search for: datums

Hello R users I recently learned how to use this command: lapply(datum,"[[","ColumnName") Unfortunately, I don't know how exactly it works, what it's called (in particular the "[[" part], and what other things you can do with it (retrieve multiple columns?). Given datum is a list of dataframes with the same column, but different number of rows,

If you redefine your NAs as below to be detected as some arbitrary large number, then the code should work through. Any 5's left in your dataset can be replaced just as easily by NAs again. Not elegant, but effective. site <- c("s1", "s1", "s1", "s2","s2", "s2") pref <- c(1, 2, 3, 1, 2, 3) R1 <- c(NA, NA, 1, NA,NA,NA) R2

Anna Crepes: Traubenzucker + Feldsalat spezielles Dressing (bringt selbst mit?) -------- Weitergeleitete Nachricht -------- Betreff: Fwd: Fwd: Fwd: Fwd: Fwd: Fwd: Fwd: 26. Datum: Thu, 11 Dec 2014 15:34:39 +0100 Von: Markus <universe at truemetal.org> An: universe at truemetal.org Geschenke Moritz: dunkle Schokolade. Geschenke Anna: normale Schokolade. -------- Weitergeleitete

Hello, I have data like the following: datum <- structure(list(Y = c(415.5, 3847.83333325, 1942.833333325, 1215.22222233333, 950.142857325, 2399.5833335, 804.75, 579.5, 841.708333325, 494.053571425 ), X = c(1.081818182, 0.492727273, 0.756363636, 0.896363636, 1.518181818, 0.499166667, 1.354545455, 1.61, 1.706363636, 1.063636364 )), .Names = c("Y", "X"), row.names = c(NA,

Dear all I try to make hourly average by cut() function, which almost works as *I* expected. What puzled me is that if there is only one item at the end of your data it results in NA. Example will explain what I mean datum<-seq(ISOdate(2004,8,31), ISOdate(2004,9,1), "min") cut(datum[1370:1381],"hour", labels=F) [1] 1 1 1 1 1 1 1 1 1 1 1 NA

Hello list! I have a data.frame which looks like this: > serv datum op.read op.write read write 1 2011-01-29 10:00:00 0 0 0 0 2 2011-01-29 10:00:01 0 0 0 0 3 2011-01-29 10:00:02 0 0 0 0 4 2011-01-29 10:00:03 0 4 0 647168 5 2011-01-29 10:00:04 0 0 0 0 6 2011-01-29

Dear All, is there a simple way to print a data.frame without its row.names? example: datum <- as.Date(c("2004-01-01", "2004-01-06", "2004-04-12")) content <- c('Neujahr', 'Hl 3 K.', 'Ostern') df1 <- data.frame(datum, content) print(df1) datum content 1 2004-01-01 Neujahr 2 2004-01-06 Hl 3 K. 3 2004-04-12 Ostern Can I get

# I am sure that I could be more efficient than this but how? Thanks in advance. #GPS in Decimal Degrees in the form longitude latitude RM215 <- matrix(c(-82.1461363, 33.5959109), nrow=1) SC <- matrix(c(-82.025888, 33.606454), nrow=1) RM202 <- matrix(c(-81.9906723, 33.5027653), nrow=1) RM198 <- matrix(c(-81.926823, 33.4634678), nrow=1) HC <- matrix(c(-81.920505, 33.46192), nrow=1)

Dear all, I am trying to project my LongLat-maps to a plane. The ultimate purpose is to do a search of points in vicinity of other points using overlay-commands (sp) with radius in km. I am applying spTransform (package rgdal) and it gives my some curious results. An example. Let's take a point lying somewhere in Germany, zone=32U x <- 8.968735 y <- 49.454735 After conversion I

I have a data frame ("res$data" in the example below), and I'd like to walk through it, doing an approx() to interpolate each element in turn based on the value of the item named "pressure" that is in the data.frame. I've put some code below. I had hoped that the commented-out "length() <-" line would let me adjust the lengths of the items, but that has

