similar to: how to transform m/d/yyyy to yyyymmdd?

Dear R users, I am building a hierarchical model on a large data set. It can take quite some time to finish one fit, I was just wondering whether it is possible to store the fit object (the result) to a file for later (offline) analysis. thanks Julia -- View this message in context: http://www.nabble.com/how-to-store-lme-lmer-fit-result-tp19910951p19910951.html Sent from the R help mailing

Hello, I have a data file (.csv) that has a size of about 2.6 GB. I am not able to read in the whole data set because of the memory limit. I actually only need some columns (3 columns) of the data set, is there a way to read in specified columns? I am using windows. Thanks, Julia -- View this message in context:

Hello, I have a data set that looks like this. ID value 111 5 111 6 111 2 178 7 178 3 138 3 138 8 138 7 138 6 . . . I'd like to calculate the mean and var for each object identified by the ID. I can in theory just loop through the whole thing..., but is there a easier way/command which let me calculate the mean/var by ID? Thanks, Julia -- View this

Dear R users, I need to compare two scatter plots, plot(x1, y1) plot(x2, y2) and would like to plot them in the same figure. How do I do it? Thank you. -- View this message in context: http://www.nabble.com/two-scatter-plots-in-one-tp21963375p21963375.html Sent from the R help mailing list archive at Nabble.com.

Hello, I have a vector: x <- c("A", "A", "A", "B", "A", "A", "C") I'd like to compare each of elements of vector x from its previous element (except for the 1st element which does not have previous element). So I'd like to get a vector y (of same length) that looks something like (0, 0, 0, 1, 1, 0, 1) or (F, F, F,

Dear R users: I have a data set that looks something like this: ID time y 1 2/01/2008 4 1 2/09/2008 12 1 9/01/2008 8 2 1/06/2008 3 2 3/01/2008 4 2 3/09/2008 9 2 6/03/2008 4 3 1/02/2008 3 3 1/10/2008 8 3 2/02/2008 7 3 2/10/2008 3 I'd like to sum up the weekly data to the monthly level, so that it looks something

Dear R users: I am running lmer(), and I had the following warning message: "In mer_finalize(ans): iteration limit reached without convergence (9) What is the default limit for lmer()? And how do I increase the limit? Many thanks, Julia -- View this message in context: http://www.nabble.com/How-to-increase-iteration-limit-for-lmer%28%29--tp20252974p20252974.html Sent from the R help

Hello, I have a time series plot drawn using 3 different colored lines, each line corresponds to different category group. I'd like to put legends on the plot. I am using "legend" to do this, however, I can either specify lty or col in legend. Will I be able to do the following in the legend box: a solid black line then g1, a solid red line then g2, a solid blue line then g3?

After v1.0 is released, I can finally get back to sane version numbers. But any comments on which one is better: a) Postfix-style: "1.1.UNSTABLE.YYYYMMDD" -> 1.1.0 (stable) b) Odd-even numbering: 1.1.x (unstable) -> 1.2.0 (stable) With a) style the releases could be done by simply copying a nightly snapshot to releases/ directory and announcing the changes since the last

Dear R-users, -I am new to R, and I am struggling with the following problem. -I am repeating the following operations hundreds of times, within a loop: I want to subset a data frame by columns. I am interested in the columns names that are given by the rows of another data frame that was built in parallel. The solution I have so far works well as long as the elements of the second data frame

How do I parse a date "yyyymmdd"? I tried asking chron(s, "ymd") but that didn't work. Would the date parsing routines of the Date class of 1.9 grok this? -- Ajay Shah Consultant ajayshah at mayin.org Department of Economic Affairs http://www.mayin.org/ajayshah Ministry of Finance, New Delhi

I know this: > library(date) > x="1979-04-04" > try=as.date(x, "ymd") > print(try) [1] 4Apr79 and that `x' here has to be a string, e.g.: > x=1979-04-04 > print(x) [1] 1971 I'm stuck in reading from a file. I say: > A <- read.table(file="try") > print(A) V1 V2 1 1979-04-04

Hi, I have several files with data in this format: 20070102 20070102 20070106 20070201 ... The data is sorted and each line represents a date (YYYYMMDD). I would like to analyze this data using R. For instance, I would like to have a histogram by year, month or day. I've already made a simple Perl script that aggregates this data but I believe that R can be much more powerful and easy on

Dear R-help group, I have a dataset with daily closing prices from a stock exchange (consecutive 5 trading days) from a firm trading a specific commodity. The date variable looks like: quote_date 20080411 With the format; yyyymmdd. Moreover, I have another data set with a (average) weekly price of the underlying commodity. The date variables in this dataset are only year and a week number. I

Is there any way to have rsync compare against a file with a *different* name? What I'd like to do is compare a vm - whatsit.img against a backup - whatsit.current, that's a symlink to whatsit.timestamp, and if different, *then* make the copy (which my script will then move to whatsit.newtimestamp, and change the symlink to point to the new one, then remove the old (or maybe save an older

Ferret 0.10.11 & AAF: the time seems to be stored in a format that can''t be sorted, the order doesn''t make any sense. Workaround: use to_i on the Time object before putting it into the index. -- Posted via http://www.ruby-forum.com/.

Hello, I have a simple question. I am trying to load data into a zoo object. I have the data in CSV format as follows SYMBOL DATE TIME PRICE XX YYYYMMDD HH:MM:SS n.nn and there are multiple symbols in this one data frame. My question is, do I need to merge DATE and TIME before loading them or can I specify multiple index.column or index.name fields?

I am unsure how to sort a column by date if it is currently in the form: YYYYMMDD For example the months: 20071031 20071130 20071231 Etc. Regards, Leigh [[alternative HTML version deleted]]

Hi, I am relatively new to R. Have scourged the help files and the www but havent been able to get a solution. I have around 250 csv files, one file for each date. They have columns of all types, numeric, string etc. The name of each file is the date in the form of 'yyyymmdd'. There is no column within the file which helps me identify the date on which the file was generated, only the

On Wed, Aug 12, 2015 at 03:12:21PM +0200, Pino Toscano wrote: > Add support for non-integer revisions of entries, comparing them as > integer when possible. Have you got an example of revisions? It may be possible to use Common_utils.compare_version which does RPM revision-style comparison. Rich. -- Richard Jones, Virtualization Group, Red Hat http://people.redhat.com/~rjones Read my