过滤函数的简单示例，特别是递归选项 [英] simple examples of filter function, recursive option specifically
问题描述
我正在为 R 中的 filter 函数寻找一些简单的(即 - 没有数学符号，长格式可重现代码)示例我想我对卷积方法有一定的了解，但我一直在推广递归选项.我已经阅读并与各种文档作斗争，但对我来说帮助有点不透明.
I am seeking some simple (i.e. - no maths notation, long-form reproducible code) examples for the filter function in R
I think I have my head around the convolution method, but am stuck at generalising the recursive option. I have read and battled with various documentation, but the help is just a bit opaque to me.
以下是我目前发现的示例:
Here are the examples I have figured out so far:
# Set some values for filter components
f1 <- 1; f2 <- 1; f3 <- 1;
我们继续:
# basic convolution filter
filter(1:5,f1,method="convolution")
[1] 1 2 3 4 5
#equivalent to:
x[1] * f1
x[2] * f1
x[3] * f1
x[4] * f1
x[5] * f1
# convolution with 2 coefficients in filter
filter(1:5,c(f1,f2),method="convolution")
[1] 3 5 7 9 NA
#equivalent to:
x[1] * f2 + x[2] * f1
x[2] * f2 + x[3] * f1
x[3] * f2 + x[4] * f1
x[4] * f2 + x[5] * f1
x[5] * f2 + x[6] * f1
# convolution with 3 coefficients in filter
filter(1:5,c(f1,f2,f3),method="convolution")
[1] NA 6 9 12 NA
#equivalent to:
NA * f3 + x[1] * f2 + x[2] * f1 #x[0] = doesn't exist/NA
x[1] * f3 + x[2] * f2 + x[3] * f1
x[2] * f3 + x[3] * f2 + x[4] * f1
x[3] * f3 + x[4] * f2 + x[5] * f1
x[4] * f3 + x[5] * f2 + x[6] * f1
现在我正在伤害我可怜的小脑干.我设法在这篇文章中使用信息找出了最基本的示例:https://stackoverflow.com/a/11552765/496803
Now's when I am hurting my poor little brain stem. I managed to figure out the most basic example using info at this post: https://stackoverflow.com/a/11552765/496803
filter(1:5, f1, method="recursive")
[1] 1 3 6 10 15
#equivalent to:
x[1]
x[2] + f1*x[1]
x[3] + f1*x[2] + f1^2*x[1]
x[4] + f1*x[3] + f1^2*x[2] + f1^3*x[1]
x[5] + f1*x[4] + f1^2*x[3] + f1^3*x[2] + f1^4*x[1]
有人可以为递归版本的卷积示例提供与上面类似的代码 filter = c(f1,f2) 和 filter = c(f1,f2,f3)?
Can someone provide similar code to what I have above for the convolution examples for the recursive version with filter = c(f1,f2) and filter = c(f1,f2,f3)?
答案应该与函数的结果相匹配:
Answers should match the results from the function:
filter(1:5, c(f1,f2), method="recursive")
[1] 1 3 7 14 26
filter(1:5, c(f1,f2,f3), method="recursive")
[1] 1 3 7 15 30
编辑
使用@agstudy 的简洁答案完成:
EDIT
To finalise using @agstudy's neat answer:
> filter(1:5, f1, method="recursive")
Time Series:
Start = 1
End = 5
Frequency = 1
[1] 1 3 6 10 15
> y1 <- x[1]
> y2 <- x[2] + f1*y1
> y3 <- x[3] + f1*y2
> y4 <- x[4] + f1*y3
> y5 <- x[5] + f1*y4
> c(y1,y2,y3,y4,y5)
[1] 1 3 6 10 15
还有……
> filter(1:5, c(f1,f2), method="recursive")
Time Series:
Start = 1
End = 5
Frequency = 1
[1] 1 3 7 14 26
> y1 <- x[1]
> y2 <- x[2] + f1*y1
> y3 <- x[3] + f1*y2 + f2*y1
> y4 <- x[4] + f1*y3 + f2*y2
> y5 <- x[5] + f1*y4 + f2*y3
> c(y1,y2,y3,y4,y5)
[1] 1 3 7 14 26
还有……
> filter(1:5, c(f1,f2,f3), method="recursive")
Time Series:
Start = 1
End = 5
Frequency = 1
[1] 1 3 7 15 30
> y1 <- x[1]
> y2 <- x[2] + f1*y1
> y3 <- x[3] + f1*y2 + f2*y1
> y4 <- x[4] + f1*y3 + f2*y2 + f3*y1
> y5 <- x[5] + f1*y4 + f2*y3 + f3*y2
> c(y1,y2,y3,y4,y5)
[1] 1 3 7 15 30
推荐答案
在递归的情况下，我认为不需要用xi来展开表达式.递归"的关键是用前一个 y 来表达右手表达式.
In the recursive case, I think no need to expand the expression in terms of xi. The key with "recursive" is to express the right hand expression in terms of previous y's.
我更喜欢考虑过滤器大小.
I prefer thinking in terms of filter size.
过滤器大小=1
y1 <- x1
y2 <- x2 + f1*y1
y3 <- x3 + f1*y2
y4 <- x4 + f1*y3
y5 <- x5 + f1*y4
过滤器大小 = 2
y1 <- x1
y2 <- x2 + f1*y1
y3 <- x3 + f1*y2 + f2*y1 # apply the filter for the past value and add current input
y4 <- x4 + f1*y3 + f2*y2
y5 <- x5 + f1*y4 + f2*y3
这篇关于过滤函数的简单示例，特别是递归选项的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
