similar to: ggplot: dodge positions

Displaying 20 results from an estimated 2000 matches similar to: "ggplot: dodge positions"

2012 Jul 03
2
ggplot2: legend
Dear all, I produced the following graph with ggplot which is almost fine, yet I don't like that the legend for "Means" and "Observations" includes a line, though no line is used in the plot for those two (the line for "Overall Mean" on the other hand is wanted): library(ggplot2) ddf <- data.frame(x = factor(rep(LETTERS[1:2], 5)), y = rnorm(10)) p <-
2012 Oct 31
1
aggregate.formula: formula from string
Dear all, I want to use aggregate.formula to conveniently summarize a data.frame. I have quiet some variables in the data.frame and thus I don't want to write all these names by hand, but instead create them on the fly. This approach has the advantage that if there will be even more columns in the data.frame I don't have to change the code. I've hence tried to construct a formula
2012 Dec 10
1
Sweep out control
Dear all, Assume that I have the following data structure: d <- expand.grid(subj=1:5, time=1:3, treatment=LETTERS[1:3]) d$value <- 10 ^ (as.numeric(d$treatment) + 1) + 10 * d$subj + d$time d$value2 <- 100000 + d$value where d$treatment == "C" stands for my control group. What I want to achieve now is to subtract the values corresponding to d$treatment == "C" from
2011 Jan 05
1
Minimum of an ordered factor
Hi everybody, Is there a particular reason, why this code does not work as intended: z <- factor(LETTERS[1:3], ordered = TRUE) u <- 4:6 min(z[u > 4]) Error in Summary.factor(2:3, na.rm = FALSE) : min not meaningful for factors I agree that min is indeed not meaningful for not ordered factors, but it makes sense for ordered factors. Especially since z[3] <
2010 Dec 13
1
predict.lm[e] with formula passed as a variable
Dear all, In a function I paste a string and convert it to a formula which I pass to lm[e]. The idea is to write a function which takes the name of the response variable and the explanatory variable and the data frame as an argument and calculates an lm[e]. (see example below) This works fine, but if I want to make a prediction on this model, R complains that the object holding the formula
2010 Aug 30
1
lattice: limits in reversed order with relation="same"
Hi everybody, I want an x-axis which has xlim=c(max, min) rather than xlim=c(min, max) in order to reflect the type of the process (cooling): library(lattice) myprepanel <- function(x,y,...) list(xlim=rev(range(x))) x <- rep(1:10, 100) z <- factor(sample(10, 1000, T)) y <- rnorm(1000, x, as.numeric(z)) xyplot(y~x|z, scales=list(x="free"), prepanel=myprepanel) This works
2011 Oct 21
1
droplevels: drops contrasts as well
Dear all, Today I figured out that there is a neat function called droplevels, which, well, drops unused levels in a data frame. I tried the function with some of my data sets and it turned out that not only the unused levels were dropped but also the contrasts I set via "C". I had a look into the code, and this behaviour arises from the fact that droplevels uses simply factor to drop
2010 May 04
1
Lazy evaluation in function call
Hi everybody, how is it possible to refer to an argument passed to a function in the function call? What I like to do, is something like f <- function(x,y) x+y f(2, x) # should give 4 The problem is of course that x is only known inside the function. Of course I could specify something like f(z<-2,z) but I'm just curious whether it is possible to use a fancy combination of eval,
2010 Aug 23
2
Sum a list of tables
Hi all, In R it is possible to sum tables: > (a <- table(rep(1:3, sample(10,3)))) 1 2 3 2 5 7 > a+a 1 2 3 4 10 14 Now suppose that I have a list of tables, where each table counts the same things > k <- list(a,a,a) How can I sum all tables in k? > do.call(sum, k) [1] 42 does not work since it sums over each table. > do.call(`+`, list(a,a)) 1 2 3 4 10
2010 Jul 20
1
apply: return list of matrices
