Question on Reduce + rollmean

x

Dear M.Rahiz,

Unfortunatelly I can't reproduce your example.

aperm(aa, c(2, 3, 1))

Let me rephrase;

Given x as

x

[[1]]
?V1 V2 V3
[1,] ?1 ?1 ?1
[2,] ?2 ?2 ?2
[3,] ?3 ?3 ?3

[[2]]
?V1 V2 V3
[1,] ?4 ?4 ?4
[2,] ?5 ?5 ?5
[3,] ?6 ?6 ?6

[[3]]
?V1 V2 V3
[1,] ?7 ?7 ?7
[2,] ?8 ?8 ?8
[3,] ?9 ?9 ?9

I'd like to calculate the moving average (interval = 2) i.e.

( x[[1]] + x[[2]] ) / 2
( x[[2]] + x[[3]] ) / 2
... and so on.

The desired output will return

2.5 2.5 2.5
3.5 3.5 3.5
4.5 4.5 4.5

5.5 5.5 5.5
6.5 6.5 6.5
7.5 7.5 7.5




Muhammad Rahiz ?| ?Doctoral Student in Regional Climate Modeling
Climate Research Laboratory, School of Geography & the Environment
Oxford University Centre for the Environment, University of Oxford
South Parks Road, Oxford, OX1 3QY, United Kingdom
Tel: +44 (0)1865-285194 ?Mobile: +44 (0)7854-625974
Email: muhammad.rahiz at ouce.ox.ac.uk



milton ruser wrote:

Dear M.Rahiz,

Unfortunatelly I can't reproduce your example.

Try apply:

library(zoo) # rollmean

# test data
m <- matrix(1:3, 3, 3)
x <- list(m, m+3, m+6)

# convert to array
a <- array(unlist(x), c(3, 3, 3)); a

# apply rollmean and permute to desired form
aa <- apply(a, 1:2, rollmean, k = 2)
aperm(aa, c(2, 3, 1))

The last line outputs:

  

aperm(aa, c(2, 3, 1))
    

, , 1

     [,1] [,2] [,3]
[1,]  2.5  2.5  2.5
[2,]  3.5  3.5  3.5
[3,]  4.5  4.5  4.5

, , 2

     [,1] [,2] [,3]
[1,]  5.5  5.5  5.5
[2,]  6.5  6.5  6.5
[3,]  7.5  7.5  7.5


On Sat, Jan 2, 2010 at 10:00 AM, Muhammad Rahiz
<muhammad.rahiz at ouce.ox.ac.uk> wrote:
  

Let me rephrase;

Given x as

    

x
      

[[1]]
 V1 V2 V3
[1,]  1  1  1
[2,]  2  2  2
[3,]  3  3  3

[[2]]
 V1 V2 V3
[1,]  4  4  4
[2,]  5  5  5
[3,]  6  6  6

[[3]]
 V1 V2 V3
[1,]  7  7  7
[2,]  8  8  8
[3,]  9  9  9

I'd like to calculate the moving average (interval = 2) i.e.

( x[[1]] + x[[2]] ) / 2
( x[[2]] + x[[3]] ) / 2
... and so on.

The desired output will return

2.5 2.5 2.5
3.5 3.5 3.5
4.5 4.5 4.5

5.5 5.5 5.5
6.5 6.5 6.5
7.5 7.5 7.5




Muhammad Rahiz  |  Doctoral Student in Regional Climate Modeling
Climate Research Laboratory, School of Geography & the Environment
Oxford University Centre for the Environment, University of Oxford
South Parks Road, Oxford, OX1 3QY, United Kingdom
Tel: +44 (0)1865-285194  Mobile: +44 (0)7854-625974
Email: muhammad.rahiz at ouce.ox.ac.uk



milton ruser wrote:
    

Dear M.Rahiz,

Unfortunatelly I can't reproduce your example.

Thanks Gabor,

It works alright. But is there alternative ways to perform rollmean apart
from permutating the data?
Possibly combining the Reduce and rollmean functions?

The easiest but traditional way is

(x[[1]]+x[[2]]) / 2
(x[[2]]+x[[3]]) / 2

Muhammad Rahiz ?| ?Doctoral Student in Regional Climate Modeling

Climate Research Laboratory, School of Geography & the Environment
Oxford University Centre for the Environment
South Parks Road, Oxford, OX1 3QY, United Kingdom Tel: +44 (0)1865-285194
?Mobile: +44 (0)7854-625974
Email: muhammad.rahiz at ouce.ox.ac.uk






Gabor Grothendieck wrote:

Try apply:

library(zoo) # rollmean

# test data
m <- matrix(1:3, 3, 3)
x <- list(m, m+3, m+6)

# convert to array
a <- array(unlist(x), c(3, 3, 3)); a

# apply rollmean and permute to desired form
aa <- apply(a, 1:2, rollmean, k = 2)
aperm(aa, c(2, 3, 1))

The last line outputs:

aperm(aa, c(2, 3, 1))

, , 1

? ? [,1] [,2] [,3]
[1,] ?2.5 ?2.5 ?2.5
[2,] ?3.5 ?3.5 ?3.5
[3,] ?4.5 ?4.5 ?4.5

, , 2

? ? [,1] [,2] [,3]
[1,] ?5.5 ?5.5 ?5.5
[2,] ?6.5 ?6.5 ?6.5
[3,] ?7.5 ?7.5 ?7.5


On Sat, Jan 2, 2010 at 10:00 AM, Muhammad Rahiz
<muhammad.rahiz at ouce.ox.ac.uk> wrote:

Let me rephrase;

Given x as

x

[[1]]
?V1 V2 V3
[1,] ?1 ?1 ?1
[2,] ?2 ?2 ?2
[3,] ?3 ?3 ?3

[[2]]
?V1 V2 V3
[1,] ?4 ?4 ?4
[2,] ?5 ?5 ?5
[3,] ?6 ?6 ?6

[[3]]
?V1 V2 V3
[1,] ?7 ?7 ?7
[2,] ?8 ?8 ?8
[3,] ?9 ?9 ?9

I'd like to calculate the moving average (interval = 2) i.e.

( x[[1]] + x[[2]] ) / 2
( x[[2]] + x[[3]] ) / 2
... and so on.

The desired output will return

2.5 2.5 2.5
3.5 3.5 3.5
4.5 4.5 4.5

5.5 5.5 5.5
6.5 6.5 6.5
7.5 7.5 7.5




Muhammad Rahiz ?| ?Doctoral Student in Regional Climate Modeling
Climate Research Laboratory, School of Geography & the Environment
Oxford University Centre for the Environment, University of Oxford
South Parks Road, Oxford, OX1 3QY, United Kingdom
Tel: +44 (0)1865-285194 ?Mobile: +44 (0)7854-625974
Email: muhammad.rahiz at ouce.ox.ac.uk



milton ruser wrote:

Dear M.Rahiz,

Unfortunatelly I can't reproduce your example.

Thanks Gabor,

It works alright. But is there alternative ways to perform rollmean apart
from permutating the data?
Possibly combining the Reduce and rollmean functions?

The easiest but traditional way is

(x[[1]]+x[[2]]) / 2
(x[[2]]+x[[3]]) / 2

Muhammad Rahiz ?| ?Doctoral Student in Regional Climate Modeling

Climate Research Laboratory, School of Geography & the Environment
Oxford University Centre for the Environment
South Parks Road, Oxford, OX1 3QY, United Kingdom Tel: +44 (0)1865-285194
?Mobile: +44 (0)7854-625974
Email: muhammad.rahiz at ouce.ox.ac.uk






Gabor Grothendieck wrote:

Try apply:

library(zoo) # rollmean

# test data
m <- matrix(1:3, 3, 3)
x <- list(m, m+3, m+6)

# convert to array
a <- array(unlist(x), c(3, 3, 3)); a

# apply rollmean and permute to desired form
aa <- apply(a, 1:2, rollmean, k = 2)
aperm(aa, c(2, 3, 1))

The last line outputs:

aperm(aa, c(2, 3, 1))

, , 1

? ? [,1] [,2] [,3]
[1,] ?2.5 ?2.5 ?2.5
[2,] ?3.5 ?3.5 ?3.5
[3,] ?4.5 ?4.5 ?4.5

, , 2

? ? [,1] [,2] [,3]
[1,] ?5.5 ?5.5 ?5.5
[2,] ?6.5 ?6.5 ?6.5
[3,] ?7.5 ?7.5 ?7.5


On Sat, Jan 2, 2010 at 10:00 AM, Muhammad Rahiz
<muhammad.rahiz at ouce.ox.ac.uk> wrote:

Let me rephrase;

Given x as

x

[[1]]
?V1 V2 V3
[1,] ?1 ?1 ?1
[2,] ?2 ?2 ?2
[3,] ?3 ?3 ?3

[[2]]
?V1 V2 V3
[1,] ?4 ?4 ?4
[2,] ?5 ?5 ?5
[3,] ?6 ?6 ?6

[[3]]
?V1 V2 V3
[1,] ?7 ?7 ?7
[2,] ?8 ?8 ?8
[3,] ?9 ?9 ?9

I'd like to calculate the moving average (interval = 2) i.e.

( x[[1]] + x[[2]] ) / 2
( x[[2]] + x[[3]] ) / 2
... and so on.

The desired output will return

2.5 2.5 2.5
3.5 3.5 3.5
4.5 4.5 4.5

5.5 5.5 5.5
6.5 6.5 6.5
7.5 7.5 7.5




Muhammad Rahiz ?| ?Doctoral Student in Regional Climate Modeling
Climate Research Laboratory, School of Geography & the Environment
Oxford University Centre for the Environment, University of Oxford
South Parks Road, Oxford, OX1 3QY, United Kingdom
Tel: +44 (0)1865-285194 ?Mobile: +44 (0)7854-625974
Email: muhammad.rahiz at ouce.ox.ac.uk



milton ruser wrote:

Dear M.Rahiz,

Unfortunatelly I can't reproduce your example.

For this example try this:

lapply(lapply(list('head', 'tail'), do.call, list(x, n = -1)),
function(x)Reduce('+', x)/2)


On Mon, Jan 4, 2010 at 1:54 PM, Muhammad Rahiz
<muhammad.rahiz at ouce.ox.ac.uk> wrote:

Thanks Gabor,

It works alright. But is there alternative ways to perform rollmean apart
from permutating the data?
Possibly combining the Reduce and rollmean functions?

The easiest but traditional way is

(x[[1]]+x[[2]]) / 2
(x[[2]]+x[[3]]) / 2

Muhammad Rahiz ?| ?Doctoral Student in Regional Climate Modeling

Climate Research Laboratory, School of Geography & the Environment
Oxford University Centre for the Environment
South Parks Road, Oxford, OX1 3QY, United Kingdom Tel: +44 (0)1865-285194
?Mobile: +44 (0)7854-625974
Email: muhammad.rahiz at ouce.ox.ac.uk






Gabor Grothendieck wrote:

Try apply:

library(zoo) # rollmean

# test data
m <- matrix(1:3, 3, 3)
x <- list(m, m+3, m+6)

# convert to array
a <- array(unlist(x), c(3, 3, 3)); a

# apply rollmean and permute to desired form
aa <- apply(a, 1:2, rollmean, k = 2)
aperm(aa, c(2, 3, 1))

The last line outputs:

aperm(aa, c(2, 3, 1))

