I have two vectors, x and y.
x is a vector where each entry represents a month for a period of several years, so I have (let's say) 10 years of data, then length(x) = 120 and so on.
(I have used the "posix.ct" command so they really are "months" in that sense, but couldn't I just have x as a numerical vector like c(1:n) or something, since I already know which month and which year a certain element of c(1:n) corresponds to? i.e if x = c(1:n), I know that x[13] is february of the second year and so on..)
y is a vector where each elements is an observation of a particular variable at a certain month.
So the observed data is grouped like this (january,0.123), (february,2.125) and so on.
I have two vectors for the months;
x1 = seq(as.POSIXct("YYYY-MM-DD", tz="GMT"),
as.POSIXct("YYYY-MM-DD", tz="GMT"),
by="month")
x2 = c(1:length(x1))
What I want to do is to run ksmooth:
plot(x1,y)
smooth = ksmooth(x2,y,"normal")
lines(smooth)
The reason that I use x1 in the plot() command is that I don't know how to otherwise get the x-axis in time.
R should automatically find a decent smoothing parameter when I haven't specified anything. The result is that ksmooth$y is equal to the input vector y! Also, a vertical bar is produced in the plot. If I replace x2 by x1 in the code above, ksmooth$y is NA for all values except for the first and last, which equal those of the input y.
So i try some bandwidths:
h = 0.1: now smooth$y = y, as before. A vertical bar is produced (it is the same color as I specified in the lines() command, so it must have to do with the ksmooth command.)
h = 10: get some non-strange results for smooth$y, however, a vertical bar is produced as before.
Then, I tried the crazy idea of very large bandwidths;
h = 1e+06: This produced nothing when I used x1 and x2 as in the code above. When I changed x2 to x1 however, I get some good results. For h = 1e+09 (that's huge!!) I get a very nice result. (I get a curve that fits the data and looks nice)
But h = 1e+09, is that reasonable? in all the examples I have looked h is something betweeen 0.1 and 10, give or take. heard something about a rule of thumb: h should equal n^(-1/5) where n is the number of data points.
I think the one thing that you are missing is that R doesn't find a decent smoothing parameter when you haven't specified anything, it just uses a bandwidth of 0.5, which is totally useless in your case.
The other thing you might be missing is that in ksmooth the bandwidth parameter is in terms of x. When ksmooth takes an x value of Date, it converts it to a numeric, which is the number of seconds. Therefore, your bandwidth will be measured in seconds, an undesirable result. When ksmooth takes an x value of months, it will default to a bandwidth of 0.5 months, also undesirable.
What you want to do is specify a reasonable bandwidth for the x that you are using. Here is an example:
x1 = seq(as.POSIXct("2000-01-01", tz="GMT"),
as.POSIXct("2010-12-31", tz="GMT"),
by="month")
x2 = c(1:length(x1))
set.seed(1)
y = runif(length(x1))
plot(x1,y,type='l')
smooth = ksmooth(x2,y,"normal")
lines(x1,smooth$y,col='blue',lwd=2)
lines(x1,ksmooth(x2,y,'normal',bandwidth=2)$y,col='red',lwd=2)
lines(x1,ksmooth(x2,y,'normal',bandwidth=10)$y,col='green',lwd=2)
lines(x1,ksmooth(x2,y,'normal',bandwidth=20)$y,col='orange',lwd=2)
Related
Taking the ideas from the following links:
the local minimum between the two peaks
How to explain ...
I look for the local minimum or minimums, avoiding the use of functions already created for this purpose [max / min locale or global].
Our progress:
#DATA
simulate <- function(lambda=0.3, mu=c(0, 4), sd=c(1, 1), n.obs=10^5) {
x1 <- rnorm(n.obs, mu[1], sd[1])
x2 <- rnorm(n.obs, mu[2], sd[2])
return(ifelse(runif(n.obs) < lambda, x1, x2))
}
data <- simulate()
hist(data)
d <- density(data)
#
#https://stackoverflow.com/a/25276661/8409550
##Since the x-values are equally spaced, we can estimate dy using diff(d$y)
d$x[which.min(abs(diff(d$y)))]
#With our data we did not obtain the expected value
#
d$x[which(diff(sign(diff(d$y)))>0)+1]#pit
d$x[which(diff(sign(diff(d$y)))<0)+1]#peak
#we check
#1
optimize(approxfun(d$x,d$y),interval=c(0,4))$minimum
optimize(approxfun(d$x,d$y),interval=c(0,4),maximum = TRUE)$maximum
#2
tp <- pastecs::turnpoints(d$y)
summary(tp)
ind <- (1:length(d$y))[extract(tp, no.tp = FALSE, peak = TRUE, pit = TRUE)]
d$x[ind[2]]
d$x[ind[1]]
d$x[ind[3]]
My questions and request for help:
Why did the command lines fail:
d$x[which.min(abs(diff(d$y)))]
It is possible to eliminate the need to add one to the index in the command lines:
d$x[which(diff(sign(diff(d$y)))>0)+1]#pit
d$x[which(diff(sign(diff(d$y)))<0)+1]#peak
How to get the optimize function to return the two expected maximum values?
