How to find inflection points in a Kernel density plot in R? - r

I am trying to find the x-values of the inflection points in the curve of a Kernel density plot that I computed with the density() function.
I found the following answered question helpful in finding the turning points:
How to find all the turning points on a kernel density curve when window width varies.
So I would think there must be a way to fnd the x-values of the inflection points, too.
Would be great if somene has a tipp.

By definition, an inflection point is the point where the second derivative of the function equals zero. In the practice, this means that an inflection point will be a point where the slope passes from increasing to decreasing, or v.v. Using this definition, I came with this approximate and non-automatic approach:
Let's say that you have a dataframe, that I will call all, which contains the x-values in the first column, and the result of the density computation in the second one. From this dataframe, we can calculate the slope of two consecutive points like this :
slopes <- vector()
for(i in (2:nrow(all))){
x1 <- all[i-1, 1]
x2 <- all[i, 1]
y1 <- all[i-1, 2]
y2 <- all[i, 2]
slope_i <- (y2-y1)/(x2-x1)
slopes <- append(slopes, slope_i)
}
By the definition of inflection point, we can now calculate if, from one point to another, the slope gets larger or smaller:
increment <- vector()
for(j in 2:length(slopes)){
increment_j <- slopes[j] - slopes[j-1]
increment <- append(increment, increment_j)
}
The inflection points will be those points were this increment passes from positive to negative, or v.v.
Now, let's separate these increments in positive and negative:
pos <- which(increment>0)
neg <- which(increment<0
Now, whenever there is a jump in these pos or neg vectors, it means we have an inflection point. So, once again:
steps_p <- vector()
for(k in 2:length(pos)){
steps_k <- pos[k] - pos[k-1]
steps_p <- append(steps_p, steps_k)
}
steps_n <- vector()
for(k in 2:length(neg)){
steps_k <- neg[k] - neg[k-1]
steps_n <- append(steps_n, steps_k)
}
Now, just ask R:
which(steps_p>1)
which(steps_n>1)
This are the indices of your inflection points, now just go to your original dataframe and ask for the x value:
all[pos[which(steps_p>1)],1]
all[neg[which(steps_n>1)],1]
Take in mind that the x value will be close to exact, but not quite, as during every loop we lose one index, but it will still be a very close solution.