, , 1

? ? [,1] [,2] [,3]
[1,] ?2.5 ?2.5 ?2.5
[2,] ?3.5 ?3.5 ?3.5
[3,] ?4.5 ?4.5 ?4.5

, , 2

? ? [,1] [,2] [,3]
[1,] ?5.5 ?5.5 ?5.5
[2,] ?6.5 ?6.5 ?6.5
[3,] ?7.5 ?7.5 ?7.5


On Sat, Jan 2, 2010 at 10:00 AM, Muhammad Rahiz
<muhammad.rahiz at ouce.ox.ac.uk> wrote:

Let me rephrase;

Given x as

x

[[1]]
?V1 V2 V3
[1,] ?1 ?1 ?1
[2,] ?2 ?2 ?2
[3,] ?3 ?3 ?3

[[2]]
?V1 V2 V3
[1,] ?4 ?4 ?4
[2,] ?5 ?5 ?5
[3,] ?6 ?6 ?6

[[3]]
?V1 V2 V3
[1,] ?7 ?7 ?7
[2,] ?8 ?8 ?8
[3,] ?9 ?9 ?9

I'd like to calculate the moving average (interval = 2) i.e.

( x[[1]] + x[[2]] ) / 2
( x[[2]] + x[[3]] ) / 2
... and so on.

The desired output will return

2.5 2.5 2.5
3.5 3.5 3.5
4.5 4.5 4.5

5.5 5.5 5.5
6.5 6.5 6.5
7.5 7.5 7.5




Muhammad Rahiz ?| ?Doctoral Student in Regional Climate Modeling
Climate Research Laboratory, School of Geography & the Environment
Oxford University Centre for the Environment, University of Oxford
South Parks Road, Oxford, OX1 3QY, United Kingdom
Tel: +44 (0)1865-285194 ?Mobile: +44 (0)7854-625974
Email: muhammad.rahiz at ouce.ox.ac.uk



milton ruser wrote:

Dear M.Rahiz,

Unfortunatelly I can't reproduce your example.

--
Henrique Dallazuanna
Curitiba-Paran?-Brasil
25? 25' 40" S 49? 16' 22" O

Here is a variation which also uses head and tail:

mapply(function(x, y) (x + y)/2, tail(x, -1), head(x, -1), SIMPLIFY = FALSE)


On Mon, Jan 4, 2010 at 12:37 PM, Henrique Dallazuanna <wwwhsd at gmail.com> wrote:
  

For this example try this:

lapply(lapply(list('head', 'tail'), do.call, list(x, n = -1)),
function(x)Reduce('+', x)/2)


On Mon, Jan 4, 2010 at 1:54 PM, Muhammad Rahiz
<muhammad.rahiz at ouce.ox.ac.uk> wrote:
    

Thanks Gabor,

It works alright. But is there alternative ways to perform rollmean apart
from permutating the data?
Possibly combining the Reduce and rollmean functions?

The easiest but traditional way is

      

(x[[1]]+x[[2]]) / 2
(x[[2]]+x[[3]]) / 2
        

Muhammad Rahiz  |  Doctoral Student in Regional Climate Modeling

Climate Research Laboratory, School of Geography & the Environment
Oxford University Centre for the Environment
South Parks Road, Oxford, OX1 3QY, United Kingdom Tel: +44 (0)1865-285194
 Mobile: +44 (0)7854-625974
Email: muhammad.rahiz at ouce.ox.ac.uk






Gabor Grothendieck wrote:
      

Try apply:

library(zoo) # rollmean

# test data
m <- matrix(1:3, 3, 3)
x <- list(m, m+3, m+6)

# convert to array
a <- array(unlist(x), c(3, 3, 3)); a

# apply rollmean and permute to desired form
aa <- apply(a, 1:2, rollmean, k = 2)
aperm(aa, c(2, 3, 1))

The last line outputs:


        

aperm(aa, c(2, 3, 1))

          

