How to add label in table() in R - r

I want to print 2*2 confusion matrix and label it. I am using table() in r.
I want to add predicted and Reality label . Can anybody suggest me , how can I do it?

This is a similar problem to the one in this question. You can follow that approach,
but simplify things a bit by just using a matrix to hold the values, and
just setting its dimension names to "predicted" and "observed":
# create some fake data (2x2, since we're building a confusion matrix)
dat <- matrix(data=runif(n=4, min=0, max=1), nrow=2, ncol=2,
dimnames=list(c("pos", "neg"), c("pos", "neg")))
# now set the names *of the dimensions* (not the row/colnames)
names(dimnames(dat)) <- c("predicted", "observed")
# and we get what we wanted
dat
# output:
# observed
# predicted pos neg
# pos 0.8736425 0.7987779
# neg 0.2402080 0.6388741
Update: #thelatemail made the nice point in the comments that you can specify dimension names when creating tables. The same is true of matrices, except you supply them as names of the dimnames list elements when calling matrix(). So here's an even more compact way:
matrix(data=runif(n=4, min=0, max=1), nrow=2, ncol=2,
dimnames=list(predicted=c("pos", "neg"), observed=c("pos", "neg")))

Related

Change title: mcmc_trace function with ggplot

I used mcmc_trace function from the bayesplot package to plot traceplot with mcmc list, which is a ggplot item so it can be further edited by ggplot function.
Follows is the plot that produced by the function. I needed to change the title k1...k[20] to subject 1... subject 20. Are there any approaches I can achieve this with ggplot function?
Follows is a simple reproducible model.
library (r2jags)
library (bayesplot)
library (ggplot2)
# data
dlist <- list(
NSubjects = 20,
k = rep (5,20),
n = rep (10,20)
)
# monitor
parameter <- 'theta'
# model
minimodel <- function(){
for (i in 1:NSubjects){
theta [i] ~ dbeta (1,1)
k[i] ~ dbin(theta[i],n[i])
}
}
samples <- jags(dlist, inits=NULL, parameter,
model.file = minimodel,
n.chains=1, n.iter=10, n.burnin=1, n.thin=1, DIC=T)
# mcmc list
codaSamples = as.mcmc.list(samples$BUGSoutput)
# select subjects
colstheta <- sprintf("theta[%d]",1:20)
# plot (here is where I need to change title, in this example: theta[1]...theta[20] to subject [1].. subject [20]
mcmc_trace(codaSamples[,colstheta]) +
labs (x='Iteration',y='theta value',
title='Traceplot - theta')
Use colnames<- to modify the column names. Since the object is a 1-element list containing a matrix-like object, you need to use [[1]]; if you have multiple chains you'll need to lapply() (or use a for loop) to apply the solution to every chain (i.e., every element in the list).
cc <- codaSamples[,colstheta]
colnames(cc[[1]]) <- gsub("theta\\[([0-9]+)\\]","subject \\1",colnames(cc[[1]]))
mcmc_trace(cc, ...)
The code above finds the numerical element in each name and inserts it into the new name; since you happen to know in this case that these are elements 1:20, you could simplify considerably, e.g.
colnames(cc[[1]]) <- paste("subject",seq(ncol(cc[[1]])))

hide the NA values when using display_numbers function in pheatmap

I am plotting heatmap by pheatmap package in r.
I applied the display_numbers function to display the values in a matrix into the heatmap, and I got:
heatmap
I got so many NA in my matrix and I would like to hide them in the heatmap, how can I do that?
First off, it is a lot easier for people to help you if you were to provide reproducible and minimal sample data. Please consider reviewing how to provide a minimal reproducible example/attempt for future posts.
As to your question:
Let's generate some sampe data
set.seed(2018)
mat <- matrix(runif(20), 4, 5)
We use a second matrix to display values via the argument display_numbers of pheatmap. Here we simply copy the original matrix and randomly generate some NA values:
mat2 <- mat
mat2[mat2 < 0.5] <- NA
We now replace NA values with empty strings.
mat2[is.na(mat2)] <- ""
Let's show the heatmap
pheatmap(mat, display_numbers = mat2)

How to add labels to original data given clustering result using hclust

Just say I have some unlabeled data which I know should be clustered into six catergories, like for example this dataset:
library(tidyverse)
ts <- read_table(url("http://kdd.ics.uci.edu/databases/synthetic_control/synthetic_control.data"), col_names = FALSE)
If I create an hclust object with a sample of 60 from the original dataset like so:
n <- 10
s <- sample(1:100, n)
idx <- c(s, 100+s, 200+s, 300+s, 400+s, 500+s)
ts.samp <- ts[idx,]
observedLabels <- c(rep(1,n), rep(2,n), rep(3,n), rep(4,n), rep(5,n), rep(6,n))
# compute DTW distances
library(dtw)#Dynamic Time Warping (DTW)
distMatrix <- dist(ts.samp, method= 'DTW')
# hierarchical clustering
hc <- hclust(distMatrix, method='average')
I know that I can then add the labels to the dendrogram for viewing like this:
observedLabels <- c(rep(1,), rep(2,n), rep(3,n), rep(4,n), rep(5,n), rep(6,n))
plot(hc, labels=observedLabels, main="")
However, I would like to the correct labels to the initial data frame that was clustered. So for ts.samp I would like to add a extra column with the correct label that each observation has been clustered into.
It would seems that ts.samp$cluster <- hc$label should add the cluster to the data frame, however hc$label returns NULL.
Can anyone help with extracting this information?
You need to define a level where you cut your dendrogram, this will form the groups.
Use:
labels <- cutree(hc, k = 3) # you set the number of k that's more appropriate, see how to read a dendrogram
ts.samp$grouping <- labels
Let's look at the dendrogram in order to find the best number for k:
plot(hc, main="")
abline(h=500, col = "red") # cut at height 500 forms 2 groups
abline(h=300, col = "blue") # cut at height 300 forms 3/4 groups
It looks like either 2 or 3 might be good. You need to find the highest jump in the vertical lines (Height).
Use the horizontal lines at that height and count the cluster "formed".

