Re: [R] rbinom
This makes sense. Guess I should have put a pencil to it. Further investigation revealed that it is indeed a possibility that the relation between x and y is nonlinear: ax+bx^2+c where a, b and c are to be determined. My question is how to code this in my simulated data. I could do something like this after appropriately defining beta. meanpred and varpred: x[,2]-rnorm(length,meanpred[2],sqrt(varpred[2])) x[,3]-rnorm(length,meanpred[3],sqrt(varpred[3])) fixpart-x%*%beta binomprob-exp(fixpart)/(1+exp(fixpart)) data$y-rbinom(n1,1,binomprob) but I'd need to square my x[,3] values before multiplying them by beta. Can I say: x[,3]-(rnorm(length,meanpred[3],sqrt(varpred[3])))^2 in lieu of x[,3]-rnorm(length,meanpred[3],sqrt(varpred[3]))? - Original Message - From: peter dalgaard pda...@gmail.com To: Scott Raynaud scott.rayn...@yahoo.com Cc: r-help@r-project.org r-help@r-project.org Sent: Tuesday, December 27, 2011 9:15 AM Subject: Re: [R] rbinom On Dec 27, 2011, at 15:47 , Scott Raynaud wrote: I have the following code (which I did not write) that generates data based on a logistic model. I'm only getting a single record with y=1. It seems implausible that in 50k cases that have a single y=1. Does that ring alarm bells for anyone else? Not really. As far as I can tell, fixpart is roughly -10.5 (= -1.5 - .25*36), so binomprob is around 2.75e-5, which - nonlinearity notwithstanding - suggests that the expected number of positives out of 50K is something like 1.4. To do this more precisely, just compute and print sum(binomprob) in the code you gave. beta-c(-1.585600,-0.246900) betasize-length(beta) meanpred-c(0,35.90) varpred-c(0,1.00) #loop code x-matrix(1,length,betasize) #length set to 50k #loop code x[,2]-rnorm(length,meanpred[2],sqrt(varpred[2])) #length set to 50k fixpart-x%*%beta binomprob-exp(fixpart)/(1+exp(fixpart)) data$y-rbinom(n1,1,binomprob) #more loop 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. -- Peter Dalgaard, Professor, Center for Statistics, Copenhagen Business School Solbjerg Plads 3, 2000 Frederiksberg, Denmark Phone: (+45)38153501 Email: pd@cbs.dk Priv: pda...@gmail.com __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] rbinom
On Dec 27, 2011, at 15:47 , Scott Raynaud wrote: I have the following code (which I did not write) that generates data based on a logistic model. I'm only getting a single record with y=1. It seems implausible that in 50k cases that have a single y=1. Does that ring alarm bells for anyone else? Not really. As far as I can tell, fixpart is roughly -10.5 (= -1.5 - .25*36), so binomprob is around 2.75e-5, which - nonlinearity notwithstanding - suggests that the expected number of positives out of 50K is something like 1.4. To do this more precisely, just compute and print sum(binomprob) in the code you gave. beta-c(-1.585600,-0.246900) betasize-length(beta) meanpred-c(0,35.90) varpred-c(0,1.00) #loop code x-matrix(1,length,betasize) #length set to 50k #loop code x[,2]-rnorm(length,meanpred[2],sqrt(varpred[2])) #length set to 50k fixpart-x%*%beta binomprob-exp(fixpart)/(1+exp(fixpart)) data$y-rbinom(n1,1,binomprob) #more loop code __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Peter Dalgaard, Professor, Center for Statistics, Copenhagen Business School Solbjerg Plads 3, 2000 Frederiksberg, Denmark Phone: (+45)38153501 Email: pd@cbs.dk Priv: pda...@gmail.com __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] rbinom
I have the following code (which I did not write) that generates data based on a logistic model. I'm only getting a single record with y=1. It seems implausible that in 50k cases that have a single y=1. Does that ring alarm bells for anyone else? beta-c(-1.585600,-0.246900) betasize-length(beta) meanpred-c(0,35.90) varpred-c(0,1.00) #loop code x-matrix(1,length,betasize) #length set to 50k #loop code x[,2]-rnorm(length,meanpred[2],sqrt(varpred[2])) #length set to 50k fixpart-x%*%beta binomprob-exp(fixpart)/(1+exp(fixpart)) data$y-rbinom(n1,1,binomprob) #more loop 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] rbinom and probability
