How I use numerical methods to calculate roots in R - r
A machine learning model predicted probability p using input x. It is unknown how model calculates the probability.
In the example below,
We have 100 xand p values.
Can someone please show an algorithm to find all values of x for which p is 0.5.
There are two challenges
I don't know the function p = f(x). I don't wish to fit some smooth polynomial curves which will remove the noise. The noises are important.
x values are discrete. So, we need to interpolate to find the desired values of x.
library(tidyverse)
x <- c(0.1,0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9,1.0,1.1,1.2,1.3,1.4,1.5,1.6,1.7,1.8,1.9,2.0, 2.1,2.2,2.3,2.4,2.5,2.6,2.7,2.8,2.9,3.0,3.1,3.2,3.3,3.4,3.5,3.6,3.7,3.8,3.9,4.0,4.1, 4.2,4.3,4.4,4.5,4.6,4.7,4.8,4.9,5.0,5.1,5.2,5.3,5.4,5.5,5.6,5.7,5.8,5.9,6.0,6.1,6.2, 6.3,6.4,6.5,6.6,6.7,6.8,6.9,7.0,7.1,7.2,7.3,7.4,7.5,7.6,7.7,7.8,7.9,8.0,8.1,8.2,8.3, 8.4,8.5,8.6,8.7,8.8,8.9,9.0,9.1,9.2,9.3,9.4,9.5,9.6,9.7,9.8,9.9, 10.0)
p <- c(0.69385203,0.67153592,0.64868391,0.72205029,0.64917218,0.66818861,0.55532616,0.58631660,0.65013198,0.53695673,0.57401464,0.57812980,0.39889101,0.41922821,0.44022287,0.48610191,0.34235438,0.30877592,0.20408235,0.17221558,0.23667792,0.29237938,0.10278049,0.20981142,0.08563396,0.12080935,0.03266140,0.12362265,0.11210208,0.08364931,0.04746024,0.14754152,0.09865584,0.16588175,0.16581508,0.14036209,0.20431540,0.19971309,0.23336415,0.12444293,0.14120138,0.21566896,0.18490258,0.34261082,0.38338941,0.41828079,0.34217964,0.38137610,0.41641546,0.58767796,0.45473784,0.60015956,0.63484702,0.55080768,0.60981219,0.71217369,0.60736818,0.78073246,0.68643671,0.79230105,0.76443958,0.74410139,0.63418201,0.64126278,0.63164615,0.68326471,0.68154362,0.75890922,0.72917978,0.55839943,0.55452549,0.69419777,0.64160572,0.63205751,0.60118916,0.40162340,0.38523375,0.39309260,0.47021037,0.33391614,0.22400555,0.20929558,0.20003229,0.15848124,0.11589228,0.13326047,0.11848593,0.17024106,0.11184393,0.12506915,0.07740497,0.02548386,0.07381765,0.02610759,0.13271803,0.07034573,0.02549706,0.02503864,0.11621910,0.08636754)
tbl <- tibble(x, p)
# plot for visualization
ggplot(data = tbl,
aes(x = x,
y = p)) +
geom_line() +
geom_point() +
geom_hline(yintercept = 0.5) +
theme_bw() +
theme(aspect.ratio = 0.4)
The figure below shows that there are five roots.
This question is clearer than your previous one: How I use numerical methods to calculate roots in R.
I don't know the function p = f(x)
So you don't have a predict function to calculate p for new x values. This is odd, though. Many statistical models have methods for predict. As BenBolker mentioned, the "obvious" solution is to use uniroot or more automated routines to find a or all roots, for the following template function:
function (x, model, p.target) predict(model, x) - p.target
But this does not work for you. You only have a set of (x, p) values that look noisy.
I don't wish to fit some smooth polynomial curves which will remove the noise. The noises are important.
So we need to interpolate those (x, p) values for a function p = f(x).
So, we need to interpolate to find the desired values of x.
Exactly. The question is what interpolation method to use.
The figure below shows that there are five roots.
This line chart is actually a linear interpolation, consisting of piecewise line segments. To find where it crosses a horizontal line, you can use function RootSpline1 defined in my Q & A back in 2018: get x-value given y-value: general root finding for linear / non-linear interpolation function
RootSpline1(x, p, 0.5)
#[1] 1.243590 4.948805 5.065953 5.131125 7.550705
Thank you very much. Please add the information of how to install the required package. That will help everyone.
This function is not in a package. But this is a good suggestion. I am now thinking of collecting all functions I wrote on Stack Overflow in a package.
The linked Q & A does mention an R package on GitHub: https://github.com/ZheyuanLi/SplinesUtils, but it focuses on splines of higher degree, like cubic interpolation spline, cubic smoothing spline and regression B-splines. Linear interpolation is not dealt with there. So for the moment, you need to grab function RootSpline1 from my Stack Overflow answer.
