Re: [R] how can I read a xlsx file

Hi Albert, Another option would be the following: install.packages('gdata') require(gdata) ?read.xlsx HTH, Jorge On Mon, Aug 15, 2011 at 11:19 AM, albert coster wrote: Hello, How can I read a xlsx file using xlsx package? Thanks Albert [[alternative HTML version deleted]]

Re: [R] Smooth ecdf

Hi Jeff, Take a look at ?density HTH, Jorge On Mon, Aug 15, 2011 at 2:38 PM, Jeffrey Joh wrote: Is it possible to smooth an ecdf plot and get a probability density plot? I have about 8000 points and I was hoping to get a density curve instead of a histogram. Jeff

Re: [R] How do I subset a dataframe

Hi eric, See R ?%in% and try the following (untested): subset(zeespan, !customer %in% c(ibm , exxon , sears) ) HTH, Jorge On Sat, Aug 13, 2011 at 7:44 PM, eric wrote: I have a dataframe zeespan. One of the columns has the name customer. The data in the customer column is text. I would

Re: [R] conditional filter resulting in 2 new dataframes

Hi, Try ifelse(initial 5, initial, 0) ifelse(initial = 5, initial, 0) and take a look at ?ifelse HTH, Jorge On Sat, Aug 13, 2011 at 9:02 PM, andrewjt wrote: This is what I am starting with: initial- matrix(c(1,5,4,8,4,4,8,6,4,2,7,5,4,5,3,2,4,6), nrow=6,

Re: [R] Not sure how to use aggregate, colSums, by

Hi eric, Try lapply(with(x, split(x, e2)), function(l){ r - with(l, aggregate(list(y, f), list(e1), sum)) colnames(r) - c('e1', 'y', 'f') r }) HTH, Jorge On Sun, Aug 14, 2011 at 1:20 PM, eric wrote: I have a data frame called test shown below that i would like to summarize in a

Re: [R] Correlation Matrix - p value?

?cor.test cor.test(x, y, method = spearman)$p.value HTH, Jorge On Tue, Aug 9, 2011 at 8:44 AM, ScottM wrote: Hello all, I've run a Spearman's Rank test to discern relationships between landscape characteristics and a specific aspect of river behaviour. I've executed a correlation matrix

Re: [R] Printing data frame with million rows

?write.table HTH, Jorge On Sun, Aug 7, 2011 at 2:08 PM, Bansal, Vikas wrote: Dear all, I was working on number of files and at the end I got a data frame with approx. million rows.To prin this data frame in output, I used capture.output(print.data.frame(end,row.names=F), file = summary,

Re: [R] Gamma distribution parameter estimation

Hi Alex, Try require(MASS) Loading required package: MASS b - c(2039L, 2088L, 5966L, 2353L, 1966L, 2312L, 3305L, 2013L, 3376L, + 3363L, 3567L, 4798L, 2032L, 1699L, 3001L, 2329L, 3944L, 2568L, + 1699L, 4545L) fitdistr(b, 'gamma') shape rate 6.4528939045 0.0021887943

Re: [R] summing columns with NAs present

Try test$Sum - rowSums(test, na.rm = TRUE) test HTH, Jorge On Fri, Aug 5, 2011 at 2:01 PM, Dimitri Liakhovitski wrote: Hello! I have a data frame with some NAs. test-data.frame(a=c(1,2,NA),b=c(10,NA,20)) I need to sum up values in 2 variables. However: test$a+test$b procudes NAs in

Re: [R] SNP Tables

Hi Jim, Here is one way: # data x - structure(list(category = structure(c(1L, 1L, 1L, 2L, 2L, 2L, 1L, 1L, 2L), .Label = c(case, control), class = factor), SNP1 = c(1L, 0L, 1L, 0L, 1L, 0L, 1L, 0L, 1L), SNP2 = c(0L, 1L, 2L, 1L, 2L, 0L, 0L, 1L, 0L), SNP3 = c(2L, 1L, 2L, 0L, 1L, 0L, 2L,

Re: [R] Deleting rows and store the deleted rows in new data frame

Or just subset(df, V5 = 10) See ?subset. HTH, Jorge On Sun, Jul 24, 2011 at 10:01 PM, Bansal, Vikas wrote: Dear Jeff, Thanks a lot for your reply. I was just curious about this thing about grep that it can perform this kind of thing or not. Otherwise with numerical comparison I know it

Re: [R] DROP OBSEVATION IN A GROUP

Hi Peter, Try this: r - with(z, tapply(y, n, function(x) sum(x == 0) 2)) z[!rep(r, with(z, table(n))), ] HTH, Jorge On Thu, Jun 30, 2011 at 1:05 AM, Peter Maclean wrote: I tried this but did not work: z0- by(z, z[,n], function(x) subset(x, sum(n==0)2)) Peter Maclean Department of

Re: [R] Saving fExtremes estimates and k-block return level with confidence intervals.