, , 1

    [,1] [,2] [,3]
[1,]  2.5  2.5  2.5
[2,]  3.5  3.5  3.5
[3,]  4.5  4.5  4.5

, , 2

    [,1] [,2] [,3]
[1,]  5.5  5.5  5.5
[2,]  6.5  6.5  6.5
[3,]  7.5  7.5  7.5


On Sat, Jan 2, 2010 at 10:00 AM, Muhammad Rahiz
<muhammad.rahiz at ouce.ox.ac.uk> wrote:

        

Let me rephrase;

Given x as


          

x

            

[[1]]
 V1 V2 V3
[1,]  1  1  1
[2,]  2  2  2
[3,]  3  3  3

[[2]]
 V1 V2 V3
[1,]  4  4  4
[2,]  5  5  5
[3,]  6  6  6

[[3]]
 V1 V2 V3
[1,]  7  7  7
[2,]  8  8  8
[3,]  9  9  9

I'd like to calculate the moving average (interval = 2) i.e.

( x[[1]] + x[[2]] ) / 2
( x[[2]] + x[[3]] ) / 2
... and so on.

The desired output will return

2.5 2.5 2.5
3.5 3.5 3.5
4.5 4.5 4.5

5.5 5.5 5.5
6.5 6.5 6.5
7.5 7.5 7.5




Muhammad Rahiz  |  Doctoral Student in Regional Climate Modeling
Climate Research Laboratory, School of Geography & the Environment
Oxford University Centre for the Environment, University of Oxford
South Parks Road, Oxford, OX1 3QY, United Kingdom
Tel: +44 (0)1865-285194  Mobile: +44 (0)7854-625974
Email: muhammad.rahiz at ouce.ox.ac.uk



milton ruser wrote:

          

Dear M.Rahiz,

Unfortunatelly I can't reproduce your example.

--
Henrique Dallazuanna
Curitiba-Paran?-Brasil
25? 25' 40" S 49? 16' 22" O

Dear Gabor & Henrique,

Thanks for your suggestions on the above problem. The following is more
traditional and unfanciful but it works. Suitable for a prodige like
myself...

m <- matrix(1:3,3,3)
x1 <- list(m, m+1, m+2, m+3, m+4)
out <- list()
for (i in 1:4){
t[[i]] <- Reduce("+", x1[c(i:i+1)])
}

Muhammad

Muhammad Rahiz ?| ?Doctoral Student in Regional Climate Modeling

Climate Research Laboratory, School of Geography & the Environment
Oxford University Centre for the Environment
South Parks Road, Oxford, OX1 3QY, United Kingdom Tel: +44 (0)1865-285194
?Mobile: +44 (0)7854-625974
Email: muhammad.rahiz at ouce.ox.ac.uk






Gabor Grothendieck wrote:

Here is a variation which also uses head and tail:

mapply(function(x, y) (x + y)/2, tail(x, -1), head(x, -1), SIMPLIFY =
FALSE)


On Mon, Jan 4, 2010 at 12:37 PM, Henrique Dallazuanna <wwwhsd at gmail.com>
wrote:

For this example try this:

lapply(lapply(list('head', 'tail'), do.call, list(x, n = -1)),
function(x)Reduce('+', x)/2)


On Mon, Jan 4, 2010 at 1:54 PM, Muhammad Rahiz
<muhammad.rahiz at ouce.ox.ac.uk> wrote:

Thanks Gabor,

It works alright. But is there alternative ways to perform rollmean
apart
from permutating the data?
Possibly combining the Reduce and rollmean functions?

