similar to: Error: no 'dimnames' attribute for array

Hi All, I have a scatter plot produced using ggplot2 and I want to add the regression line to this scatter plot. I suppose I can use geom_smooth() to do this, but for the sake of learning ( I am new both to R and ggplot2), I want to try and add it as a function (something that curve() does in the standard R plotting). I did some search and found that stat_function() can be used for this. But

Hi, Try this: #mat1 is the data res<-do.call(cbind,lapply(seq_len(nrow(mat1)),function(i) {new1<-do.call(rbind,lapply(seq_len(nrow(mat1[-i,])),function(j) {x1<-rbind(mat1[i,],mat1[j,]); x2<-(abs(x1[1,1]-x1[2,1])*abs(x1[1,5]-x1[2,5]))+(abs(x1[1,2]-x1[2,2])*abs(x1[1,6]-x1[2,6]))+(abs(x1[1,3]-x1[2,3])*abs(x1[1,7]-x1[2,7]))+(abs(x1[1,4]-x1[2,4])*abs(x1[1,8]-x1[2,8]))}));new1}))

Hi, I'm trying to understand some details about an example maintened in [1]. According that link, I have total.data as a data set (am I right?). But I don't understand how is built that table. I saved the dataset in a file, with dput(), and had something like this: structure(list(df.all = structure(list(V1 = structure(c(1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L, 11L, 12L, 13L, 14L, 15L,

Hi R-Help, data at bottom I've been struggling with a problem where I need to order based on 1) the Frequency "Freq" and 2) keeping each group of 3 of the same type together "Var2" but I want across all groups it to go "high to low" based on the earn factor. Thank you! structure(list(Var1 = structure(c(1L, 3L, 2L, 1L, 2L, 3L, 1L, 3L, 2L, 3L, 1L, 2L, 3L, 1L,

what am I doing wrong? chron(as.character(f), format=c(dates="%m/%d/%y", times="%h:%m")) f <- structure(c(51L, 60L, 66L, 87L, 90L, 115L, 23L, 35L, 37L, 6L, 12L, 55L, 84L, 96L, 109L, 17L, 29L, 41L, 3L, 74L, 94L, 102L, 30L, 8L, 46L, 69L, 107L, 15L, 25L, 39L, 1L, 71L, 95L, 19L, 56L, 62L, 76L, 85L, 99L, 111L, 42L, 4L, 52L, 61L, 67L, 91L, 13L, 24L, 36L, 38L, 7L, 81L, 82L, 57L,

Dear xpeRts, I would like to ask a question about plotting in R. I have four columns in which first column contains x-coordinate and second column contains y-coordinate of certain points, whereas the third column "z" contains classification number of those points from 1 to 4, it should be noted that this column doesn't quantify anything, but just classify the points. More precisely

Hello All... I gave a task to my students that involved using mclust to look for clusters in some bivariate data of isotopes vs various mining locations. They discovered something I didn?t expect; the data (called tur) is appended below. p <- qplot(x = dD, y = dCu65, data = tur, color = mine) print(p) # simple bivariate plot of the data; looks fine mod1 <- Mclust(tur[,2:3]) mod1$G mod2

Hi all: I am trying to use the ddply function to estimate the mean of 'Total','Fry','Smolt' and 'Fry.Eq' columns without success. I have the dput of my dataset below. I wonder if someone can give me a hand with this function. # dput(winter) winter <-structure(list(IDDate = structure(c(37L, 48L, 59L, 62L, 63L, 64L, 65L, 66L, 67L, 38L, 39L, 40L, 41L, 42L, 43L,

It is very wierd... Can some of you confirm the following behavior ? It is a new bug (feature ?) which was not yet in 0.49 ... noquote <- function(obj) { ## constructor for a useful "minor" class if(!inherits(obj,"noquote")) class(obj) <- c(class(obj),"noquote") obj } "[.noquote" <- function (x, subs) structure(unclass(x)[subs], class =

