※ 引述《jackhzt (巴克球)》之铭言： : [问题类型]: : 程式谘询(我想用R 做某件事情，但是我不知道要怎麽用R 写出来) : [软体熟悉度]: : 入门(写过其他程式，只是对语法不熟悉) : [问题叙述]:piecewise linear approximation(PLA) : 目的:使用线性的方式，切割一个序列 ( 时间序列) : 目标方法: 简单来说连结起点和终点， 依时间依序比较点和线的距离 : 太大就将点和终点连线， 以新的线再依序和接下来的点比距离， 重复做 : 以下是原文叙述: : http://imgur.com/eUcKxg6 : http://imgur.com/BLbYeyJ : 问题: : (step1): : Input time series Q(i : j) and threshold value "error". A vector Bp : is used to restore the breakpoints. "k" records the number of : the present breakpoints. "pos" denotes the position of the : newest breakpoint. : Initially, i = 1, j = m, where m is the length of time series. : Since the first point and the last point : are the special breakpoints, let k = 2, Bp(1) = q1 and : Bp(2) = qm. : (step2): : For time series Q(i : j), create line segment L(i : j) according : to the formula (6). Set two variables l = i + 1 and : best_so_far = 0. *公式6在我的程式码中有付 : (step3): : Calculate the distance of point ql to the line segment L(i : j), : that is D(ql,L(i : j)). : (step4): : If D(ql,L(i : j)) > best_so_far, best_so_far = D(ql,L(i : j)) and : pos = l. : (step5): : l = l + 1. If l>=j, go to the step 6; otherwise, go back to step3. : (step6): : If best_so_farPe, k = k + 1, Bp(k) = q_pos, go back to the Step : 2 and let the two subsequences Q(i : pos) and Q(pos : j) : redo the step 2 to step 6, respectively. : (step7): : Sort the element of vector Bp by an ascending time and : output the sorted result. : 出自:http://tinyurl.com/hhosdmk -3.1 : 1.我的程式码看起来有点问题，尤其是step 6这地方不太会表示，有高手可以解惑吗? : 2.有没有比较正常的打法?我的打法好像问题很大 : 3.package方面有试过一些，但是有办法表达和上面叙述一样的package目前好像没找到 : 程式码可贴於以下网站： http://ideone.com/TOEISf : 求高手救援 我不确定我有做对，看图应该是对了 基本上这个要套用递回才能解决 程式如下： 好读版：http://pastebin.com/95ATSHHV q<-c(18, 15, 24, 23, 18, 22, 19, 29, 22, 25, 20, 19, 18, 20, 26, 32, 26, 26, 34, 29, 23, 34, 22, 19, 21, 19, 34, 23, 23, 23, 30, 21, 15, 29, 32, 19, 21, 28, 22, 32, 29, 25, 28, 28, 23, 12, 26, 24, 27, 14, 38, 27, 28, 25, 38, 34, 25, 37, 15, 28, 15, 23, 23, 28, 15, 15, 19, 25, 28, 16, 19, 17, 23, 19, 16, 18, 18, 17, 20, 18, 21, 13, 11, 12, 13, 16, 13, 16, 10, 13, 14, 6, 19, 18, 19, 15, 17, 6, 14, 28, 15, 20, 16, 12) # distance function dis_f <- function(t, q, i, j){ a <- (q[j]-q[i])/(j-i) abs((q[i]*j-q[j]*i)/(j-i) + a * t - q[t]) / sqrt(a^2 + 1) } pla <- function(q, i, j, time, eplison){ if (i > j || j - i <= 1) return(sort(time)) # find the maximum distance (Following two lines represents the Step 3~5) dis_t <- dis_f((i+1):(j-1), q, i, j) # calculate distance of qi~qj loc <- which.max(dis_t) # find the position pos <- i + loc # record the position best_so_far <- dis_t[loc] # print the segment cat(sprintf("segment: %i, %i, %.2f\n", i, j, best_so_far)) # Step 6: find more segments if (best_so_far >= eplison) { # record the time time <- c(time, pos) if (pos < j) { time <- pla(q, i, pos, time, eplison) time <- pla(q, pos, j, time, eplison) } } return(sort(time)) } # calculate eplison eplison <- sd(dis_f(1:length(q), q, 1, length(q))) time <- pla(q, 1, length(q), c(1, length(q)), eplison) plot(1:length(q), q,type="o") lines(time, q[time], col = 2) --

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