52, 0.74, 0.26, 0.01, 0.06, 0.02, 0.33, 5.95, 85.93, 5.30, 1.17, 1.12, 0.18, 0.03, 0.14, 0.67, 7.73, 80.53, 8.84, 1.00, 1.06, 0.01, 0.11, 0.24, 0.43, 6.48, 83.46, 4.07, 5.20, 0.21, 0, 0.22, 1.30, 2.38, 11.24, 64.86, 19.79, 0, 0, 0, 0, 0, 0, 0, 100 )/100, 8, 8, dimnames = list(rc, rc), byrow = TRUE) then followed through with the steps: nth_root - X %*% L_star %*% X_inv Despite my (distant) physics training, I am no matrix mechanic, so I cannot comment on that method. I would instead search for an nth-root matrix function http://search.r-project.org/cgi-bin/namazu.cgi?query=nth+root+of+matrixmax=100result=normalsort=scoreidxname=functionsidxname=vignettesidxname=views And finding one in package 'pracma', see if it succeeds: nthroot(M, 12) AAAAA A BBBBB B CCC D AAA 0.9919988 0.8129311 0.6597444 0.5389055 0.5519810 0.4917592 0.4641589 0.4641589 AA 0.6613401 0.9918530 0.8084034 0.6564198 0.5389055 0.5747716 0.4917592 0.4641589 A 0.5574256 0.7294611 0.9922170 0.7855279 0.6644097 0.6089493 0.4641589 0.5389055 BBB 0.4917592 0.6211686 0.7904537 0.9874431 0.7828700 0.6902644 0.6877567 0.5905718 BB 0.5086590 0.5783322 0.6589304 0.8078827 0.9821168 0.8169667 0.6812921 0.6846083 B 0.4641589 0.5668255 0.6049010 0.6350224 0.7960944 0.9850460 0.7658309 0.7816283 CCC 0.5982072 0.000 0.6005307 0.6963517 0.7323434 0.8334839 0.9645648 0.8737164 D 0.000 0.000 0.000 0.000 0.000 0.000 0.000 1.000 nthroot(M, 12)^12 AAA AA ABBB BB BCCC D AAA 0.9081 0.0833 0.0068 0.0006 0.0008 0.0002 0.0001 0.0001 AA 0.0070 0.9065 0.0779 0.0064 0.0006 0.0013 0.0002 0.0001 A 0.0009 0.0227 0.9105 0.0552 0.0074 0.0026 0.0001 0.0006 BBB 0.0002 0.0033 0.0595 0.8593 0.0530 0.0117 0.0112 0.0018 BB 0.0003 0.0014 0.0067 0.0773 0.8053 0.0884 0.0100 0.0106 B 0.0001 0.0011 0.0024 0.0043 0.0648 0.8346 0.0407 0.0520 CCC 0.0021 0. 0.0022 0.0130 0.0238 0.1124 0.6486 0.1979 D 0. 0. 0. 0. 0. 0. 0. 1. all.equal(M , nthroot(M, 12)^12) [1] TRUE Success! -- David. But the check (going back 12 to the power again) doesn't yield the original matrix. Now some rounding errors can be expected, but I didn't expect a perfectly diagonal matrix, when the initial matrix isn't diagonal at all. round(nth_root^12,4) [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [1,] 0.9078 0. 0. 0. 0. 0. 0.0 [2,] 0. 0.9053 0. 0. 0. 0. 0.0 [3,] 0. 0. 0.9079 0. 0. 0. 0.0 [4,] 0. 0. 0. 0.8553 0. 0. 0.0 [5,] 0. 0. 0. 0. 0.7998 0. 0.0 [6,] 0. 0. 0. 0. 0. 0.8285 0.0 [7,] 0. 0. 0. 0. 0. 0. 0.64570 [8,] 0. 0. 0. 0. 0. 0. 0.1 Any takers -- David Winsemius, MD Heritage Laboratories West Hartford, CT __ 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] 12th Root of a Square (Transition) Matrix
Is it just me or are you confusing the 12th root of a matrix with taking the 12th root of each entry? Because your formula involving the eigenvectors and eigenvalues calculates the 12th root of the matrix, while round(nth_root^12,4) will print out a matrix whose components are powers of 12 of the nth_root, which is very different. To find the 12th power of a matrix, you can either search for an appropriate function, or do res = nth_root; for (p in 1:11) res = res %*% nth_root then compare res to your original matrix M. Peter On Fri, Nov 4, 2011 at 3:34 AM, clangkamp christian.langk...@gmxpro.de wrote: I have tried this method, but the result is not working, at least not as I expect: I used the CreditMetrics package transition matrix rc - c(AAA, AA, A, BBB, BB, B, CCC, D) M - matrix(c(90.81, 8.33, 0.68, 0.06, 0.08, 0.02, 0.01, 0.01, 0.70, 90.65, 7.79, 0.64, 0.06, 0.13, 0.02, 0.01, 0.09, 2.27, 91.05, 5.52, 0.74, 0.26, 0.01, 0.06, 0.02, 0.33, 5.95, 85.93, 5.30, 1.17, 1.12, 0.18, 0.03, 0.14, 0.67, 7.73, 80.53, 8.84, 1.00, 1.06, 0.01, 0.11, 0.24, 0.43, 6.48, 83.46, 4.07, 5.20, 0.21, 0, 0.22, 1.30, 2.38, 11.24, 64.86, 19.79, 0, 0, 0, 0, 0, 0, 0, 100 )/100, 8, 8, dimnames = list(rc, rc), byrow = TRUE) then followed through with the steps: nth_root - X %*% L_star %*% X_inv But the check (going back 12 to the power again) doesn't yield the original matrix. Now some rounding errors can be expected, but I didn't expect a perfectly diagonal matrix, when the initial matrix isn't diagonal at all. round(nth_root^12,4) [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [1,] 0.9078 0. 0. 0. 0. 0. 0. 0 [2,] 0. 0.9053 0. 0. 0. 0. 0. 0 [3,] 0. 0. 0.9079 0. 0. 0. 0. 0 [4,] 0. 0. 0. 0.8553 0. 0. 0. 0 [5,] 0. 0. 0. 0. 0.7998 0. 0. 0 [6,] 0. 0. 0. 0. 0. 0.8285 0. 0 [7,] 0. 0. 0. 0. 0. 0. 0.6457 0 [8,] 0. 0. 0. 0. 0. 0. 0. 1 Any takers - Christian Langkamp christian.langkamp-at-gmxpro.de -- View this message in context: http://r.789695.n4.nabble.com/12th-Root-of-a-Square-Transition-Matrix-tp2259736p3989618.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. -- Sent from my Linux computer. Way better than iPad :) __ 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] nproc parameter in efpFunctional
On Fri, 4 Nov 2011, bonda wrote: The 2006 CSDA paper is really very informative, perhaps, I'm trying to understand the things lying beyond. If we have e.g. k=3, then taking nproc=3 for the functional maxBB we get a critical value (boundary) maxBB$computeCritval(0.05,nproc=3) [1] 1.544421, and this for nproc=NULL (Bonferroni approximation) will be maxBB$computeCritval(0.05) [1] 1.358099. No. In the latter case no Bonferroni approximation is applied. If you want to use it, you can do so via the rule of thumb R maxBB$computeCritval(0.05/3, nproc = 1) [1] 1.547175 which essentially matches the critical value computed for nproc = 3. If you use the more precise value 1 - (1 - 0.05)^(1/3) instead of 0.05/3, you get a match (up to some small numerical differences). Setting nproc=NULL is only possible in efpFunctional(): efpFunctional() sets up the computeCritval() and computePval() functions via simulation methods (unless closed form solutions are supplied). For the simulation two strategies are available: Simulate nproc = 1, 2, 3, ... explicitly. Simulate only nproc = 1 and apply a Bonferroni correction. The last option is chosen if you set nproc=NULL -- it makes only sense if you aggregate via the maximum across the components. The resulting computeCritval() and computePval() function always need to have the correct nproc supplied (i.e., nproc=NULL makes no sense). Aggregating 3 Brownian bridges first over components, we obtain time series process. Now, we wonder if maximum value of the process (aggregation over time) lies over boundary. Which boundary - 1.544421 or 1.358099 - should one take? They look too different and, for instance, lead to unfair computing of empirical size (as rejection rate of null hypothesis) or empirical power (as acception rate of alternative). -- View this message in context: http://r.789695.n4.nabble.com/nproc-parameter-in-efpFunctional-tp3972419p3989598.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.
Re: [R] Lattice plots and missing x-axis labels on second page
On Nov 2, 2011, at 8:23 PM, Evans, David G (DFG) wrote: I should say I'm using Windows-7, R version 2.13.0 and lattice version 0.19-33. I've pared down my code to this : pdat = read.table(RGRAPHSDGE.csv,header=T,sep=,,fill=T) print(xyplot(pdat$NITRATE~pdat$DATEYR|pdat$WELL, I generally try to avoid building lattice plots this way. It is better to use data=pdat and shorten the formula specification. That way the functions can use the aggregate information in the dataframe. Try: xyplot( NITRATE~ DATEYR| WELL, data=pdat, as.table=TRUE, layout=c(3,4), You may want to try layout=c(3,4,2) See help(xyplot) subsection layout, where Sarkar says that pages are somethimes incorrectly calculated. xlab=Year, ylab=Nitrate mg / litre, strip=FALSE )) First 3 lines of pdat looks like this: WELL DATEYR NITRATE 1 ALASKA CHILDRENS SERVICES 1993.8360.81 2 ALASKA CHILDRENS SERVICES 1994.8500.91 3 ALASKA CHILDRENS SERVICES 1995.8030.94 Thanks again. -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org ] On Behalf Of Evans, David G (DFG) Sent: Wednesday, November 02, 2011 3:24 PM To: r-help@r-project.org Subject: [R] Lattice plots and missing x-axis labels on second page Hello, I'm trying to make a lattice plot (using xyplot()). I have included a layout=c(3,4) statement, giving me 12 plots per page and an as.table=TRUE statement, directing the way the plots are laid out. I have 18 plots altogether and so 6 of them end up on the second page. Everything looks fine for the first page, but the x-axis labels (e.g. 1993, 1994...) are all missing on the second page. The x-axis variable name (Year) is there at the bottom, however. Any help is appreciated. Thanks. David G. Evans Biometrician Division of Sport Fish Alaska Dept . of Fish and Game Anchorage, Ak 99518 __ 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. David Winsemius, MD Heritage Laboratories West Hartford, CT __ 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] Lattice plots and missing x-axis labels on second page
On Nov 2, 2011, at 8:23 PM, Evans, David G (DFG) wrote: I should say I'm using Windows-7, R version 2.13.0 and lattice version 0.19-33. I've pared down my code to this : pdat = read.table(RGRAPHSDGE.csv,header=T,sep=,,fill=T) print(xyplot(pdat$NITRATE~pdat$DATEYR|pdat$WELL, I generally try to avoid building lattice plots this way. It is better to use data=pdat and shorten the formula specification. That way the functions can use the aggregate information in the dataframe. Try: xyplot( NITRATE~ DATEYR| WELL, data=pdat, as.table=TRUE, layout=c(3,4), You may want to try layout=c(3,4,2) See help(xyplot) subsection layout, where Sarkar says that pages are somethimes incorrectly calculated. xlab=Year, ylab=Nitrate mg / litre, strip=FALSE )) First 3 lines of pdat looks like this: WELL DATEYR NITRATE 1 ALASKA CHILDRENS SERVICES 1993.8360.81 2 ALASKA CHILDRENS SERVICES 1994.8500.91 3 ALASKA CHILDRENS SERVICES 1995.8030.94 Thanks again. -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org ] On Behalf Of Evans, David G (DFG) Sent: Wednesday, November 02, 2011 3:24 PM To: r-help@r-project.org Subject: [R] Lattice plots and missing x-axis labels on second page Hello, I'm trying to make a lattice plot (using xyplot()). I have included a layout=c(3,4) statement, giving me 12 plots per page and an as.table=TRUE statement, directing the way the plots are laid out. I have 18 plots altogether and so 6 of them end up on the second page. Everything looks fine for the first page, but the x-axis labels (e.g. 1993, 1994...) are all missing on the second page. The x-axis variable name (Year) is there at the bottom, however. Any help is appreciated. Thanks. David G. Evans Biometrician Division of Sport Fish Alaska Dept . of Fish and Game Anchorage, Ak 99518 __ 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. David Winsemius, MD Heritage Laboratories West Hartford, CT __ 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] 12th Root of a Square (Transition) Matrix
On Nov 4, 2011, at 4:04 PM, Peter Langfelder wrote: Is it just me or are you confusing the 12th root of a matrix with taking the 12th root of each entry? I think I got confused as well. Thanks for clarifying. Because your formula involving the eigenvectors and eigenvalues calculates the 12th root of the matrix, while round(nth_root^12,4) will print out a matrix whose components are powers of 12 of the nth_root, which is very different. To find the 12th power of a matrix, you can either search for an appropriate function, or do res = nth_root; for (p in 1:11) res = res %*% nth_root The 12th (matrix) root of M: e^( 1/n * log(M) ) require(Matrix) M1.12 - expm( (1/12)*logm(M) ) res = M1.12; nth_root=M1.12 for (p in 1:11) + res = res %*% nth_root # Check accuracy round(res, 4) [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [1,] 0.9081 0.0833 0.0068 0.0006 0.0008 0.0002 0.0001 0.0001 [2,] 0.0070 0.9065 0.0779 0.0064 0.0006 0.0013 0.0002 0.0001 [3,] 0.0009 0.0227 0.9105 0.0552 0.0074 0.0026 0.0001 0.0006 [4,] 0.0002 0.0033 0.0595 0.8593 0.0530 0.0117 0.0112 0.0018 [5,] 0.0003 0.0014 0.0067 0.0773 0.8053 0.0884 0.0100 0.0106 [6,] 0.0001 0.0011 0.0024 0.0043 0.0648 0.8346 0.0407 0.0520 [7,] 0.0021 0. 0.0022 0.0130 0.0238 0.1124 0.6486 0.1979 [8,] 0. 0. 0. 0. 0. 0. 0. 1. M AAA AA ABBB BB BCCC D AAA 0.9081 0.0833 0.0068 0.0006 0.0008 0.0002 0.0001 0.0001 AA 0.0070 0.9065 0.0779 0.0064 0.0006 0.0013 0.0002 0.0001 A 0.0009 0.0227 0.9105 0.0552 0.0074 0.0026 0.0001 0.0006 BBB 0.0002 0.0033 0.0595 0.8593 0.0530 0.0117 0.0112 0.0018 BB 0.0003 0.0014 0.0067 0.0773 0.8053 0.0884 0.0100 0.0106 B 0.0001 0.0011 0.0024 0.0043 0.0648 0.8346 0.0407 0.0520 CCC 0.0021 0. 0.0022 0.0130 0.0238 0.1124 0.6486 0.1979 D 0. 0. 0. 0. 0. 0. 0. 1. Rather good agreement to 4 decimal places anyway. then compare res to your original matrix M. Peter On Fri, Nov 4, 2011 at 3:34 AM, clangkamp christian.langk...@gmxpro.de wrote: I have tried this method, but the result is not working, at least not as I expect: I used the CreditMetrics package transition matrix rc - c(AAA, AA, A, BBB, BB, B, CCC, D) M - matrix(c(90.81, 8.33, 0.68, 0.06, 0.08, 0.02, 0.01, 0.01, 0.70, 90.65, 7.79, 0.64, 0.06, 0.13, 0.02, 0.01, 0.09, 2.27, 91.05, 5.52, 0.74, 0.26, 0.01, 0.06, 0.02, 0.33, 5.95, 85.93, 5.30, 1.17, 1.12, 0.18, 0.03, 0.14, 0.67, 7.73, 80.53, 8.84, 1.00, 1.06, 0.01, 0.11, 0.24, 0.43, 6.48, 83.46, 4.07, 5.20, 0.21, 0, 0.22, 1.30, 2.38, 11.24, 64.86, 19.79, 0, 0, 0, 0, 0, 0, 0, 100 )/100, 8, 8, dimnames = list(rc, rc), byrow = TRUE) then followed through with the steps: nth_root - X %*% L_star %*% X_inv But the check (going back 12 to the power again) doesn't yield the original matrix. Now some rounding errors can be expected, but I didn't expect a perfectly diagonal matrix, when the initial matrix isn't diagonal at all. round(nth_root^12,4) [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [1,] 0.9078 0. 0. 0. 0. 0. 0.0 [2,] 0. 0.9053 0. 0. 0. 0. 0.0 [3,] 0. 0. 0.9079 0. 0. 0. 0.0 [4,] 0. 0. 0. 0.8553 0. 0. 0.0 [5,] 0. 0. 0. 0. 0.7998 0. 0.0 [6,] 0. 0. 0. 0. 0. 0.8285 0.0 [7,] 0. 0. 0. 0. 0. 0. 0.64570 [8,] 0. 0. 0. 0. 0. 0. 0.1 Any takers - Christian Langkamp christian.langkamp-at-gmxpro.de -- View this message in context: http://r.789695.n4.nabble.com/12th-Root-of-a-Square-Transition-Matrix-tp2259736p3989618.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. -- Sent from my Linux computer. Way better than iPad :) __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. David Winsemius, MD Heritage Laboratories West Hartford, CT __ 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] 12th Root of a Square (Transition) Matrix