Hello compadRes, I'm developing a script that selects cells over a certain metabolic rate to kill them. A rate between 9 and 12 means that the cells are candidates for death. I'll show you what I mean: # a would be a vector of cell metabolic rates. a-c(8, 7, 9, 6, 10, 11, 4, 5, 6) #now identify which cells will be candidates for death, namely those cells with metabolic rates9 b-a[a6] #This would select all cells with rates 6 rbinom(length(b), size=1, prob=1) #This will select cells using a an exponential probability distribution rbinom(length(b), size=1, prob=(seq(0, 1, by=0.1)^2)) I have two questions: 1) Am I correct in my interpretations? 2) I'd actually like to have this probability scaled so that cells with rates of 9 are unlikely to be selected and those near 12 are highly likely. How can I code this? Thanks in advance, Wyatt sessionInfo() R version 2.11.1 (2010-05-31) i386-pc-mingw32 locale: [1] LC_COLLATE=English_United States.1252 [2] LC_CTYPE=English_United States.1252 [3] LC_MONETARY=English_United States.1252 [4] LC_NUMERIC=C [5] LC_TIME=English_United States.1252 attached base packages: [1] stats graphics grDevices utils datasets methods base loaded via a namespace (and not attached): [1] tools_2.11.1 K. Wyatt McMahon, Ph.D. Postdoctoral Research Associate Department of Internal Medicine 3601 4th St. - Lubbock, TX - 79430 P: 806-743-4072 F: 806-743-3148 [[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] rbinom and probability
On Feb 4, 2011, at 4:24 PM, Mcmahon, Kwyatt wrote: Hello compadRes, I'm developing a script that selects cells over a certain metabolic rate to kill them. A rate between 9 and 12 means that the cells are candidates for death. I'll show you what I mean: # a would be a vector of cell metabolic rates. a-c(8, 7, 9, 6, 10, 11, 4, 5, 6) #now identify which cells will be candidates for death, namely those cells with metabolic rates9 b-a[a6] #This would select all cells with rates 6 rbinom(length(b), size=1, prob=1) No, you have an overly complex way of generating a vector of all 1's with length(b). And it's not clear what you mean by select. I see no caode that would do any selection or indexing. #This will select cells using a an exponential probability distribution rbinom(length(b), size=1, prob=(seq(0, 1, by=0.1)^2)) Again no selection appears to be taking place. I have two questions: 1) Am I correct in my interpretations? 2) I'd actually like to have this probability scaled so that cells with rates of 9 are unlikely to be selected and those near 12 are highly likely. How can I code this? a[ sample(1:length(a), prob=0.1+(a9) ,replace=TRUE) ] [1] 10 10 11 10 10 11 10 11 10 a[sample(1:length(a), prob=0.1+(a9) ,replace=TRUE) ] [1] 10 10 11 11 11 11 5 10 10 The rates over 9 are 11 times (= 1.1/0.1) as likely to be selected. You can adjust the ratio by altering the 0.1 value.+ with the first argument numeric act to coerce the logical vector to 1's and 0's. -- David. Thanks in advance, Wyatt sessionInfo() R version 2.11.1 (2010-05-31) i386-pc-mingw32 locale: [1] LC_COLLATE=English_United States.1252 [2] LC_CTYPE=English_United States.1252 [3] LC_MONETARY=English_United States.1252 [4] LC_NUMERIC=C [5] LC_TIME=English_United States.1252 attached base packages: [1] stats graphics grDevices utils datasets methods base loaded via a namespace (and not attached): [1] tools_2.11.1 K. Wyatt McMahon, Ph.D. Postdoctoral Research Associate Department of Internal Medicine 3601 4th St. - Lubbock, TX - 79430 P: 806-743-4072 F: 806-743-3148 [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. David Winsemius, MD 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] rbinom for a matrix
ACroske Audy3272 at yahoo.com writes: Ben: Thanks for the reply. One further question, and this is where my novice status at R shows through. The code makes sense, but what would I put it for m? Is it the same number for all three (that was my first thought since it was the same placeholder for all three). Number of rows in the matrix is 56. There are 2576 total cells (observations). Each cell has its own probability. Thanks for your help! matrix(rbinom(length(m),prob=m,size=1),nrow=nrow(m)) or (perhaps marginally more efficiently?) y - (runif(m)m) storage.mode(y) - double m is your matrix of probabilities; for example, you might do m = as.matrix(read.table(mystuff.txt)) (or import it from ArcGIS, or whatever). length(m) will be computed automatically (=2576), as will nrow(m) (=56). Erik Iverson's solution should work too, but I think mine will be much more efficient -- picking a whole slug of random numbers at once is much faster than looping and picking them one at a time. Ben Bolker __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] rbinom for a matrix
I have a large matrix full of probabilities; I would like to convert each probability to a 1 or a 0 using rbinom. How can I do this on the entire matrix? The matrix was converted from a raster ArcMap dataset, so the matrix is essentially a map. Because of this, I have no column headings. Thanks! -- View this message in context: http://www.nabble.com/rbinom-for-a-matrix-tp18366867p18366867.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] rbinom for a matrix