Related

Shrink convex hull

I have a bunch of points in 2D space and have calculated a convex hull for them. I would now like to "tighten" the hull so that it no longer necessarily encompasses all points. In the typical nails-in-board-with-rubber-band analogy, what I'd like to achieve is to be able to tune the elasticity of the rubber band and allow nails to bend at pressure above some limit. That's just an analogy, there is no real physics here. This would kind-of be related to the reduction in hull area if a given point was removed, but not quite because there could be two points that are very close to each-other. This is not necessarily related to outlier detection, because you could imagine a pattern where a large fractions of the nails would bend if they are on a narrow line (imagine a hammer shape for example). All of this has to be reasonably fast for thousands of points. Any hints where I should look in terms of algorithms? An implementation in R would be perfect, but not needed.
EDIT AFTER COMMENT: The three points I've labelled are those with largest potential for reducing the hull area if they are excluded. In the plot there is no other set of three points that would result in a larger area reduction. A naive implementation of what I'm looking for would maybe be to randomly sample some fraction of the points, calculate the hull area, remove each point on the hull iteratively, recalculate the area, repeat many times and remove points that tend to lead to high area reduction. Maybe this could be implemented in some random forest variant? It's not quite right though, because I would like the points to be removed one by one so that you get the following result. If you looked at all points in one go it would possibly be best to trim from the edges of the "hammer head".
Suppose I have a set of points like this:
set.seed(69)
x <- runif(20)
y <- runif(20)
plot(x, y)
Then it is easy to find the subset points that sit on the convex hull by doing:
ss <- chull(x, y)
This means we can plot the convex hull by doing:
lines(x[c(ss, ss[1])], y[c(ss, ss[1])], col = "red")
Now we can randomly remove one of the points that sits on the convex hull (i.e. "bend a nail") by doing:
bend <- ss[sample(ss, 1)]
x <- x[-bend]
y <- y[-bend]
And we can then repeat the process of finding the convex hull of this new set of points:
ss <- chull(x, y)
lines(x[c(ss, ss[1])], y[c(ss, ss[1])], col = "blue", lty = 2)
To get the point which will, on removal, cause the greatest reduction in area, one option would be the following function:
library(sp)
shrink <- function(coords)
{
ss <- chull(coords[, 1], coords[, 2])
outlier <- ss[which.min(sapply(seq_along(ss),
function(i) Polygon(coords[ss[-i], ], hole = FALSE)#area))]
coords[-outlier, ]
}
So you could do something like:
coords <- cbind(x, y)
new_coords <- shrink(coords)
new_chull <- new_coords[chull(new_coords[, 1], new_coords[, 2]),]
new_chull <- rbind(new_chull, new_chull[1,])
plot(x, y)
lines(new_chull[,1], new_chull[, 2], col = "red")
Of course, you could do this in a loop so that new_coords is fed back into shrink multiple times.
Calculate a robust center and variance using mcd.cov in MASS and the mahalanobis distance of each point from it (using mahalanobis in psych). We then show a quantile plot of the mahalanobis distances using PlotMD from modi and also show the associated outliers in red in the second plot. (There are other functions in modi that may be of interest as well.)
library(MASS)
library(modi)
library(psych)
set.seed(69)
x <- runif(20)
y <- runif(20)
m <- cbind(x, y)
mcd <- cov.mcd(m)
md <- mahalanobis(m, mcd$center, mcd$cov)
stats <- PlotMD(md, 2, alpha = 0.90)
giving:
(continued after screenshot)
and we show the convex hull using lines and the outliers in red:
plot(m)
ix <- chull(m)
lines(m[c(ix, ix[1]), ])
wx <- which(md > stats$halpha)
points(m[wx, ], col = "red", pch = 20)
Thank you both! I've tried various methods for outlier detection, but it's not quite what I was looking for. They have worked badly due to weird shapes of my clusters. I know I talked about convex hull area, but I think filtering on segment lengths yields better results and is closer to what I really wanted. Then it would look something like this:
shrink <- function(xy, max_length = 30){
to_keep <- 1:(dim(xy)[1])
centroid <- c(mean(xy[,1]), mean(xy[,2]))
while (TRUE){
ss <- chull(xy[,1], xy[,2])
ss <- c(ss, ss[1])
lengths <- sapply(1:(length(ss)-1), function(i) sum((xy[ss[i+1],] - xy[ss[i],])^2))
# This gets the point with the longest convex hull segment. chull returns points
# in clockwise order, so the point to remove is either this one or the one
# after it. Remove the one furthest from the centroid.
max_point <- which.max(lengths)
if (lengths[max_point] < max_length) return(to_keep)
if (sum((xy[ss[max_point],] - centroid)^2) > sum((xy[ss[max_point + 1],] - centroid)^2)){
xy <- xy[-ss[max_point],]
to_keep <- to_keep[-ss[max_point]]
}else{
xy <- xy[-ss[max_point + 1],]
to_keep <- to_keep[-ss[max_point + 1]]
}
}
}
It's not optimal because it factors in the distance to the centroid, which I would have liked to avoid, and there is a max_length parameter that should be calculated from the data instead of being hard-coded.
No filter:
It looks like this because there are 500 000 cells in here, and there are many that end up "wrong" when projecting from ~20 000 dimensions to 2.
Filter:
Note that it filters out points at tips of some clusters. This is less-than-optimal but ok. The overlap between some clusters is true and should be there.