On Fri, Nov 4, 2011 at 2:37 PM, David Winsemius dwinsem...@comcast.net wrote: The 12th (matrix) root of M: e^( 1/n * log(M) ) require(Matrix) M1.12 - expm( (1/12)*logm(M) ) I like this - haven't thought of the matrix algebra functions in Matrix. Thanks, Peter __ 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] Decision tree model using rpart ( classification
aajit75 aajit75 at yahoo.co.in writes: fit - rpart(decile ~., method=class, control=rpart.control(minsplit=min_obs_split, cp=c_c_factor), data=dtm_ip) In A and B target variable 'segment' is from the clustering data using same set of input variables , while in C target variable 'decile' is derived from behavioural variables and input variables are from profile data. Number of rows in the input table in all three cases are same. What is the value of modeling the deciles as the target? They are a lower resolution version of information you already have, and without this model that doesn't finish fitting you should already be able to assign a decile to every customer. Andrew __ 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] Determining r2 values for a SEM
Hello, I have been using the SEM package and developed 4 models that all have an adequate fit. I require r2 values for the variables in my SEM and cannot find a way to get r2 values for a SEM. I have attempted using the smc (squared multiple correlation) function, but this only takes the initial covariance matrix and not the fitted SEM as arguements, so the r2 values cannot be correct. Is there a way to determine r2 values for an SEM in the SEM package or another way to get these values in R? Thank you for your assistance, Katherine Stewart -- View this message in context: http://r.789695.n4.nabble.com/Determining-r2-values-for-a-SEM-tp3990855p3990855.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] How to delete only those rows in a dataframe in which all records are missing
It does! Thanks, José -Original Message- From: R. Michael Weylandt [mailto:michael.weyla...@gmail.com] Sent: 04 November 2011 15:18 To: Jose Iparraguirre Cc: r-help@r-project.org Subject: Re: [R] How to delete only those rows in a dataframe in which all records are missing Perhaps something like this will work. df[!(rowSums(is.na(df))==NCOL(df)),] Michael On Fri, Nov 4, 2011 at 9:27 AM, Jose Iparraguirre jose.iparragui...@ageuk.org.uk wrote: Hi, Imagine I have the following data frame: a - c(1,NA,3) b - c(2,NA,NA) c - data.frame(cbind(a,b)) c a b 1 1 2 2 NA NA 3 3 NA I want to delete the second row. If I use na.omit, that would also affect the third row. I tried to use a loop and an ifelse clause with is.na to get R identify that row in which all records are missing, as opposed to the first row in which no records are missing or the third one, in which only one record is missing. How can I get R identify the row in which all records are missing? Or, how can I get R delete/omit only this row? Thanks in advance, José José Iparraguirre Chief Economist Age UK T 020 303 31482 E jose.iparragui...@ageuk.org.ukmailto:jose.iparragui...@ageuk.org.uk Tavis House, 1- 6 Tavistock Square London, WC1H 9NB www.ageuk.org.ukhttp://www.ageuk.org.uk | ageukblog.org.ukhttp://ageukblog.org.uk/ | @AgeUKPAhttp://twitter.com/ageukpa Age UK Improving later life www.ageuk.org.uk --- Age UK is a registered charity and company limited by guarantee, (registered charity number 1128267, registered company number 6825798). Registered office: Tavis House, 1-6 Tavistock Square, London WC1H 9NA. For the purposes of promoting Age UK Insurance, Age UK is an Appointed Representative of Age UK Enterprises Limited, Age UK is an Introducer Appointed Representative of JLT Benefit Solutions Limited and Simplyhealth Access for the purposes of introducing potential annuity and health cash plans customers respectively. Age UK Enterprises Limited, JLT Benefit Solutions Limited and Simplyhealth Access are all authorised and regulated by the Financial Services Authority. -- This email and any files transmitted with it are confidential and intended solely for the use of the individual or entity to whom they are addressed. If you receive a message in error, please advise the sender and delete immediately. Except where this email is sent in the usual course of our business, any opinions expressed in this email are those of the author and do not necessarily reflect the opinions of Age UK or its subsidiaries and associated companies. Age UK monitors all e-mail transmissions passing through its network and may block or modify mails which are deemed to be unsuitable. Age Concern England (charity number 261794) and Help the Aged (charity number 272786) and their trading and other associated companies merged on 1st April 2009. Together they have formed the Age UK Group, dedicated to improving the lives of people in later life. The three national Age Concerns in Scotland, Northern Ireland and Wales have also merged with Help the Aged in these nations to form three registered charities: Age Scotland, Age NI, Age Cymru. [[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. Age UK Improving later life www.ageuk.org.uk --- Age UK is a registered charity and company limited by guarantee, (registered charity number 1128267, registered company number 6825798). Registered office: Tavis House, 1-6 Tavistock Square, London WC1H 9NA. For the purposes of promoting Age UK Insurance, Age UK is an Appointed Representative of Age UK Enterprises Limited, Age UK is an Introducer Appointed Representative of JLT Benefit Solutions Limited and Simplyhealth Access for the purposes of introducing potential annuity and health cash plans customers respectively. Age UK Enterprises Limited, JLT Benefit Solutions Limited and Simplyhealth Access are all authorised and regulated by the Financial Services Authority. -- This email and any files transmitted with it are confidential and intended solely for the use of the individual or entity to whom they are addressed. If you receive a message in error, please advise the sender and delete immediately. Except where this email is sent in the usual course of our business, any opinions expressed in this email are those of the author and do not necessarily reflect the opinions of Age UK or its subsidiaries and associated companies. Age UK monitors all e-mail transmissions passing through its
[R] HoltWinters in R 2.14.0
Hey All, First time on these forums. Thanks in advance. S... I have a process that was functioning well before the 2.14 update. Now the HoltWinters function is throwing an error whereby I get the following: Error in HoltWinters(sales.ts) : optimization failure I've been looking around to determine why this happens (see if I can test the data beforehand) but I haven't come across anything. Any help appreciated! -- View this message in context: http://r.789695.n4.nabble.com/HoltWinters-in-R-2-14-0-tp3991247p3991247.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] Reading parameters from dataframe and loading as objects
Hi, assign is your friend here: apply(data,1,function(x)assign(x[1],x[2],envir = .GlobalEnv)) As a note, you probably don't want to use data as a variable because it overwrites the data function, leading to unwanted side-effects if you ever use it. Cheers, François Pepin Scientist Sequenta, Inc. 400 E. Jamie Court, Suite 301 South San Francisco, CA 94080 650 243 3929 p francois.pe...@sequentainc.com www.sequentainc.com The contents of this e-mail message and any attachments are intended solely for the addressee(s) named in this message. This communication is intended to be and to remain confidential and may be subject to applicable attorney/client and/or work product privileges. If you are not the intended recipient of this message, or if this message has been addressed to you in error, please immediately alert the sender by reply e-mail and then delete this message and its attachments. Do not deliver, distribute or copy this message and/or any attachments and if you are not the intended recipient, do not disclose the contents or take any action in reliance upon the information contained in this communication or any attachments. On Nov 3, 2011, at 23:24 , Aher wrote: Hi List, I want to read several parameters from data frame and load them as object into R session, Is there any package or function in R for this?? Here is example param -c(clust_num, minsamp_size, maxsamp_size, min_pct, max_pct) value -c(15, 2, 20, 0.001, .999) data - data.frame ( cbind(param , value)) data param value 1clust_num 15 2 minsamp_size2 3 maxsamp_size 2e+05 4 min_pct 0.001 5 max_pct 0.999 My data contains many such parameters, I need to read each parameter and its value from the data and load it as objects in R session as below: clust_num - 15 minsamp_size -2 maxsamp_size -2e+05 min_pct -0.001 max_pct -0.999 The way right now I am doing it is as creating as many variables as parameters in the data frame and one observation for value of each parameter. example: clust_num minsamp_sizemaxsamp_sizemin_pct max_pct 152 20 0.001 0.999 data$ clust_num , data$minsamp_size, . Is there any better way for doing this? -- View this message in context: http://r.789695.n4.nabble.com/Reading-parameters-from-dataframe-and-loading-as-objects-tp3989150p3989150.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.
Re: [R] Determining r2 values for a SEM
On Nov 04, 2011 at 6:55pm Katherine Stewart wrote: Is there a way to determine r2 values for an SEM in the SEM package or another way to get these values in R? Katherine, rsquare.sem() in package sem.additions will do it for you. Regards, Mark. - Mark Difford (Ph.D.) Research Associate Botany Department Nelson Mandela Metropolitan University Port Elizabeth, South Africa -- View this message in context: http://r.789695.n4.nabble.com/Determining-r2-values-for-a-SEM-tp3990855p3991279.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] error message
Hello, I am a PhD candidate at University of Sao Paulo, finishing the thesis and know nothing about R language... I am trying to analyze a structure data set (STRs) with 82 samples and 10 loci (missing data = -9) and I keep getting the same Error in text[(last line - n + 1):last line] : only 0's may be mixed with negative subscripts message. I read this thread and tried the traceback () function, but all I got after I typed it was 1: read.structure(file = SimStru3.str)... When I import the data using the read.table function, it works well, but then I am not being able to transform the data frame in a genind object... I know I may be asking very stupid things, but I really could use the help... Thank you so much! Juliana -- View this message in context: http://r.789695.n4.nabble.com/error-message-tp3223412p3991100.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] representing wind date using windrose
hello,
i am new to R and want to use it for a small project to draw a wind
data from a microclimate datasource, can someone give me an example of
how i can represent this in a neat way?
for example, i have:
speed, direction
0.3,NNE
0.45,NNE
0.32,NE
0.28,N
0.30,NE
how do i put this data to get a windrose graph?
many thanks
norman
--
% .join( [ {'*':'@','^':'.'}.get(c,None) or
chr(97+(ord(c)-83)%26) for c in ,adym,*)uzq^zqf ] )
__
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 use 'prcomp' with CLUSPLOT?
Hello Michael, Thank you for replying to my post! That was an interesting solution - good to know, but I am now getting a different error: /Error in if (length(clus) != n) stop(The clustering vector is of incorrect length) : argument is of length zero/ which brought me here: https://svn.r-project.org/R-packages/trunk/cluster/R/plotpart.q I am trying to figure that out now... FYI, as a test set, one could just delete columns until they are = to the number of rows... clusplot has some nice extras, but I am also looking at just plotting w/pca... Thank you again, Jo -- View this message in context: http://r.789695.n4.nabble.com/How-to-use-prcomp-with-CLUSPLOT-tp3989022p3991868.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] Unused Arguments Error Even Though I'm Using Them?
Greetings, I am running into an error I can't seem to get past; something tells me I am making an obvious mistake as I am new to R, but everything looks fine to me. Basically, I'm getting the message unused arguments even though my arguments are all being used. I have posted my code at the following URL: http://textsnip.com/8af5b2 Thanks for any help, CLS __ 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] ANCOVA with many levels of one factor
I am trying to do and ANCOVA with ten sites that I want to compare condition at, with Length as a covariate. Most examples I have found only deal with two levels and I am unsure if the same code applies for more than two levels. Here is what I have, and I just wanted to double check that I am on the right track ancova-lm(Condition~site+Length+site:Length) summary(ancova) anova(ancova) ancova1-update(ancova,~.-site:Length) anova(ancova,ancova1) summary(ancova1) Any help/suggestions welcome! Thanks! -- View this message in context: http://r.789695.n4.nabble.com/ANCOVA-with-many-levels-of-one-factor-tp3991474p3991474.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] how to interpret vglm output?