Hi Peter, Try data.frame(n = names(res2), t(sapply(res2, function(l) l@fit$par.ests))) for the first part. HTH, Jorge On Thu, Jun 30, 2011 at 12:16 AM, Peter Maclean wrote: I am estimating a large model by groups. How do you save the results and returns the associated quantiles? For

Re: [R] 2d rndom walk

Hi Komal, Try this: walk2d-function(n){ rw - matrix(0, ncol = 2, nrow = n) # generate the indices to set the deltas indx - cbind(seq(n), sample(c(1, 2), n, TRUE)) # now set the values rw[indx] - sample(c(-1, 1), n, TRUE) # cumsum the columns rw[,1] - cumsum(rw[, 1]) rw[,2] - cumsum(rw[, 2])

Re: [R] Simple simulations

Hi robcinm, You might also consider: # data x - c(rep(0, 20), 1:37) # number of simulations B - 1000 # result: TRUE/FALSE out - replicate(B, { y - sample(x, 3, replace = FALSE) all(y == 0) }) mean(out) HTH, Jorge On Mon, Jun 27, 2011 at 5:08 PM, robcinm wrote: I am

Re: [R] i want to unsubscribe

Check R-help Subscribers at https://stat.ethz.ch/mailman/listinfo/r-help HTH, Jorge On Sun, Jun 26, 2011 at 2:17 AM, elisheva corn wrote: how do i unsubscribe [[alternative HTML version deleted]] __ R-help@r-project.org mailing list

Re: [R] Fw:

Hi Ungku, Check ?persp ?volcano in the R console. HTH, Jorge On Sun, Jun 26, 2011 at 9:22 PM, Ungku Akashah wrote: - Forwarded Message - From: Ungku Akashah To: r-help@r-project.org Sent: Friday, June 24, 2011 3:15 PM Subject: hello. I need some help about this R

Re: [R] ddply to count frequency of combinations

Hi Idris, I do not know what the correct use of ddply would be, but here is another option using paste() and table(); data.frame(with(df, table(paste((, x, ,, y, ), sep = HTH, Jorge On Tue, Jun 21, 2011 at 2:30 PM, Idris Raja wrote: I have a dataframe df with two columns x and y. I

Re: [R] omitting columns from a data frame

Hi Erin, One option woild be subset(), especially the select parameter. HTH, Jorge On Mon, Jun 20, 2011 at 11:45 PM, Erin Hodgess wrote: Dear R People: I have a data frame, xm1, which has 12 rows and 4 columns. If I put is xm1[,-4], I get all rows, and columns 1 - 3, which is as it

Re: [R] Likelihood ratio test

Hi Diviya, Take a look at the lrtest function in the lmtest package: install.packages('lmtest) require(lmtest) ?lrtest HTH, Jorge On Sun, Jun 12, 2011 at 1:16 PM, Diviya Smith wrote: Hello there, I want to perform a likelihood ratio test to check if a single exponential or a sum of 2

Re: [R] Linear model - coefficients

Hi Robert, Try this: reg2 - lm( Y ~ factor(x1) + factor(x2) + factor(x3) + factor(x4) + factor(x5) - 1, data = X ) cof(ref2) HTH, Jorge On Sun, Jun 12, 2011 at 4:40 PM, Robert Ruser wrote: Prof. Ripley, thank you very much for the answer but wanted to get something else. There is an

Re: [R] Counting the Number of Letters in a row

Hi Abraham, Try foo - function(x){ x - as.character(x) sapply(strsplit(x, ), function(s) sum(nchar(s))) } foo(f1$keyword) HTH, Jorge On Fri, Jun 10, 2011 at 12:48 PM, Abraham Mathew wrote: I'm trying to find the total number of letters in a row of a data frame. Let's say I have the

Re: [R] How to subset based on column name that is a number ?

Hi Mauricio, Try the following: # using the iris data require(MASS) x - iris x[x[, 5] == 'setosa',] # and your data dframe[dframe[, 1] == FY11_Q4, ] HTH, Jorge On Thu, Jun 9, 2011 at 3:34 PM, Mauricio Cornejo wrote: Hi, I have a data frame with column names 1, 2, 3, ... and I'd like to

Re: [R] Problem in R documentation

Hi Siddharth, adf.test() is part of the tseries package, so you need to download and install it before using that function. Try the following and let us now what you get: install.packages('tseries') require(tseries) ?adf.test HTH, Jorge On Mon, Jun 6, 2011 at 2:41 AM, siddharth arun wrote:

Re: [R] How to convert a factor column into a numeric one?

