Does rarecurve() (vegan) accept standard error for plotting?
If so, how can I plot such a curve?
I am following a classical script for this, with the BCI dataset:
S <- specnumber(BCI)
(raremax <- min(rowSums(BCI)))
Srare <- rarefy(BCI, raremax)
plot(S, Srare, xlab = "Observed No. of Species", ylab = "Rarefied No. of Species")
abline(0, 1)
rarecurve(BCI, step = 20, sample = raremax, col = "blue", cex = 0.6)
Statistically speaking, facilitating a function as this one would be helpful to most vegan users.
Thank you!
André
rarecurve does not give you SE. The reason is obvious and already given to you: there is enough clutter without extra curves. If you really want to do this, you must do it manually. That is not too complicated, because rarefy function accepts a vector sample sizes and gives you all the numbers you need. The following draws a basic plot using one site of Barro Colorado data set:
library(vegan)
data(BCI)
sum(BCI[1,]) # site 1, 448 tree stems
N <- seq(2, 448, by=8)
S <- rarefy(BCI[1,], N, se = TRUE)
plot(N, S[1,], type="l", lwd=3)
lines(N, S[1,] + 2*S[2,]) ## 2*SE is good enough for 95% CI
lines(N, S[1,] - 2*S[2,])
Statistically speaking, this gives you only the error caused by the subsampling process assuming that the observed data have no random variation. To me this makes little sense, and I find the rarefaction SE's misleading and meaningless. That does not stop me providing them in vegan.
Related
I am having trouble with plotting a forest plot based on a multi-level model, in which I'd also like to display pooled effects of subgroups, as well as the results for subgroup differences.
So far, I have managed to produce a plot of the data where clusters are grouped together. I would like to extend this plot by adding pooled effects of subgroups at the right positions, without losing the grouping of the clusters. (As it is explained here, but also while keeping what is shown in the last example of this).
This is the code I have used so far to produce the "normal" forest plot for my model (sorry, it's pretty long):
# ma_data => my data
# main_3L => my multi-level model
# Prepare row argument for separation by study
dd <- c(0, diff(ma_data$ID))
dd[dd > 0] <- 1
rows <- (1:main_3L$k) + cumsum(dd)
par(tck=-.01, mgp = c(1.6,.2,0), cex=1)
# refactor ID var
ma_data$ID_plot <- substr(ma_data$short_cite, 1, nchar(ma_data$short_cite))
ma_data$ID_plot <- paste(sub(" ||) ","",substr(ma_data$ID_plot,0,2)), substr(ma_data$ID_plot,3,nchar(ma_data$ID_plot)), sep="")
tiff("./figures/forestFull_ext1.tiff", width=3200,height=4500, res=300)
# Plot the forest!
metafor::forest(main_3L,
addpred = TRUE, # adds prediction interval
cex=0.5,
header="Author(s) and Year",
rows=rows, # uses the vector created above
order=order(ma_data$ID, ma_data$es_adj),
ylim=c(0.5,max(rows)+3),
xlim=c(-5,3),
xlab="Hedges' G",
ilab=cbind(as.character(ma_data$setup),as.character(ma_data$target_1), as.character(ma_data$measure_type), ma_data$task, as.character(ma_data$cogdom_pooled), ma_data$sample_size_exp),
ilab.xpos=c(-3.9,-3.6,-3.3,-2.8,-2.2,-1.7),
slab=ma_data$ID_plot,
mlab = mlabfun("Overall RE Modell", main_3L, main_3L.I2)) # Adds Q,Qp, I² and sigma² values.
abline(h = rows[c(1,diff(rows)) == 2] - 1, lty="dotted")
# adds a second polygon with robust estimates for standard error
addpoly(coeftest.main_3L$beta, sei = coeftest.main_3L$SE,
rows = -2.5,
cex = 0.5,
mlab = "Robust RE Model estimate",
col = "darkred")
par(cex=0.5, font=2)
# text(c(-4,-3.7,-3.2,-2.5, -2), 150.5, pos=3, c("Target", "Measure","Task","Cognitive Domain", "N"))
text(c(-3.9,-3.6,-3.3,-2.8,-2.2,-1.7), 150.5, pos=3, c("Setup", "Target", "Measure","Task","Cognitive Domain", "N"))
dev.off()
Specifically, I need to know how to "make space" for the additional rows and polygons.
