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R Error In Stripchart.default

What types are the plot commands you call expecting? –Glen_b♦ Jul 17 '14 at 10:19 @Glen_b, the variables are all coming as "factor" when I cheek their class type. up vote 2 down vote favorite I have a simple, single data file test.txt which contains: 1 5 7 9 11 I want to plot this file with index numbers. Here is how to extract those columns from the data x=mydata['avg_ms'] y=mydata['MB.s'] Now plot > plot(x,y) Error in stripchart.default(x1, ...) : invalid plotting method huh ... If we look at x and/or y, they are actually columns from mydata and plot() wants rows (actually vectors but we'll get there). > x    avg_ms 1    0.02 2    0.04 have a peek here

What to do with my pre-teen daughter who has been out of control since a severe accident? Also scale <- data[1] creates an element from a data.frame data[1] 1 5 2 10 3 12 4 15 whereas scale from the read.table is a vector 5 10 12 15 Let's look at the original data and then the transformed data x=mydata['avg_ms']     #  column of data tx=t(mydata['avg_ms']) #  transform the column of data into a row Look at the datatypes I am using R, so things I can implement in R would be relevant - and guidance on how to do so would be especially useful.

Looks like I have ALOT of reading to do. –B.Miller Jul 17 '14 at 10:33 It sounds like you think a stripchart does something other than what it actually share|improve this answer edited May 18 '09 at 9:44 answered May 18 '09 at 9:25 luapyad 3,1811519 So what should his plot() call look like? –anon May 18 '09 Reload to refresh your session. Thanks for the help! –Chris Bunch May 18 '09 at 14:57 add a comment| up vote 7 down vote In your example, plot(scale, serial) won't work because scale and serial are

  • The apply() takes 3 args input variable 1=row,2=col,1:2=both function to apply see http://nsaunders.wordpress.com/2010/08/20/a-brief-introduction-to-apply-in-r/ > ra=apply(row,2,as.numeric) > class(ra) [1] "numeric" > ra  IOsize threads  avg_ms    MB.s   0.000  64.000   0.200   0.025 The
  • Below is an illustration of some of your options.
  • Not the answer you're looking for?
  • They don't mean line of data, they mean line on the plot.
  • By adding a "," with no value before it, we are giving a wild card to the row identifier and saying give me all the values for all rows in the
  • Following advice in luapyad's answer, I came up with this.

mydata <- seq(1,5) # generate some data sq <- seq(1,5) plot(sq, mydata) # Happy (two vectors) x <- data.frame(sq) # Put x into data.frame plot(x, mydata) # Unhappy (one data.frame, one Donc plot(x,y) renvoie le message : Code : Tout sélectionnerErreur dans stripchart.default(x1, ...) : méthode graphique incorrecteQue faire ? I had thought it'd place time along the x axis and then there wouldnt be a Y axis as such rather (based on the x axis showing time) it'd show me I am wanting to make plots to show what is happening over time .

r plot share|improve this question edited Jul 23 '09 at 13:09 theycallmemorty 6,13583864 asked May 18 '09 at 8:16 Chris Bunch 43.1k28101118 add a comment| 7 Answers 7 active oldest votes Put it in the verbose mode (-v) to see the corresponding R script. I'd be grateful for any suggestions as to how I should go about this. > I think what you are doing wrong is misinterpreting what people mean when they say "line". what's that Error?

Flom, PhD Statistical Consultant www DOT peterflomconsulting DOT com ______________________________________________ [hidden email] mailing list https://stat.ethz.ch/mailman/listinfo/r-helpPLEASE do read the posting guide http://www.R-project.org/posting-guide.htmland provide commented, minimal, self-contained, reproducible code. How to explain leaving a job for a huge ethical/moral issue to a potential employer - without REALLY explaining it Why don't cameras offer more than 3 colour channels? (Or do share|improve this answer answered Jul 17 '14 at 10:30 Glen_b♦ 151k19248517 ahh right . First pick what to plot.