The easiest but traditional way is

(x[[1]]+x[[2]]) / 2
(x[[2]]+x[[3]]) / 2

Muhammad Rahiz ?| ?Doctoral Student in Regional Climate Modeling

Climate Research Laboratory, School of Geography & the Environment
Oxford University Centre for the Environment
South Parks Road, Oxford, OX1 3QY, United Kingdom Tel: +44
(0)1865-285194
?Mobile: +44 (0)7854-625974
Email: muhammad.rahiz at ouce.ox.ac.uk






Gabor Grothendieck wrote:

Try apply:

library(zoo) # rollmean

# test data
m <- matrix(1:3, 3, 3)
x <- list(m, m+3, m+6)

# convert to array
a <- array(unlist(x), c(3, 3, 3)); a

# apply rollmean and permute to desired form
aa <- apply(a, 1:2, rollmean, k = 2)
aperm(aa, c(2, 3, 1))

The last line outputs:

aperm(aa, c(2, 3, 1))

, , 1

? ?[,1] [,2] [,3]
[1,] ?2.5 ?2.5 ?2.5
[2,] ?3.5 ?3.5 ?3.5
[3,] ?4.5 ?4.5 ?4.5

, , 2

? ?[,1] [,2] [,3]
[1,] ?5.5 ?5.5 ?5.5
[2,] ?6.5 ?6.5 ?6.5
[3,] ?7.5 ?7.5 ?7.5


On Sat, Jan 2, 2010 at 10:00 AM, Muhammad Rahiz
<muhammad.rahiz at ouce.ox.ac.uk> wrote:

Let me rephrase;

Given x as

x

[[1]]
?V1 V2 V3
[1,] ?1 ?1 ?1
[2,] ?2 ?2 ?2
[3,] ?3 ?3 ?3

[[2]]
?V1 V2 V3
[1,] ?4 ?4 ?4
[2,] ?5 ?5 ?5
[3,] ?6 ?6 ?6

[[3]]
?V1 V2 V3
[1,] ?7 ?7 ?7
[2,] ?8 ?8 ?8
[3,] ?9 ?9 ?9

I'd like to calculate the moving average (interval = 2) i.e.

( x[[1]] + x[[2]] ) / 2
( x[[2]] + x[[3]] ) / 2
... and so on.

The desired output will return

2.5 2.5 2.5
3.5 3.5 3.5
4.5 4.5 4.5

5.5 5.5 5.5
6.5 6.5 6.5
7.5 7.5 7.5




Muhammad Rahiz ?| ?Doctoral Student in Regional Climate Modeling
Climate Research Laboratory, School of Geography & the Environment
Oxford University Centre for the Environment, University of Oxford
South Parks Road, Oxford, OX1 3QY, United Kingdom
Tel: +44 (0)1865-285194 ?Mobile: +44 (0)7854-625974
Email: muhammad.rahiz at ouce.ox.ac.uk



milton ruser wrote:

Dear M.Rahiz,

Unfortunatelly I can't reproduce your example.

--
Henrique Dallazuanna
Curitiba-Paran?-Brasil
25? 25' 40" S 49? 16' 22" O

That gives an error when I run it.  I think you want this:

m <- matrix(1:3,3,3)
x1 <- list(m, m+1, m+2, m+3, m+4)

out <- list()
for(i in 1:4) out[[i]] <- (x1[[i]] + x1[[i+1]]) / 2


On Thu, Jan 7, 2010 at 12:28 PM, Muhammad Rahiz
<muhammad.rahiz at ouce.ox.ac.uk> wrote:
  

Dear Gabor & Henrique,

Thanks for your suggestions on the above problem. The following is more
traditional and unfanciful but it works. Suitable for a prodige like
myself...

m <- matrix(1:3,3,3)
x1 <- list(m, m+1, m+2, m+3, m+4)
out <- list()
for (i in 1:4){
t[[i]] <- Reduce("+", x1[c(i:i+1)])
}

Muhammad

Muhammad Rahiz  |  Doctoral Student in Regional Climate Modeling

Climate Research Laboratory, School of Geography & the Environment
Oxford University Centre for the Environment
South Parks Road, Oxford, OX1 3QY, United Kingdom Tel: +44 (0)1865-285194
 Mobile: +44 (0)7854-625974
Email: muhammad.rahiz at ouce.ox.ac.uk






