Question

有一系列8,7,6,5,6,7,8,7,6,7,8在5和6处有局部最小值。

如何使用R？

Answer 1

library(zoo)

x <- c(8,7,6,5,6,7,8,7,6,7,8)

# eliminate runs
r <- rle(x)
r$lengths[] <- 1
xx <- inverse.rle(r)        

xx[ rollapply(c(Inf, xx, Inf), 3, function(x) x[2] < min(x[-2])) ]
## [1] 5 6

如果我们知道x没有运行，就像问题示例中的情况一样，我们可以省略消除运行的三行，并在最后一个语句中替换xx x。如果您不想考虑端点，请使用-Inf代替Inf两次出现。

假设输入c(2, 1, 1, 3)应该导致输出1次。

Answer 2

如果没有运行，我们可以简单地检查滞后和前导数字的区别：

library(dplyr)
x[x - lag(x) < 0 & x - lead(x) < 0]

给出：

[1] 5 6

Answer 3

我尝试使用没有任何包装并获得结果。

series<-c(8,7,6,5,6,7,8,7,6,7,8)
#We have local minima at 5 and 6.
#Check if series is decreasing in the begining itself
series[2]<series[1]
#If so let us check for minima.
#First check if you have multiple minima
less<-numeric()
#Fill first position and write a loop to check rest of positions.
less<-1
for(i in 2:length(series)){
less[i]<-ifelse(series[i-1]>=series[i],1,0)
}
#First occurance of 0 after 1
checkless<-numeric()
#Fill first position and write a loop to check rest of positions.
checkless<-1
for(i in 2:length(less)){
  checkless[i]<-ifelse(less[i-1]==0,1,less[i])
}
#Indexes where 0 exists indicates the change in signs.
which(checkless==0)
#Local minima exists once index before change
which(checkless==0)-1
series[which(checkless==0)-1]
#Check how many cases exists
length(series[which(checkless==0)])
#If number of such cases greater than 1 then we have multiple minima in the series
length(series[which(checkless==0)])>1

找到系列的局部最小值

3 个答案: