Find scatterplot area where ~50% of points have one of 2 values - r

I have a data frame that has 3 values for each point in the form: (x, y, boolean). I'd like to find an area bounded by values of (x, y) where roughly half the points in the area are TRUE and half are FALSE.
I can scatterplot then data and color according to the 3rd value of each point and I get a general idea but I was wondering if there would be a better way. I understand that if you take a small enough area where there are only 2 points and one if TRUE and the other is FALSE then you have 50/50 so I was thinking there has to be a better way of deciding what size area to look for.
Visually I see this has drawing a square on the scatter plot and moving it around the x and y axis each time checking the number of TRUE and FALSE points in the area, but is there a way to determine what a good size for the area is based on the values?
Thanks
EDIT: G5W's answer is a step in the right direction but based on their scatterplot, I'm looking to create a square / rectangle idea in which ~ half the points are green and half are red. I understand that there is potentially an infinite amount of those areas but thinking there might be a good way to determine an optimal size for the area (maybe it should contain at least a certain percentage of the points or something)

Note update below
You do not provide any sample data, so I have created some bogus data like this:
TestData = data.frame(x = c(rnorm(100, -1, 1), rnorm(100, 1,1)),
y = c(rnorm(100, -1, 1), rnorm(100, 1,1)),
z = rep(c(TRUE,FALSE), each=100))
I think that what you want is how much area is taken up by each of the TRUE and FALSE points. A way to interpret that task is to find the convex hull for each group and take its area. That is, find the minimum convex polygon that contains a group. The function chull will compute the convex hull of a set of points.
plot(TestData[,1:2], pch=20, col=as.numeric(TestData$z)+2)
CH1 = chull(TestData[TestData$z,1:2])
CH2 = chull(TestData[!TestData$z,1:2])
polygon(TestData[which(TestData$z)[CH1],1:2], lty=2, col="#00FF0011")
polygon(TestData[which(!TestData$z)[CH2],1:2], lty=2, col="#FF000011")
Once you have the polygons, the polyarea function from the pracma package will compute the area. Note that it computes a "signed" area so you either need to be careful about which direction you traverse the polygon or take the absolute value of the area.
library(pracma)
abs(polyarea(TestData[which(TestData$z)[CH1],1],
TestData[which(TestData$z)[CH1],2]))
[1] 16.48692
abs(polyarea(TestData[which(!TestData$z)[CH2],1],
TestData[which(!TestData$z)[CH2],2]))
[1] 15.17897
Update
This is a completely different answer based on the updated question. I am leaving the old answer because the question now refers to it.
The question now gives a little more information about the data ("There are about twice as many FALSE than TRUE") so I have made an updated bogus data set to reflect that.
set.seed(2017)
TestData = data.frame(x = c(rnorm(100, -1, 1), rnorm(200, 1, 1)),
y = c(rnorm(100, 1, 1), rnorm(200, -1,1)),
z = rep(c(TRUE,FALSE), c(100,200)))
The problem is now to find regions where the density of TRUE and FALSE are approximately equal. The question asked for a rectangular region, but at least for this data, that will be difficult. We can get a good visualization to see why.
We can use the function kde2d from the MASS package to get the 2-dimensional density of the TRUE points and the FALSE points. If we take the difference of these two densities, we need only find the regions where the difference is near zero. Once we have this difference in density, we can visualize it with a contour plot.
library(MASS)
Grid1 = kde2d(TestData$x[TestData$z], TestData$y[TestData$z],
lims = c(c(-3,3), c(-3,3)))
Grid2 = kde2d(TestData$x[!TestData$z], TestData$y[!TestData$z],
lims = c(c(-3,3), c(-3,3)))
GridDiff = Grid1
GridDiff$z = Grid1$z - Grid2$z
filled.contour(GridDiff, color = terrain.colors)
In the plot it is easy to see the place that there are far more TRUE than false near (-1,1) and where there are more FALSE than TRUE near (1,-1). We can also see that the places where the difference in density is near zero lie in a narrow band in the general area of the line y=x. You might be able to get a box where a region with more TRUEs is balanced by a region with more FALSEs, but the regions where the density is the same is small.
Of course, this is for my bogus data set which probably bears little relation to your real data. You could perform the same sort of analysis on your data and maybe you will be luckier with a bigger region of near equal densities.