Gabor Grothendieck wrote:
    

Here is a variation which also uses head and tail:

mapply(function(x, y) (x + y)/2, tail(x, -1), head(x, -1), SIMPLIFY =
FALSE)


On Mon, Jan 4, 2010 at 12:37 PM, Henrique Dallazuanna <wwwhsd at gmail.com>
wrote:

      

For this example try this:

lapply(lapply(list('head', 'tail'), do.call, list(x, n = -1)),
function(x)Reduce('+', x)/2)


On Mon, Jan 4, 2010 at 1:54 PM, Muhammad Rahiz
<muhammad.rahiz at ouce.ox.ac.uk> wrote:

        

Thanks Gabor,

It works alright. But is there alternative ways to perform rollmean
apart
from permutating the data?
Possibly combining the Reduce and rollmean functions?

The easiest but traditional way is


          

(x[[1]]+x[[2]]) / 2
(x[[2]]+x[[3]]) / 2

            

Muhammad Rahiz  |  Doctoral Student in Regional Climate Modeling

Climate Research Laboratory, School of Geography & the Environment
Oxford University Centre for the Environment
South Parks Road, Oxford, OX1 3QY, United Kingdom Tel: +44
(0)1865-285194
 Mobile: +44 (0)7854-625974
Email: muhammad.rahiz at ouce.ox.ac.uk






Gabor Grothendieck wrote:

          

Try apply:

library(zoo) # rollmean

# test data
m <- matrix(1:3, 3, 3)
x <- list(m, m+3, m+6)

# convert to array
a <- array(unlist(x), c(3, 3, 3)); a

# apply rollmean and permute to desired form
aa <- apply(a, 1:2, rollmean, k = 2)
aperm(aa, c(2, 3, 1))

The last line outputs:



            

aperm(aa, c(2, 3, 1))


              

, , 1

   [,1] [,2] [,3]
[1,]  2.5  2.5  2.5
[2,]  3.5  3.5  3.5
[3,]  4.5  4.5  4.5

, , 2

   [,1] [,2] [,3]
[1,]  5.5  5.5  5.5
[2,]  6.5  6.5  6.5
[3,]  7.5  7.5  7.5


On Sat, Jan 2, 2010 at 10:00 AM, Muhammad Rahiz
<muhammad.rahiz at ouce.ox.ac.uk> wrote:


            

Let me rephrase;

Given x as



              

x


                

[[1]]
 V1 V2 V3
[1,]  1  1  1
[2,]  2  2  2
[3,]  3  3  3

[[2]]
 V1 V2 V3
[1,]  4  4  4
[2,]  5  5  5
[3,]  6  6  6

[[3]]
 V1 V2 V3
[1,]  7  7  7
[2,]  8  8  8
[3,]  9  9  9

I'd like to calculate the moving average (interval = 2) i.e.

( x[[1]] + x[[2]] ) / 2
( x[[2]] + x[[3]] ) / 2
... and so on.

The desired output will return

2.5 2.5 2.5
3.5 3.5 3.5
4.5 4.5 4.5

5.5 5.5 5.5
6.5 6.5 6.5
7.5 7.5 7.5




Muhammad Rahiz  |  Doctoral Student in Regional Climate Modeling
Climate Research Laboratory, School of Geography & the Environment
Oxford University Centre for the Environment, University of Oxford
South Parks Road, Oxford, OX1 3QY, United Kingdom
Tel: +44 (0)1865-285194  Mobile: +44 (0)7854-625974
Email: muhammad.rahiz at ouce.ox.ac.uk



milton ruser wrote:


              

Dear M.Rahiz,

Unfortunatelly I can't reproduce your example.

--
Henrique Dallazuanna
Curitiba-Paran?-Brasil
25? 25' 40" S 49? 16' 22" O