R Function to Find Derivative of Every Point in Time Series

I have a smoothed time series and want to find the instantaneous velocity of the function at any point along the line.
What I want to do is take a series of values: ex(1,6,5,4,3,5,6,7,1)
and return the derivative of each relative to the function of the entire series, such that at every point in time, I know what direction the line is trending.
I am new to R, but know there must be a way.
Any tips?
Ex:
library(smoother)
data(BJsales)
m <- data.frame(BJsales)
x.smth <- as.data.frame(smth.gaussian(m$BJsales,tails=TRUE,alpha = 5))
x.smth.ts <- cbind(seq(1:nrow(m)),x.smth)
colnames(x.smth.ts) <- c("x","y")
x.smth.ts
plot(x.smth.ts$y~x.smth.ts$x)
Desired output:
df with 2 columns: x, deriv.of.y
Edit: Final Result thanks to G5W
TS with Color by Derivative
Your proposed example using the BJSales data is decidedly not differentiable,
so instead I will show the derivative of a much smoother function. If your real data is smooth, this should work for you.
The simplest way to approximate the derivative is simply to use finite differences.
f'(x) ≈ (f(x+h) - f(x))/h
## Smooth sample function
x = seq(0,10,0.1)
y = x/2 + sin(x)
plot(x,y, pch=20)
## Simplest - first difference
d1 = diff(y)/diff(x)
d1 = c(d1[1],d1)
Let's use it to plot a tangent line as an error check. I picked a place to draw the tangent line arbitrarily: the 18th point, x=1.7
plot(x,y, type="l")
abline(y[18]-x[18]*d1[18], d1[18])
To get the data.frame that you requested, you just need
Derivative = data.frame(x, d1)

Coordinates of greatest kernel density for trivariate KDE

I have calculated the kernel density of a 3-column matrix in R using the following code:
ss<-read.table("data.csv",header=TRUE,sep=",")
x<-ss[,1]
y<-ss[,2]
z<-ss[,3]
ssdata<-c(x,y,z)
ssmat<-matrix(ssdata,,3)
rp<-kde(ssmat)
plot(rp)
What I need now are the (x,y,z) coordinates of the point of maximum kernel density. Based on the answer provided at on the R-help list, I understand that the kde() function plots the joint density of the three variables in a fourth dimension which is represented in the 3d plot by shading to indicate areas of greater point density. So in effect I am trying to locate the maximum value of this "fourth" dimension. I suspect that this is a relatively simple problem but I haven't been able to find the answer. Any ideas?
You can extract the max value from the info returned from kde. To see all the stuff returned, use str(rp).
## Get the indices
inds <- which(abs(rp$estimate - max(rp$estimate)) < 1e-10, arr.ind=T)
xyz <- mapply(function(a, b) a[b], rp$eval.points, inds)
## Add it to plot
plot(rp)
points3d(x=xyz[1], y=xyz[2], z=xyz[3], size=20, col="blue")

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.