Related

How to get the best polygon point pattern data in spatstat analysis in R

I have a dataset of spatial locations data. I want to do a point pattern analysis using the spatstat package in R using this data. I want the best polygon area for the analysis instead of the rectangle area. The code I have is
original_data = read.csv("/home/hudamoh/PhD_Project_Moh_Huda/Dataset_files/my_coordinates.csv")
plot(original_data$row, original_data$col)
which results in a plot that looks like this
Setting the data for point pattern data
point_pattern_data = ppp(original_data$row, original_data$col, c(0, 77), c(0, 116))
plot(point_pattern_data)
summary(point_pattern_data)
resulting in a plot that looks like this
#The observed data has considerably wide white spaces, which I want to remove for a better analysis area. Therefore, I want to make the point pattern a polygon instead of a rectangle. The vertices for the polygon are the pairs of (x,y) below to avoid white space as much as possible.
x = c(3,1,1,0.5,0.5,1,2,2.5,5.5, 16,21,28,26,72,74,76,75,74,63,58,52,47,40)
y = c(116,106,82.5,64,40,35,25,17.5,5,5,5,10,8,116,100,50,30,24,17,10,15,15,8)
I find these vertices above manually by considering the plot below (with the grid lines)
plot(original_data$row,original_data$col)
grid(nx = 40, ny = 25,
lty = 2, # Grid line type
col = "gray", # Grid line color
lwd = 2) # Grid line width
So I want to make the point pattern polygon. The code is
my_data_poly = owin(poly = list(x = c(3,1,1,0.5,0.5,1,2,2.5,5.5, 16,21,28,26,72,74,76,75,74,63,58,52,47,40), y = c(116,106,82.5,64,40,35,25,17.5,5,5,5,10,8,116,100,50,30,24,17,10,15,15,8)))
plot(my_data_poly)
but it results in an error. The error is
I fix it by
my_data_poly = owin(poly = list(x = c(116,106,82.5,64,40,35,25,17.5,5,5,5,10,8,116,100,50,30,24,17,10,15,15,8), y = c(3,1,1,0.5,0.5,1,2,2.5,5.5, 16,21,28,26,72,74,76,75,74,63,58,52,47,40)))
plot(my_data_poly)
It results in a plot
However, this is not what I want. How to get the observed area as a polygon in point pattern data analysis?
This should be a reasonable solution to the problem.
require(sp)
poly = Polygon(
cbind(original_data$col,
original_data$row)
))
This will create a polygon from your points. You can use this document to understand the sp package better
We don’t have access to the point data you read in from file, but if you just want to fix the polygonal window that is not a problem.
You need to traverse the vertices of your polygon sequentially and anti-clockwise.
The code connects the first point you give to the next etc. Your vertices are:
library(spatstat)
x = c(3,1,1,0.5,0.5,1,2,2.5,5.5, 16,21,28,26,72,74,76,75,74,63,58,52,47,40)
y = c(116,106,82.5,64,40,35,25,17.5,5,5,5,10,8,116,100,50,30,24,17,10,15,15,8)
vert <- ppp(x, y, window = owin(c(0,80),c(0,120)))
plot.ppp(vert, main = "", show.window = FALSE, chars = NA)
text(vert)
Point number 13 is towards the bottom left and 14 in the top right, which gives the funny crossing in the polygon.
Moving the order around seems to help:
xnew <- c(x[1:11], x[13:12], x[23:14])
ynew <- c(y[1:11], y[13:12], y[23:14])
p <- owin(poly = cbind(xnew, ynew))
plot(p, main = "")
It is unclear from your provided plot of the data that you really should apply point pattern analysis.
The main assumption underlying point process modelling as implemented in spatstat
is that the locations of events (points) are random and the process that
generated the random locations is of interest.
Your points seem to be on a grid and maybe you need another tool for your analysis.
Of course spatstat has a lot of functionality for simply handling and summarising data like this so you may still find useful tools in there.

In ggplot, how to draw a circle/disk with a line that divides its area according to a given ratio and colored points inside?

I want to visualize proportions using points inside a circle. For example, let's say that I have 100 points that I wish to scatter (somewhat randomly jittered) in a circle.
Next, I want to use this diagram to represent the proportions of people who voted Biden/Harris in 2020 US presidential elections, in each state.
Example #1 -- Michigan
Biden got 50.62% of Michigan's votes. I'm going to draw a horizontal diameter that splits the circle to two halves, and then color the points under the diameter in blue (Democrats' color).
Example #2 -- Wyoming
Unlike Michigan, in Wyoming Biden got only 26.55% of the votes, which is approximately a quarter of the vote. In this case I'd draw a horizontal chord that divides the circle such that the disk's area under the chord is 25% of the entire disk area. Then I'll color the respective points in that area in blue. Since I have 100 points in total, 25 points represent the 25% who voted Biden in Wyoming.
My question: How can I do this with ggplot? I researched this issue, and there's a lot of geometry going on here. First, the kind of area I'm talking about is called a "circular segment". Second, there are many formulas to calculate its area, if we know some other parameters about the shape (such as the radius length, etc.). See this nice demo.
However, my goal isn't to solve geometry problems, but just to represent proportions in a very specific way:
draw a circle
sprinkle X number of points inside
draw a (real or invisible) horizontal line that divides the circle/disk area according to a given proportion
ensure that the points are arranged respective to the split. That is, if we want to represent a 30%-70% split, then have 30% of the points under the line that divides the disk.
color the points under the line.
I understand that this is somewhat an exotic visualization, but I'll be thankful for any help with this.
EDIT
I've found a reference to a JavaScript package that does something very similar to what I'm asking.
I took a crack at this for fun. There's a lot more that could be done. I agree that this is not a great way to visualize proportions, but if it's engaging your audience ...
Formulas for determining appropriate heights are taken from Wikipedia. In particular we need the formulas
a/A = (theta - sin(theta))/(2*pi)
h = 1-cos(theta/2)
where a is the area of the segment; A is the whole area of the circle; theta is the angle described by the arc that defines the segment (see Wikipedia for pictures); and h is the height of the segment.
Machinery for finding heights.
afun <- function(x) (x-sin(x))/(2*pi)
## curve(afun, from=0, to = 2*pi)
find_a <- function(a) {
uniroot(
function(x) afun(x) -a,
interval=c(0, 2*pi))$root
}
find_h <- function(a) {
1- cos(find_a(a)/2)
}
vfind_h <- Vectorize(find_h)
## find_a(0.5)
## find_h(0.5)
## curve(vfind_h(x), from = 0, to= 1)
set up a circle
dd <- data.frame(x=0,y=0,r=1)
library(ggforce)
library(ggplot2); theme_set(theme_void())
gg0 <- ggplot(dd) + geom_circle(aes(x0=x,y0=y,r=r)) + coord_fixed()
finish
props <- c(0.2,0.5,0.3) ## proportions
n <- 100 ## number of points to scatter
cprop <- cumsum(props)[-length(props)]
h <- vfind_h(cprop)
set.seed(101)
r <- runif(n)
th <- runif(n, 0, 2 * pi)
dd <-
data.frame(x = sqrt(r) * cos(th),
y = sqrt(r) * sin(th))
dd2 <- data.frame(x=r*cos(2*pi*th), y = r*sin(2*pi*th))
dd2$g <- cut(dd2$y, c(1, 1-h, -1))
gg0 + geom_point(data=dd2, aes(x, y, colour = g), size=3)
There are a bunch of tweaks that would make this better (meaningful names for the categories; reverse the axis order to match the plot; maybe add segments delimiting the sections, or (more work) polygons so you can shade the sections.
You should definitely check this for mistakes — e.g. there are places where I may have used a set of values where I should have used their first differences, or vice versa (values vs cumulative sum). But this should get you started.

how to get point set (x,y) in a desired area in r

The figure is the plot of x,y set in a excel file, total 8760 pair of x and y. I want to remove the noise data pair in red circle area and output a new excel file with remain data pair. How could I do it in R?
Using #G5W's example:
Make up data:
set.seed(2017)
x = runif(8760, 0,16)
y = c(abs(rnorm(8000, 0, 1)), runif(760,0,8))
XY = data.frame(x,y)
Fit a quantile regression to the 90th percentile:
library(quantreg)
library(splines)
qq <- rq(y~ns(x,20),tau=0.9,data=XY)
Compute and draw the predicted curve:
xvec <- seq(0,16,length.out=101)
pp <- predict(qq,newdata=data.frame(x=xvec))
plot(y~x,data=XY)
lines(xvec,pp,col=2,lwd=2)
Keep only points below the predicted line:
XY2 <- subset(XY,y<predict(qq,newdata=data.frame(x)))
plot(y~x,data=XY2)
lines(xvec,pp,col=2,lwd=2)
You can make the line less wiggly by lowering the number of knots, e.g. y~ns(x,10)
Both R and EXCEL read and write .csv files, so you can use those to transfer the data back and forth.
You do not provide any data so I made some junk data to produce a similar problem.
DATA
set.seed(2017)
x = runif(8760, 0,16)
y = c(abs(rnorm(8000, 0, 1)), runif(760,0,8))
XY = data.frame(x,y)
One way to identify noise points is by looking at the distance to the nearest neighbors. In dense areas, nearest neighbors will be closer. In non-dense areas, they will be further apart. The package dbscan provides a nice function to get the distance to the k nearest neighbors. For this problem, I used k=6, but you may need to tune for your data. Looking at the distribution of distances to the 6th nearest neighbor we see that most points have 6 neighbors within a distance of 0.2
XY6 = kNNdist(XY, 6)
plot(density(XY6[,6]))
So I will assume that point whose 6th nearest neighbor is further away are noise points. Just changing the color to see which points are affected, we get
TYPE = rep(1,8760)
TYPE[XY6[,6] > 0.2] = 2
plot(XY, col=TYPE)
Of course, if you wish to restrict to the non-noise points, you can use
NonNoise = XY[XY6[,6] > 0.2,]

Find correct 2D translation of a subset of coordinates

I have a problem I wish to solve in R with example data below. I know this must have been solved many times but I have not been able to find a solution that works for me in R.
The core of what I want to do is to find how to translate a set of 2D coordinates to best fit into an other, larger, set of 2D coordinates. Imagine for example having a Polaroid photo of a small piece of the starry sky with you out at night, and you want to hold it up in a position so they match the stars' current positions.
Here is how to generate data similar to my real problem:
# create reference points (the "starry sky")
set.seed(99)
ref_coords = data.frame(x = runif(50,0,100), y = runif(50,0,100))
# generate points take subset of coordinates to serve as points we
# are looking for ("the Polaroid")
my_coords_final = ref_coords[c(5,12,15,24,31,34,48,49),]
# add a little bit of variation as compared to reference points
# (data should very similar, but have a little bit of noise)
set.seed(100)
my_coords_final$x = my_coords_final$x+rnorm(8,0,.1)
set.seed(101)
my_coords_final$y = my_coords_final$y+rnorm(8,0,.1)
# create "start values" by, e.g., translating the points we are
# looking for to start at (0,0)
my_coords_start =apply(my_coords_final,2,function(x) x-min(x))
# Plot of example data, goal is to find the dotted vector that
# corresponds to the translation needed
plot(ref_coords, cex = 1.2) # "Starry sky"
points(my_coords_start,pch=20, col = "red") # start position of "Polaroid"
points(my_coords_final,pch=20, col = "blue") # corrected position of "Polaroid"
segments(my_coords_start[1,1],my_coords_start[1,2],
my_coords_final[1,1],my_coords_final[1,2],lty="dotted")
Plotting the data as above should yield:
The result I want is basically what the dotted line in the plot above represents, i.e. a delta in x and y that I could apply to the start coordinates to move them to their correct position in the reference grid.
Details about the real data
There should be close to no rotational or scaling difference between my points and the reference points.
My real data is around 1000 reference points and up to a few hundred points to search (could use less if more efficient)
I expect to have to search about 10 to 20 sets of reference points to find my match, as many of the reference sets will not contain my points.
Thank you for your time, I'd really appreciate any input!
EDIT: To clarify, the right plot represent the reference data. The left plot represents the points that I want to translate across the reference data in order to find a position where they best match the reference. That position, in this case, is represented by the blue dots in the previous figure.
Finally, any working strategy must not use the data in my_coords_final, but rather reproduce that set of coordinates starting from my_coords_start using ref_coords.
So, the previous approach I posted (see edit history) using optim() to minimize the sum of distances between points will only work in the limited circumstance where the point distribution used as reference data is in the middle of the point field. The solution that satisfies the question and seems to still be workable for a few thousand points, would be a brute-force delta and comparison algorithm that calculates the differences between each point in the field against a single point of the reference data and then determines how many of the rest of the reference data are within a minimum threshold (which is needed to account for the noise in the data):
## A brute-force approach where min_dist can be used to
## ameliorate some random noise:
min_dist <- 5
win_thresh <- 0
win_thresh_old <- 0
for(i in 1:nrow(ref_coords)) {
x2 <- my_coords_start[,1]
y2 <- my_coords_start[,2]
x1 <- ref_coords[,1] + (x2[1] - ref_coords[i,1])
y1 <- ref_coords[,2] + (y2[1] - ref_coords[i,2])
## Calculate all pairwise distances between reference and field data:
dists <- dist( cbind( c(x1, x2), c(y1, y2) ), "euclidean")
## Only take distances for the sampled data:
dists <- as.matrix(dists)[-1*1:length(x1),]
## Calculate the number of distances within the minimum
## distance threshold minus the diagonal portion:
win_thresh <- sum(rowSums(dists < min_dist) > 1)
## If we have more "matches" than our best then calculate a new
## dx and dy:
if (win_thresh > win_thresh_old) {
win_thresh_old <- win_thresh
dx <- (x2[1] - ref_coords[i,1])
dy <- (y2[1] - ref_coords[i,2])
}
}
## Plot estimated correction (your delta x and delta y) calculated
## from the brute force calculation of shifts:
points(
x=ref_coords[,1] + dx,
y=ref_coords[,2] + dy,
cex=1.5, col = "red"
)
I'm very interested to know if there's anyone that solves this in a more efficient manner for the number of points in the test data, possibly using a statistical or optimization algorithm.

