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Editorial Team
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Editorial Team
Asked: June 17, 20262026-06-17T10:02:35+00:00 2026-06-17T10:02:35+00:00

I am seeking some simple (i.e. – no maths notation, long-form reproducible code) examples

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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
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  1. Editorial Team
    Editorial Team
    2026-06-17T10:02:36+00:00Added an answer on June 17, 2026 at 10:02 am

    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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