Dr. LaBudde, Perhaps as.numeric(as.character(x)) is what you are looking for. HTH, Jorge On Sun, Jun 5, 2011 at 12:31 AM, Robert A. LaBudde wrote: I have a data frame: head(df) Time Temp Conc ReplLog10 10 -20H1 6.406547 22 -20H1 5.738683 37 -20

Re: [R] Transforming a data matrix into a vector

Hi Charles, Try rep(c(tmp), each = 3) HTH, Jorge On Mon, May 30, 2011 at 4:22 PM, Charles Ellis wrote: Hi, I am trying to transform a data matrix into a vector and have not be able to accomplish want I am looking for. The setup is as follows. I start with a 3 x 3 matrix: 5 1 3 3 3

Re: [R] Simulation from discrete uniform

Hi Serdar, Take a look at the following: sample(0:9, 100, replace = FALSE) Error in sample(0:9, 100, replace = FALSE) : cannot take a sample larger than the population when 'replace = FALSE' sample(0:9, 100, replace = TRUE) [1] 5 6 5 7 3 0 8 4 8 2 2 4 7 6 0 7 0 0 0 7 5 6 3 6 0 9 6 1 2 6 9

Re: [R] how to compute the inverse percentile of a given observation w.r.t. a reference distribution

Hi Rudi, Take a look at ?ecdf HTH, Jorge On Wed, May 25, 2011 at 3:42 PM, rudi wrote: Hi, can anyone help me to figure out how to compute the percentile of an individual observation with respect to a reference distribution. What I mean is. Let's assume I have a vector consisting of 10

Re: [R] convert binary vector to time series of sum(x)/minute

Hi mac, Try N - 6000 x - sample(1:0, N, TRUE) tapply(x, rep(1:(N/60), each = 60), sum) HTH, Jorge On Sat, May 21, 2011 at 1:59 PM, andyjmac wrote: Dear members, I apologize for the relatively simple request, but I couldn't find exactly what I was looking for. I have a binary vector

Re: [R] Separating boot results

Hi Patrick, How about this (untested)? a - codboot[c(4)] round(a$bca[4, 5], 2) HTH, Jorge On Thu, May 19, 2011 at 7:20 AM, Patrick Santoso wrote: Good Morning, I'm having what I hope to be a simple problem. I am generating bootstrap confidence intervals using package (boot) - which

Re: [R] How to get rid of the index in result file

Hi Yighua, Try m - 2.35343 m [1] 2.35343 cat(m) 2.35343 HTH, Jorge On Thu, May 19, 2011 at 2:35 PM, Hu, Yinghua wrote: Hi, I am running some function in Ubuntu command line and get some problem. I used some command like below $ R --slave -vanilla my_infile my_outfile The return

Re: [R] Separating boot results

tweak the parameters? I've never worked with that command. Pat On Thu, May 19, 2011 at 2:12 PM, Jorge Ivan Velez wrote: Hi Patrick, How about this (untested)? a - codboot[c(4)] round(a$bca[4, 5], 2) HTH, Jorge On Thu, May 19, 2011 at 7:20 AM, Patrick Santoso wrote: Good Morning

Re: [R] Adding a numeric to all values in the dataframe

Hi, Does the following work for you? set.seed(123) d - data.frame(x = rpois(10, 4), y = rnorm(10)) d d + 2 See ?within for one more option. HTH, Jorge On Thu, May 19, 2011 at 11:54 PM, Ramya wrote: Hi there I just want to add 2 to all the values in dataframe. I tried using sapply but

Re: [R] Finding row in matrix with minimum column

Hi Worik, See ?which.min x - matrix(c(7, 7, 9, 7, 7, 9, 2, 9), ncol = 2, byrow = FALSE) which.min(x[,2]) x[which.min(x[,2]), ] HTH, Jorge On Wed, May 18, 2011 at 10:38 PM, Worik R wrote: Friends If I have a matrix such as... [,1] [,2] [1,]77 [2,]79 [3,]9 2

Re: [R] subsetting a list of dataframes

Hi Lara, You might try the following (untested): yourlistofdataframes[sapply(yourlistofdataframes, function(d) nrow(d) 1)] HTH, Jorge On Tue, May 17, 2011 at 4:24 PM, Lara Poplarski wrote: Hello All, I have a list of dataframes, and I need to subset it by keeping only those dataframes

Re: [R] Simulating correlations with varying sample sizes

Hi Holger, Replace N by N[i]. HTH, Jorge On Mon, May 16, 2011 at 9:42 AM, Holger Steinmetz wrote: Hi there, I would like to draw 10 correlations from a bivariate population - but every draw should be done with a different sample size. I thought I could to this with a loop:

Re: [R] Quick question: Omitting rows and cols with certain percents of missing values