Question 1
The answer to the first question is straighforward. The line d$x[which.min(abs(diff(d$y)))] asks for the x value at which there was the smallest change in y between two consecutive points. The answer is that this happened at the extreme right of the plot where the density curve is essentially flat:
which.min(abs(diff(d$y)))
#> [1] 511
length(abs(diff(d$y)))
#> [1] 511
This is not only smaller than the difference at your local maxima /minima points; it is orders of magnitude smaller. Let's zoom in to the peak value of d$y, including only the peak and the point on each side:
which.max(d$y)
#> [1] 324
plot(d$x[323:325], d$y[323:325])
We can see that the smallest difference is around 0.00005, or 5^-5, between two consecutive points. Now look at the end of the plot where it is flattest:
plot(d$x[510:512], d$y[510:512])
The difference is about 1^-7, which is why this is the flattest point.
Question 2
The answer to your second question is "no, not really". You are taking a double diff, which is two elements shorter than x, and if x is n elements long, a double diff will correspond to elements 2 to (n - 1) in x. You can remove the +1 from the index, but you will have an off-by-one error if you do that. If you really wanted to, you could concatenate dummy zeros at each stage of the diff, like this:
d$x[which(c(0, diff(sign(diff(c(d$y, 0))))) > 0)]
which gives the same result, but this is longer, harder to read and harder to justify, so why would you?
Question 3
The answer to the third question is that you could use the "pit" as the dividing point between the minimum and maximum value of d$x to find the two "peaks". If you really want a single call to get both at once, you could do it inside an sapply:
pit <- optimize(approxfun(d$x,d$y),interval=c(0,4))$minimum
peaks <- sapply(1:2, function(i) {
optimize(approxfun(d$x, d$y),
interval = c(min(d$x), pit, max(d$x))[i:(i + 1)],
maximum = TRUE)$maximum
})
pit
#> [1] 1.691798
peaks
#> [1] -0.02249845 3.99552521
I have about 50 datasets that include all trades within a timeframe of 30 days for about 10 pairs on 5 exchanges. All pairs are of the same asset class, meaning they are strongly correlated and expect to have similar properties, but are on different scales. An example of this data would be
set.seed(1)
n <- 1000
dates <- seq(as.POSIXct("2019-08-05 00:00:00", tz="UTC"), as.POSIXct("2019-08-05 23:59:00", tz="UTC"), by="1 min")
x <- data.frame("t" = sort(sample(dates, 1000)),"p" = cumsum(sample(c(-1, 1), n, TRUE)))
Roughly, I need to identify the relevant local minima and maxima, which happen daily. The yellow marks are my points of interest. Unlike this example, there is usually only one such point per day and I consider each day separately. However, it is hard to filter out noise from my actual points of interest.
My actual goal is to find the exact point, at which the pair started to make a jump and the exact point, at which the jump is over. This needs to be as accurate as possible, as I want to observe which asset moved first and which asset followed at which point in time (as said, they are highly correlated).
Between two extreme values, I want to minimize the distance and maximize the relative/absolute change, as my points of interest are usually close to each other and their difference is quite large.
I already looked at other questions like
Finding local maxima and minima and Algorithm to locate local maxima and also this algorithm that has the same goal. However, my dataset is extremely noisy. I already reduced the dataset to 5-minute intervals, however, this has led to omitting the relevant points in the functions to identify local minima & maxima. Therefore, this was a not good solution given my goal.
How can I achieve my goal with a quite accurate algorithm? Manually skimming through all the time-series is not an option, since this would require me to evaluate 50 * 30 time-series manually, which is too time-consuming. I'm really puzzled and trying to find a suitable solution for a week.
