R help - Jan 2011 - [zoo] - Individual zoo or data frames from non-continuous zoo series

#Is there a way to break the below zoo object into non-NA data frames
algorithmically
#this is a small example of a much larger problem.
#It is really no even necessary to have the continuous chunks
#end up as zoo objects but it is important to have them end
#up with the index column.
#thanks for all of your help in advance, and
#if you need anything else please let me know

library(zoo)

ind. <- 1:200
data <- c(1:50, rep(NA, 50), 1:50, rep(NA, 50))

z <- zoo(data, ind.)

-- 
Stephen Sefick
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Let's not spend our time and resources thinking about things that are
so little or so large that all they really do for us is puff us up and
make us feel like gods.? We are mammals, and have not exhausted the
annoying little problems of being mammals.

? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? -K. Mullis

"A big computer, a complex algorithm and a long time does not equal
science."

? ? ? ? ? ? ? ? ? ? ? ? ? ? ? -Robert Gentleman

On Thu, Jan 6, 2011 at 5:41 PM, stephen sefick <ssefick at gmail.com>
wrote:
> #Is there a way to break the below zoo object into non-NA data frames
> algorithmically
> #this is a small example of a much larger problem.
> #It is really no even necessary to have the continuous chunks
> #end up as zoo objects but it is important to have them end
> #up with the index column.
> #thanks for all of your help in advance, and
> #if you need anything else please let me know
>
> library(zoo)
>
> ind. <- 1:200
> data <- c(1:50, rep(NA, 50), 1:50, rep(NA, 50))
>
> z <- zoo(data, ind.)
>
Below c(TRUE, diff(is.na(z)) != 0) gives a logical vector which is
TRUE at the first position of any run of NAs or non-NAs.  The cumsum
of that gives a vector the same length as z such that each position of
the first run is 1, each position of the second run is 2, etc.  The NA
runs are then set to 0 in the second line.  In the third line we split
z on g and drop the portion that corresponds to the NAs.
> g <- cumsum(c(TRUE, diff(is.na(z)) != 0))
> g[is.na(z)] <- 0
> split(z, g)[-1]
$`1`
 1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26
 1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26
27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50

$`3`
101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120
  1   2   3   4   5   6   7   8   9  10  11  12  13  14  15  16  17  18  19  20
121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140
 21  22  23  24  25  26  27  28  29  30  31  32  33  34  35  36  37  38  39  40
141 142 143 144 145 146 147 148 149 150
 41  42  43  44  45  46  47  48  49  50

This could be combined into a multivariate series like this:

do.call("merge", split(z, g)[-1])

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