Hi Vickie, You might try the following: # some data set.seed(123) X - matrix(rnorm(1000), ncol = 20) X[sample(1000, 100)] - NA # excluding rows with NA 20% X[!rowMeans(is.na(X)) 0.2, ] # excluding columns with NA 10% X[, !colMeans(is.na(X)) 0.1] See ?is.na, ?rowMeans and ?colMeans for more

Re: [R] Generating a best fit line for non linear data

Hi, Try d - data.frame(samples, species) fit = nls(species ~ a *(1 - exp(-b*samples)), start = list(a = 27, b = .15), data = d) summary(fit) Formula: species ~ a * (1 - exp(-b * samples)) Parameters: Estimate Std. Error t value Pr(|t|) a 35.824723.02073 11.860 6.10e-10 *** b 0.07168

Re: [R] Question on list object

Hi Christofer, You might try sapply(listObj, function(l) l[1:max(sapply(listObj, length))] ) HTH, Jorge On Wed, Apr 27, 2011 at 1:23 PM, Bogaso Christofer wrote: Dear all, let say, I have following list object: listObj - vector(list, length = 3) listObj[[1]] - rnorm(3) listObj[[2]]

Re: [R] MASS fitdistr with plyr or data.table?

Hi Justin, One way of doing it is using a combination of tapply() and sapply() as follows: # data set.seed(144) weib.dist-rweibull(1,shape=3,scale=8) weib.test.too-data.frame(cbind(1:10,weib.dist)) names(weib.test.too)-c('site','wind_speed') # results require(MASS) out - with(weib.test.too,

Re: [R] Tell the difference between characters

Hi Lisa, Is this what you have in mind? temp - c(aa, aA, ab) temp == temp[1] [1] TRUE FALSE FALSE HTH, Jorge On Tue, Apr 26, 2011 at 2:09 PM, Lisa wrote: Dear all, I just want to determine if the characters in a character string are the same or not. For example, temp - c(aa, aA,

Re: [R] Normality tests

Hi Bruce, One way is via apply() # some data set.seed(123) X - matrix(rnorm(100), ncol = 5) X # tests t(apply(X, 2, function(x){ sw - shapiro.test(x) c(sw$statistic, P = sw$p.value) })) See ?apply and ?str and ?shapiro.test for more information. HTH, Jorge

Re: [R] indexing list elements with lapply?

Dear Simon, Try any of the following: sapply(r, function(l) l[,2] / l[1, 2]) lapply(r, function(l) l[,2] / l[1, 2]) HTH, Jorge On Fri, Apr 22, 2011 at 5:52 PM, Simon Kiss wrote: Dear colleagues, I have a list that looks like what the code below produces. I need a function to go through

Re: [R] Problem having tick marks aligned when plotting three graphs on top of one another.

Hi Dr. Sorkin, One way of doing what you want is via matplot(): with(d, matplot(Slope, d[, -1], type = 'l', lty = 1)) where d is your data. HTH, Jorge On Fri, Apr 22, 2011 at 11:55 PM, John Sorkin wrote: R 2.10 Windows 7 I am trying to plot three graphs on top of each other. I need to

Re: [R] line type lty

See ?par Best, Jorge On Thu, Apr 21, 2011 at 12:22 PM, Hui Du wrote: Hi All, Does somebody know how to know the detail of the line types? For example, lty = 1, means what kind of line?, lty = 2, means what kind of line? Thanks. HXD

Re: [R] all combinations with replacement

Hi Kehl, How large are n and k in your case? Using Dimitris' approach and I got the following timings for 1000 replicates: # function based on Dimitri's reply foo - function(n, k){ r - expand.grid(rep(list(0:n), k)) subset(r, rowSums(r) == n) } # a second try foo2 - function(n,

Re: [R] Dump the source code of data frame

Hi CH, Take a look at ?dput HTH, Jorge On Wed, Apr 13, 2011 at 3:09 AM, C.H. wrote: Dear R experts, I remember a similar function existed and have been mentioned in R-help before. I tried my best to search but I really can't find it out. suppose I have an data frame like this:

Re: [R] Find characters in a matrix and return a TRUE-FALSE vector

Hi Nina, You might try sapply(yourdata, function(x) any(x == C)) See ?sapply for more details. HTH, Jorge On Wed, Apr 13, 2011 at 7:17 AM, Vitrifizierung wrote: I have the following problem: My data is a matrix of multiple columns and rows. The column I am interested in looks like

Re: [R] list to data frame

Hi Franklin, Try do.call(rbind, ineffFilesList) See ?do.call for more details. HTH, Jorge On Sun, Apr 10, 2011 at 2:01 PM, Franklin Tamborello II wrote: I need to make a data frame out of the data that I currently have in a list. This works, but is ugly:

Re: [R] Getting number of students with zeroes in long format

, Apr 6, 2011 at 10:25 PM, Jorge Ivan Velez wrote: Hi Chris, Sorry I did not see your email before ;-) Here is one option: r - with(d, tapply(sus, id_r, function(x) any(x 0))) r 111516181920212224252630 3132 FALSE TRUE FALSE FALSE