Hi all, I would like to ask you for help. I did a dotplot - using dotchart function. There are two localites (loc) with values 75 or 56 in my data ZZ. The f column has 4 levels: P1, S1, S8, R6. The dataframe is ordered by N value, pchloc value is assign to use "pch" in plot. > head(ZZ) loc f N color ordered pchloc 98 75 S1 6.39 green 1 16 99 75 S8 6.44 blue

The source to the noquote() function looks like this: noquote <- function(obj, right = FALSE) { ## constructor for a useful "minor" class if(!inherits(obj,"noquote")) class(obj) <- c(attr(obj, "class"), if(right) c(right = "noquote") else "noquote") obj } Notice what happens with right =

On Fri, 2017-01-06 at 12:54 -0500, m.roth at 5-cent.us wrote: > James B. Byrne wrote: > > > > On Thu, January 5, 2017 17:23, Always Learning wrote: > >> > >> Cyber attacks are gradually replacing armed conflicts. > > > > Better fight with bits than blood. > > Yes, but... attacks on the friggin' IoT could result in lots of blood. Or, >

I currently have a program that automates 2-way ANOVA on a series of endpoints, but before the ANOVA is carried out I want the code to test the assumptions of normality and equal variance and report along with each anova result in the output file.  How can I do this? I have pasted below the code that I currently use.   library(car) numFiles = x #

Dear R-xperts I have just written a little hypergeometric function, included below [the hypergeometric function crops up when solving a common type of ODE]. It works fine on single values of the primary argument z, but vectorizing it is getting confusing. The best I have come up with so far just tests for z being longer than 1 and if so, uses sapply() recursively. This is fine, except that it

Hello, I'm having difficulty passing an object name to a lapply function. Can somebody tell me the trick to make this work? #Works T13702 <- TRACKDATA[["13702.xls"]][["data"]] min(unlist(lapply(list(T13702), function(x) mdy.date(x[1, 2], x[1, 1], x[1, 3])))) 16553 #Works d<-2 assign(paste("T",substr(names(TRACKDATA)[d],1,(nchar(names(TRACKDATA)[d]

Dear R users, A have a dataframe (matrix) with two collumns (plot, and diameter (d)). I want all diameters values for different combination of plots. For example I want all d values for all posible combination, 100C2 (all d values for plot 1 with all d values in the plot 2.......with all d values from plot 1 with all d values from plot 100, ...... with all d values from plot 99 with all d values

I haven't been able to get anywhere tracking this down. It seems that c.noquote() does something strange with its third (and subsequent) parameters: R-2.7.0 under NetBSD, R-2.6.0 under Solaris, and R-2.8.0 (unstable) (2008-06-10 r45893) under WinXP, I get: > c(noquote('z'), 'y', 'x', '*') [1] z y x * x * > or: > c(noquote('z'),

G'day all, I am trying to use a string as an argument in a selection but things are not working as I expect, seems the selection is not seeing the expanded string and I do not know how to make it. Perhaps the noquote class value that is returned is the problem. Here is an example. > selection #this is my string [1] "attackprogress$Se=='Toona ciliata [19825: JMM35]'"

Hi R, When I use the below to write the text file try=data.frame(rep("a",5), rep("b",5)) write.table(try,"z:\\try.txt",row.names=F,col.names=F,sep="\t") the output contains two columns with quotes! Is there a way to write without quotes? I tried try[,1]=noquote(try[,1]) try[,2]=noquote(try[,2]) Thank you, Regards, Ravi Shankar

Hello, It's an alternative to use SAS by function in R? I want to plot d histograms by plot.from example bellow: Thank you! plot d 1 1 16.3 2 1 25.0 3 1 57.8 4 1 17.0 5 2 10.8 13 2 96.4 17 3 76.0 18 3 32.0 19 3 11.0 20 3 11.0 24 3 106.0 25 3 12.5 21 4 19.3 22 4 12.0 26 4 15.0 27 5 99.3 32 7 11.0 36