What techniques exists in R to visualize a "distance matrix"?

I wish to present a distance matrix in an article I am writing, and I am looking for good visualization for it.
So far I came across balloon plots (I used it here, but I don't think it will work in this case), heatmaps (here is a nice example, but they don't allow to present the numbers in the table, correct me if I am wrong. Maybe half the table in colors and half with numbers would be cool) and lastly correlation ellipse plots (here is some code and example - which is cool to use a shape, but I am not sure how to use it here).
There are also various clustering methods but they will aggregate the data (which is not what I want) while what I want is to present all of the data.
Example data:
nba <- read.csv("http://datasets.flowingdata.com/ppg2008.csv")
dist(nba[1:20, -1], )
I am open for ideas.
You could also use force-directed graph drawing algorithms to visualize a distance matrix, e.g.
nba <- read.csv("http://datasets.flowingdata.com/ppg2008.csv")
dist_m <- as.matrix(dist(nba[1:20, -1]))
dist_mi <- 1/dist_m # one over, as qgraph takes similarity matrices as input
library(qgraph)
jpeg('example_forcedraw.jpg', width=1000, height=1000, unit='px')
qgraph(dist_mi, layout='spring', vsize=3)
dev.off()
Tal, this is a quick way to overlap text over an heatmap. Note that this relies on image rather than heatmap as the latter offsets the plot, making it more difficult to put text in the correct position.
To be honest, I think this graph shows too much information, making it a bit difficult to read... you may want to write only specific values.
also, the other quicker option is to save your graph as pdf, import it in Inkscape (or similar software) and manually add the text where needed.
Hope this helps
nba <- read.csv("http://datasets.flowingdata.com/ppg2008.csv")
dst <- dist(nba[1:20, -1],)
dst <- data.matrix(dst)
dim <- ncol(dst)
image(1:dim, 1:dim, dst, axes = FALSE, xlab="", ylab="")
axis(1, 1:dim, nba[1:20,1], cex.axis = 0.5, las=3)
axis(2, 1:dim, nba[1:20,1], cex.axis = 0.5, las=1)
text(expand.grid(1:dim, 1:dim), sprintf("%0.1f", dst), cex=0.6)
A Voronoi Diagram (a plot of a Voronoi Decomposition) is one way to visually represent a Distance Matrix (DM).
They are also simple to create and plot using R--you can do both in a single line of R code.
If you're not famililar with this aspect of computational geometry, the relationship between the two (VD & DM) is straightforward, though a brief summary might be helpful.
Distance Matrices--i.e., a 2D matrix showing the distance between a point and every other point, are an intermediate output during kNN computation (i.e., k-nearest neighbor, a machine learning algorithm which predicts the value of a given data point based on the weighted average value of its 'k' closest neighbors, distance-wise, where 'k' is some integer, usually between 3 and 5.)
kNN is conceptually very simple--each data point in your training set is in essence a 'position' in some n-dimension space, so the next step is to calculate the distance between each point and every other point using some distance metric (e.g., Euclidean, Manhattan, etc.). While the training step--i.e., construcing the distance matrix--is straightforward, using it to predict the value of new data points is practically encumbered by the data retrieval--finding the closest 3 or 4 points from among several thousand or several million scattered in n-dimensional space.
Two data structures are commonly used to address that problem: kd-trees and Voroni decompositions (aka "Dirichlet tesselation").
A Voronoi decomposition (VD) is uniquely determined by a distance matrix--i.e., there's a 1:1 map; so indeed it is a visual representation of the distance matrix, although again, that's not their purpose--their primary purpose is the efficient storage of the data used for kNN-based prediction.
