I have created plots of heart rate recovery against time, using ggplot2.
I have noticed them seem to have an inflection point where the heart rate levels off. I was looking to ask if there is a way of determining this inflection point for each subject, using R studio instead of doing it manually myself?
This is the script for plots:
ggplot(data = f, aes(x=Seconds, y=Heart.Rate, group=ID, colour=ID + geom_point() + geom_line()
You should write a script to calculate the change in Y divided by change in X. This is basically derivative of y wrt x. In a distance versus time curve this gives you velocity. Often a sudden change will create an easily discernable peak or inflection in the derivative curve. This should be simple to calculate as well. It's just the difference in 2 consecutive y points divided by the difference in their corresponding x values. You can then plot this new curve and see what it looks like. You will have to start at point 2 since there is no earlier value for point 1.
Related
I am back at college learning maths and I want to try and use some this knowledge to create some svg with d3.js.
If I have a function f(x) = x^3 - 3x^2 + 3x - 1
I would take the following steps:
Find the x intercepts for when y = 0
Find the y intercept when x = 0
Find the stationary points when dy\dx = 0
I would then have 2 x values from point 3 to plug into the original equation.
I would then draw a nature table do judge the flow of the graph or curve.
Plot the known points from the above and sketch the graph.
Translating what I would do on pen and paper into code instructions is what I really could do with any sort of advice on the following:
How can I programmatically factorise point 1 of the above to find the x-intercepts for when y = 0. I honestly do not know where to even start.
How would I programmatically find dy/dx and the values for the stationary points.
If I actually get this far then what should I use in d3 to join the points on the graph.
Your other "steps" have nothing to do with d3 or plotting.
Find the x intercepts for when y = 0
This is root finding. Look for algorithms to help with this.
Find the y intercept when x = 0
Easy: substitute to get y = 1.
Find the stationary points when dy\dx = 0
Take the first derivative to get 3x^2 - 12x + 9 and repeat the root finding step. Easy to get using quadratic equation.
I would then have 2 x values from point 3 to plug into the original
equation. I would then draw a nature table do judge the flow of the
graph or curve. Plot the known points from the above and sketch the
graph.
I would just draw the curve. Pick a range for x and go.
It's great to learn d3. You'll end up with something like this:
https://maurizzzio.github.io/function-plot/
For a cubic polynomial, there are closed formulas available to find all the particular points that you want (https://en.wikipedia.org/wiki/Cubic_function), and it is a sound approach to determine them.
Anyway, you will have to plot the smooth curve, which means that you will need to compute close enough points and draw a polyline that joins them.
Doing this, you are actually performing the first steps of numerical root isolation, with such an accuracy that the approximate and exact roots will be practically undistinguishable.
So an easy combined solution is to draw the curve as a polyline and find the intersections with the X axis as well as extrema using this polyline representation, rather than by means of more sophisticated methods.
This approach works for any continuous curve and is very easy to implement. So you actually draw the curve to find particular points rather than conversely as is done by analytical methods.
For best results on complicated curves, you can adapt the point density based on the local curvature, but this is another story.
The only equation to calculate this that I can find involves t in the range [0, 1], but I have no idea how long it will take to travel the entire path, so I can't calculate (1 - t).
I know the speed at which I'm traveling, but it seems to be a heavy idea to calculate the total time beforehand (nor do I actually know how to do that calculation). What is an equation to figure out the position without knowing the total time?
Edit To clarify on the cubic bezier curve: I have four control points (P0 to P1), and to get a value on the curve with t, I need to use the four points as such:
B(t) = (1-t)^3P0 + 3t(1-t)^2P1 + 3t^2(1-t)P2 + t^3P3
I am not using a parametric equation to define the curve. The control points are what define the curve. What I need is an equation that does not require the use of knowing the range of t.
I think there is a misunderstanding here. The 't' in the cubic Bezier curve's definition does not refer to 'time'. It is parameter that the x, y or even z functions based on. Unlike the traditional way of representing y as a function of x, such as y=f(x), an alternative way of representing a curve is by the parametric form that represents x, y and z as functions of an additional parameter t, C(t)=(x(t), y(t), z(t)). Typically the t value will range from 0 to 1, but this is not a must. The common representation for a circle as x=cos(t) and y=sin(t) is an example of parametric representation. So, if you have the parametric representation of a curve, you can evaluate the position on the curve for any given t value. It has nothing to do with the time it takes to travel the entire path.
