similar to: assessing data variation

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

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

Hi all, I am trying to filter the element of a df that start with "TF", like this: alfa = c(123221,"TF13124",41243,"TF1234",32414,"TF13124","TF14333",2134123,"TF1234") beta =

An undergraduate here, so do not hesitate to let me know if you feel that I''m heading in a wrong direction. I have a data frame containing panel data across 10 years (hence 120 months). I want to be able to run regression separately for each month (or year). The below shows how I ran the regression for each month, but I need to know how I would combine the regression results

Hi, I've been struggling most of the morning with an IF ELSE problem, and I wonder if someone might be able to sort me out. Here's what I need to do (dummy example, my data are more complicated): If type = A or B or C and status = a then count = 1 and status = b then count = 2 and status = c then count = 3 Else if type = D or E or F and status = a then count = 9

Dear R-helpers, I am helping a SAS user run some analyses in R that she cannot do in SAS and she is complaining about R's peculiar (to her!) way of recoding variables. In particular, she is wondering if there is an R package that allows this kind of SAS recoding: IF TYPE='TRUCK' and count=12 THEN VEHICLES=TRUCK+((CAR+BIKE)/2.2); Thanks for any help or suggestions you might be able

-------- Original-Nachricht -------- Datum: Wed, 27 Oct 2010 15:03:48 +0200 Von: "Sibylle St?ckli" <sibylle.stoeckli at gmx.ch> An: Rhelp <r-help at r-project.org> Betreff: as.list Dear R-users sorry, here Rcode included (attachment has been removed) I would like to read a txt file as list, to select rows and columns, and to create barplots. (1) selection of parameter

Thanks! I just figured it out (thanks to "Beyond Compare") and was coming here to post back. The boxM test doesn't work with that (now, finally working) REXP structure, but I probably now need to create a table or something and parse that structure. So much fun! :) Thanks again. - M Sent from [ProtonMail](https://protonmail.com), Swiss-based encrypted email. > --------

Dear Group: I am working with a data frame containing 316 rows of individuals with 79 variables. Each of these 79 variables have values that range between -4 to +4, and I want to subset this data frame so that in the resulting new dataframe, values of _all_ of these variables should range between -3 and +3. Let's say I have the following dataframe (it's a toy example with 4 individuals

-------- Original-Nachricht -------- Datum: Wed, 27 Oct 2010 15:09:59 +0200 Von: "Sibylle St?ckli" <sibylle.stoeckli at gmx.ch> An: Rhelp <r-help at r-project.org> Betreff: Fwd: as.list -------- Original-Nachricht -------- Datum: Wed, 27 Oct 2010 15:03:48 +0200 Von: "Sibylle St?ckli" <sibylle.stoeckli at gmx.ch> An: Rhelp <r-help at r-project.org>

All - Not sure if this is a real programming question, but here goes: I have data that looks like Lake Length Weight 1 158 45 1 179 70 1 200 125 1 202 150 1 206 145 1 209 165 1 210 140 1 215 175 1 216 152 1 220 150 1 221 165 ... where lake goes from 1 - 84 and the number of rows for each lake is variable (but > ~20). I'm trying to do two things: 1) build a simple linear model of the

Hi, How can I get the values of mean and median (not only points but values too) on the boxplot. I am using boxplot function from graphics package. Following is my data set > df [1] 5 1 1 0 0 10 38 47 2 5 0 28 5 8 81 21 12 9 1 12 2 4 22 3 > mean.val <- sapply(df,mean) > boxplot(df,las = 1,col = "light blue") > points(seq(df), mean.val,

Dear All, I've a 10 by 5 data frame like this set.seed(1001) a <- rnorm(10) b <- rnorm(10) c <- rnorm(10) d <- rnorm(10) e <- rnorm(10) A <- cbind(a,b,c,d,e) Each row is one datum. I want to resample within A[,5]. Fit the regression lines lm(A[,5] ~ A[,1] + A[,2]) and lm(A[,5] ~ A[,3] + A[,4]) to each bootstrap sample. I don't know how to write the statistics part. Can

Hi Guys, Do you any one know how to convert a long format table to an adjacency matrix used in sna? The long table looks like p1 p2 counts a b 100 a c 200 a d 100 b c 80 b d 90 b e 100 c d 100 c e 40 d e 60 and I want to convert it to an adjacency matrix which can be used in sna? Any methods will be appreciated! btw, besides sna package, is there any better package can be used in social

I would like to remove columns of a df which have too many NAs. I think that summary() should give me the information, I just don't know how to access it. Advice? _____________________________ Professor Michael Kubovy University of Virginia Department of Psychology USPS: P.O.Box 400400 Charlottesville, VA 22904-4400 Parcels: Room 102 Gilmer Hall McCormick Road

Running R 1.8.1 on a Windows machine In dotchart, I would like to shrink the labels on the tick marks (that is, the numbers) without shrinking anything else. I could not find this in either the Rhelp archives or in ?dotchart, which recmmends cex to avoid 'label overlap', but cex shrinks all the characters in the plot. Is there a way to do this? Thanks Peter Peter L. Flom, PhD

Dear list, I have a data.frame with nine categorical variables (0,1,2 and NAs) that I would like to get the number of events for each of them. I can extract this using summary() for each variable at a time with the as.factor()argument (otherwise it will get me the mean value): >summary(as.factor(mydf[,3])) 0 1 2 NA's 194 67 4 2 Trying to use apply() to get this for

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

# 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)