Also, is there an option in the forest() function to display only the pooled effects of subgroups and main effect, bot not the individual effect sizes? I know that it is possible in the meta package, but have not found anything similar in metafor.
Any help is greatly appreciated!
This should be easy to fix, I genuinely don't know what is wrong.
Suppose I wanted to perform the EM algorithm for the Old Faithful data in R and plot the results:
install.packages('mixtools')
library('mixtools')
test<-normalmixEM(faithful$waiting, k=2)
plot(test, which=2, xlim= c(30, 100))
lines(density(faithful$waiting), lty=2, lwd=2)
This works.
But if I wanted to change the x-label or y-axis I get an error message:
plot(test, which=2, xlim= c(30, 100), xlab="", ylim= c(0, 0.06))
lines(density(faithful$waiting), lty=2, lwd=2)
The message is:
argument 4 matches multiple formal arguments
Can someone please help me out? What am I doing wrong? I'm really puzzled.
Thanks!
From the documentation you need to follow this form:
plot(x, whichplots = 1,
loglik = 1 %in% whichplots,
density = 2 %in% whichplots,
xlab1="Iteration", ylab1="Log-Likelihood",
main1="Observed Data Log-Likelihood", col1=1, lwd1=2,
xlab2=NULL, ylab2=NULL, main2=NULL, col2=NULL,
lwd2=2, alpha = 0.05, marginal = FALSE, ...)
you'll need to use xlab2 = ...
plot() is a generic function that actually calls a more specific function (called a "method") depending on what you are trying to plot (see this chapter from Hadley Wickham's Advanced R book for details). In this case, you are feeding-in an object of class "mixEM" to plot(). You can see this by running, e.g.:
class(test)
The generic function plot() is calling the method plot.mixEM() because you are feeding in an object of type "mixEM". To see which parameters of plot.mixEM() you can control, check out that function's help page
?plot.mixEM
The helpfile makes it clear that you need xlab2 as an argument instead of xlab. However, I don't immediately see how to change ylim, so you should view the source code for plot.mixEM to see if there's a way to adjust it other components of the graph:
getAnywhere(plot.mixEM)
I tried to visualize a Power Law p(x)=x^(-2.5) with following R code. When you use an log-scale in the end you get a lot of vibrations what is okay as can be seen here
But know, and this is my Problem, I read an article where the author says I have to use a cumulative distribution function to remove this vibrations at the end. But for me it doesn't work, as can be seen here
library(ggplot2)
chol_r <- read.table("C:\\Users\\me\\Desktop\\1M_just_random_py.txt",
header = FALSE)
chol <- (chol_r)**(-2.5) #this p(x)
chol2 = (1/1.5)*chol_r**(-1.5) # the cumulative distribution function
qplot(chol2,
geom="histogram",
binwidth = 0.001, #0.001 oder 0.38
main = "Histogram",
xlab = "Numbers",
fill=I("blue"),
col=I("red"),
log="xy")
So does anybody know what I am doing wrong? Or how i can get a straight line falling without that vibrations? I really don't know what I am doing wrong
I can not figure out how the lattice levelplot works. I have played with this now for some time, but could not find reasonable solution.
Sample data:
Data <- data.frame(x=seq(0,20,1),y=runif(21,0,1))
Data.mat <- data.matrix(Data)
Plot with levelplot:
rgb.palette <- colorRampPalette(c("darkgreen","yellow", "red"), space = "rgb")
levelplot(Data.mat, main="", xlab="Time", ylab="", col.regions=rgb.palette(100),
cuts=100, at=seq(0,1,0.1), ylim=c(0,2), scales=list(y=list(at=NULL)))
This is the outcome:
Since, I do not understand how this levelplot really works, I can not make it work. What I would like to have is the colour strips to fill the whole window of the corresponding x (Time).
Alternative solution with other method.
Basically, I'm trying here to plot the increasing risk over time, where the red is the highest risk = 1. I would like to visualize the sequence of possible increase or clustering risk over time.
