On Wed, 30 Nov 2011, Riccardo (Jack) Lucchetti wrote:
On Wed, 30 Nov 2011, Sven Schreiber wrote:
> On 11/30/2011 10:55 PM, Henrique Andrade wrote:
>> Dear Hansl experts,
>>
>> I would like to write a Hansl code but unfortunately I'm out of
>> creativity :(
>> I have a binary series with blocks of 0 and 1. Something like
>>
>> X=(0,0,0,1,1,1,1,1,0,0,0,0,0,1,1,1,1,1,1,1,1,1,1,1).
>>
>> Here are the steps I need do follow:
>>
>> (1) Find the number of 1-blocks;
>> (2) Calculate the average number of observations inside these blocks.
>>
>> In my hypothetical example, the X series, I have two blocks, and these
>> blocks have an average of 8 observations (five observations in the first
>> block and eleven observations in the second block).
>>
>
> Just a quick thought: you could take the (serial/first) difference of
> your series and count the occurrences of -1, which indicates how many
> times you have a 01 pattern. And each 1 in the difference indicates the
> position of a 10 pattern (or vice versa, depending on whether you take
> left-to-right or right-to-left differences). The position indices help
> you determine how many ones you have in between, i.e. in each block.
>
> But your problem in general does look like unpleasant details to code.
Building on Sven's idea: call your binary series y. Then,
<hansl>
series chg = diff(y)
chg[1] = (y[1] == 1)
scalar n_blocks = sum(chg==1)
scalar avg_len = sum(y) / n_blocks
</hansl>
I think this ought to work.
Here's another approach -- not so elegant in that it involves a
loop, but it prepares a vector of block lengths, in case that might
be wanted.
<hansl>
matrix X={0,0,0,1,1,1,1,1,0,0,0,0,0,1,1,1,1,1,1,1,1,1,1,1}'
matrix B = {}
scalar blen = 0
scalar n = rows(X)
loop i=1..n -q
if X[i] == 1
blen++
if i == n || X[i+1] == 0
B |= {blen}
blen = 0
endif
endif
endloop
printf "Found %d blocks, average length %g\n",
rows(B), meanc(B)
</hansl>
Allin Cottrell