Converting columns into vectors When originally selecting out the columns of the data, we could have selected out vectors directly instead of selecting a column and transforming the column to a What to do with my pre-teen daughter who has been out of control since a severe accident? The scale variable will be set from the read.table, and it should not interfere with the "global" scale() function. –luapyad May 18 '09 at 9:50 Yup, that did it. Haut Afficher les messages postés depuis : Tous1 jour7 jours2 semaines1 mois3 mois6 mois1 an Trier par AuteurDateSujet CroissantDécroissant Répondre 4 messages • Page 1 sur 1 Retourner vers « Questions en cours »

Peter Flom-2 Threaded Open this post in threaded view ♦ ♦ | Report Content as Inappropriate ♦ ♦ Re: Problems producing a simple plot Steve Murray <[hidden email]> wrote > The brackets take the equivalent of x and y coordinates or row and column position. I'm trying this (from reading this site): data <- read.table("foo.csv", header=T,sep=",") attach(data) scale <- data[1] serial <- data[2] plot(scale,serial) But I get this error back: Error in stripchart.default(x1, ...) : invalid To make it more intuitive add the argument       "byrow=TRUE," and add a       "NULL" for the rowname position in the row and columns name section m=matrix(c(      0 ,      1

more hot questions question feed default about us tour help blog chat data legal privacy policy work here advertising info mobile contact us feedback Technology Life / Arts Culture / Recreation Note, I used the ylim parameter in plot to accommodate the range in the third column. My original data is from a CSV file that I have exported from another program. –B.Miller Jul 17 '14 at 21:39 Jittering the points would be much more informative Haut Nicolas Péru Messages : 1405 Enregistré le : 07 Aoû 2006, 08:13 Citation Messagepar Nicolas Péru » 12 Sep 2008, 08:56 et lire les réponses qui sont faites sur d'autres post comportant

I renamed the header "scale": scaling, serial, spawn, for, worker 5, 0.000178, 0.000288, 0.000292, 0.000300 10, 0.156986, 0.297926, 0.064509, 0.066297 12, 2.658998, 6.059502, 0.912733, 0.923606 15, 188.023411, 719.463264, 164.111459, 161.687982 then: all other columns in one plot, easy and with some elegance :) for printing or presentation, you may use Prof. The system returned: (22) Invalid argument The remote host or network may be down.

Get 2 lines yanked or 1 line yanked confirmation more hot questions question feed lang-r about us tour help blog chat data legal privacy policy work here advertising info mobile contact

I'd be grateful for any suggestions as to how I should go about this. Instead, an error message is shown: Error in stripchart.default(x1, ...) : invalid plotting method Can you point out what I'm doing wrong, or suggest a better solution? Join them; it only takes a minute: Sign up Plotting Simple Data in R up vote 18 down vote favorite 9 I have a comma separated file named foo.csv containing the First, plot all dots of a variable on a line > stripchart(Duncan$income, method="stack") #stack means the dots of the same value are stacked #how about on the "type" column? > stripchart(Duncan$type,

Many thanks, Steve Previous message: [R] Trouble installing packages on Macintosh OS-X 10.5.6 Next message: [R] Problems producing a simple plot Messages sorted by: [ date ] [ thread ] [ My assumption was wrong though. –B.Miller Jul 17 '14 at 10:24 I've edited the question to come closer to what I was trying to suggest in chat; I suspect share|improve this answer answered Jun 14 '09 at 17:36 zzr 167312 add a comment| up vote 2 down vote I am far from being an R expert, but I think you Counterintuitive polarizing filters How neutrons interact if not through an electromagnetic interaction?

Print some JSON Would it be ok to eat rice using spoon in front of Westerners? asked 1 year ago viewed 4341 times active 1 year ago Blog Stack Overflow Podcast #92 - The Guerilla Guide to Interviewing Related 317Plot two graphs in same plot in R76Shading it show me time along x axis and at what the corresponding Name value was at that time. Next number in sequence, understand the 1st mistake to avoid the 2nd Using multiple custom meta data keyword Criteria in a single query as LIKE operators Is there an adverb meaning