Labeling outliers on boxplot in R

I would like to plot each column of a matrix as a boxplot and then label the outliers in each boxplot as the row name they belong to in the matrix. To use an example:
vv=matrix(c(1,2,3,4,8,15,30),nrow=7,ncol=4,byrow=F)
rownames(vv)=c("one","two","three","four","five","six","seven")
boxplot(vv)
I would like to label the outlier in each plot (in this case 30) as the row name it belongs to, so in this case 30 belongs to row 7. Is there an easy way to do this? I have seen similar questions to this asked but none seemed to have worked the way I want it to.
There is a simple way. Note that b in Boxplot in following lines is a capital letter.
library(car)
Boxplot(y ~ x, id.method="y")
Or alternatively, you could use the "Boxplot" function from the {car} package which labels outliers for you.
See the following link: https://CRAN.R-project.org/package=car
In the example given it's a bit boring because they are all the same row. but here is the code:
bxpdat <- boxplot(vv)
text(bxpdat$group, # the x locations
bxpdat$out, # the y values
rownames(vv)[which(vv == bxpdat$out, arr.ind=TRUE)[, 1]], # the labels
pos = 4)
This picks the rownames that have values equal to the "out" list (i.e., the outliers) in the result of boxplot. Boxplot calls and returns the values from boxplot.stats. Take a look at:
str(bxpdat)
#DWin's solution works very well for a single boxplot, but will fail for anything with duplicate values, like the dataset I have created:
#Create data
set.seed(1)
basenums <- c(1,2,3,4,8,15,30)
vv=matrix(c(basenums, sample(basenums), 1-basenums,
c(0, 29, 30, 31, 32, 33, 60)),nrow=7,ncol=4,byrow=F)
dimnames(vv)=list(c("one","two","three","four","five","six","seven"), 1:4)
On this dataset, #DWin's solution gives:
Which is false, because in the 4th example, it is not possible for the minimum and maximum to be in the same row.
This solution is monstrous (and I hope can be simplified), but effective.
#Reshape data
vv_dat <- as.data.frame(vv)
vv_dat$row <- row.names(vv_dat)
library(reshape2)
new_vv <- melt(vv_dat, id.vars="row")
#Get boxplot data
bxpdat <- as.data.frame(boxplot(value~variable, data=new_vv)[c("out", "group")])
#Get matches with boxplot data
text_guide <- do.call(rbind, apply(bxpdat, 1,
function(x) new_vv[new_vv$value==x[1]&new_vv$variable==x[2], ]))
#Add labels
with(text_guide, text(x=as.numeric(variable)+0.2, y=value, labels=row))
Or you can simply run the code from this blog post:
source("https://raw.githubusercontent.com/talgalili/R-code-snippets/master/boxplot.with.outlier.label.r") # Load the function
set.seed(6484)
y <- rnorm(20)
x1 <- sample(letters[1:2], 20,T)
lab_y <- sample(letters, 20)
# plot a boxplot with interactions:
boxplot.with.outlier.label(y~x1, lab_y)
(which handles multiple outliers which are close to one another)
#sebastian-c
This is a slight modification of DWin solution that seem to work with more generality
bx1<-boxplot(pb,las=2,cex.axis=.8)
if(length(bx1$out)!=0){
## get the row of each outlier
out.rows<-sapply(1:length(bx1$out),function(i) which(vv[,bx1$group[i]]==bx1$out[i]))
text(bx1$group,bx1$out,
rownames(vv)[out.rows],
pos=4
)
}

How to include variable values in histogram titles in R - using by()

I want to produce histograms using by(), how can I access the values of the factors, to include in histogram headings, for example...
a <- runif(500, 0, 10)
b <- LETTERS[1:5]
c <- c("Condition1", "Condition2")
x <- data.frame("Variable1" = b, "Variable2"= c, "Value"=a)
head(x)
by(x$Value, x$Variable2, hist)
or using two variables
by(x$Value, list(x$Variable2, x$Variable1), hist)
Is there a way of passing the variable value (eg Condition1) to the title of the histogram using the options within hist(), eg putting function(x) hist(x, main=...) into by()?
Pass the split up dataframe rather than just the Values. Then you will have more to work with:
by(x, x$Variable2, function(x) hist(x$Value, main=unique(x$Variable2) ) )
Produced two plots labled Condition1, Condition2
This doesn't really answer your question, since you're specifying the use of by(), but I usually use split() and lapply() for these types of problems. My approach is usually along the lines of:
temp <- split(x$Value, list(x$Variable2, x$Variable1))
lapply(names(temp), function(x) hist(temp[[x]], main = x, xlab = "Value"))

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