Re: [R] Selecting data from list object

Hi Santosh, One way would be sapply(d, [, 1) [1] 20110405 20110405 20110405 20110405 20110405 20110405 HTH, Jorge On Thu, Apr 7, 2011 at 2:14 AM, santosh wrote: Hello Group, Is there a simpler way to get data out of a list object? (like in data frame without using the apply functions)

Re: [R] Correlation Matrix

Hi Dmitry, You might try with(MyDataFrame[, 1:3]) is variable1, variable2 and variable3 correspond to the first three columns of your data, or with( MyDataFrame( cor( cbind( variable1, variable2, variable3) ) ) ) otherwise. HTH, Jorge On Thu, Apr 7, 2011 at 3:09 PM, Dmitry Berman wrote:

Re: [R] Correlation Matrix

I am sorry for the noise, but with(MyDataFrame[, 1:3]) should have been with(cor(MyDataFrame[, 1:3])) Best, Jorge On Thu, Apr 7, 2011 at 3:21 PM, Jorge Ivan Velez wrote: Hi Dmitry, You might try with(MyDataFrame[, 1:3]) is variable1, variable2 and variable3 correspond to the first

Re: [R] Getting number of students with zeroes in long format

Hi Chris, Is this what you have in mind? sum(with(yourdata, tapply(sus, id_r, function(x) any(x==0 [1] 13 HTH, Jorge On Wed, Apr 6, 2011 at 4:44 PM, Christopher Desjardins wrote: Hi, I have longitudinal school suspension data on students. I would like to figure out how many students

Re: [R] force output dimension of table function

Hi, You might try table(factor(s, levels = 0:5)) 0 1 2 3 4 5 0 1 1 1 0 2 HTH, Jorge On Wed, Apr 6, 2011 at 11:37 PM, fisken wrote: I have a small annoying problem. When I use the 'table' function on a simple vector it counts the number of occurences. So depending on the values of my

Re: [R] round number

Hi Alfredo, Try noquote(sprintf(%.2f, a*.2)) HTH, Jorge On Sat, Mar 26, 2011 at 2:05 PM, Alfredo Alessandrini wrote: Hi, a - 4 a*0.2 [1] 0.8 ok!! Is there a method to obtain this: a*0.2 [1] 0.80 I need to round the number also with the zero. Thanks in advance,

Re: [R] for loop assistance -

Hi Steven, One would be with(yourdataset, aggregate(x, list(lc1, id), mean)) Group.1 Group.2x 1 85 ga1 45.47261 2 95 ga1 53.38831 3 105 ga1 58.18282 4 115 ga1 63.77469 5 125 ga1 66.98222 6 85 ga2 47.55711 7 95 ga2

Re: [R] apply mean to a three-dimension data

Sorry, the above should have been mymean - function(x) Reduce(+, x)/length(x) mymean(b) [,1] [,2] [1,] 10.3 12.3 [2,] 11.3 13.3 Apologies for the noise. Best, Jorge On Thu, Mar 24, 2011 at 3:54 PM, Jorge Ivan Velez wrote: Hi Hui, Ssee ?Reduce for more

Re: [R] apply mean to a three-dimension data

Hi Hui, Ssee ?Reduce for more details. You might try something along the lines of mymean - function(x) Reduce(+, x)/length(x) add(b) [,1] [,2] [1,] 10.3 12.3 [2,] 11.3 13.3 HTH, Jorge On Thu, Mar 24, 2011 at 11:07 AM, Hui Du wrote: Hi All,

Re: [R] How create vector that sums correct responses for multiple subjects?

Hi Kevin, Try this (untested): sapply(split(pretestdata, Subject), function(l) with(l$Correct == C)) HTH, Jorge On Thu, Mar 24, 2011 at 3:24 PM, Kevin Burnham wrote: I have a data file with indicates pretest scores for a linguistics experiment. The data are in long form so for each of 33

Re: [R] tapply with specific quantile value

Hi Steven, See the prob argument under ?quantile. The following should be what you want: tapply(x, l.c.1, quantile, prob = 0.75) HTH, Jorge * * On Thu, Mar 24, 2011 at 7:18 PM, Steven Ranney wrote: All - I have an example data frame x l.c.1 43.38812035 085 47.55710661

Re: [R] error in bargraph.CI {sciplot}

Hi Barbara, Works just fine for me: require(sciplot) Loading required package: sciplot data(ToothGrowth) # Two-way design with options bargraph.CI(dose, len, group = supp, data = ToothGrowth, xlab = Dose, ylab = Growth, cex.lab = 1.5, x.leg = 1, col = black, angle = 45,

Re: [R] Urgent query about R!

