tapply(t(data.matrix(sim_sub)),rep( rep(1:4, each=5), each=10),mean) 
   #  1      2      3      4 
#0.0086 0.0074 0.0082 0.0108 

unlist(lapply(split(sim_sub,((seq_len(nrow(sim_sub))-1)%/%5)+1),function(x) 
mean(unlist(x))))
#    1      2      3      4 
#0.0086 0.0074 0.0082 0.0108 
A.K.

----- Original Message -----
From: David Winsemius <dwinsem...@comcast.net>
To: Keith S Weintraub <kw1...@gmail.com>
Cc: "r-help@r-project.org" <r-help@r-project.org>
Sent: Wednesday, April 17, 2013 4:05 PM
Subject: Re: [R] Best way to calculate averages of Blocks in an matrix?


On Apr 17, 2013, at 9:54 AM, Keith S Weintraub wrote:

> Folks,
>  I recently was given a simulated data set like the following subset:
> 
> sim_sub<-structure(list(V11 = c(0.01, 0, 0, 0.01, 0, 0.01, 0, 0, 0, 0, 
> 0, 0, 0, 0, 0, 0, 0, 0, 0, 0), V12 = c(0, 0, 0, 0.01, 0.03, 0, 
> 0, 0, 0, 0, 0, 0.01, 0, 0.01, 0, 0, 0, 0, 0, 0.04), V13 = c(0, 
> 0, 0, 0.01, 0, 0, 0, 0, 0, 0.01, 0, 0, 0, 0, 0.01, 0, 0, 0, 0, 
> 0.01), V14 = c(0, 0.01, 0.01, 0.01, 0.01, 0, 0, 0, 0, 0.03, 0, 
> 0, 0.01, 0.01, 0.04, 0.01, 0.02, 0, 0.01, 0.03), V15 = c(0, 0.01, 
> 0, 0, 0.01, 0, 0, 0, 0.01, 0.02, 0.01, 0, 0, 0.01, 0, 0, 0, 0.01, 
> 0.01, 0.04), V16 = c(0, 0, 0, 0.03, 0.02, 0.01, 0, 0, 0.02, 0.02, 
> 0, 0.02, 0.02, 0, 0.01, 0.01, 0, 0, 0.03, 0.01), V17 = c(0, 0.01, 
> 0, 0.01, 0, 0, 0, 0.01, 0.05, 0.03, 0, 0.01, 0, 0.02, 0.02, 0, 
> 0, 0.01, 0.02, 0.04), V18 = c(0, 0.01, 0, 0.03, 0.03, 0, 0, 0, 
> 0.02, 0.01, 0, 0.02, 0.01, 0.02, 0.03, 0.02, 0, 0, 0.04, 0.04
> ), V19 = c(0, 0.01, 0.01, 0.02, 0.07, 0, 0, 0, 0.04, 0.01, 0.02, 
> 0, 0, 0, 0.04, 0, 0, 0, 0, 0.05), V20 = c(0, 0, 0, 0.01, 0.04, 
> 0.01, 0, 0, 0.02, 0.04, 0.01, 0, 0.02, 0, 0.03, 0, 0.02, 0.01, 
> 0.03, 0.03)), .Names = c("V11", "V12", "V13", "V14", "V15", "V16", 
> "V17", "V18", "V19", "V20"), row.names = c(NA, 20L), class = "data.frame")
> 
>> sim_sub
>    V11  V12  V13  V14  V15  V16  V17  V18  V19  V20
> 1  0.01 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00
> 2  0.00 0.00 0.00 0.01 0.01 0.00 0.01 0.01 0.01 0.00
> 3  0.00 0.00 0.00 0.01 0.00 0.00 0.00 0.00 0.01 0.00
> 4  0.01 0.01 0.01 0.01 0.00 0.03 0.01 0.03 0.02 0.01
> 5  0.00 0.03 0.00 0.01 0.01 0.02 0.00 0.03 0.07 0.04
> 6  0.01 0.00 0.00 0.00 0.00 0.01 0.00 0.00 0.00 0.01
> 7  0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00
> 8  0.00 0.00 0.00 0.00 0.00 0.00 0.01 0.00 0.00 0.00
> 9  0.00 0.00 0.00 0.00 0.01 0.02 0.05 0.02 0.04 0.02
> 10 0.00 0.00 0.01 0.03 0.02 0.02 0.03 0.01 0.01 0.04
> 11 0.00 0.00 0.00 0.00 0.01 0.00 0.00 0.00 0.02 0.01
> 12 0.00 0.01 0.00 0.00 0.00 0.02 0.01 0.02 0.00 0.00
> 13 0.00 0.00 0.00 0.01 0.00 0.02 0.00 0.01 0.00 0.02
> 14 0.00 0.01 0.00 0.01 0.01 0.00 0.02 0.02 0.00 0.00
> 15 0.00 0.00 0.01 0.04 0.00 0.01 0.02 0.03 0.04 0.03
> 16 0.00 0.00 0.00 0.01 0.00 0.01 0.00 0.02 0.00 0.00
> 17 0.00 0.00 0.00 0.02 0.00 0.00 0.00 0.00 0.00 0.02
> 18 0.00 0.00 0.00 0.00 0.01 0.00 0.01 0.00 0.00 0.01
> 19 0.00 0.00 0.00 0.01 0.01 0.03 0.02 0.04 0.00 0.03
> 20 0.00 0.04 0.01 0.03 0.04 0.01 0.04 0.04 0.05 0.03
> 
> Every 5 rows represents one block of simulated data.
> 
> What would be the best way to average the blocks?