From ?levelplot we're told that if the first argument is a matrix then "'x' provides the
'z' vector described above, while its rows and columns are
interpreted as the 'x' and 'y' vectors respectively.", so
> m = Data.mat[, 2, drop=FALSE]
> dim(m)
[1] 21 1
> levelplot(m)
plots a levelplot with 21 columns and 1 row, where the levels are determined by the values in m. The formula interface might look like
> df <- data.frame(x=1, y=1:21, z=runif(21))
> levelplot(z ~ y + x, df)
(these approaches do not quite result in the same image).
Unfortunately I don't know much about lattice, but I noted your "Alternative solution with other method", so may I suggest another possibility:
library(plotrix)
color2D.matplot(t(Data[ , 2]), show.legend = TRUE, extremes = c("yellow", "red"))
Heaps of things to do to make it prettier. Still, a start. Of course it is important to consider the breaks in your time variable. In this very simple attempt, regular intervals are implicitly assumed, which happens to be the case in your example.
Update
Following the advice in the 'Details' section in ?color2D.matplot: "The user will have to adjust the plot device dimensions to get regular squares or hexagons, especially when the matrix is not square". Well, well, quite ugly solution.
par(mar = c(5.1, 4.1, 0, 2.1))
windows(width = 10, height = 2.5)
color2D.matplot(t(Data[ , 2]),
show.legend = TRUE,
axes = TRUE,
xlab = "",
ylab = "",
extremes = c("yellow", "red"))
In an effort to help populate the R tag here, I am posting a few questions I have often received from students. I have developed my own answers to these over the years, but perhaps there are better ways floating around that I don't know about.
The question: I just ran a regression with continuous y and x but factor f (where levels(f) produces c("level1","level2"))
thelm <- lm(y~x*f,data=thedata)
Now I would like to plot the predicted values of y by x broken down by groups defined by f. All of the plots I get are ugly and show too many lines.
My answer: Try the predict() function.
##restrict prediction to the valid data
##from the model by using thelm$model rather than thedata
thedata$yhat <- predict(thelm,
newdata=expand.grid(x=range(thelm$model$x),
f=levels(thelm$model$f)))
plot(yhat~x,data=thethedata,subset=f=="level1")
lines(yhat~x,data=thedata,subset=f=="level2")
Are there other ideas out there that are (1) easier to understand for a newcomer and/or (2) better from some other perspective?
The effects package has good ploting methods for visualizing the predicted values of regressions.
thedata<-data.frame(x=rnorm(20),f=rep(c("level1","level2"),10))
thedata$y<-rnorm(20,,3)+thedata$x*(as.numeric(thedata$f)-1)
library(effects)
model.lm <- lm(formula=y ~ x*f,data=thedata)
plot(effect(term="x:f",mod=model.lm,default.levels=20),multiline=TRUE)
Huh - still trying to wrap my brain around expand.grid(). Just for comparison's sake, this is how I'd do it (using ggplot2):
thedata <- data.frame(predict(thelm), thelm$model$x, thelm$model$f)
ggplot(thedata, aes(x = x, y = yhat, group = f, color = f)) + geom_line()
The ggplot() logic is pretty intuitive, I think - group and color the lines by f. With increasing numbers of groups, not having to specify a layer for each is increasingly helpful.
I am no expert in R. But I use:
xyplot(y ~ x, groups= f, data= Dat, type= c('p','r'),
grid= T, lwd= 3, auto.key= T,)
This is also an option:
interaction.plot(f,x,y, type="b", col=c(1:3),
leg.bty="0", leg.bg="beige", lwd=1, pch=c(18,24),
xlab="",
ylab="",
trace.label="",
main="Interaction Plot")
Here is a small change to the excellent suggestion by Matt and a solution similar to Helgi but with ggplot. Only difference from above is that I have used the geom_smooth(method='lm) which plots regression lines directly.
set.seed(1)
y = runif(100,1,10)
x = runif(100,1,10)
f = rep(c('level 1','level 2'),50)
thedata = data.frame(x,y,f)
library(ggplot2)
ggplot(thedata,aes(x=x,y=y,color=f))+geom_smooth(method='lm',se=F)