If more code snippets are demanded, I'm happy to share, however they didn't give me meaningful results, which would be opposed to the idea of providing a minimum working example, therefore I decided to leave them out for now.
EDIT:
First off, I updated the plot and added timestamps to the dataset to give you an idea (the actual resolution). Ideally, the algorithm would detect both jumps on the left. The inner two dots because they're closer together and jump without interception, and the outer dots because they're more extreme in values. In fact, this maybe answers the question whether the algorithm is allowed to look into the future. Yes, if there's another local extrema in the range of, say, 30 observations (or 30 minutes), then ignore the intermediate local extrema.
In my data, jumps have been from 2% - ~ 15%, such that a jump needs to be at least 2% to be considered. And only if a threshold of 15 (this might be adaptable) consecutive steps in the same direction before / after the peaks and valleys is reached.
A very naive approach was to subset the data around the global minimum and maximum of a day. In most cases, this has denoised data and worked as an indicator. However, this is not robust when the global extrema are not in the range of the jump.
Hope this clarifies why this isn't a statistical question (there are some tests to determine whether a jump has happened, but not for jump arrival time afaik).
In case anyone wants a real example:
this is a corresponding graph, this is the raw data of the relevant period and this is the reduced dataset.
Perhaps as a starting point, look at function streaks
in package PMwR (which I maintain). A streak is
defined as a move of a specified size that is
uninterrupted by a countermove of the same size. The
function works with returns, not differences, so I add
100 to your data.
For instance:
set.seed(1)
n <- 1000
x <- 100 + cumsum(sample(c(-1, 1), n, TRUE))
plot(x, type = "l")
s <- streaks(x, up = 0.12, down = -0.12)
abline(v = s[, 1])
abline(v = s[, 2])
The vertical lines show the starts and ends of streaks.
Perhaps you can then filter the identified streaks by required criteria such as length. Or
you may play around with different thresholds for up
and down moves (though this is not really recommended
in the current implementation, but perhaps the results
are good enough). For instance, up streaks might look as follows. A green vertical shows the start of a streak; a red line shows its end.
plot(x, type = "l")
s <- streaks(x, up = 0.12, down = -0.05)
s <- s[!is.na(s$state) & s$state == "up", ]
abline(v = s[, 1], col = "green")
abline(v = s[, 2], col = "red")
Suppose you have two lines, L1 and L2, which for each x value (x1 and x2 for example) they have known points at L1={(x1,L1_y1), (x2,L1_y2)}, and L2={(x1,L2_y1), (x2,L2_y2)}. By joining these points they may or may not have an intersection at some x3 where x1
Now suppose you want to know the maximum at any x value (not restricted to just x1, x2 etc, but anywhere along the axis) of both of these lines. Obviously it is often trivial to calculate for just a few lines, and a few different x value, but in my case I have several tens of thousand x values and a few lines to check it against, so it can't be done manually.
In R, is there some code which will calculate the maximum at any given point x3?
An example of this can be seen here with L1={(1,1), (2,4)}, and L2={(1,4),(2,1)}, illustrated by:
Here the intersection of these lines is at (1.5, 2.5). L2 is the maximum before this, and L1 after. This maximum line is shown in red below.
As you can see, it isn't enough just to take the max at every point and join these up, and so it will need to consider the lines as some form of function, and then take the maximum of this.
Also, as mention before as there are several thousand x values it will need to generalise to larger data.
To test the code further if you wish you can randomly generate y values for some x values, and it will be clear to see from a plot if it works correctly or not.
Thanks in advance!
Defining points constituting your lines from the example
L1 <- list(x = c(1, 2), y = c(1, 4))
L2 <- list(x = c(1, 2), y = c(4, 1))
defining a function taking a pointwise maximum of two functions corresponding to the lines
myMax <- function(x)
pmax(approxfun(L1$x, L1$y)(x), approxfun(L2$x, L2$y)(x))
This gives
plot(L1$x, L1$y, type = 'l')
lines(L2$x, L2$y, col = 'red')
curve(myMax(x), from = 1, to = 2, col = 'blue', add = TRUE)
Clearly this extends to more complex L1 and L2 as approxfun is just a piecewise-linear approximation. Also, you may add L3, L4, and so on.
