过滤功能的简单例子,具体递归选项 [英] simple examples of filter function, recursive option specifically

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问题描述

我正在为R
中的过滤器函数寻找一些简单的(即 - 没有数学符号,长形式可重复的代码)的例子我想我有我的头绕卷积方法,但我坚持推广递归选项。我已经阅读并与各种文档作斗争,但对我的帮助只是有点不透明。



以下是我到目前为止所了解的例子:

 #为滤波器组件设置一些值
f1 < - 1; f2 <-1; f3 <-1;

然后我们去:

<$ p $基本卷积滤波器
滤波器(1:5,f1,method =convolution)
[1] 1 2 3 4 5

#eqivalent to:
x [1] * f1
x [2] * f1
x [3] * f1
x [4] * f1
x [5] * f1

#滤波器中的2个系数的卷积
filter(1:5,c(f1,f2),method =convolution)
[1] 3 5 7 9 NA

#eqivalent 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

# (1:5,c(f1,f2,f3),method =convolution)
[1] NA 6 9 12 NA

#等同于:
NA * f3 + x [1] * f2 + x [2] * f1 #x [0] =不存在/ 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

现在当我受伤的时候我可怜的小脑干。
我设法弄明白这个职位信息最基本的例子: https://stackoverflow.com/a/11552765 / 496803
$ b

 过滤器(1:5,f1,method =recursive)
[1 ] 1 3 6 10 15

#eqivalent 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]

有人可以提供类似的代码,以上的递归版本的卷积实例具有 filter = c(f1,f2 ) filter = c(f1,f2,f3)?



(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的整齐答案:

 >过滤器(1:5,f1,method =递归)
时间序列:
开始= 1
结束= 5
频率= 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

和...

 > filter(1:5,c(f1,f2),method =recursive)
时间序列:
开始= 1
结束= 5
频率= 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

和...

 > filter(1:5,c(f1,f2,f3),method =recursive)
时间序列:
开始= 1
结束= 5
频率= 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


解决方案

在递归的情况下,我认为不需要用xi来扩展表达式。
递归的关键是用前面的y来表示右手表达式。

我更喜欢根据过滤器大小来思考。

filter size = 1

  y1 < -  x1 
y2 < - x2 + f1 * y1
y3 < - x3 + f1 * y2
y4 < - x4 + f1 * y3
y5 < - x5 + f1 * y4

filter size = 2

  y1 < -  x1 
y2 < - x2 + f1 * y1
y3 < - x3 + f1 * y2 + f2 * y1#将过滤器应用于过去值并添加当前输入
y4 < - x4 + f1 * y3 + f2 * y2
y5 < - x5 + f1 * y4 + f2 * y3
/ pre>

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.

Here are the examples I have figured out so far:

# Set some values for filter components
f1 <- 1; f2 <- 1; f3 <- 1;

And on we go:

# 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

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]

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)?

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

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

and...

> 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

and...

> 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

解决方案

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.

I prefer thinking in terms of filter size.

filter size =1

y1 <- x1                                            
y2 <- x2 + f1*y1      
y3 <- x3 + f1*y2 
y4 <- x4 + f1*y3 
y5 <- x5 + f1*y4 

filter size = 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

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