Related
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R - poly.calc not stable when using many points
I've been trying to solve a problem using Lagrange interpolation, which is implemented in poly.calc method (polynom package) in R language. Basically, my problem is to predict the population of a certain country using Lagrange Interpolation. I have the population from the past years (1961 - 2014). The csv file is here w1 = read.csv(file="country.csv", sep=",", head=TRUE) array_x = w1$x array_y = w1$y #calls Lagrange Method p = poly.calc(array_x, array_y) #create a function to evaluate the polynom prf <- as.function(p) #create some points to plot myx = seq(1961, 2020, 0.5) #y's to plot myy = prf(myx) #plot plot(myx, myy,col='blue') After that, the plotted curve is declining and the y-axis is (very big) negative (power of 134). It does not make sense. However, if I use like five points, it is correct.
This is not really an SO question but rather a numerical analysis question. R is doing everything you want it to, it's not a programming error. It's just that what you want it to do is notoriously bad. Lagrange polynomials are notorious for being incredibly unstable, especially when a large number of points are fit. A much more stable alternative is the use of splines, such as B-splines. They can be fit very easily with R's default spline library into any regression model, i.e. you could fit a least squares model with library(splines) x <- sort(runif(500, -3,3) ) #sorting makes for easier plotting ahead y <- sin(x) splineFit <- lm(y ~ bs(x, df = 5) ) est_y <- predict(splineFit) plot(x, y, type = 'l') lines(x, est_y, col = 'blue') You can see from the above model that the splines can do a good job of fitting non-linear relations.
How to plot a log curve in R?
I have the following set of data: x = c(8,16,64,128,256) y = c(7030.8, 3624.0, 1045.8, 646.2, 369.0) Which, when plotted, looks like an exponential decay or negative ln function. I'm trying to fit a smooth curve to this data, but I don't know how. I've tried nls and lm functions, but I can't seem to get it right. The online examples have too many steps for the simple data I have, and I can't understand well enough to modify the examples for what I need. Any help or advice would be appreciated. Thank you. Edit: When I say I tried nls and lm functions, I mean that the lines produced were linear, no matter what parameters I tried. And when I say too many steps, I mean the examples I found were for predicting with 2 independent variables, or for creating multiple fit lines. What I'm asking is what is the best way to fit a simple smooth line to data that, when graphed, looks like an exponential decay or negative ln. What the equation of the line is isn't important, it's meant to be a reference for the shape of the data.
A good way to fit a curve to a function is the built-in nls function, which performs non-linear least squares optimization. For example, if you wanted to fit the model y = b * x^e, you could do: n <- nls(y ~ b * x ^ e, data = data.frame(x, y), start = c(b = 1000, e = -1)) (?nls, or this walkthrough, can tell you more about these options). You could then plot the curve on top of your points: plot(x, y) curve(predict(n, newdata = data.frame(x = x)), add = TRUE) You can try a few other models (specified by that formula in nls) that may fit your data.
Maybe 'lowess' is what you're looking for? Try: plot(y ~ x) lines(lowess(y ~ x)) That function just connects the dots. It sounds like you would prefer something that smooths out the elbows. In principle, 'loess' is useful for that, but you don't have enough data points here for that to work.
Plotting fitted values vs observed ones in R or winbugs
I want to plot the fitted values versus the observed ones and want to put straight line showing the goodness of fit. However, I do not want to use abline() because I did not calculate the fitted values using lm command as my I used a model that R does not cover. I calculated the coefficients and used them to calculate the fitted values. So, what can I do to obtain such a plot in R or in winbugs? Here is what I want
Still no data provided, but maybe this simple example using the curve function will inform the process: x <- 1:10 y <- 2+ 3*(1:10) + rnorm(10) plot(1:10, y) curve( 2+3*x, 0, 10, add=TRUE) Note to new R users. the expression y_i = 1 - xbeta + delta_i + e_i would fail in R in part because the x and beta are not separated by an operator. But if you do understand R's matrix syntax it might be a very compact expression even if "X" were multidimensional. All of htis depends on the specifics which we are so far lacking.
finding functions that match dot plots
ggplot(test,aes(x=timepoints,y= mean,ymax = mean + sde, ymin = mean - sde)) +
geom_errorbar(width=2) +
geom_point() +
geom_line() +
stat_smooth(method='loess') +
xlab('Time (min)') +
ylab('Fold Induction') +
opts(title = 'yo')
I can plot the blue 'loess'-ed line. But is there a way to find the mathematical function of the blue 'loess'-ed line?
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Rule Number One: not all distributions have a (closed-form) function which generates them. Yes, you can create a close fit by way of splines, or calculating moments (mean, variance, skew, etc) and building the series, so your choice depends on whether you intend to interpolate, extrapolate, or just "view" the resultant function. In the scientific world, it's more common to have a theory, or premise, about the behavior behind your data. You can then do standard (e.g. nls) fitting methods to see how well the proposed fit function can be made to match your data.
To understand how the loess line is computed see the loess.demo function in the TeachingDemos package. This is an interactive graphical demonstration that will show how the y-value at each point is computed for each x-value based on the data and bandwidth parameter (it also shows the difference in the raw loess fit and the spline that is often fit to the loess estimates).