I'm new to R and I am attempting to take a set of time series and run them through a Conditional Inference Tree to help classify the shape of the time series. The problem is that not all of the time sereis are of the same number of periods. I am trying to expand each time series to be 30 periods long, but still maintain the same "shape". This is as far as I have got
Require(zoo)
test<-c(606,518,519,541,624,728,560,512,777,728,1014,1100,930,798,648,589,680,635,607,544,566)
accordion<-function(A,N){
x<-ts(scale(A), start=c(1,1), frequency=1)
X1 <- zoo(x,seq(from = 1, to = N, by =(N-1)/(length(x)-1) ))
X2<-merge(X1, zoo(order.by=seq(start(X1), end(X1)-1, by=((N-1)/length(x))/(N/length(x)))))
X3<-na.approx(X2)
return(X3)}
expand.test<-accordion(test,30)
plot(expand.test); lines(scale(test))
length(expand.test)
The above code, scales the time series and then evenly spaces it out to 30 periods and interpolates the missing values. However, the length of the returned series is 42 units and not 30, however it retains the same "shape" as the orignal time series. Does anyone know how to modify this so that the results produced by the function accordian are 30 periods long and the time series shape remains relatively unchanged?
I think there's a base R solution here. Check out approx(), which does linear (or constant) interpolation with as many points n as you specify. Here I think you want n = 30.
test2 <- approx(test, n=30)
plot(test2)
points(test, pch="*")
This returns a list test2 where the second element y is your interpolated values. I haven't yet used your time series object, but it seems that was entirely interior to your function, correct?
I have a following matrix [500,2], so we have 500 rows and 2 columns, the left one gives us the index of X observations, and the right one gives the probability with which this X comes true, so - a typical probability density relationship.
So, my question is, how to plot the histogram the right way, so that the x-axis is the x-index, and the y-axis is the density(0.01-1.00). The bandwidth of the estimator is 0.33.
Thanks in advance!
the end of the whole data looks like this: just for a little orientation
[490,] 2.338260830 0.04858685
[491,] 2.347839477 0.04797310
[492,] 2.357418125 0.04736149
[493,] 2.366996772 0.04675206
[494,] 2.376575419 0.04614482
[495,] 2.386154067 0.04553980
[496,] 2.395732714 0.04493702
[497,] 2.405311361 0.04433653
[498,] 2.414890008 0.04373835
[499,] 2.424468656 0.04314252
[500,] 2.434047303 0.04254907
#everyone,
yes, I have made the estimation before, so.. the bandwith is what I mentioned, the data is ordered from low to high values, so respecively the probability at the beginning is 0,22, at the peak about 0,48, at the end 0,15.
The line with the density is plotted like a charm but I have to do in addition is to plot a histogram! So, how I can do this, ordering the blocks properly(ho the data to be splitted in boxes etc..)
Any suggestions?
Here is a part of the data AFTER the estimation, all values are discrete, so I assume histogram can be created.., hopefully.
[491,] 4.956164 0.2618131
[492,] 4.963014 0.2608723
[493,] 4.969863 0.2599309
[494,] 4.976712 0.2589889
[495,] 4.983562 0.2580464
[496,] 4.990411 0.2571034
[497,] 4.997260 0.2561599
[498,] 5.004110 0.2552159
[499,] 5.010959 0.2542716
[500,] 5.017808 0.2533268
[501,] 5.024658 0.2523817
Best regards,
appreciate the fast responses!(bow)
What will do the job is to create a histogram just for the indexes, grouping them in a way x25/x50 each, for instance...and compute the average probability for each 25 or 50/100/150/200/250 etc as boxes..?
Assuming the rows are in order from lowest to highest value of x, as they appear to be, you can use the default plot command, the only change you need is the type:
plot(your.data, type = 'l')
EDIT:
Ok, I'm not sure this is better than the density plot, but it can be done:
x = dnorm(seq(-1, 1, length = 500))
x.bins = rep(1:50, each = 10)
bars = aggregate(x, by = list(x.bins), FUN = sum)[,2]
barplot(bars)
In your case, replace x with the probabilities from the second column of your matrix.
EDIT2:
On second thought, this only makes sense if your 500 rows represent discrete events. If they are instead points along a continuous distribution function adding them together as I have done is incorrect. Mathematically I don't think you can produce the binned probability for a range using only a few points from within that range.
Assuming M is the matrix. wouldn't this just be :
plot(x=M[ , 1], y = M[ , 2] )
You have already done the density estimation since this is not the original data.