ACroske Audy3272 at yahoo.com writes: I have a large matrix full of probabilities; I would like to convert each probability to a 1 or a 0 using rbinom. How can I do this on the entire matrix? The matrix was converted from a raster ArcMap dataset, so the matrix is essentially a map. Because of this, I have no column headings. Thanks! How about matrix(rbinom(length(m),prob=m,size=1),nrow=nrow(m)) or (perhaps marginally more efficiently?) y - (runif(m)m) storage.mode(y) - double Ben Bolker __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] rbinom for a matrix
Is this what you're looking for? test - matrix(runif(100, 0, 1), nrow = 20) nr - nrow(test) matrix(sapply(test, rbinom, n = 1, size = 1), nrow = nr) ACroske wrote: I have a large matrix full of probabilities; I would like to convert each probability to a 1 or a 0 using rbinom. How can I do this on the entire matrix? The matrix was converted from a raster ArcMap dataset, so the matrix is essentially a map. Because of this, I have no column headings. Thanks! __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] rbinom for a matrix
On Wednesday 09 July 2008, Ben Bolker wrote: ACroske Audy3272 at yahoo.com writes: I have a large matrix full of probabilities; I would like to convert each probability to a 1 or a 0 using rbinom. How can I do this on the entire matrix? The matrix was converted from a raster ArcMap dataset, so the matrix is essentially a map. Because of this, I have no column headings. Thanks! How about matrix(rbinom(length(m),prob=m,size=1),nrow=nrow(m)) or (perhaps marginally more efficiently?) y - (runif(m)m) storage.mode(y) - double Ben Bolker Wait a second. Are you trying to convert each probability into the most likely '1' or '0' through rounding? The code example above will give you a different answer every time you run it. Is that what you are looking for? Just curious, Dylan -- Dylan Beaudette Soil Resource Laboratory http://casoilresource.lawr.ucdavis.edu/ University of California at Davis 530.754.7341 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] rbinom for a matrix
-BEGIN PGP SIGNED MESSAGE- Hash: SHA1 Dylan Beaudette wrote: | On Wednesday 09 July 2008, Ben Bolker wrote: | ACroske Audy3272 at yahoo.com writes: | I have a large matrix full of probabilities; I would like to convert each | probability to a 1 or a 0 using rbinom. | How can I do this on the entire matrix? The matrix was converted from a | raster ArcMap dataset, so the matrix is essentially a map. Because of | this, I have no column headings. | Thanks! | How about | | matrix(rbinom(length(m),prob=m,size=1),nrow=nrow(m)) | | or (perhaps marginally more efficiently?) | | y - (runif(m)m) | storage.mode(y) - double | | Ben Bolker | | | Wait a second. Are you trying to convert each probability into the most | likely '1' or '0' through rounding? The code example above will give you a | different answer every time you run it. Is that what you are looking for? | | Just curious, | | Dylan | ~ I assumed that since the original poster said convert each probability to a 1 or 0 using rbinom that they did indeed want a random assignment, rather than rounding ... -BEGIN PGP SIGNATURE- Version: GnuPG v1.4.6 (GNU/Linux) Comment: Using GnuPG with Mozilla - http://enigmail.mozdev.org iD8DBQFIdRiZc5UpGjwzenMRAuLkAKCA6ZhjrVsg5RJGYFGwq+6vz2pWxgCfetYk 9V/ZPGi2jfVGV5jx+LGPs8s= =d8c9 -END 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] rbinom for a matrix
On Wednesday 09 July 2008, Ben Bolker wrote: Dylan Beaudette wrote: | On Wednesday 09 July 2008, Ben Bolker wrote: | ACroske Audy3272 at yahoo.com writes: | I have a large matrix full of probabilities; I would like to convert each | probability to a 1 or a 0 using rbinom. | How can I do this on the entire matrix? The matrix was converted from a | raster ArcMap dataset, so the matrix is essentially a map. Because of | this, I have no column headings. | Thanks! | | How about | | matrix(rbinom(length(m),prob=m,size=1),nrow=nrow(m)) | | or (perhaps marginally more efficiently?) | | y - (runif(m)m) | storage.mode(y) - double | | Ben Bolker | | Wait a second. Are you trying to convert each probability into the most | likely '1' or '0' through rounding? The code example above will give you a | different answer every time you run it. Is that what you are looking for? | | Just curious, | | Dylan ~ I assumed that since the original poster said convert each probability to a 1 or 0 using rbinom that they did indeed want a random assignment, rather than rounding ... Sorry about not including a reference to whom I was addressing. Meant that comment to go to the original poster. -- Dylan Beaudette Soil Resource Laboratory http://casoilresource.lawr.ucdavis.edu/ University of California at Davis 530.754.7341 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] rbinom for a matrix
