如何在R中获得线性插值的逆

Question

I went through many solutions for that problem, but could not find a reliable automated one. Please find below a detailed self-suficient description.

These are the data: data.txt

x y
1 1
2 2
3 3
4 2
5 1

Plotting as a scatter plot:

t=read.table("data.txt",header=T)
plot(t$x,t$y,"l")

You will see a peak, my question is now: assuming I am happy with the linear interpolation, what is the width at half maximum of that "curve"? So for which values x0 of x I have f(x0)=max(y)/2, where f is the linear interpolation. I tried with approxfun and some kernel density, but I do not want to smooth my data.

Any input is very welcome.

Answer 1

There are probably a lot of better ways of doing this, but here is one solution. It would be easier if we knew how you did the interpolation in the first place.

# Extend your line for testing
y<-c(1,2,3,2,1,2,3,4,5,4,3)

# Split into linear segments.
segments<-c(1,diff(diff(y)))!=0
seg.points<-which(c(segments,TRUE))

# Calculate desired value
val<-max(y)/2

# Loop through and find value
res<-c()
for(i in 1:(length(seg.points)-1)) {
  start<-y[seg.points[i]]
  end<-y[seg.points[i+1]]
  if ((val>=start & val<=end) | (val<=start & val >=end)) {
    slope=(end-start)/(seg.points[i+1] - seg.points[i])
    res<-c(res,seg.points[i] + ((val - start) / slope))
  }
}
res
# [1] 2.5 3.5 6.5

Answer 2

Here is a simple function to do what you want, assuming your data are unimodal, here using linear interpolation:

# FUNCTION TO INFER FWHM USING LINEAR INTERPOLATION
fwhm = function(x, y) { 
  halfheight = max(y)/2
  id.maxy = which.max(y)
  y1 = y[1:id.maxy]
  y2 = y[id.maxy:length(y)]
  x1 = x[1:id.maxy]
  x2 = x[id.maxy:length(y)]
  x.start = approx(x=y1,y=x1, xout=halfheight, method="linear")$y # or without interpolation: x[which.min(abs(y1-halfheight))]
  x.stop = approx(x=y2,y=x2, xout=halfheight, method="linear")$y # or without interpolation: x[which.min(abs(y2-halfheight))+id.maxy]
  fwhm = x.stop-x.start
  width = fwhm/(2*sqrt(2*log(2)))
  return(list(fwhm=fwhm, width=width))
  }

# GAUSSIAN PEAK FUNCTION WITH MODE AT u, WIDTH AT INFLECTION POINT w (FWHM=2.355*w) AND PEAK HEIGHT h
gauspeak=function(x, u, w, h=1) h*exp(((x-u)^2)/(-2*(w^2)))

# EXAMPLE
x = seq(0,200,length.out=1000)
y = gauspeak(x=x, u=100, w=10, h=100)
fwhm(x=x, y=y)
# $`fwhm`
# [1] 23.54934
# 
# $width
# [1] 10.00048

You could also use the spline() function instead of approx() to do cubic spline interpolation instead of linear interpolation. And if your data were unimodal but noisy you could smooth your data first using smooth.spline() (maybe on a log(y+1E-20) scale if your data are strictly positive, after which you can backtransform to the original scale).