如何计算R中移动窗口内的平均斜率 [英] How to calculate the average slope within a moving window in R

问题描述

我的数据集包含两个变量 y和t [05s] .每隔05秒测量一次y.

My dataset contains 2 variables y and t [05s]. y was measured every 05 seconds.

我正在尝试计算移动的 20秒窗口内的平均斜率，即在计算出第一个20秒的斜率值后，窗口向前移动一个时间单位(05秒)并计算接下来的20秒窗口，以 05秒的增量产生连续的20秒斜率值.

I am trying to calculate the average slope within a moving 20-second-window, i.e. after calculating the first 20-second slope value the window moves forward one time unit (05 seconds) and calculates the next 20-second-window, producing successive 20-second slope values at 05-second increments.

我认为，使用rollapply(zoo包)计算滚动回归可以达到目的，但是我一次又一次地获得每个窗口相同的截距和斜率值.我该怎么办?

I thought that calculating a rolling regression with rollapply (zoo package) would do the trick, but I get the same intercept and slope values for each window over and over again. What can I do?

我的数据:

dput(DataExample)
structure(list(t = c(0, 0.05, 0.1, 0.15, 0.2, 0.25, 0.3, 0.35, 
0.4, 0.45, 0.5, 0.55, 0.6, 0.65, 0.7, 0.75, 0.8, 0.85, 0.9, 0.95, 
1, 1.05, 1.1, 1.15, 1.2, 1.25, 1.3, 1.35, 1.4, 1.45, 1.5, 1.55, 
1.6, 1.65, 1.7, 1.75, 1.8, 1.85, 1.9, 1.95, 2, 2.05, 2.1, 2.15, 
2.2, 2.25, 2.3, 2.35, 2.4, 2.45, 2.5, 2.55, 2.6, 2.65, 2.7, 2.75, 
2.8, 2.85, 2.9, 2.95, 3, 3.05, 3.1, 3.15, 3.2, 3.25, 3.3, 3.35, 
3.4, 3.45, 3.5, 3.55, 3.6, 3.65, 3.7, 3.75, 3.8, 3.85, 3.9, 3.95, 
4, 4.05, 4.1, 4.15, 4.2, 4.25, 4.3, 4.35, 4.4, 4.45, 4.5, 4.55, 
4.6, 4.65, 4.7, 4.75, 4.8, 4.85, 4.9, 4.95, 5, 5.05, 5.1, 5.15, 
5.2, 5.25, 5.3, 5.35, 5.4, 5.45, 5.5, 5.55, 5.6, 5.65, 5.7, 5.75, 
5.8, 5.85, 5.9, 5.95, 6, 6.05, 6.1, 6.15, 6.2, 6.25, 6.3, 6.35, 
6.4, 6.45, 6.5, 6.55, 6.6, 6.65, 6.7, 6.75, 6.8, 6.85, 6.9, 6.95, 
7, 7.05, 7.1, 7.15, 7.2, 7.25, 7.3, 7.35, 7.4, 7.45, 7.5, 7.55, 
7.6, 7.65, 7.7, 7.75, 7.8, 7.85, 7.9, 7.95, 8, 8.05, 8.1, 8.15, 
8.2, 8.25, 8.3, 8.35, 8.4, 8.45, 8.5, 8.55, 8.6, 8.65, 8.7, 8.75, 
8.8, 8.85, 8.9, 8.95, 9, 9.05, 9.1, 9.15, 9.2, 9.25, 9.3, 9.35, 
9.4, 9.45, 9.5, 9.55, 9.6, 9.65, 9.7, 9.75, 9.8, 9.85, 9.9, 9.95, 
10, 10.05, 10.1, 10.15, 10.2, 10.25, 10.3), y = c(3.05, 3.04, 
3.02, 3.05, 3.01, 3.02, 3.02, 3.05, 3.02, 3.01, 3.04, 3.04, 3.03, 
3.03, 3.03, 3.02, 3.02, 3.03, 3.03, 3.03, 3.04, 3.03, 3.03, 3.03, 
3.03, 3.02, 3.02, 3.02, 3.01, 3.03, 3.03, 3.03, 3.03, 3.03, 3.02, 
3.01, 3.02, 3.02, 3.01, 3.02, 3.02, 3.02, 3.03, 3.02, 3.02, 3.01, 
3.01, 3.02, 3.01, 3.02, 3.02, 3.02, 3.02, 3.01, 3.01, 3.01, 3.01, 
3.02, 3, 3.01, 3.02, 3.02, 3.02, 3.01, 3.01, 3.01, 3.01, 3.02, 
3, 3.01, 3.01, 3.01, 3.01, 3.01, 3.01, 3, 3, 3.01, 3, 3, 3.01, 
3.01, 3.01, 3.01, 3, 3, 3, 3.01, 3, 3, 3.01, 3.01, 3.01, 3.01, 
3.01, 3.01, 3, 3.02, 3, 3.01, 3.02, 3.04, 3.05, 3.08, 3.04, 3.06, 
3.08, 3.06, 3.08, 3.09, 3.04, 3.05, 3.07, 3.08, 3.06, 3.08, 3.08, 
3.07, 3.08, 3.08, 3.05, 3.06, 3.07, 3.07, 3.06, 3.08, 3.08, 3.08, 
3.08, 3.08, 3.05, 3.06, 3.08, 3.08, 3.06, 3.09, 3.07, 3.08, 3.08, 
3.08, 3.06, 3.07, 3.07, 3.07, 3.06, 3.09, 3.07, 3.07, 3.08, 3.08, 
3.06, 3.07, 3.07, 3.07, 3.06, 3.09, 3.07, 3.07, 3.07, 3.08, 3.07, 
3.07, 3.07, 3.07, 3.06, 3.08, 3.07, 3.07, 3.06, 3.08, 3.07, 3.07, 
3.07, 3.07, 3.06, 3.08, 3.07, 3.07, 3.06, 3.08, 3.06, 3.07, 3.06, 
3.07, 3.06, 3.08, 3.07, 3.07, 3.06, 3.07, 3.06, 3.07, 3.06, 3.07, 
3.06, 3.07, 3.06, 3.06, 3.06, 3.07, 3.04, 3.04, 3.04, 3.06, 3.06, 
3.04, 3.04)), class = c("tbl_df", "tbl", "data.frame"), row.names = c(NA, 
-207L), .Names = c("t", "y"))

R代码:

require(zoo)
library("zoo", lib.loc="~/R/win-library/3.3")
rollapply(zoo(DataExample),
          width=5,
          FUN = function(Z) 
          { 
            z = lm(formula=y~t, data = as.data.frame(DataExample)); 
            return(z$coef) 
          }, by=1,
          by.column=FALSE, align="right")

推荐答案

该注释似乎已删除，但指出问题代码中rollapply中的函数未使用传递给它的参数.解决此问题并进行其他一些小的改进后，这将分别返回第1列和第2列中的截距和斜率.

The comment seems to have been deleted but it was pointed out that the function in rollapply in the code in the question was not using the argument passed to it. After fixing that and making some other minor improvements, this returns the intercept and the slope in columns 1 and 2 respectively.

library(zoo)

Coef <- function(Z) coef(lm(y ~ t, as.data.frame(Z)))    
rollapplyr(zoo(DataExample), 5, Coef, by.column = FALSE)

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