Hi everybody, Suppose we have the following data structure: ddf <-data.frame(a=rep(1:4,3), b=rep(paste("p", 1:3, sep=""), each=4), c=c(1,0,0,1,1,1,0,1,1,1,1,1)) I want now to make a contingency table for each pair of values of p, i.e. a contingency table for each of the pairs (p1,p2), (p1,p3) and (p2,p3). The result should be given as a list of matrices: # [[1]]
2010 Jun 21
2
list() assigning the same value to two items
Hi everybody, I'd like to have a list with two elements, where both elements have the same value: z <- list(a=1, b=1) If it happens, that I've to change the value, I've to assure that I change both. This is error prone. Hence, a better way to achieve this is to define: tmp <- 1 z <- list(a=tmp, b=tmp) Now, I'm wondering if it is possible to make this more compact: z
2011 Jan 25
2
ggplot geom_boxplot and stat_smooth
Dear all I would like to superpose some smoothing line through boxplot in ggplot > dput(ad) structure(list(konc.f = structure(c(1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 6L, 6L, 6L, 7L, 7L, 7L, 7L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 6L, 6L, 6L, 7L, 7L, 7L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 4L, 4L, 4L,
2002 Sep 23
2
R crash with internet2.dll
Hi, I'm using: platform i386-pc-mingw32 arch i386 os mingw32 system i386, mingw32 status major 1 minor 5.1 year 2002 month 06 day 17 language R and I would like to apply: > update.packages() trying URL
2013 Feb 03
1
ggplot2 plotting errorbars.
Hi, i'm using this lines of code: dodge <-position_dodge(width=0.9) ggplot(dfm,aes(x = X,y = value)) + geom_bar(aes(fill = variable), position=dodge, stat="identity") + geom_errorbar(aes(ymin=value-er, ymax=value+er),width=0.25, position=dodge,stat="identity") to plot this data frame X variable value er 1 A X4 58.74 9.44 2 B X4
2010 Mar 12
1
simple plot in ggplot2, wrong error bars
I was wondering if anyone could help me with this, simple problem. I am essentially following the example on Hadley's webpage (http://had.co.nz/ggplot2/geom_errorbar.html), but it still doesn't make any sense to me. df <- data.frame(trt = factor(c("intact", "intact", "removed", "removed")), coon = c(0.093, 0.06, 0.057, 0.09), group =
2011 May 27
2
help with barplot
Hi, I'm really struggling with barplot I have a data.frame with 3 columns. The first column represents an "incident" type The second column represents a "month" The third column represents a "time" Code for a sample data.frame incidents <- rep(c('a','b','d','e'), each =25) months <- rep(c(1,2), each =10) times
2011 Dec 21
3
black and white in qplot? layout 4 graphs in one screen
Hello, I am trying to plot means and standard errors conditioned by a factor, using qplot. I am successful at getting the bar graph I want with a error bar, however I have tried many things and cannot get the bars to change colors. Currently showing as red and blue, but need it to be black and white for publication. Any suggestions please? Using a data set June, which is str:
2002 Nov 07
4
Preferable contrasts?
Dear all, I'm working with Cox-regression, because data could be censored. But in this particular case not. Now I have a simple example: PRO and PRE are (0,1) coded. The response is not normal distributed. We are interested in a model which could describe interaction. But my results are depending strongly in the choose of the contrast option. It is clear that there is some dependence in
2009 Jul 06
3
Speed up code, profiling, optimization, lapply vs. loops
High everybody, currently I'm writinig a package that, for a given family of variance functions depending on a parameter theta, say, computes the extended quasi likelihood (eql) function for different values of theta. The computation involves a couple of calls of the 'glm' routine. What I'm doing now is to call 'lapply' for a list of theta values and a function, that
2013 Nov 04
1
ggplot2: Add '+' operator for aes (uneval) objects
Dear all, Is there a reason, why there is no +-operator for aes (i.e. uneval) objects (as there is for themes and gg objects)? I had a couple of cases where such an operator would be useful, for instance to combine the result of aes and aes_string in functions. Any flaws with the following proposition: `+.uneval` <- function(e1, e2) { dup <- names(e1) %in% names(e2) if (any(dup)) {