similar to: Problem adjusting x-labels with bargraphCI

Hi there, I wonder if anyone can help me. I'm trying to use bargraph.CI in the Sciplot package when there is a missing combination of the factor levels. Unfortunately the standard errors on the plot do not appear to be correct. Consider an analysis consisting of two factors A and B. When all factor level combinations are present all appears fine: library(sciplot) #all data

Hello list: I am using "cut" and "table" to obtain a frequency table from a numeric sample vector.  The idea is to calculate mean and standard deviation on grouped data.  However, I can't extract the midpoints of the class intervals, which seem to be strings treated as factors.  How do i extract the midpoint? Thanks, jose loreto [[alternative HTML version deleted]]

Hello, When using bargraph.CI in package sciplot can the bars for each group be different colors? How do I select the color for each group? When I use this instead of the default (SD vs SE): bargraph.CI(x.factor = dose, response = len, data = ToothGrowth, ci.fun= function(x) c(mean(x)-sd(x), mean(x) + sd(x)) ) Am I getting 95% CI bars? Thank you kindly, Michael Just

Dear R-users, I need some help concerning the plotting of dendrograms for hierarchical agglomerative clustering. The agglomeration niveau of each step should be displayed at the branches of the dendrogram. For this I need the x/y coordinates of the branch-agglomerations of the dendrogram. The y-values are known (the heights of the agglomeration), but how can I get the x-values? > mydata

Hi to all, Does anybody knows why this is giving an error? data(ToothGrowth) # Two-way design with options bargraph.CI(dose, len, group = supp, data = ToothGrowth, xlab = "Dose", ylab = "Growth", cex.lab = 1.5, x.leg = 1, col = "black", angle = 45, cex.names = 1.25, density = c(0,20), legend = TRUE) Error in dn.call[[1]] :

Hello r-help, I have a lengthy vector of data (with values anywhere from 1-200), and another index vector of 'groups' representing values 0-2, 3-5, 6-8, ... of length 67. The index vector has the structure (1, 4, 7, ... , 196, 199), where each value is the midpoint of each respective group. I'm trying to convert the data vector such that values falling into each group are changed to

I'm trying to create a bargraph of means with standard error bars using the function bargraph.CI (in the sciplot package). Like this: bargraph.CI(x.factor, response,data,xlab, ylab, par(family="serif"),font=11) However, an error message comes up when I try to use the par funtion. Does the character font/style need to be changed in a different way for this kind of plot? Or am I

Hi there, I am a beginner. I would like to change the error bars in the bargraph.CI function from the default (se) to (sd). The help file says ci.fun= function(x) c(fun(x)-se(x), fun(x)+se(x)) Is there a simple way of telling the function what (x) precisely is - I already define in in the of the bargraph.CI function and assume that is should be able to use that information. cheers, Herwig --

May I raise the issue of the behaviour of image()? In image() dim(z) is supposed to be c(length(x), length(y)). This suggests that the user is supplying the midpoints of the rectangles. For all but the outer rectangles this is true. The outer rectangles, however, are only displayed in half width. So it is not possible to show correctly an image plot of equally-spaced abscissa. I would hazard

#If I have two lists as follows a1<- array(1:6, dim=c(2,3)) a2<- array(7:12, dim=c(2,3)) l1<- list(a1,a2) a3<- array(1:4, dim=c(2,2)) a4<- array(5:8, dim=c(2,2)) l2<- list(a3,a4) #how can I create a new list with the mean across all arrays within the list, so all components are included? As an example for [[1]]; cbind((l1[[1]][,1]+l2[[1]][,1])/2,

Starting with the head of a 499 element matrix whose column names are now the labels trom a cut() operation, I needed to get to a vector of midpoints to serve as the basis for plotting a calibration curve ( exp(linear predictor) vs. : > dput(head(dimnames(mtcal)[2][[1]])) # was starting point testvec <- c("(-8.616,-3.084]", "(-3.084,-2.876]",

I have a data set that I display using barplot. I don't know what you call it, but when I look at it, it looks like this: > lsu (0,0.1] (0.1,0.2] (0.2,0.3] (0.3,0.4] (0.4,0.5] (0.5,0.6] A 0.052631579 0.000000000 0.000000000 0.000000000 0.000000000 0.000000000 B 0.000000000 0.000000000 0.001007049 0.003021148 0.000000000 0.000000000 E 0.200000000 0.000000000

Hi, Can anyone help with plotting vertical error bars on a bar graph. I have tried following examples online and in the big R book and writing my own function but I have been unsuccessful and don’t really have an understanding of what it is I am doing. I have calculated my standard errors so basically just need to draw the bars on the graph but just don’t have a clue!!! I don’t even know what

Dear R helpers, I think that this may be a bit of a math question as the more I consider it, the harder it seems. I am trying to come up with a way to work out the minimum distance between line segments. For instance, consider 20 random line segments: x1 <- runif(20) y1 <- runif(20) x2 <- runif(20) y2 <- runif(20) plot(x1, y1, type = "n") segments(x1, y1, x2, y2)

Dear R-lister I have a question about how to create multiple error bar plots. I found that I can use an "errbar" function from Hmisc package to create one error bar plot in which there are multiple data points (x, y) with the error bars. Thus, I know that I can get "one" error bar plot which consists of many data points. However, I did not find a way to put

Hello, I'd appreciate a suggestion on how to construct plots (barplots?) that use means on the Y axis instead of density/count. I'd also like to use groups and plot error or confidence interval bars on these graphs. I know this is a read the manual situation. I'd appreciate help with what to read, or your benevolence with some sample code. I've looked at lattice and gplots2, but

I have a very standard barplot. My labels are too long to be printed horizontally under each bar, so I am using text to put the labels on a 45 degree slant. However, the labels are spaced more narrowly than the bars, so on an 8 vertical bar plot, the end of the eighth label is lined up with the seventh bar. Preferably I don't want to do every text label separately (I'm having this

Greetings, A couple general questions regarding the use of splines to interpolate depth profile data. Here is an example of a set of depths, with associated attributes for a given soil profile, along with a function for calculating midpoints from a set of soil horizon boundaries: #calculate midpoints: mid <- function(x) { for( i in 1:length(x)) { if( i > 1) { a[i] = (x[i] -

Hello, I'm trying to sort my bargraph.CI plot (function like barplot in the SCIPLOT package) by the mean of the response variable. Does somebody have a trick for it. Thank you. Romain Mayor, PHD student.

Hello again I'm trying to put error bars onto a barplot. I've tried something that Bill Simpson suggested a while ago, ie: x<-c(1,2,3,4,5) y<-c(1.1, 2.3, 3.0, 3.9, 5.1) ucl<-c(1.3, 2.4, 3.5, 4.1, 5.3) lcl<-c(.9, 1.8, 2.7, 3.8, 5.0) plot(x,y, ylim=range(c(lcl,ucl))) arrows(x,ucl,x,lcl,length=.05,angle=90,code=3) #or segments(x,ucl,x,lcl) but I can't get it to work on a