This answers the posed question:

  > tapply( data.matrix(sim_sub),  rep( rep(1:4, each=5), each=10) ,mean)
     1      2      3      4 
0.0030 0.0070 0.0106 0.0144 


Your code following suggests that you do not want the average values within 
blocks but within blocks AND ALSO within columns (although how you get 5 rows 
of 5 blocks from a 20 row input object is unclear to me)

> data.frame( lapply(sim_sub, function(col) tapply(col, rep(1:4, each=5), mean) 
>  ) )
    V11   V12   V13   V14   V15  V16   V17   V18   V19   V20
1 0.004 0.008 0.002 0.008 0.004 0.01 0.004 0.014 0.022 0.010
2 0.002 0.000 0.002 0.006 0.006 0.01 0.018 0.006 0.010 0.014
3 0.000 0.004 0.002 0.012 0.004 0.01 0.010 0.016 0.012 0.012
4 0.000 0.008 0.002 0.014 0.012 0.01 0.014 0.020 0.010 0.018

>From your code I am guessing a typo of 5 for 4?

> 
> My way was to reshape sim_sub, average over the columns and then reshape back 
> like so:
> 
>> matrix(colSums(matrix(t(sim_sub), byrow = TRUE, ncol = 50)), byrow = TRUE, 
>> ncol = 10)/4
>       [,1]   [,2]   [,3]   [,4]   [,5]  [,6]   [,7]   [,8]   [,9]  [,10]
> [1,] 0.0050 0.0000 0.0000 0.0025 0.0025 0.005 0.0000 0.0050 0.0050 0.0050
> [2,] 0.0000 0.0025 0.0000 0.0075 0.0025 0.005 0.0050 0.0075 0.0025 0.0050
> [3,] 0.0000 0.0000 0.0000 0.0050 0.0025 0.005 0.0050 0.0025 0.0025 0.0075
> [4,] 0.0025 0.0050 0.0025 0.0075 0.0075 0.020 0.0250 0.0275 0.0150 0.0150
> [5,] 0.0000 0.0175 0.0075 0.0275 0.0175 0.015 0.0225 0.0275 0.0425 0.0350
> 
> 
> How bad is "t(sim_sub)" in the above?

The whole matrix( matrix( t(.), ... )) approach seems kind of tortured, but to 
your question, t() is a fairly efficient function.

-- 

David Winsemius
Alameda, CA, USA

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