Hello , I need to estimate the parameters of generalized poisson regression model. I found that I could use : vglm(formula, family,.., data), but I dont know how to interpret the output!!! min 1Q Median 3Q Max elogit (lambda)-.66 -.61-.46 -.06 35.33 log(theta) -10.2-.02 .11 .64 1.2 coefficients: value stdt.value intercept:1 .76 .019 39.32 intercept:2 1.02 .0239.88 sex .55 1.216.00 log-likelihood -990.83, I dont know what is the exactly meaning of these two intercept? Thanks in advance for your help. Akram [[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] Reading parameters from dataframe and loading as objects
On Nov 4, 2011, at 1:50 PM, Francois Pepin francois.pe...@sequentainc.com wrote: Hi, assign is your friend here: apply(data,1,function(x)assign(x[1],x[2],envir = .GlobalEnv)) As a note, you probably don't want to use data as a variable because it overwrites the data function, leading to unwanted side-effects if you ever use it. While it is true that using data as an object name is a bad choice, the specific reason offered is incorrect. Functions are kept in a different list from other named objects and creating a data data.frame will NOT overwrite the data function. -- David. Cheers, François Pepin Scientist Sequenta, Inc. 400 E. Jamie Court, Suite 301 South San Francisco, CA 94080 650 243 3929 p francois.pe...@sequentainc.com www.sequentainc.com The contents of this e-mail message and any attachments are intended solely for the addressee(s) named in this message. This communication is intended to be and to remain confidential and may be subject to applicable attorney/client and/or work product privileges. If you are not the intended recipient of this message, or if this message has been addressed to you in error, please immediately alert the sender by reply e-mail and then delete this message and its attachments. Do not deliver, distribute or copy this message and/or any attachments and if you are not the intended recipient, do not disclose the contents or take any action in reliance upon the information contained in this communication or any attachments. On Nov 3, 2011, at 23:24 , Aher wrote: Hi List, I want to read several parameters from data frame and load them as object into R session, Is there any package or function in R for this?? Here is example param -c(clust_num, minsamp_size, maxsamp_size, min_pct, max_pct) value -c(15, 2, 20, 0.001, .999) data - data.frame ( cbind(param , value)) data param value 1clust_num 15 2 minsamp_size2 3 maxsamp_size 2e+05 4 min_pct 0.001 5 max_pct 0.999 My data contains many such parameters, I need to read each parameter and its value from the data and load it as objects in R session as below: clust_num - 15 minsamp_size -2 maxsamp_size -2e+05 min_pct -0.001 max_pct -0.999 The way right now I am doing it is as creating as many variables as parameters in the data frame and one observation for value of each parameter. example: clust_numminsamp_sizemaxsamp_sizemin_pctmax_pct 152200.0010.999 data$ clust_num , data$minsamp_size, . Is there any better way for doing this? -- View this message in context: http://r.789695.n4.nabble.com/Reading-parameters-from-dataframe-and-loading-as-objects-tp3989150p3989150.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-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] Creating a sequence from two samples with several constraints (frequency and repeats)
I'm attempting to create a sequence for an experiment and am hoping I can use R to create it. It has several constraints: (1) It is made up of two sequences (red and green) that have 4 different repeating triplets (e.g. T1=ABC T2=DEF T3=GHI JKL) (2) Each sequence has the following constraints: (a) there cannot be repeating triplets (e.g. T1 T1), (b) there cannot be repeating triplet pairs (e.g. T1 T2 T1 T2) (3) Triplets occur with the following frequency: T1=20, T2=23, T3=26, T4=36 (same for red and green sequences) (4) Red and green sequences are then interleaved, such that you never get more than 6 in a row of one color. (For those that are interested, I'm trying to replicate Turke-Browne, Junge, Scholl, 2005, *JEP*, but with frequency manipulations to the triplets) Thanks in advance for any help. [[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] XLConnect Error
Hi Daniel, you can get Java from http://www.java.com/en/download/manual.jsp?locale=en Simply download and follow the instructions. Hope that helps. Martin -- View this message in context: http://r.789695.n4.nabble.com/XLConnect-Error-tp3628528p3991681.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] bug calculating ROC with caret and earth?
Does caret have a bug calculating ROC with earth? When using caret and earth
on any of my data sets, caret's ROC never varies. This could mean earth is
finding the same model (for example, because of using an nprune parameter that
is too high). However, if that were true, sensitivity and specificity would
also not vary, but they do vary. Also, I verified nprune is not too high.
I am attaching sample output from R 2.14.0 on Windows 7 64-bit with earth 3.2
and caret 5.07.
I don't have this problem with caret and ctree.
Andrew
R version 2.14.0 (2011-10-31)
Copyright (C) 2011 The R Foundation for Statistical Computing
ISBN 3-900051-07-0
Platform: x86_64-pc-mingw32/x64 (64-bit)
R is free software and comes with ABSOLUTELY NO WARRANTY.
You are welcome to redistribute it under certain conditions.
Type 'license()' or 'licence()' for distribution details.
R is a collaborative project with many contributors.
Type 'contributors()' for more information and
'citation()' on how to cite R or R packages in publications.
Type 'demo()' for some demos, 'help()' for on-line help, or
'help.start()' for an HTML browser interface to help.
Type 'q()' to quit R.
# install and load packages, as needed
for (pkg in c('caret','earth','mlbench', 'e1071')) {
+ if (!require(pkg, character.only=T)) {install.packages(pkg)}
+ require(pkg, character.only=T)
+ }
Loading required package: caret
Loading required package: lattice
Loading required package: reshape
Loading required package: plyr
Attaching package: reshape
The following object(s) are masked from package:plyr:
rename, round_any
Loading required package: cluster
Loading required package: foreach
Loading required package: iterators
Loading required package: codetools
foreach: simple, scalable parallel programming from Revolution Analytics
Use Revolution R for scalability, fault tolerance and more.
http://www.revolutionanalytics.com
Loading required package: earth
Loading required package: leaps
Loading required package: plotmo
Loading required package: plotrix
Loading required package: mlbench
Loading required package: e1071
Loading required package: class
Attaching package: class
The following object(s) are masked from package:reshape:
condense
# system information
installed.packages()[c('earth','caret'),'Version']
earth caret
3.2-1 5.07-001
# prepare data
data(etitanic)
mydata - etitanic
mydata$survived - as.factor(ifelse(etitanic$survived==1, 'T', 'F'))
summary(mydata)
pclasssurvived sex age sibspparch
1st:284 F:619female:388 Min. : 0.1667 Min. :0. Min.
:0.
2nd:261 T:427male :658 1st Qu.:21. 1st Qu.:0. 1st
Qu.:0.
3rd:501 Median :28. Median :0. Median
:0.
Mean :29.8811 Mean :0.5029 Mean
:0.4207
3rd Qu.:39. 3rd Qu.:1. 3rd
Qu.:1.
Max. :80. Max. :8. Max.
:6.