Ben: Thanks for the reply. One further question, and this is where my novice status at R shows through. The code makes sense, but what would I put it for m? Is it the same number for all three (that was my first thought since it was the same placeholder for all three). Number of rows in the matrix is 56. There are 2576 total cells (observations). Each cell has its own probability. Thanks for your help! Ben Bolker wrote: ACroske Audy3272 at yahoo.com writes: I have a large matrix full of probabilities; I would like to convert each probability to a 1 or a 0 using rbinom. How can I do this on the entire matrix? The matrix was converted from a raster ArcMap dataset, so the matrix is essentially a map. Because of this, I have no column headings. Thanks! How about matrix(rbinom(length(m),prob=m,size=1),nrow=nrow(m)) or (perhaps marginally more efficiently?) y - (runif(m)m) storage.mode(y) - double Ben Bolker __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- View this message in context: http://www.nabble.com/rbinom-for-a-matrix-tp18366867p18369250.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] rbinom for a matrix
Yes I do want a random assignment, instead of rounding. (From what I understand of the rbinom command, it will randomly assign 1 or 0, and the higher the given probability, the higher the likelihood of a 1... Feel free to correct me if I'm wrong!) Ben Bolker wrote: -BEGIN PGP SIGNED MESSAGE- Hash: SHA1 Dylan Beaudette wrote: | On Wednesday 09 July 2008, Ben Bolker wrote: | ACroske Audy3272 at yahoo.com writes: | I have a large matrix full of probabilities; I would like to convert each | probability to a 1 or a 0 using rbinom. | How can I do this on the entire matrix? The matrix was converted from a | raster ArcMap dataset, so the matrix is essentially a map. Because of | this, I have no column headings. | Thanks! | How about | | matrix(rbinom(length(m),prob=m,size=1),nrow=nrow(m)) | | or (perhaps marginally more efficiently?) | | y - (runif(m)m) | storage.mode(y) - double | | Ben Bolker | | | Wait a second. Are you trying to convert each probability into the most | likely '1' or '0' through rounding? The code example above will give you a | different answer every time you run it. Is that what you are looking for? | | Just curious, | | Dylan | ~ I assumed that since the original poster said convert each probability to a 1 or 0 using rbinom that they did indeed want a random assignment, rather than rounding ... -BEGIN PGP SIGNATURE- Version: GnuPG v1.4.6 (GNU/Linux) Comment: Using GnuPG with Mozilla - http://enigmail.mozdev.org iD8DBQFIdRiZc5UpGjwzenMRAuLkAKCA6ZhjrVsg5RJGYFGwq+6vz2pWxgCfetYk 9V/ZPGi2jfVGV5jx+LGPs8s= =d8c9 -END 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. -- View this message in context: http://www.nabble.com/rbinom-for-a-matrix-tp18366867p18370010.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] rbinom not using probability of success right
Message: 24 Date: Wed, 28 May 2008 05:53:26 -0700 (PDT) From: Philip Twumasi-Ankrah [EMAIL PROTECTED] Subject: [R] rbinom not using probability of success right To: r-help@r-project.org Message-ID: [EMAIL PROTECTED] Content-Type: text/plain I am trying to simulate a series of ones and zeros (1 or 0) and I am using rbinom but realizing that the number of successes expected is not accurate. Any advice out there. This is the example: N-500 status-rbinom(N, 1, prob = 0.15) count-sum(status) 15 percent of 500 should be 75 but what I obtain from the count variable is 77 that gives the probability of success to be 0.154. Not very good. Is there another way beyond using sample and rep together? I understand you correctly you want there to be exactly 75 ones. If this is what you are trying to do then using pseudorandom variables is the incorrect way of going about it. Your suggestion of sample(c(rep (0,545),rep(1,75))) seems to me to be the best way of going about it since conceptually this is what you are doing: taking permutation of a fixed set of numbers. Best, Kyle __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] rbinom not using probability of success right
I am trying to simulate a series of ones and zeros (1 or 0) and I am using rbinom but realizing that the number of successes expected is not accurate. Any advice out there. This is the example: N-500 status-rbinom(N, 1, prob = 0.15) count-sum(status) 15 percent of 500 should be 75 but what I obtain from the count variable is 77 that gives the probability of success to be 0.154. Not very good. Is there another way beyond using sample and rep together? A Smile costs Nothing But Rewards Everything Happiness is not perfected until it is shared -Jane Porter [[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] rbinom not using probability of success right
You asked for each of 500 to be included with probability 0.15, not for 15% of 500. If you want the latter, use sample, e.g. sample(c(rep(1,75), rep(0,425))) And to see if your 77 is reasonable for binomial sampling: binom.test(77, 500, 0.15) Exact binomial test data: 77 and 500 number of successes = 77, number of trials = 500, p-value = 0.8022 alternative hypothesis: true probability of success is not equal to 0.15 95 percent confidence interval: 0.1234860 0.1886725 sample estimates: probability of success 0.154 so it certainly is. On Wed, 28 May 2008, Philip Twumasi-Ankrah wrote: I am trying to simulate a series of ones and zeros (1 or 0) and I am using rbinom but realizing that the number of successes expected is not accurate. Any advice out there. This is the example: N-500 status-rbinom(N, 1, prob = 0.15) count-sum(status) 15 percent of 500 should be 75 but what I obtain from the count variable is 77 that gives the probability of success to be 0.154. Not very good. Is there another way beyond using sample and rep together? [...] -- Brian D. Ripley, [EMAIL PROTECTED] 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] rbinom not using probability of success right
On 28-May-08 12:53:26, Philip Twumasi-Ankrah wrote: I am trying to simulate a series of ones and zeros (1 or 0) and I am using rbinom but realizing that the number of successes expected is not accurate. Any advice out there. This is the example: N-500 status-rbinom(N, 1, prob = 0.15) count-sum(status) 15 percent of 500 should be 75 but what I obtain from the count variable is 77 that gives the probability of success to be 0.154. Not very good. The difference (77 - 75 =2) is well within the likely sampling variation when 500 values are sampled independently with P(1)=0.15: The standard deviation of the resulting number of 1s is sqrt(500*0.15*0.85) = 7.98, so the difference of 2 is only 1/4 of a standard deviation, hence very likely to be equalled or exceeded. Your chance of getting exactly 75 by this method is quite small: dbinom(75,500,0.15) [1] 0.04990852 and your chance of being 2 or more off your target is 1 - sum(dbinom((74:76),500,0.15)) [1] 0.8510483 Is there another way beyond using sample and rep together? It looks as though you are seeking to obtain exactly 75 1s, randomly situated, the rest being 0s, so in effect you do need to do something on the lines of sample and rep. Hence, something like status - rep(0,500) status[sample((1:500),75,replace=FALSE)] - 1 Hoping this helps, Ted. E-Mail: (Ted Harding) [EMAIL PROTECTED] Fax-to-email: +44 (0)870 094 0861 Date: 28-May-08 Time: 14:19:24 -- XFMail -- __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] rbinom not using probability of success right
Do it again. What did you get this time? Then do it another time. Do you see what is happening? Charles Annis, P.E. [EMAIL PROTECTED] phone: 561-352-9699 eFax: 614-455-3265 http://www.StatisticalEngineering.com -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Philip Twumasi-Ankrah Sent: Wednesday, May 28, 2008 8:53 AM To: r-help@r-project.org Subject: [R] rbinom not using probability of success right I am trying to simulate a series of ones and zeros (1 or 0) and I am using rbinom but realizing that the number of successes expected is not accurate. Any advice out there. This is the example: N-500 status-rbinom(N, 1, prob = 0.15) count-sum(status) 15 percent of 500 should be 75 but what I obtain from the count variable is 77 that gives the probability of success to be 0.154. Not very good. Is there another way beyond using sample and rep together? A Smile costs Nothing But Rewards Everything Happiness is not perfected until it is shared -Jane Porter [[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] rbinom : Does randomness preclude precision?