You have the given curve and you have your speed. To calculate what you're asking for you need to divide the total distance by the speed you traveled given that time. That will give you the parametric (t) you need. So if the total curve has a distance of 72.2 units and your speed is 1 unit then your t is 1/72.2.
Your only missing bit is calculating the length of a given curve. This is typically done by subdividing it into line segments small enough that you don't care, and then adding up the total distance of those line segments. You could likely combine those two steps as well if you were so inclined. If you have your given speed, just iteration like 1000th of the curve add the line segment between the start and point 1000th of the way through the curve, and subtract that from how far you need to travel (given that you have speed and time, you have distance you need to travel), and keep that up until you've gone as far as you need to go.
The range for t is between 0 and 1.
x = (1-t)*(1-t)*(1-t)*p0x + 3*(1-t)*(1-t)*t*p1x + 3*(1-t)*t*t*p2x + t*t*t*p3x;
y = (1-t)*(1-t)*(1-t)*p0y + 3*(1-t)*(1-t)*t*p1y + 3*(1-t)*t*t*p2y + t*t*t*p3y;
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.
I am capturing some points on an X, Y plane to represent some values. I ask the user to select a few points and I want the system to then generate a curve following the trend that the user creates. How do I calculate this? So say it is this:
Y = dollar amount
X = unit count
user input: (2500, 200), (4500, 500), (9500, 1000)
Is there a way I can calculate some sort of curve to follow those points so I would know based off that selection what Y = 100 would be on the same scale/trend?
EDIT: People keep asking for the nature of the curve, yes logarithmic. But I'd also like to check out some other options. It's for pricing the the restraint is that the as X increases Y should always be higher. However the rate of change of the curve should change related to the two adjacent points that the user selected, we could probably require a certain number of points. Does that help?
EDIT: Math is hard.
EDIT: Maybe a parabola then?
The problem is that there are multiple curves that you can fit to the same data. To borrow an example from my old stats book, here is the same data set (1, 1, 1, 10, 1, 1, 1) with four curves:
You need to specify the overall trend to get a meaningful result.
First, you are going to have to have an idea of what your line is or better said, what type of line fits your data the best. Is it linear (straight line) or does it curve (x-squared). Sounds like this is a curve.
If your curve is a parabola, then you will need to solve y = Ax(2) + Bx + c using your three points that the user has chosen. You will need at least 3 points to solve for 3 unknowns.
200 = A(2500)(2) + B(2500) + C
500 = A(4500)(2) + B(4500) + C
1000 = A(9500)(2) + B(9500) + C
Given these three equations, you should be able to solve for A, B and C, then use these to plot a new curve.
The Least Square Fit would give you a nice data matching curve.
This is a rather general extrapolation problem. In your case, fitting a quadric (parabola) is probably the most reasonable course of action. Depending on how well your data fits a quadric, you may want to fit it to more than 3 points (the noisier and weirder the data, the more points you should use).
Depending on the amount and type of data you have, you may want to try LOESS regression.
However, this may not be a good option if you only have 3 points as in your example (but keep in mind that you will not be able to have good extrapolation with 3 points no matter the algorithm you use)
Another option would be B-splines
I'm trying to design a program that draws graphs given a set of points (x, y), and it also should recognize the curve (straight line, hyperbole, parabola), with only the help of the points.
Is there an algorithm to do that?
You'll need five points if the curve could be a straight line or a conic (hyperbola, parabola, ellipse, circle).
If the five points are collinear, you have a straight line. (Or a degenerate conic? But if you're expecting straight lines, this should indicate a straight line.)
If four are collinear, you have a degenerate conic, given by the line through the four collinear points and any line through the fifth point that is not parallel to the first line.
If three are collinear, you have a degenerate conic, given by the line through the three collinear points and the line through the two other points. (Unless these two lines are parallel, in which case this isn't a conic.)
If no three points are collinear, you have a unique, non-degenerate conic.
To find the equation for this conic (Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0), take a look at this page, specifically the formula in the DETAILS section. Put in your five x and y values, calculate the determinant of the matrix in terms of x and y, and this will give you the formula of your conic. Then see here to figure out what kind of conic you have based on the values of A, B and C.
If you have more than 5 points, pick five points (preferably so no three are collinear), find the conic, and then check that the remaining points lie on the conic.
You can do it by watching the function extreme but is maybe not optimal solution for this problem (i mean a problem in parabola function like that y = sqrt(x*x-1)).
For straight line you can calc y = ax + b by the 2 random point's and check what other points equal this condition, if yes. This is straight line if no you may check a 2 other exceptions or nothing from it. Maybe somone else get solution for next 2 cases?