# show natural maximum pruning is 9
fit - earth(survived ~ ., data=mydata)
summary(fit, style=max)
Call: earth(formula=survived~., data=mydata)
T =
1.094732
- 0.2113713 * max(0, pclass2nd - 0)
- 0.3413489 * max(0, pclass3rd - 0)
- 0.4851343 * max(0, sexmale - 0)
- 0.004222467 * max(0, age -10)
+ 0.02569032 * max(0,10 - age)
- 0.09699376 * max(0, sibsp - 1)
- 0.06266133 * max(0, parch - 1)
- 0.09015484 * max(0, 1 - parch)
Selected 9 of 10 terms, and 6 of 6 predictors
Importance: sexmale, pclass3rd, age, pclass2nd, sibsp, parch
Number of terms at each degree of interaction: 1 8 (additive model)
GCV 0.1519922RSS 153.8581GRSq 0.3720351RSq 0.3911174
# custom metric
twoClassSummaryPlus - function (data,
+ lev = NULL,
+ model = NULL)
+
+ {
+ out1 - twoClassSummary(data, lev, model)
+ out2 - defaultSummary(data, lev, model)
+ #browser() # debug
+ #print(out1)
+ #print(dim(data))
+ c(out1, out2)
+ }
# tne
train_earth - function(nprune)
+ {
+ # prepare tuning parameters
+ grid - expand.grid(.degree=c(1), .nprune=nprune)
+
+ trControl- trainControl(summaryFunction = twoClassSummaryPlus,
+ classProbs = T,
+ verboseIter=T)
+
+ # tune
+ mydata.best - train(survived ~ .,
+ data = mydata,
+ method = earth,
+ trControl = trControl,
+ metric=Sens,
+ tuneGrid=grid)
+
+ # show tuned
+ print(mydata.best)
+ }
train_earth(c(1:9)) # ROC is constant
Fitting: degree=1, nprune=9
Fitting: degree=1, nprune=9
Fitting: degree=1, nprune=9
Fitting: degree=1, nprune=9
Fitting: degree=1, nprune=9
Fitting: degree=1, nprune=9
Fitting: degree=1, nprune=9
Fitting: degree=1, nprune=9
Fitting: degree=1, nprune=9
Fitting: degree=1, nprune=9
Fitting:
Re: [R] representing wind date using windrose
There are several windrose functions in various packages. Try this
RSiteSearch(winrose)
--
David
On Nov 4, 2011, at 6:06 PM, Norman Khine nor...@khine.net wrote:
hello,
i am new to R and want to use it for a small project to draw a wind
data from a microclimate datasource, can someone give me an example of
how i can represent this in a neat way?
for example, i have:
speed, direction
0.3,NNE
0.45,NNE
0.32,NE
0.28,N
0.30,NE
how do i put this data to get a windrose graph?
many thanks
norman
--
% .join( [ {'*':'@','^':'.'}.get(c,None) or
chr(97+(ord(c)-83)%26) for c in ,adym,*)uzq^zqf ] )
__
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] representing wind date using windrose
Try this: http://rss.acs.unt.edu/Rdoc/library/climatol/html/rosavent.html
Michael
On Fri, Nov 4, 2011 at 6:06 PM, Norman Khine nor...@khine.net wrote:
hello,
i am new to R and want to use it for a small project to draw a wind
data from a microclimate datasource, can someone give me an example of
how i can represent this in a neat way?
for example, i have:
speed, direction
0.3,NNE
0.45,NNE
0.32,NE
0.28,N
0.30,NE
how do i put this data to get a windrose graph?
many thanks
norman
--
% .join( [ {'*':'@','^':'.'}.get(c,None) or
chr(97+(ord(c)-83)%26) for c in ,adym,*)uzq^zqf ] )
__
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] Determining r2 values for a SEM
Dear Mark and Katherine, As well, version 2.0-0 of the sem package on R-Forge, and soon to be on CRAN, includes R^2 values as part of the summary() output (along with many other enhancements). Best, John -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-bounces@r- project.org] On Behalf Of Mark Difford Sent: November-04-11 3:06 PM To: r-help@r-project.org Subject: Re: [R] Determining r2 values for a SEM On Nov 04, 2011 at 6:55pm Katherine Stewart wrote: Is there a way to determine r2 values for an SEM in the SEM package or another way to get these values in R? Katherine, rsquare.sem() in package sem.additions will do it for you. Regards, Mark. - Mark Difford (Ph.D.) Research Associate Botany Department Nelson Mandela Metropolitan University Port Elizabeth, South Africa -- View this message in context: http://r.789695.n4.nabble.com/Determining-r2-values-for-a-SEM- tp3990855p3991279.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.
Re: [R] HoltWinters in R 2.14.0
I believe there were some changes to Holt-Winters, specifically in re optimization that probably lead to your problem, but you'll have to provide more details. See the NEWS file for citations about the change. If you put example code/data others may be able to help you -- I haven't updated yet so I can't be of much help. Michael On Fri, Nov 4, 2011 at 2:55 PM, TimothyDalbey tmdal...@gmail.com wrote: Hey All, First time on these forums. Thanks in advance. S... I have a process that was functioning well before the 2.14 update. Now the HoltWinters function is throwing an error whereby I get the following: Error in HoltWinters(sales.ts) : optimization failure I've been looking around to determine why this happens (see if I can test the data beforehand) but I haven't come across anything. Any help appreciated! -- View this message in context: http://r.789695.n4.nabble.com/HoltWinters-in-R-2-14-0-tp3991247p3991247.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.
Re: [R] Unused Arguments Error Even Though I'm Using Them?
I can't see anything in your code at first glance that would lead to that error: could you perhaps provide some self-contained code that reproduces that error? Michael On Fri, Nov 4, 2011 at 3:21 PM, Christopher Simons christopherleesim...@gmail.com wrote: Greetings, I am running into an error I can't seem to get past; something tells me I am making an obvious mistake as I am new to R, but everything looks fine to me. Basically, I'm getting the message unused arguments even though my arguments are all being used. I have posted my code at the following URL: http://textsnip.com/8af5b2 Thanks for any help, CLS __ 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] representing wind date using windrose
I'm also very impressed with openair
http://www.openair-project.org/,
also
cran.r-project.org/package=openair
Clint
--
Clint BowmanINTERNET: cl...@ecy.wa.gov
Air Quality Modeler INTERNET: cl...@math.utah.edu
Department of Ecology VOICE: (360) 407-6815
PO Box 47600FAX:(360) 407-7534
Olympia, WA 98504-7600
USPS: PO Box 47600, Olympia, WA 98504-7600
Parcels:300 Desmond Drive, Lacey, WA 98503-1274
On Fri, 4 Nov 2011, R. Michael Weylandt wrote:
Try this: http://rss.acs.unt.edu/Rdoc/library/climatol/html/rosavent.html
Michael
On Fri, Nov 4, 2011 at 6:06 PM, Norman Khine nor...@khine.net wrote:
hello,
i am new to R and want to use it for a small project to draw a wind
data from a microclimate datasource, can someone give me an example of
how i can represent this in a neat way?
for example, i have:
speed, direction
0.3,NNE
0.45,NNE
0.32,NE
0.28,N
0.30,NE
how do i put this data to get a windrose graph?
many thanks
norman
--
% .join( [ {'*':'@','^':'.'}.get(c,None) or
chr(97+(ord(c)-83)%26) for c in ,adym,*)uzq^zqf ] )
__
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.
