2017-02-27 48 views
2

我有一些我想分析的噪音數據。以這裏的數據爲例。確定當地最小值後的局部最大值

set.seed(98765) 
A <- seq(0, 3, 0.01) 
B <- round(exp(A),digits = 2) 
B <-round(ifelse(B>1.5, jitter(B, factor = 200),B),digits = 2) 
# 
C <-seq(3,3.5,0.01) 
D <- rep(B[301],51) 
D <-round(jitter(D, factor = 0.8),digits = 2) 
# 
E <-seq(3.5,4,0.01) 
Ff <- rep(D[51],51) 
Ff <-round(jitter(Ff, factor = 1.3),digits = 2) 
# 
G <- seq(4,5, 0.01) 
H <- (-10*G)+60 
H <- round(jitter(H, factor = 50),digits = 2) 
# 
I <- seq(5,6,0.01) 
J <- 5*I-15 
J <- round(jitter(J, factor = 50),digits = 2) 
# 
K <- seq(6,8,0.01) 
L <- (-2*K)+27 
L <- round(jitter(L, factor = 40),digits = 2) 
# 
X <- c(A,C,E,G,I,K) 
Y <- c(B,D,Ff,H,J,L) 
# 
df1 <- data.frame(X,Y) 
ggplot(df1, aes(X,Y))+geom_point() 

於是密謀後,我們看到後我們如何達到局部最大值周圍x=5,不久發生失水最小。

如何獲得這個局部最大值的兩個最大Y值?

我可以通過指定x約束(對應於最小值)來獲得最大值,但在我的數據中,實際位置發生了變化,但形狀的形式保持不變。

## max value 
## max(df1$Y[df1$X>5]) 
+0

僅供參考,你可以看到在尋找局部最大值和最小值HTTP這樣的問題://計算器.COM /問題/ 6836409 /查找本地極大值,和最小值 – Djork

回答

1

這是不完整的答案,但這裏是一個辦法,可以提供一些方向

#Fit a smooth spline to the data and extract y-value 
dty = smooth.spline(df1$X, df1$Y)$y 

#Obtain delta values for dty 
delta = c(0, diff(dty)) 

#Plot the data as a line 
plot(df1$X, df1$Y, type = "l") 

#Points corresponding to zero in diff for dty are local maxima and minima (in theory) 
#In practice, you may have to tweak the tolerance 
points(df1$X[which(delta > -0.01 & delta < 0.01)], 
     df1$Y[which(delta > -0.01 & delta < 0.01)], 
     pch = 19, col = 'red') 

enter image description here

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