Hello, again I'm on my weblog-script... having problems... This code: =========================== weblog <- read_weblog("web.log") weblog_by_date <- split(weblog, weblog$date) #for ( i in names(weblog_by_day) ) { print(i); print(weblog_by_day$i) } for ( datum in names(weblog_by_date) ) { print(datum) selected <- weblog_by_date[[datum]] res_size_by_host <-

Readers, I have been reading the r book (Crawley) and tried to use the influence measures function for linear regression, as described. I have one datum that I wish to show in the graph but exclude from the regression and ab line. x y 0 5 10 9 20 10 30 19 40 4 With the influence measures function I plot the graph but linear regression includes the datum (40,4), which I want to show on the graph

Readers, I have data with different time stamps that I wish to plot (for example): data set 1 time(hh:mm:ss),datum 01:00:00,500 01:00:15,600 01:00:30,750 01:00:45,720 01:01:00,700 01:01:15,725 01:01:30,640 01:01:45,710 data set 2 time,datum 01:00:12,20 01:01:01,55 01:01:55,22 The time interval in data set 1 does not change, but the time interval in data set 2 does change, such that for a

Dear useRs, I'm struggling again with date-related stuff: I am using R to draw water levels at certain measuring stations. My data comes as a tab-delimited text file and looks like this: DATUM P1 P2 ... 2006-11-16 425.21 423.99 2006-12-15 425.12 423.97 2007-01-16 425.16 424.06 ... (measurements started in July 2004 and still continue on a monthly or bi-weekly basis) This

OMG.. how embarassing.. that was my personal reminder E-Mail for x-mas dinner. Not meant for this list. Please ignore. Shame on me.. *blushing* LOL. Am 12.12.2014 um 21:19 schrieb Markus: > Anna Crepes: Traubenzucker > + Feldsalat spezielles Dressing (bringt selbst mit?) > > > > -------- Weitergeleitete Nachricht -------- > Betreff: Fwd: Fwd: Fwd: Fwd: Fwd: Fwd: Fwd: 26.

How to do the attachment in mail like the yahoo and gmail does? --~--~---------~--~----~------------~-------~--~----~ You received this message because you are subscribed to the Google Groups "Ruby on Rails: Talk" group. To post to this group, send email to rubyonrails-talk-/JYPxA39Uh5TLH3MbocFFw@public.gmane.org To unsubscribe from this group, send email to

Hey, I am trying to create barplot of abundances over time (in days). The period is over 171 days, so I don't want to have all labels there but only the first day of the month. I couldn't find anything like this on the forum yet. Mostrly it's about year to year data. This is the relevant part of my table: datum month abundance 26/03/11 March 1 27/03/11 March 0 28/03/11 March 1

Since the getelementptrs were implicitly generated by the CreateStore/Load I'm not sure how to get access to them. So I hacked the assignment to be done thrice: once using a manual decomposition into two GEPs and stores, once using the "big" CreateStore, once via the setGlobal function, printing addresses and memory contents at each point to the degree that I have access to them.

Hi, I am using in my workflow gIntersection from sp package. Part of my relevant sessionInfo is: R version 4.0.3 (2020-10-10) Platform: x86_64-w64-mingw32/x64 (64-bit) Running under: Windows 10 x64 (build 17763) attached base packages: [1] stats graphics grDevices utils datasets methods base other attached packages: [1] CircStats_0.2-6 boot_1.3-25 MASS_7.3-53 stringr_1.4.0

Hello, all. I am working on a project with a large (~350Mb, about 5800 rows) insurance claims dataset. It was supplied in a tilde(~)-delimited format. I imported it into a data frame in R by setting memory.limit to maximum (4Gb) for my computer and using read.table. The resulting data frame had 10 bad rows. The errors appear due to read.table missing delimiter characters, with multiple data