Re: [R] Creating a sequence from two samples with several constraints (frequency and repeats)
I believe the permute package is set up to generate restricted permutations. Michael On Fri, Nov 4, 2011 at 4:59 PM, Jeremy Wang wang0...@umn.edu wrote: I'm attempting to create a sequence for an experiment and am hoping I can use R to create it. It has several constraints: (1) It is made up of two sequences (red and green) that have 4 different repeating triplets (e.g. T1=ABC T2=DEF T3=GHI JKL) (2) Each sequence has the following constraints: (a) there cannot be repeating triplets (e.g. T1 T1), (b) there cannot be repeating triplet pairs (e.g. T1 T2 T1 T2) (3) Triplets occur with the following frequency: T1=20, T2=23, T3=26, T4=36 (same for red and green sequences) (4) Red and green sequences are then interleaved, such that you never get more than 6 in a row of one color. (For those that are interested, I'm trying to replicate Turke-Browne, Junge, Scholl, 2005, *JEP*, but with frequency manipulations to the triplets) Thanks in advance for any help. [[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] error message
Could you provide an example of your code? The error is coming up because lastLine - n + 1 0 but obviously I can't tell you why it's happening in your code without seeing it. Michael On Fri, Nov 4, 2011 at 2:06 PM, JulianaMF j...@ib.usp.br wrote: Hello, I am a PhD candidate at University of Sao Paulo, finishing the thesis and know nothing about R language... I am trying to analyze a structure data set (STRs) with 82 samples and 10 loci (missing data = -9) and I keep getting the same Error in text[(last line - n + 1):last line] : only 0's may be mixed with negative subscripts message. I read this thread and tried the traceback () function, but all I got after I typed it was 1: read.structure(file = SimStru3.str)... When I import the data using the read.table function, it works well, but then I am not being able to transform the data frame in a genind object... I know I may be asking very stupid things, but I really could use the help... Thank you so much! Juliana -- View this message in context: http://r.789695.n4.nabble.com/error-message-tp3223412p3991100.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] set seed for random draws
Hello, all!
I need help on these two problems:
1) If I want to randomly draw numbers from standard normal (or other
distributions) in loops e.g.:
ty=0; ks=0
for (i in 1:5) {
set.seed(14537+i)
k-rnorm(1)
ks[i]-.3*k+.9
if (ty==0) {
while ((ks.2)||(ks3)) {
#set.seed(13237+i*100)
k-rnorm(1)
ks[i]-.3*k+.9 }
}
}
}
Question: Here I draw initial a, then if the drawn initial a satisfied 2
conditions I redraw a. I set.seed(13237) in the first draw of a, should I
set.seed() in the redraw part?
2) I also have more loops after this i loop that also draw from normal(0,1). I
want to randomly draws from normal(0,1) for loop j (inside loop j I draw
another random numbers from N(0,1))
My question: Should I or shouldn't I set seed again and again for each loop?
Why or why not.
I guess this problem concerned about setting seed as I want to have different
number for each i.
Thanks!
Deana
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Re: [R] set seed for random draws
This might be more fundamental, but why do you feel the need to reset
the seed each loop? There's nothing that suggests you need to...
Michael
On Fri, Nov 4, 2011 at 8:38 PM, Md Desa, Zairul Nor Deana Binti
znde...@ku.edu wrote:
Hello, all!
I need help on these two problems:
1) If I want to randomly draw numbers from standard normal (or other
distributions) in loops e.g.:
ty=0; ks=0
for (i in 1:5) {
set.seed(14537+i)
k-rnorm(1)
ks[i]-.3*k+.9
if (ty==0) {
while ((ks.2)||(ks3)) {
#set.seed(13237+i*100)
k-rnorm(1)
ks[i]-.3*k+.9 }
}
}
}
Question: Here I draw initial a, then if the drawn initial a satisfied 2
conditions I redraw a. I set.seed(13237) in the first draw of a, should I
set.seed() in the redraw part?
2) I also have more loops after this i loop that also draw from normal(0,1).
I want to randomly draws from normal(0,1) for loop j (inside loop j I draw
another random numbers from N(0,1))
My question: Should I or shouldn't I set seed again and again for each loop?
Why or why not.
I guess this problem concerned about setting seed as I want to have different
number for each i.
Thanks!
Deana
__
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[R] Matrix element-by-element multiplication
is there a way to do element-by-element multiplication as in Gauss and MATLAB, as shown below? Thanks. --- a 1.000 2.000 3.000 x 1.0002.0003.000 2.0004.0006.000 3.0006.0009.000 a.*x 1.0002.0003.000 4.0008.00012.00 9.00018.0027.00 -- Steven T. Yen, Professor of Agricultural Economics The University of Tennessee http://web.utk.edu/~syen/ [[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 avoid ifelse statement converting factor to character
most all those posts were right in sourcing the problemits just that nobody actually offered a viable solution the problem is that the new level C was not one of the original levels in $social status add C as a level and then just do the ol fashioned way and it works just fine do this: data$SOCIAL_STATUS - factor(data$SOCIAL_STATUS, levels = c(levels(data$SOCIAL_STATUS), C)) then this: data$SOCIAL_STATUS-ifelse(data$SOCIAL_STATUS==B data$MALE4, C, data$SOCIAL_STATUS) it is sooo much more helpful when someone who has addressed the specific problem being asked replies instead of people just throwing random ideas out there -- View this message in context: http://r.789695.n4.nabble.com/How-to-avoid-ifelse-statement-converting-factor-to-character-tp895726p3992067.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] nested for loops
Hi all , I have written a code with nested for loops .
The aim is to estimate the maximum likelihood by creating 3 vectors with the
same length( sequence )
and then to utilize 3 for loops to make combinations among the 3 vectors ,
which are (length)^3 in number , and find the one that maximize the
likelihood ( maximum likelihood estimator).
The code I created, runs but I think something goes wrong...because when I
change the length of the vectors but not the bounds the result is the
same!!!
I will give a simple example(irrelevant but proportional to the above) to
make it more clear...
Lets say we want to find the combination that maximize the multiplication of
the entries of some vectors.
V1-c(1,2,3)
V2-c(5, 2 , 4)
V3-c( 4, 3, 6)
The combination we look for is ( 3 , 5 , 6) that give us 3*5*6 = 90
If I apply the following in R , I won't take this result
V1-c(1,2,3)
V2-c(5, 2 , 4)
V3-c( 4, 3, 6)
for( i in V1){
for( j in V2) {
for( k in V3){
l- i*j*k
}
}
}
l
Then l- i*j*k is number and not vector(of all multiplications of all
the combinations) , and is 3*4*6 = 72.
How can I fix the code?
--
View this message in context:
http://r.789695.n4.nabble.com/nested-for-loops-tp3992089p3992089.html
Sent from the R help mailing list archive at Nabble.com.
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Re: [R] Matrix element-by-element multiplication
Did you even try? a - 1:3 x - matrix(c(1,2,3,2,4,6,3,6,9),3) a*x [,1] [,2] [,3] [1,]123 [2,]48 12 [3,]9 18 27 Michael On Fri, Nov 4, 2011 at 7:26 PM, Steven Yen s...@utk.edu wrote: is there a way to do element-by-element multiplication as in Gauss and MATLAB, as shown below? Thanks. --- a 1.000 2.000 3.000 x 1.000 2.000 3.000 2.000 4.000 6.000 3.000 6.000 9.000 a.*x 1.000 2.000 3.000 4.000 8.000 12.00 9.000 18.00 27.00 -- Steven T. Yen, Professor of Agricultural Economics The University of Tennessee http://web.utk.edu/~syen/ [[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] nested for loops
Your problem is that you redefine l each time through the loops and
don't record old values; you could do so by using c() for
concatenation, but perhaps this is what you are looking for:
exp(rowSums(log(expand.grid(V1, V2, V3
Hope this helps,
Michael
On Fri, Nov 4, 2011 at 7:49 PM, nick_pan nick_pa...@yahoo.gr wrote:
Hi all , I have written a code with nested for loops .