Hi Rachel, You might also try apply(expand.grid(rep(list(1:6), 4)), 1, paste, collapse = , sep = ) HTH, Jorge * * On Tue, Mar 22, 2011 at 6:41 PM, Rachel Chu wrote: Hi there, I am currently working on a R programming project and got stuck. I am supposed to generate a set of

Re: [R] Replacing Period in String

Hi John, Try gsub([.],,txt) See Extended Regular Expressions in ?regex. HTH, Jorge * * On Mon, Mar 21, 2011 at 12:49 AM, Sparks, John James wrote: Dear R Users, I am working with gsub for the first time. I am trying to remove some characters from a string. I have hit the problem

Re: [R] Output a table formatted with standard deviations below means

Hi Nathan, Do not know a direct way, but the following seems to work: # data means - matrix(1:10,nrow=2) sds - matrix(seq(0.1,1,by=0.1),nrow=2) colnames(means) - colnames(sds) - c(a,b,c,d,e) # adding ( ) to the SDs sdsn - t(apply(sds, 1, function(x) paste('(', x, ')', sep = ))) # formatting

Re: [R] Multiple testing corrections on very large vector

Hi terdon, Very happy to help and know it worked. Honestly, I do not know exactly where the differences are, but it is not hard to check the sources and compare both algorithms. When doing this, you can can see that p.adjust() uses vectorization whereas mt.rawp2adjp() does not. Perhaps that is

Re: [R] Loop Through Columns to Remove Rows

Hi Sam, How about this? test[apply(test, 1, function(x) !any(x == '#DIV/0!')), ] HTH, Jorge On Wed, Mar 9, 2011 at 3:29 PM, Sam Albers wrote: Hello Venerable List, I am trying to loop (I think) an operation through a list of columns in a dataframe to remove set of #DIV/0! values. I am

Re: [R] Differences per group

Hi Rens, One way would be x$difference - do.call(c, with(x, tapply(amount, customer, function(x) c(0, diff(x) x Take a look at ?tapply and ?aggregate for more information. HTH, Jorge On Wed, Mar 9, 2011 at 10:27 AM, rens_1112 wrote: Dear all, Probably a rather stupid question, but I

Re: [R] Multiple testing corrections on very large vector

Hi terdon, Take a look at ?p.adjust and its argument n. For example, you could adjust B pv values by using p.adjust(pv[1:B], method = 'BH', n = B) Then, you can continue processing other subsets of pv and concatenate the result. Here, a for() loop might be useful. HTH, Jorge On Tue, Mar 8,

Re: [R] Multiple testing corrections on very large vector

Hi terdon, You are absolutely right. I apologize for any inconvenience my lack of coffee might have caused :-) I simulated some p-values with the length of your vector and ran the p.adjust() function on them. Here is what I got: system.time(res - p.adjust(pv, method = 'fdr')) user system

Re: [R] How to reference a package in academical paper

Hi Jan, R citation('RODBC') To cite package ‘RODBC’ in publications use: Brian Ripley and and from 1999 to Oct 2002 Michael Lapsley (2010). RODBC: ODBC Database Access. R package version 1.3-2. http://CRAN.R-project.org/package=RODBC A BibTeX entry for LaTeX users is @Manual{,

Re: [R] MCMC_glm models

Hi Nick, The following would be one way of doing what you want: # function to estimate one model foo - function(mu = 9.244655, n = 50){ X - rpois(n, mu) glm(X ~ Y, family = poisson) # note I am using family = poisson } B - 1000 # number of samples -- change accordingly Y -

Re: [R] Sorting

Hi Whitney, If I understood correctly, what you actually want is to construct a 2x2 table considering smoking and retlevel. Perhaps something along the lines of with(yourdata, table(retlevel, smoking)) could give you some insights. See ?table for more details. HTH, Jorge On Mon, Mar 7, 2011

Re: [R] Grouping data in ranges in table

Hi Jason, Something along the lines of with(Orange, table(cut(age, breaks = c(118, 664, 1004, 1372, 1582, Inf)), cut(circumference, breaks = c(30, 58, 62, 115, 145, 179, 214 should get you started. HTH, Jorge On Sat, Mar 5, 2011 at 5:38 PM, Jason Rupert wrote:

Re: [R] testing power of correlation

Hi Anna, Take a look at ?cor ?cor.test and http://www.statmethods.net/stats/power.html HTH, Jorge On Sat, Mar 5, 2011 at 3:02 PM, Anna Gretschel wrote: Dear List, does anyone know how I can test the strength of a correlation? Cheers, Anna

Re: [R] displaying label meeting condition (i.e. significant, i..e p value less than 005) in plot function