Teds reply is a bit comforting and as indicated in my post, I am resorting to using sample but as an academic issue, does randomness preclude precision? Randomness should be in the sequence of zeros and ones and how they are simulated at each iteration of the process but not in the eventual nature of the distribution. I mean if I simulated a Normal (0, 1) and got a Normal(1.5, 2) these would be very different distributions. It is the same with simulating a Binomial(1, p=0.15) and getting Binomial(1, 0.154) [EMAIL PROTECTED] wrote: On 28-May-08 12:53:26, Philip Twumasi-Ankrah wrote: I am trying to simulate a series of ones and zeros (1 or 0) and I am using rbinom but realizing that the number of successes expected is not accurate. Any advice out there. This is the example: N-500 status-rbinom(N, 1, prob = 0.15) count-sum(status) 15 percent of 500 should be 75 but what I obtain from the count variable is 77 that gives the probability of success to be 0.154. Not very good. The difference (77 - 75 =2) is well within the likely sampling variation when 500 values are sampled independently with P(1)=0.15: The standard deviation of the resulting number of 1s is sqrt(500*0.15*0.85) = 7.98, so the difference of 2 is only 1/4 of a standard deviation, hence very likely to be equalled or exceeded. Your chance of getting exactly 75 by this method is quite small: dbinom(75,500,0.15) [1] 0.04990852 and your chance of being 2 or more off your target is 1 - sum(dbinom((74:76),500,0.15)) [1] 0.8510483 Is there another way beyond using sample and rep together? It looks as though you are seeking to obtain exactly 75 1s, randomly situated, the rest being 0s, so in effect you do need to do something on the lines of sample and rep. Hence, something like status - rep(0,500) status[sample((1:500),75,replace=FALSE)] - 1 Hoping this helps, Ted. E-Mail: (Ted Harding) Fax-to-email: +44 (0)870 094 0861 Date: 28-May-08 Time: 14:19:24 -- XFMail -- A Smile costs Nothing But Rewards Everything Happiness is not perfected until it is shared -Jane Porter [[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] rbinom : Does randomness preclude precision?
What do you mean by ... *eventual* nature of the distribution? If you simulated 100 samples, would you expect to see 1.5 successes? Or 1? Or 2? How many, in your thinking, is eventual? Charles Annis, P.E. [EMAIL PROTECTED] phone: 561-352-9699 eFax: 614-455-3265 http://www.StatisticalEngineering.com -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Philip Twumasi-Ankrah Sent: Wednesday, May 28, 2008 9:52 AM To: [EMAIL PROTECTED] Cc: r-help@r-project.org Subject: Re: [R] rbinom : Does randomness preclude precision? Teds reply is a bit comforting and as indicated in my post, I am resorting to using sample but as an academic issue, does randomness preclude precision? Randomness should be in the sequence of zeros and ones and how they are simulated at each iteration of the process but not in the eventual nature of the distribution. I mean if I simulated a Normal (0, 1) and got a Normal(1.5, 2) these would be very different distributions. It is the same with simulating a Binomial(1, p=0.15) and getting Binomial(1, 0.154) [EMAIL PROTECTED] wrote: On 28-May-08 12:53:26, Philip Twumasi-Ankrah wrote: I am trying to simulate a series of ones and zeros (1 or 0) and I am using rbinom but realizing that the number of successes expected is not accurate. Any advice out there. This is the example: N-500 status-rbinom(N, 1, prob = 0.15) count-sum(status) 15 percent of 500 should be 75 but what I obtain from the count variable is 77 that gives the probability of success to be 0.154. Not very good. The difference (77 - 75 =2) is well within the likely sampling variation when 500 values are sampled independently with P(1)=0.15: The standard deviation of the resulting number of 1s is sqrt(500*0.15*0.85) = 7.98, so the difference of 2 is only 1/4 of a standard deviation, hence very likely to be equalled or exceeded. Your chance of getting exactly 75 by this method is quite small: dbinom(75,500,0.15) [1] 0.04990852 and your chance of being 2 or more off your target is 1 - sum(dbinom((74:76),500,0.15)) [1] 0.8510483 Is there another way beyond using sample and rep together? It looks as though you are seeking to obtain exactly 75 1s, randomly situated, the rest being 0s, so in effect you do need to do something on the lines of sample and rep. Hence, something like status - rep(0,500) status[sample((1:500),75,replace=FALSE)] - 1 Hoping this helps, Ted. E-Mail: (Ted Harding) Fax-to-email: +44 (0)870 094 0861 Date: 28-May-08 Time: 14:19:24 -- XFMail -- A Smile costs Nothing But Rewards Everything Happiness is not perfected until it is shared -Jane Porter [[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] rbinom : Does randomness preclude precision?