The aim is to estimate the maximum likelihood by creating 3 vectors with the
same length( sequence )
and then to utilize 3 for loops to make combinations among the 3 vectors ,
which are (length)^3 in number , and find the one that maximize the
likelihood ( maximum likelihood estimator).
The code I created, runs but I think something goes wrong...because when I
change the length of the vectors but not the bounds the result is the
same!!!
I will give a simple example(irrelevant but proportional to the above) to
make it more clear...
Lets say we want to find the combination that maximize the multiplication of
the entries of some vectors.
V1-c(1,2,3)
V2-c(5, 2 , 4)
V3-c( 4, 3, 6)
The combination we look for is ( 3 , 5 , 6) that give us 3*5*6 = 90
If I apply the following in R , I won't take this result
V1-c(1,2,3)
V2-c(5, 2 , 4)
V3-c( 4, 3, 6)
for( i in V1){
for( j in V2) {
for( k in V3){
l- i*j*k
}
}
}
l
Then l- i*j*k is number and not vector(of all multiplications of all
the combinations) , and is 3*4*6 = 72.
How can I fix the code?
--
View this message in context:
http://r.789695.n4.nabble.com/nested-for-loops-tp3992089p3992089.html
Sent from the R help mailing list archive at Nabble.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.
__
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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] How to avoid ifelse statement converting factor to character
On Nov 4, 2011, at 7:33 PM, clint cl...@ucsd.edu wrote: most all those posts were right in sourcing the problemits just that nobody actually offered a viable solution Very few problems that are posed with a test dataset go unanswered. the problem is that the new level C was not one of the original levels in $social status add C as a level and then just do the ol fashioned way and it works just fine do this: data$SOCIAL_STATUS - factor(data$SOCIAL_STATUS, levels = c(levels(data$SOCIAL_STATUS), C)) then this: data$SOCIAL_STATUS-ifelse(data$SOCIAL_STATUS==B data$MALE4, C, data$SOCIAL_STATUS) it is sooo much more helpful when someone who has addressed the specific problem being asked replies instead of people just throwing random ideas out there Just a little Friday night trolling? This was a posting from 2 years ago. -- David -- View this message in context: http://r.789695.n4.nabble.com/How-to-avoid-ifelse-statement-converting-factor-to-character-tp895726p3992067.html Sent from the R help mailing list archive at Nabble.com. This is a mailing list. Nabble users who fail to include context who replying to ancient threads should not be throwing stones. __ 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] 12th Root of a Square (Transition) Matrix
This is just one of many 12-th roots. (Peter knows this i'm sure.) The negative of this would also be an nth root, and I read that there are quite few others that arise from solutions based on permuting negatives of eigen values of a triangularized form. But as I said , I'm not a matrix mechanic, so no code for that. -- David. On Nov 4, 2011, at 6:10 PM, Peter Langfelder peter.langfel...@gmail.com wrote: On Fri, Nov 4, 2011 at 2:37 PM, David Winsemius dwinsem...@comcast.net wrote: The 12th (matrix) root of M: e^( 1/n * log(M) ) require(Matrix) M1.12 - expm( (1/12)*logm(M) ) I like this - haven't thought of the matrix algebra functions in Matrix. Thanks, Peter __ 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] HoltWinters in R 2.14.0
On Fri, 4 Nov 2011, R. Michael Weylandt wrote: I believe there were some changes to Holt-Winters, specifically in re optimization that probably lead to your problem, but you'll have to provide more details. See the NEWS file for citations about the change. If you put example code/data others may be able to help you -- I haven't updated yet so I can't be of much help. Michael On Fri, Nov 4, 2011 at 2:55 PM, TimothyDalbey tmdal...@gmail.com wrote: Hey All, First time on these forums. Thanks in advance. S... I have a process that was functioning well before the 2.14 update. Now the HoltWinters function is throwing an error whereby I get the following: Error in HoltWinters(sales.ts) : optimization failure Most likely it was incorrect before. You cannot assume that it was actually 'functioning well': all the cases where we have seen this message it was giving incorrect answers before and not detecting them. And in all those cases the model was a bad fit and using starting values for the optimization helped. I've been looking around to determine why this happens (see if I can test the data beforehand) but I haven't come across anything. Any help appreciated! -- 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.
Re: [R] barplot as histogram
On 11/05/2011 05:04 AM, Jesse Brown wrote: Hello: I'm dealing with an issue currently that I'm not sure the best way to approach. I've got a very large (10G+) dataset that I'm trying to create a histogram for. I don't seem to be able to use hist directly as I can not create an R vector of size greater than 2.2G. I considered condensing the data previous to loading it into R and just plotting the frequencies as a barplot; unfortunately, barplot does not support plotting the values according to a set of x-axis positions. What I have is something similar to: ys - c(12,3,7,22,10) xs - c(1,30,35,39,60) and I'd like the bars (ys) to appear at the positions described by xs. I can get this to work on smaller sets by filling zero values in for missing ys for the entire range of xs but in my case this would again create a vector too large for R. Is there another way to use the two vectors to create a simulated frequency histogram? Is there a way to create a histogram object (as returned by hist) from the condensed data so that plot would handle it correctly? Hi Jesse, I think that barp (plotrix) will get you out of trouble. Jim __ 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] acf?
I started to check what I thought I knew with autocovariance and it doesnt jive with the the calculations given by R. I was wondering if there is some scaling or something that I am not aware of. Take the example Ø d - 1:10 Ø (a - acf(d, type=covariance, demean=FALSE, plot=FALSE)) Autocovariances of series d, by lag 0123456789 38.5 33.0 27.6 22.4 17.5 13.0 9.0 5.6 2.9 1.0 But when I calculate it manually (for lag of 1) like: Ø y1 - d mean(d) Ø dl - c(d[-1], d[1]) Ø y2 - dl mean(d) Ø mean(y1*y2) [1] 3.75 What am I missing to get this basic concept? Isnt it E[(Yt ut)(Ys us)]? Thank you. Kevin [[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] acf?
On Sat, Nov 5, 2011 at 12:26 AM, Kevin Burton rkevinbur...@charter.net wrote: I started to check what I thought I knew with autocovariance and it doesn’t jive with the the calculations given by ‘R’. I was wondering if there is some scaling or something that I am not aware of. Take the example Ø d - 1:10 Ø (a - acf(d, type=covariance, demean=FALSE, plot=FALSE)) Autocovariances of series ‘d’, by lag 0 1 2 3 4 5 6 7 8 9 38.5 33.0 27.6 22.4 17.5 13.0 9.0 5.6 2.9 1.0 But when I calculate it manually (for lag of 1) like: Ø y1 - d – mean(d) Ø dl - c(d[-1], d[1]) Ø y2 - dl – mean(d) Ø mean(y1*y2) [1] 3.75 What am I missing to get this basic concept? Isn’t it E[(Yt – ut)(Ys – us)]? Try this: d - 1:10 dm - d - mean(d) sum(dm[-1] * dm[-10]) / 10 [1] 5.775 acf(d, type = cov, plot = FALSE)[1] Autocovariances of series ‘d’, by lag 1 5.78 -- Statistics Software Consulting GKX Group, GKX Associates Inc. tel: 1-877-GKX-GROUP email: ggrothendieck at 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.
