Question

我试图创建一个函数来查找＆＃34; maxima＆＃34;和＆＃34; minima＆＃34;。我有以下数据：

我试过这个功能来找到＆＃34; maxima＆＃34;

localMaxima <- function(x) {
  # Use -Inf instead if x is numeric (non-integer)
  y <- diff(c(-.Machine$integer.max, x)) > 0L
  rle(y)$lengths
  y <- cumsum(rle(y)$lengths)
  y <- y[seq.int(1L, length(y), 2L)]
  if (x[[1]] == x[[2]]) {
    y <- y[-1]
  }
  y
}

maks <- localMaxima(x)

找到＆＃34; minima＆＃34;

的功能

localMinima <- function(x) {
      # Use -Inf instead if x is numeric (non-integer)
      y <- diff(c(.Machine$integer.max, x)) > 0L
      rle(y)$lengths
      y <- cumsum(rle(y)$lengths)
      y <- y[seq.int(1L, length(y), 2L)]
      if (x[[1]] == x[[2]]) {
        y <- y[-1]
      }
      y
    }

    mins <- localMinima(x)

结果不是100％正确

maks = 1  5  7 10 12 14 16 20 24 27 31 33 35
mins = 3  6  8 11 13 15 19 23 26 28 32 34 36

结果应

maks = 5  7 10 12 14 16 20 24 27 31 33 35
mins = 3  6  8 11 13 15 19 23 26 28 32 34

Finding local maxima and minima in R接近，但并不合适。

我该如何解决这个问题？

非常感谢你

Answer 1

您可以定义两个函数，如下所示，它们生成您需要的向量：

library(data.table)
#shift lags or leads a vector by a certain amount defined as the second argument
#the default is to lag a vector.
#The rationale behind the below code is that each local minimum's adjucent
#values will be greater than itself. The opposite is true for a local 
#maximum. I think this is what you are trying to achieve and one way to do 
#it is the following code
maximums <- function(x) which(x - shift(x, 1) > 0  & x - shift(x, 1, type='lead') > 0)
minimums <- function(x) which(x - shift(x, 1) < 0  & x - shift(x, 1, type='lead') < 0)

输出：

> maximums(y)
 [1]  5  7 10 12 14 16 20 24 27 31 33 35
> minimums(y)
 [1]  3  6  8 11 13 15 19 23 26 28 32 34

Answer 2

这是我前一段时间写的一个功能（它比你需要的更通用）。它在序列数据x中找到峰值，其中我将峰值定义为局部最大值，其中m点的任何一侧都具有低于它的值（因此更大m会导致更严格的标准寻找高峰）：

 find_peaks <- function (x, m = 3){
     shape <- diff(sign(diff(x, na.pad = FALSE)))
     pks <- sapply(which(shape < 0), FUN = function(i){
        z <- i - m + 1
        z <- ifelse(z > 0, z, 1)
        w <- i + m + 1
        w <- ifelse(w < length(x), w, length(x))
        if(all(x[c(z : i, (i + 2) : w)] <= x[i + 1])) return(i + 1) else return(numeric(0))
    })
     pks <- unlist(pks)
     pks
}

因此，对于您的案例m = 1：

 find_peaks(x, m = 1)
 #[1]  5  7 10 12 14 16 20 24 27 31 33 35

和最小值：

 find_peaks(-x, m = 1)
 #[1]  3  6  8 11 13 15 19 23 26 28 32 34

在R中找到局部最大值和最小值

2 个答案: