如何平滑R中的曲线？

Question

location diffrence<-c(0,0.5,1,1.5,2)
Power<-c(0,0.2,0.4,0.6,0.8,1)
plot(location diffrence,Power)

The guy which has written the paper said he has smoothed the curve using a weighted moving average with weights vector w = (0.25,0.5,0.25) but he did not explained how he did this and with which function he achieved that.i am really confused

Answer 1

Up front, as @MartinWettstein cautions, be careful in when you smooth data and what you do with it (infer from it). Having said that, a simple exponential moving average might look like this.

# replacement data
x <- seq(0, 2, len=5)
y <- c(0, 0.02, 0.65, 1, 1)

# smoothed
ysm <- 
  zoo::rollapply(c(NA, y, NA), 3,
                 function(a) Hmisc::wtd.mean(a, c(0.25, 0.5, 0.25), na.rm = TRUE),
                 partial = FALSE)

# plot
plot(x, y, type = "b", pch = 16)
lines(x, ysm, col = "red")

Notes:

• the zoo:: package provides a rolling window (3-wide here), calling the function once for indices 1-3, then again for indices 2-4, then 3-5, 4-6, etc.
• with rolling-window operations, realize that they can be center-aligned (default of zoo::rollapply ) or left/right aligned. There are some good explanations here: How to calculate 7-day moving average in R? )
• I surround the y data with NA s so that I can mimic a partial window. Normally with rolling-window ops, if k=3 , then the resulting vector is length(y) - (k-1) long. I'm inferring that you want to include data on the ends, so the first smoothed data point would be effectively (0.5*0 + 0.25*0.02)/0.75 , the second smoothed data point (0.25*0 + 0.5*0.02 + 0.25*0.65)/1 , and the last smoothed data point (0.25*1 + 0.5*1)/0.75 . That is, omitting the 0.25 times a missing data point. That's a guess and can easily be adjusted based on your real needs.
• I'm using Hmisc::wtd.mean , though it is trivial to write this weighted-mean function yourself.

This is suggestive only, and not meant to be authoritative. Just to help you begin exploring your smoothing processes.