R get and restrain number of local maxima position and value

Question

I'm using R to analyze some spectra and I'm trying to get the local maxima, namely their position and their value.

For example, with a vector:

spectrum <- c(1,1,2,3,5,3,3,2,1,1,5,6,9,5,1,1)

I would like the following result:

pos.peaks = c(5,13)
val.peaks = c(5,9)

I've already used the solution provided here: Finding local maxima and minima for the position of the peaks but how do I extract the corresponding value afterwards? Knowing that I don't have just one vector, I have several columns within several dataframes within a list, and I want to apply the function to every single column of all dataframes in the list. For example, for all the positions I did this:

example <- lapply(mylist, function (x) lapply(x, function(y) which(diff(sign(diff(y)))==-2)+1))

I didn't manage to make it work with slice or filter, because I don't need the same rows within the same dataframe...

Furthermore, I would like to know how to reduce the amount of local maxima I get because my data is very noisy.

I'd appreciate any help you can give me.

Thanks!

Nath

You can use `spectrum[which(diff(sign(diff(spectrum))) == - 2) + 1]`. — tmfmnk, Jul 17 '19 at 12:43

GKi · Accepted Answer · 2019-07-18T08:45:55.830

peakPosition <- function(x, inclBorders=TRUE) {
  if(inclBorders) {y <- c(min(x), x, min(x))
  } else {y <- c(x[1], x)}
  y <- data.frame(x=sign(diff(y)), i=1:(length(y)-1))
  y <- y[y$x!=0,]
  idx <- diff(y$x)<0
  (y$i[c(idx,F)] + y$i[c(F,idx)] - 1)/2
}

(pos.peaks  <- peakPosition(spectrum))
#[1]  5 13

(val.peaks  <- spectrum[pos.peaks])
#[1] 5 9

And for the loop to get the values something like:

example <- lapply(mylist, function(x) {x[peakPosition(x)]})

and for the positions:

lapply(mylist, peakPosition)

In the comment you say your data is very noisy and you get to many local maxima, so you may try first to smooth your data like following:

d <- predict(loess(spectrum ~ seq_along(spectrum)))
pos.peaksS  <- peakPosition(d)
(i <- pos.peaks[apply(abs(outer(pos.peaks, pos.peaksS, "-")), 1, FUN=which.min)])
#[1]  5 13
spectrum[i]
#[1] 5 9

or you make some aggregation to the index like:

set.seed(42)
x <- rnorm(1e3)

y <- peakPosition(x)
(pos.peaks <- sort(aggregate(y, list(k=kmeans(y, 7)$cluster), FUN = function(i) i[which.max(x[i])])[,2]))
#[1] 118 287 459 525 613 820 988

(val.peaks  <- x[pos.peaks])
#[1] 2.701891 2.459594 2.965865 3.229069 2.223534 3.211199 3.495304

Thank you for going all the way to the looping, I'm going to experiment with both solutions to see what fits best to my data (it's very noisy so I get to many local maxima) — nathkpa, Jul 18 '19 at 06:42
@nathkpa I have updated the Answer, but maybe you ask a new question with the topic of reducing the amount of local maxima. — GKi, Jul 18 '19 at 08:47
Thanks again, I've adapted the title and the text, is the term "restrain" appropriate for the title? — nathkpa, Jul 18 '19 at 09:28
@nathkpa I don't know if "restrain" is appropriate - no native speaker. — GKi, Jul 18 '19 at 09:30

score 1 · Answer 2 · answered Jul 17 '19 at 13:20

If the vector has at least length 3:

find_peaks <- function(x, max = TRUE) {
  if (max == FALSE) x <- x * (-1)
  res <- rep(FALSE, length(x))
  if (x[1] > x[2]) res[1] <- TRUE
  if (x[length(x)-1] < x[length(x)]) res[length(res)] <- TRUE
  for (i in (2:(length(x)-1))) {
    if ((x[i-1] < x[i]) & (x[i+1] < x[i])) res[i] <- TRUE
  }
  res
}
spectrum[find_peaks(spectrum)]
# [1] 5 9
which(find_peaks(spectrum))
# [1]  5 13

R get and restrain number of local maxima position and value

2 Answers2