Hi Umesh, You can try something along the lines of: d - dataf[dataf$p 0.05, ] # p 0.05 with(d, plot(xvar, p, col = 'white')) with(d, text(xvar, p, name, cex = .7)) HTH, Jorge On Sat, Mar 5, 2011 at 12:29 PM, Umesh Rosyara wrote: Dear R users, Here is my problem: # example data

Re: [R] Entering table with multiple columns rows

Hi Laura, May be yo meant: diet - matrix(c(24,134,9,52,23,72,12,15), ncol = 2, byrow = TRUE) # note ncol = 2 rownames(diet) - c(none, healthy, unhealthy, dangerous) colnames(diet) - c('Yes', 'No') diet HTH, Jorge On Mon, Feb 28, 2011 at 9:17 PM, Laura Clasemann wrote: Hi, I'm having

Re: [R] substract 2 data.frames

Hi Nicolas, Try popn[!rownames(popn) %in% rownames(fish), ] HTH, Jorge On Sun, Feb 27, 2011 at 3:29 PM, Nicolas Gutierrez wrote: Hi! I have 2 data.frames: fish and popn: fish xloc yloc id birth size weight energy gonad 20 15 15 54 -60 107.9 63.0 15952.9 8.0 21 15

Re: [R] R in different OS

Hi Hui, May be sessionInfo() is what you are looking for. See ?sessionInfo as well as ?version for more details. You can run the following on your R session and see what comes up: sessionInfo() sessionInfo()$R.version$platform version$platform Then, you might use ifelse() to set up the right

Re: [R] identify an element in a column

Hi Gary, Try transform(z, y = ifelse(x == 5, y-1, y)) HTH, Jorge On Tue, Feb 22, 2011 at 12:18 PM, Hongwei Dong wrote: Hi, R users, I'm wondering if I can identify an element in a column by an element in another column. For example: x-1:10 y-11:20 z-cbind(x,y) z x y [1,] 1

Re: [R] concatenate vector after strsplit()

Hi Robert, You might try do.call(rbind, lapply(yourlist, [, 1:4)) and then write the resulting file using write.table(...). Best, Jorge On Sun, Feb 20, 2011 at 11:13 AM, Robert Baer wrote: ls is a list of character vectors created by strsplit() I want to concatenate the 1st 4 character

Re: [R] Generating uniformly distributed correlated data.

Hi Soren, Take a look at http://tolstoy.newcastle.edu.au/R/help/05/07/7741.html HTH, Jorge On Sat, Feb 19, 2011 at 9:17 PM, Søren Faurby wrote: I wish to generate a vector of uniformly distributed data with a defined correlation to another vector The only function I have been able to

Re: [R] questions about counting numbers

Hi Carrie, Try x - rle(a) rep(x$lengths, x$lengths) [1] 1 2 2 1 HTH, Jorge On Sun, Feb 6, 2011 at 8:21 PM, Carrie Li wrote: Hello R-helpers, I have a question about counting numbers. Here is a simple example. a=c(2, 3, 3,4) table(a) a 2 3 4 1 2 1 so, I can to create another

Re: [R] sum the values in a vector as a complete number

Hi AD, You might try the following: # data a - c(2,3,5) b - c(8,7) # you got this wrong ;) # option 1 foo - function(x) as.numeric(paste(x, sep = , collapse = )) # examples foo(a) # [1] 235 foo(b) # [1] 87 foo(a) + foo(b) # [1] 322 # option 2 foo2 - function(x, y) foo(x) + foo(y) # example

Re: [R] data frame column name change

Hi eric, Try colnames(x) colnames(x)[1] - 'newname' colnames(x) HTH, Jorge On Sun, Jan 16, 2011 at 11:28 PM, eric wrote: How do I change the name of one column in a data frame ? Suppose I have a data frame x with 5 columns. If the names were date, col1, col2, col3, col4 and I wanted to

Re: [R] standard errors in johansen test

Hi Walter, The paper can be found at http://cran.r-project.org/web/packages/vars/vignettes/vars.pdf It seems that you need the vars library before trying the function you mention. What happens if you do the following? install.packages(vars) require(vars) ?cajorls ?ca.jo HTH, Jorge On Wed,

Re: [R] weighed mean of a data frame row-by-row

Hi Vassilis, Try test.df$y - with(test.df, x1*w + x2*(1-w)) test.df HTH, Jorge On Thu, Jan 6, 2011 at 8:33 AM, Vassilis wrote: Dear list, This must be an easy one. I have a data frame like this one: test.df - data.frame(x1=c(2,3,5), x2=c(5, 3, 4), w=c(0.8, 0.3, 0.5)) and I want to

Re: [R] How to 'explode' a matrix

Hi Kevin, Take a look at ?kronecker HTH, Jorge On Wed, Jan 5, 2011 at 7:03 AM, Kevin Ummel wrote: Hi everyone, I'm looking for a way to 'explode' a matrix like this: matrix(1:4,2,2) [,1] [,2] [1,]13 [2,]24 into a matrix like this:

Re: [R] Plotting colour-coded points

Hi Anjan, Try something along the lines of d$bb - with(d, cut(b, c(0,9,19,29))) with(d, plot(a, id, col = bb, pch = 16, las = 1)) legend('topright', as.character(levels(d$bb)), col = 1:length(levels(d$bb)), ncol = 3, pch = 16) where 'd' is your original data.frame. HTH, Jorge On Wed, Jan 5,

Re: [R] Saving objects inside a list

Hi Eduardo, Try r - ls() result - sapply(r, get) result HTH, Jorge On Mon, Jan 3, 2011 at 12:25 PM, Eduardo de Oliveira Horta wrote: Hello there, any ideas on how to save all the objects on my workspace inside a list object? For example, say my workspace is as follows ls() [1] x y z

Re: [R] Counting number of datasets and appending them

Hi Grace, Try something along the lines of do.call(rbind, lapply(1:maxi, function(x) get(paste('data', x, sep = HTH, Jorge On Wed, Dec 29, 2010 at 5:06 PM, Li, Grace wrote: Hi there, I have a question on how to read a bunch of dataset, assign each of the dataset to a matrix in the

Re: [R] subset question

Hi Anjan, Try subset(d, gene %in% c(i1, i2, i3)) HTH, Jorge On Wed, Dec 29, 2010 at 4:55 PM, ANJAN PURKAYASTHA wrote: Hi, I'm having a problem with a step that should be pretty simple. I have a dataframe, d, with column names : gene s1 s2 s3. The column gene stores an Id; the rest of

Re: [R] Non-uniformly distributed plot

Hi Eric, You can try plot(x, y, log = 'xy') fit - lm(log(y)~log(x)) abline(fit, col = 2, lty = 2) summary(fit) par(mfrow = c(2,2)) plot(fit) HTH, Jorge On Thu, Dec 23, 2010 at 5:55 PM, Eric Hu wrote: Thanks David. I am reposting the data here. Eric Hi, I would like to plot a

Re: [R] vectorised recovery of strsplit value ??

Try sapply(strsplit(sampleIDs, _), [, 1) HTH, Jorge On Wed, Dec 22, 2010 at 4:02 PM, maddox wrote: Dear Guru's My first steps with R have ground to a halt! I have a vector of sample identifiers sampleIDs [1] D1_1 D1_2 D1_3 D1_4 D1_5 D1_6 D1_7 D1_8 [9] D1_9 D1_10

Re: [R] regression

Hi Ufuk, Using Michael's data, here is one more way of doing it: allmodels - lapply(1:nrow(x), function(row) with(x, lm(y ~ ., data = x[-row,]))) allmodels To access the information contained in the model when the first row is removed, you can do summary(allmodels[[1]]) And, if you want to

Re: [R] sample() issue

Hi Cory, Check out http://cran.r-project.org/doc/FAQ/R-FAQ.html#Why-doesn_0027t-R-think-these-numbers-are-equal_003f HTH, Jorge On Mon, Dec 20, 2010 at 2:04 PM, cory n wrote: length(sample(25000, 25000*(1-.55))) [1] 11249 25000*(1-.55) [1] 11250 length(sample(25000, 11250)) [1]

Re: [R] Time Series of Histograms

Hi Enrico, Is this close to what you want to do? http://addictedtor.free.fr/graphiques/RGraphGallery.php?graph=109 HTH, Jorge On Sun, Dec 19, 2010 at 7:03 PM, Enrico R. Crema wrote: Dear List, I have a set of distributions recorded at an equal interval of time and I would like to plot

Re: [R] check for item in vector

Hi CH, Check ?is.element ?%in% HTH, Jorge On Mon, Dec 13, 2010 at 9:48 AM, C.H. wrote: Dear R users, Suppose I have an vector like this: animal - c(Tiger,Panda) I would like to know is there any function that check for the existence of certain item in a vector. e.g.

Re: [R] break

If I understand correctly, the following should do it: lag.max2 - function(object, n = 12){ matris - matrix(NA, nrow = n) for(i in 1:n){ matris[i] - ur.df(object, lags = i, type = trend)@testreg$coefficients[i+3,4] if(matris[i]0.1) break } # output c('lag' = i, 'p' = matris[i]) } a2 -

Re: [R] String to array

Try f - function(string) as.numeric(strsplit(string, - )[[1]]) f(x) f(y) f(z) HTH, Jorge On Thu, Dec 9, 2010 at 9:24 AM, Romildo Martins wrote: Hello, how convert x in xarray (numbers)? x [1] 0 - 13 y [1] 11 - 23 z [1] 220 - 9 xarray [1] 0 13 yarray [1] 11 23 zarray

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