Detecting dips in a 2D plot

I need to automatically detect dips in a 2D plot, like the regions marked with red circles in the figure below. I'm only interested in the "main" dips, meaning the dips have to span a minimum length in the x axis. The number of dips is unknown, i.e., different plots will contain different numbers of dips. Any ideas?
Update:
As requested, here's the sample data, together with an attempt to smooth it using median filtering, as suggested by vines.
Looks like I need now a robust way to approximate the derivative at each point that would ignore the little blips that remain in the data. Is there any standard approach?
y <- c(0.9943,0.9917,0.9879,0.9831,0.9553,0.9316,0.9208,0.9119,0.8857,0.7951,0.7605,0.8074,0.7342,0.6374,0.6035,0.5331,0.4781,0.4825,0.4825,0.4879,0.5374,0.4600,0.3668,0.3456,0.4282,0.3578,0.3630,0.3399,0.3578,0.4116,0.3762,0.3668,0.4420,0.4749,0.4556,0.4458,0.5084,0.5043,0.5043,0.5331,0.4781,0.5623,0.6604,0.5900,0.5084,0.5802,0.5802,0.6174,0.6124,0.6374,0.6827,0.6906,0.7034,0.7418,0.7817,0.8311,0.8001,0.7912,0.7912,0.7540,0.7951,0.7817,0.7644,0.7912,0.8311,0.8311,0.7912,0.7688,0.7418,0.7232,0.7147,0.6906,0.6715,0.6681,0.6374,0.6516,0.6650,0.6604,0.6124,0.6334,0.6374,0.5514,0.5514,0.5412,0.5514,0.5374,0.5473,0.4825,0.5084,0.5126,0.5229,0.5126,0.5043,0.4379,0.4781,0.4600,0.4781,0.3806,0.4078,0.3096,0.3263,0.3399,0.3184,0.2820,0.2167,0.2122,0.2080,0.2558,0.2255,0.1921,0.1766,0.1732,0.1205,0.1732,0.0723,0.0701,0.0405,0.0643,0.0771,0.1018,0.0587,0.0884,0.0884,0.1240,0.1088,0.0554,0.0607,0.0441,0.0387,0.0490,0.0478,0.0231,0.0414,0.0297,0.0701,0.0502,0.0567,0.0405,0.0363,0.0464,0.0701,0.0832,0.0991,0.1322,0.1998,0.3146,0.3146,0.3184,0.3578,0.3311,0.3184,0.4203,0.3578,0.3578,0.3578,0.4282,0.5084,0.5802,0.5667,0.5473,0.5514,0.5331,0.4749,0.4037,0.4116,0.4203,0.3184,0.4037,0.4037,0.4282,0.4513,0.4749,0.4116,0.4825,0.4918,0.4879,0.4918,0.4825,0.4245,0.4333,0.4651,0.4879,0.5412,0.5802,0.5126,0.4458,0.5374,0.4600,0.4600,0.4600,0.4600,0.3992,0.4879,0.4282,0.4333,0.3668,0.3005,0.3096,0.3847,0.3939,0.3630,0.3359,0.2292,0.2292,0.2748,0.3399,0.2963,0.2963,0.2385,0.2531,0.1805,0.2531,0.2786,0.3456,0.3399,0.3491,0.4037,0.3885,0.3806,0.2748,0.2700,0.2657,0.2963,0.2865,0.2167,0.2080,0.1844,0.2041,0.1602,0.1416,0.2041,0.1958,0.1018,0.0744,0.0677,0.0909,0.0789,0.0723,0.0660,0.1322,0.1532,0.1060,0.1018,0.1060,0.1150,0.0789,0.1266,0.0965,0.1732,0.1766,0.1766,0.1805,0.2820,0.3096,0.2602,0.2080,0.2333,0.2385,0.2385,0.2432,0.1602,0.2122,0.2385,0.2333,0.2558,0.2432,0.2292,0.2209,0.2483,0.2531,0.2432,0.2432,0.2432,0.2432,0.3053,0.3630,0.3578,0.3630,0.3668,0.3263,0.3992,0.4037,0.4556,0.4703,0.5173,0.6219,0.6412,0.7275,0.6984,0.6756,0.7079,0.7192,0.7342,0.7458,0.7501,0.7540,0.7605,0.7605,0.7342,0.7912,0.7951,0.8036,0.8074,0.8074,0.8118,0.7951,0.8118,0.8242,0.8488,0.8650,0.8488,0.8311,0.8424,0.7912,0.7951,0.8001,0.8001,0.7458,0.7192,0.6984,0.6412,0.6516,0.5900,0.5802,0.5802,0.5762,0.5623,0.5374,0.4556,0.4556,0.4333,0.3762,0.3456,0.4037,0.3311,0.3263,0.3311,0.3717,0.3762,0.3717,0.3668,0.3491,0.4203,0.4037,0.4149,0.4037,0.3992,0.4078,0.4651,0.4967,0.5229,0.5802,0.5802,0.5846,0.6293,0.6412,0.6374,0.6604,0.7317,0.7034,0.7573,0.7573,0.7573,0.7772,0.7605,0.8036,0.7951,0.7817,0.7869,0.7724,0.7869,0.7869,0.7951,0.7644,0.7912,0.7275,0.7342,0.7275,0.6984,0.7342,0.7605,0.7418,0.7418,0.7275,0.7573,0.7724,0.8118,0.8521,0.8823,0.8984,0.9119,0.9316,0.9512)