I think I see the rub: You would like to see the distribution of a sample be identical to the distribution from which it was sampled. But if it is random then that can happen only in the long run, not on every sample. That is why samples from a normal density are *not* themselves normal - they're t. When the sample size is large enough the differences between a random sample's density and its parent density become vanishingly small. Thus the differences you observe from repeated random samples from the binomial. Repeated sampling produces slightly different numbers of successes. How could it be otherwise? Charles Annis, P.E. [EMAIL PROTECTED] phone: 561-352-9699 eFax: 614-455-3265 http://www.StatisticalEngineering.com From: Philip Twumasi-Ankrah [mailto:[EMAIL PROTECTED] Sent: Wednesday, May 28, 2008 10:36 AM To: [EMAIL PROTECTED] Subject: RE: [R] rbinom : Does randomness preclude precision? Charles, When you simulate data from a distribution, what you effect are doing is generating a sequence of values that would correspond to that distribution. So you can generate 1000 values from a normal distribution and expect that when you check on the distribution of your sample (what you do with your qqnorm or Q-Q plot), it should be a close fit with the theoretical distribution with the assigned parameter values. It will be difficult to explain why a simulated data may be different from the distribution it is was generated from . I think you can not blame it on randomness. I hope you understand what I am trying to determine. Charles Annis, P.E. [EMAIL PROTECTED] wrote: What do you mean by ... *eventual* nature of the distribution? If you simulated 100 samples, would you expect to see 1.5 successes? Or 1? Or 2? How many, in your thinking, is eventual? Charles Annis, P.E. [EMAIL PROTECTED] phone: 561-352-9699 eFax: 614-455-3265 http://www.StatisticalEngineering.com -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Philip Twumasi-Ankrah Sent: Wednesday, May 28, 2008 9:52 AM To: [EMAIL PROTECTED] Cc: r-help@r-project.org Subject: Re: [R] rbinom : Does randomness preclude precision? Teds reply is a bit comforting and as indicated in my post, I am resorting to using sample but as an academic issue, does randomness preclude precision? Randomness should be in the sequence of zeros and ones and how they are simulated at each iteration of the process but not in the eventual nature of the distribution. I mean if I simulated a Normal (0, 1) and got a Normal(1.5, 2) these would be very different distributions. It is the same with simulating a Binomial(1, p=0.15) and getting Binomial(1, 0.154) [EMAIL PROTECTED] wrote: On 28-May-08 12:53:26, Philip Twumasi-Ankrah wrote: I am trying to simulate a series of ones and zeros (1 or 0) and I am using rbinom but realizing that the number of successes expected is not accurate. Any advice out there. This is the example: N-500 status-rbinom(N, 1, prob = 0.15) count-sum(status) 15 percent of 500 should be 75 but what I obtain from the count variable is 77 that gives the probability of success to be 0.154. Not very good. The difference (77 - 75 =2) is well within the likely sampling variation when 500 values are sampled independently with P(1)=0.15: The standard deviation of the resulting number of 1s is sqrt(500*0.15*0.85) = 7.98, so the difference of 2 is only 1/4 of a standard deviation, hence very likely to be equalled or exceeded. Your chance of getting exactly 75 by this method is quite small: dbinom(75,500,0.15) [1] 0.04990852 and your chance of being 2 or more off your target is 1 - sum(dbinom((74:76),500,0.15)) [1] 0.8510483 Is there another way beyond using sample and rep together? It looks as though you are seeking to obtain exactly 75 1s, randomly situated, the rest being 0s, so in effect you do need to do something on the lines of sample and rep. Hence, something like status - rep(0,500) status[sample((1:500),75,replace=FALSE)] - 1 Hoping this helps, Ted. E-Mail: (Ted Harding) Fax-to-email: +44 (0)870 094 0861 Date: 28-May-08 Time: 14:19:24 -- XFMail -- A Smile costs Nothing But Rewards Everything Happiness is not perfected until it is shared -Jane Porter [[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. A Smile costs Nothing But Rewards Everything Happiness is not perfected until it is shared -Jane Porter __ R-help@r-project.org mailing
Re: [R] rbinom : Does randomness preclude precision?
-Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Philip Twumasi-Ankrah Sent: Wednesday, May 28, 2008 6:52 AM To: [EMAIL PROTECTED] Cc: r-help@r-project.org Subject: Re: [R] rbinom : Does randomness preclude precision? Teds reply is a bit comforting and as indicated in my post, I am resorting to using sample but as an academic issue, does randomness preclude precision? I would say yes, at least in the manner you seem to be thinking of it. Think about it for a minute. Say you use the following code status - sample(rep(c(0,1),c(425,75))) You will get your 75 ones randomly distributed throughout the vector status. What would you expect the value to be for sum(status[1:100]) I suggest that you will see the same kind of variation in the sum as you would see for sum(rbinom(100, 1, p=.15)) So, you can randomly arrange a certain percentage of ones in a sequence, but you should not expect to see that exact percentage in any subset of the sequence. One other example, if you flip a fair coin 100 times, do you expect to get exactly 50 heads? Hope this is helpful, Dan Daniel Nordlund Bothell, WA USA Randomness should be in the sequence of zeros and ones and how they are simulated at each iteration of the process but not in the eventual nature of the distribution. I mean if I simulated a Normal (0, 1) and got a Normal(1.5, 2) these would be very different distributions. It is the same with simulating a Binomial(1, p=0.15) and getting Binomial(1, 0.154) [EMAIL PROTECTED] wrote: On 28-May-08 12:53:26, Philip Twumasi- Ankrah wrote: I am trying to simulate a series of ones and zeros (1 or 0) and I am using rbinom but realizing that the number of successes expected is not accurate. Any advice out there. This is the example: N-500 status-rbinom(N, 1, prob = 0.15) count-sum(status) 15 percent of 500 should be 75 but what I obtain from the count variable is 77 that gives the probability of success to be 0.154. Not very good. The difference (77 - 75 =2) is well within the likely sampling variation when 500 values are sampled independently with P(1)=0.15: The standard deviation of the resulting number of 1s is sqrt(500*0.15*0.85) = 7.98, so the difference of 2 is only 1/4 of a standard deviation, hence very likely to be equalled or exceeded. Your chance of getting exactly 75 by this method is quite small: dbinom(75,500,0.15) [1] 0.04990852 and your chance of being 2 or more off your target is 1 - sum(dbinom((74:76),500,0.15)) [1] 0.8510483 Is there another way beyond using sample and rep together? It looks as though you are seeking to obtain exactly 75 1s, randomly situated, the rest being 0s, so in effect you do need to do something on the lines of sample and rep. Hence, something like status - rep(0,500) status[sample((1:500),75,replace=FALSE)] - 1 Hoping this helps, Ted. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] rbinom with computed probability
G'day Sigalit,
On Fri, 23 Nov 2007 11:25:30 +0200
sigalit mangut-leiba [EMAIL PROTECTED] wrote:
Hello,
I have a loop with probability computed from a logistic model like
this: for (i in 1:300){
p[i]-exp(-0.834+0.002*x[i]+0.023*z[i])/(1+exp(-0.834+0.002*x[i]+0.023
+z[i]))
x and z generated from normal distribution.
I get 300 different probabilities And I want to generate variables
from bernulli distribution
with P for every observation:
T[i]-rbinom(1,1,p[i])
But i get missing values for T.
What I'm doing wrong?
I guess the problem is that in the numerator you have 0.023*z[i] but
0.023+z[i] in the denominator. Thus, some p[i] can be outside of
[0,1] which would produce NAs in T.
But why a for loop? This code is readily vectorised:
p - exp(-0.834+0.002*x+0.023*z)/(1+exp(-0.834+0.002*x+0.023*z))
HTH.
Cheers,
Berwin
=== Full address =
Berwin A TurlachTel.: +65 6515 4416 (secr)
Dept of Statistics and Applied Probability+65 6515 6650 (self)
Faculty of Science FAX : +65 6872 3919
National University of Singapore
6 Science Drive 2, Blk S16, Level 7 e-mail: [EMAIL PROTECTED]
Singapore 117546http://www.stat.nus.edu.sg/~statba
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[R] rbinom with computed probability
Hello,
I have a loop with probability computed from a logistic model like this:
for (i in 1:300){
p[i]-exp(-0.834+0.002*x[i]+0.023*z[i])/(1+exp(-0.834+0.002*x[i]+0.023
+z[i]))
x and z generated from normal distribution.
I get 300 different probabilities And I want to generate variables from
bernulli distribution
with P for every observation:
T[i]-rbinom(1,1,p[i])
But i get missing values for T.
What I'm doing wrong?
Thank you,
Sigalit.
[[alternative HTML version deleted]]
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] rbinom with computed probability
I don't run into problems doing this:
x=rnorm(300)
z=rnorm(300)
for (i in 1:300){
p[i]-exp(-0.834+0.002*x[i]+0.023*z[i])/(1+exp(-0.834+0.002*x[i]+0.023
+z[i])); T[i]-rbinom(1,1,p[i])
}
hth, Ingmar
On 23 Nov 2007, at 10:25, sigalit mangut-leiba wrote:
Hello,
I have a loop with probability computed from a logistic model like
this:
for (i in 1:300){
p[i]-exp(-0.834+0.002*x[i]+0.023*z[i])/(1+exp(-0.834+0.002*x[i]+0.023
+z[i]))
x and z generated from normal distribution.
I get 300 different probabilities And I want to generate variables
from
bernulli distribution
with P for every observation:
T[i]-rbinom(1,1,p[i])
But i get missing values for T.
What I'm doing wrong?
Thank you,
Sigalit.
[[alternative HTML version deleted]]
__
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PLEASE do read the posting guide http://www.R-project.org/posting-
guide.html
and provide commented, minimal, self-contained, reproducible code.
Ingmar Visser
Department of Psychology, University of Amsterdam
Roetersstraat 15
1018 WB Amsterdam
The Netherlands
t: +31-20-5256723
[[alternative HTML version deleted]]
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