Beyond that, whether it's a good idea to represent a distance matrix this way probably depends most of all on your audience. To most, the relationship between a VD and the antecedent distance matrix will not be intuitive. But that doesn't make it incorrect--if someone without any statistics training wanted to know if two populations had similar probability distributions and you showed them a Q-Q plot, they would probably think you haven't engaged their question. So for those who know what they are looking at, a VD is a compact, complete, and accurate representation of a DM.
So how do you make one?
A Voronoi decomp is constructed by selecting (usually at random) a subset of points from within the training set (this number varies by circumstances, but if we had 1,000,000 points, then 100 is a reasonable number for this subset). These 100 data points are the Voronoi centers ("VC").
The basic idea behind a Voronoi decomp is that rather than having to sift through the 1,000,000 data points to find the nearest neighbors, you only have to look at these 100, then once you find the closest VC, your search for the actual nearest neighbors is restricted to just the points within that Voronoi cell. Next, for each data point in the training set, calculate the VC it is closest to. Finally, for each VC and its associated points, calculate the convex hull--conceptually, just the outer boundary formed by that VC's assigned points that are farthest from the VC. This convex hull around the Voronoi center forms a "Voronoi cell." A complete VD is the result from applying those three steps to each VC in your training set. This will give you a perfect tesselation of the surface (See the diagram below).
To calculate a VD in R, use the tripack package. The key function is 'voronoi.mosaic' to which you just pass in the x and y coordinates separately--the raw data, not the DM--then you can just pass voronoi.mosaic to 'plot'.
library(tripack)
plot(voronoi.mosaic(runif(100), runif(100), duplicate="remove"))
You may want to consider looking at a 2-d projection of your matrix (Multi Dimensional Scaling). Here is a link to how to do it in R.
Otherwise, I think you are on the right track with heatmaps. You can add in your numbers without too much difficulty. For example, building of off Learn R :
library(ggplot2)
library(plyr)
library(arm)
library(reshape2)
nba <- read.csv("http://datasets.flowingdata.com/ppg2008.csv")
nba$Name <- with(nba, reorder(Name, PTS))
nba.m <- melt(nba)
nba.m <- ddply(nba.m, .(variable), transform,
rescale = rescale(value))
(p <- ggplot(nba.m, aes(variable, Name)) + geom_tile(aes(fill = rescale),
colour = "white") + scale_fill_gradient(low = "white",
high = "steelblue")+geom_text(aes(label=round(rescale,1))))
A dendrogram based on a hierarchical cluster analysis can be useful:
http://www.statmethods.net/advstats/cluster.html
A 2-D or 3-D multidimensional scaling analysis in R:
http://www.statmethods.net/advstats/mds.html
If you want to go into 3+ dimensions, you might want to explore ggobi / rggobi:
http://www.ggobi.org/rggobi/
In the book "Numerical Ecology" by Borcard et al. 2011 they used a function called *coldiss.r *
you can find it here: http://ichthyology.usm.edu/courses/multivariate/coldiss.R
it color codes the distances and even orders the records by dissimilarity.
another good package would be the seriation package.
Reference:
Borcard, D., Gillet, F. & Legendre, P. (2011) Numerical Ecology with R. Springer.
A solution using Multidimensional Scaling
data = read.csv("http://datasets.flowingdata.com/ppg2008.csv", sep = ",")
dst = tcrossprod(as.matrix(data[,-1]))
dst = matrix(rep(diag(dst), 50L), ncol = 50L, byrow = TRUE) +
matrix(rep(diag(dst), 50L), ncol = 50L, byrow = FALSE) - 2*dst
library(MASS)
mds = isoMDS(dst)
#remove {type = "n"} to see dots
plot(mds$points, type = "n", pch = 20, cex = 3, col = adjustcolor("black", alpha = 0.3), xlab = "X", ylab = "Y")
text(mds$points, labels = rownames(data), cex = 0.75)

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