yy <- runmed(y, 41)
plot(y, type="l", ylim=c(0,1), ylab="", xlab="", lwd=0.5)
points(yy, col="blue", type="l", lwd=2)
EDITED : function strips the regions to contain nothing but the lowest part, if wanted.
Actually, Using the mean is easier than using the median. This allows you to find regions where the real values are continuously below the mean. The median is not smooth enough for an easy application.
One example function to do this would be :
FindLowRegion <- function(x,n=length(x)/4,tol=length(x)/20,p=0.5){
nx <- length(x)
n <- 2*(n %/% 2) + 1
# smooth out based on means
sx <- rowMeans(embed(c(rep(NA,n/2),x,rep(NA,n/2)),n),na.rm=T)
# find which series are far from the mean
rlesx <- rle((sx-x)>0)
# construct start and end of regions
int <- embed(cumsum(c(1,rlesx$lengths)),2)
# which regions fulfill requirements
id <- rlesx$value & rlesx$length > tol
# Cut regions to be in general smaller than median
regions <-
apply(int[id,],1,function(i){
i <- min(i):max(i)
tmp <- x[i]
id <- which(tmp < quantile(tmp,p))
id <- min(id):max(id)
i[id]
})
# return
unlist(regions)
}
where
n determines how much values are used to calculate the running mean,
tol determines how many consecutive values should be lower than the running mean to talk about a low region, and
p determines the cutoff used (as a quantile) for stripping the regions to their lowest part. When p=1, the complete lower region is shown.
Function is tweaked to work on data as you presented, but the numbers might need to be adjusted a bit to work with other data.
This function returns a set of indices, which allows you to find the low regions. Illustrated with your y vector :
Lows <- FindLowRegion(y)
newx <- seq_along(y)
newy <- ifelse(newx %in% Lows,y,NA)
plot(y, col="blue", type="l", lwd=2)
lines(newx,newy,col="red",lwd="3")
Gives :
You have to smooth the graph in some way. Median filtration is quite useful for that purpose (see http://en.wikipedia.org/wiki/Median_filter). After smoothing, you will simply have to search for the minima, just as usual (i.e. search for the points where the 1st derivative switches from negative to positive).
A simpler answer (which also does not require smoothing) could be provided by adapting the maxdrawdown() function from the tseries. A drawdown is commonly defined as the retreat from the most-recent maximum; here we want the opposite. Such a function could then be used in a sliding window over the data, or over segmented data.
maxdrawdown <- function(x) {
if(NCOL(x) > 1)
stop("x is not a vector or univariate time series")
if(any(is.na(x)))
stop("NAs in x")
cmaxx <- cummax(x)-x
mdd <- max(cmaxx)
to <- which(mdd == cmaxx)
from <- double(NROW(to))
for (i in 1:NROW(to))
from[i] <- max(which(cmaxx[1:to[i]] == 0))
return(list(maxdrawdown = mdd, from = from, to = to))
}
So instead of using cummax(), one would have to switch to cummin() etc.
My first thought was something much cruder than filtering. Why not look for the big drops followed by long enough stable periods?
span.b <- 20
threshold.b <- 0.2
dy.b <- c(rep(NA, span.b), diff(y, lag = span.b))
span.f <- 10
threshold.f <- 0.05
dy.f <- c(diff(y, lag = span.f), rep(NA, span.f))
down <- which(dy.b < -1 * threshold.b & abs(dy.f) < threshold.f)
abline(v = down)
The plot shows that it's not perfect, but it doesn't discard the outliers (I guess it depends on your take on the data).

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