similar to: re form data for aov()?

Greeting. Dear Mr/Mrs/Miss, OTU ID Health Disease Bacterial 1 0.29 0.34 Bacterial 2 0.25 0.07 Bacterial 3 0.06 0.06 Bacterial 4 0.07 0.09 Bacterial 5 0.02 0.05 Above show the first 6 data sets, may I ask that the reason of R show the error like "Error in cor(data) : 'x' must be numeric" ? And how to solve it? Besides, isn't this data can conduct correlation matrix?

Hi Chin Yi, If you are trying to correlate "Health" with "Disease", i.e. cydf<-read.table(text="OTU ID Health Disease Bacterial 1 0.29 0.34 Bacterial 2 0.25 0.07 Bacterial 3 0.06 0.06 Bacterial 4 0.07 0.09 Bacterial 5 0.02 0.05", header=TRUE) print(cor(cydf$Health,cydf$Disease)) [1] 0.7103517 If you are getting that error, it probably means that

On 2017-07-05 11:56, Jim Lemon wrote: > Hi Chin Yi, > If you are trying to correlate "Health" with "Disease", i.e. > > cydf<-read.table(text="OTU ID Health Disease > Bacterial 1 0.29 0.34 > Bacterial 2 0.25 0.07 > Bacterial 3 0.06 0.06 > Bacterial 4 0.07 0.09 > Bacterial 5 0.02 0.05", > header=TRUE) >

Dear useRs, I'm trying to plot spray quantities against dates, and label the points on the plot. Basically quite simple, but sometimes two chemicals have been used and are listed in separate rows in the table as below; then the labels are written on top of each other. > spray SprayDate PD Trt Qwater Qai 1 2005-03-09 Spidermites Pegasus 1300 1.04 2

Full_Name: Karel Zvara Version: 1.8.1 OS: MS Winows 2000 Submission from: (NULL) (195.113.30.163) The test statistics of the fligner.test (ctest package) depends on the order of cases: > fligner.test(count~spray,data=InsectSprays) Fligner-Killeen test for homogeneity of variances data: count by spray Fligner-Killeen:med chi-squared = 14.4828, df = 5, p-value = 0.01282 >

Hey, Here is a snippet that generated a boxplot and separates points so that they do not overlap. I have a problem to avoir text overlapping. Any help would be helpful. > attach(InsectSprays) boxplot(count ~ spray, data = InsectSprays, outpch = NA) stripchart(count ~ spray, data = InsectSprays, vertical = TRUE, method = "jitter", pch = 21, col =

Full_Name: Bob O'Hara Version: 2.6.0 OS: Windows XP Submission from: (NULL) (88.112.20.250) Using horizontal=TRUE with boxplot() confuses it as to what is an x- or y-axis. At least, xlim= and ylim= are the wrong way round, log="x" (or "y") and xaxt= work as expected, I haven't looked at anything else. Some code to see if you can reproduce the bug (or discover

I am using R 2.2.1 on a Windows 2000 PC. When I do a grid() after the boxplot it overprints the boxplot: > boxplot(count ~ spray, data = InsectSprays, col = "lightgray") > grid(nx=NA, ny=NULL) > if I try the panel.first > boxplot(count ~ spray, data = InsectSprays, col = "lightgray", + panel.first=grid(nx=NA, ny=NULL)) > I can see the grid flash

Hi everybody! I recently tried out Wine, and decided to write an article about it. Well, that means I need to get some interesting windows apps working, so far it's only Notepad and Solitaire (won't make anyone cheer). The problem is, when I try to start any installer.exe file, I get some rows of nonsense ("fixme pthread...blabla") and then it exits with the message that it

I'm trying to select a subset of a dataframe while dropping some factors. While the dataset gets smaller all Factor levels remain and I need to get rid of them. Strangely enough, I am almost certain that the same code on the same data worked OK earlier today - and it is not the first time that I'm not able to replicate earlier results with this command (I know, I might just be going

I'm still having trouble getting to grips with time classes. I wish to calculate the difference in days between events. Browse[1]> insp.j$First [1] "2002-02-19 13:00:00 NZDT" Browse[1]> spray.j$Date [1] "2001-11-29 13:00:00 NZDT" Browse[1]> insp.jk - spray.j$Date Time difference of 82 days If I save insp.jk to a vector, I get a nice useful value of 82.

the data is : >table.8.3<-data.frame(expand.grid( marijuana=factor(c("Yes","No"),levels=c("No","Yes")), cigarette=factor(c("Yes","No"),levels=c("No","Yes")), alcohol=factor(c("Yes","No"),levels=c("No","Yes"))), count=c(911,538,44,456,3,43,2,279))

I am pleased to announce the newest release of Locomotive! http://locomotive.sourceforge.net Mostly a bug fix release. If you are using Locomotive on 10.4 you should definitely upgrade. New in 0.2.4 ============ * Rename application entries in status window * Run/Restart/Stop buttons now immediately update * Bundle choosing is fixed for Mac OS X 10.4 * Missing gem dependencies in Rails 0.13.1

Dear all - With the following code I get Boxplots and Stripcharts in one Plot: with(InsectSprays, boxplot(count ~ spray, boxwex = 0.3)) with(InsectSprays, stripchart(count ~ spray, col = "red", vertical = TRUE, add = TRUE)) But the dots from the stripchart are plotted over the Boxes. Is there any possibility to have Stripchart and Boxplots side-by-side, i.e. to move boxplots and/or

Dear all, I am interested in correctly testing effects of continuous environmental variables and ordered factors on bacterial abundance. Bacterial abundance is derived from counts and expressed as percentage. My problem is that the abundance data contain many zero values: Bacteria <-

Hello everyone! I downloaded CodeWeavers wine the other week and had it up and running in the instant of a sec. So, I mounted my Cd and installed StarCraft with no problems (except all this jibbrisch about missing fonts). Happy as a newborn daddy I started it up, and noticed that since I couldn't fill in a userid for playing, I couldn't play. it turned out that all my keyboard pressing

Hello! I have a suggestion that I want to present before the final release. One thing I've missed in mp3/ID3 is a field that could tell either in what language is the song/clip performed, or otherwise from what country is the artist. What do you think of that? It would be a possibility to supercede mp3. :) Regards Bo _________________________________________s_p_r_a_y_ Här börjar Internet!

Dear All I am trying to include sampling weights in multilavel regression analysis using packege lme4 using following codes print(fm1 &lt;- lmer(DC~sex+age+smoker+alcohol+fruits(1|setting), dataset,REML = FALSE), corr = FALSE) print(fm2 &lt;- lmer(DC~sex+age+smoker+alcohol+fruits(1|setting), dataset,REML = FALSE), corr = FALSE,weights=sweight) The problem is both the

This is probably very simple but I'm new to R so apologies for being stupid. I have some data with No coded as 0 and yes coded as 1. e.g. id sex alcohol smoker 1 M 0 1 2 F 1 0 3 M 0 0 I realise I can covert the numerical variable back to a factor by falcohol<-factor(alcohol,levels=0:1) levels<-c("No","Yes")

Hi, I have a factor and I would like to find the most frequent level. I think my current approach is a bit long winded and I was wondering if there was a more elegant way to do it: x <- factor(sample(1:0, 5,replace=TRUE)) levels(x)[ which( as.logical((table(x) == max(table(x)))) == TRUE ) ] (The length of x will always be an odd number, so I wont get a tie in max()) Thanks,