On Jun 20, 2010, at 5:52 PM, Schmidt Martin wrote:
Dear R users
A quite simple question: how can I get the mean duration
(persistence) of daily data, when the data set looks as follows:
[1] 7 7 7 9 9 5 7 5 5 5 9 5 5 6 6 6 1 1 1 2 2 4 4 4 4 7 7 3 3 2 4 4
7 7 7 7 7
[38] 7 7 5 5 9 9 5 5 5 1 7 9 9 9 9 9 5 5 5 5 5 3 4 4 4 4 4 4 4 4 2 2
2 2 2 2 8
[75] 8 6 6 6 8 8 8 8 8 6 6 2 8 8 8 8 8 8 8 8 8 2 2 2 2 2 6 6 6 6 6 6
6 6 6 6 3
[112] 3 2 2 2 1 5 1 1 4 2 2 2 2 2 2 2 3 4 4 4 3 3 3 6 6 6 8 8 4 4 2
4 4 1 1 6 6
.............................and so on:
Thus, I'd like to know, from each number (1-9) the mean duration
from the first to the last day (here not visible, but each number
stands for a day).
With:
length(which(test==1))
I certainly get the total number of number 1......unfortunately
that's all I got!
I am guessing you want:
?rle
> dur <- scan()
1: 7 7 7 9 9 5 7 5 5 5 9 5 5 6 6 6 1 1 1 2 2 4 4 4 4 7 7 3 3 2 4 4 7 7
7 7 7
38: 7 7 5 5 9 9 5 5 5 1 7 9 9 9 9 9 5 5 5 5 5 3 4 4 4 4 4 4 4 4 2 2 2
2 2 2 8
75: 8 6 6 6 8 8 8 8 8 6 6 2 8 8 8 8 8 8 8 8 8 2 2 2 2 2 6 6 6 6 6 6 6
6 6 6 3
112: 3 2 2 2 1 5 1 1 4 2 2 2 2 2 2 2 3 4 4 4 3 3 3 6 6 6 8 8 4 4 2 4
4 1 1 6 6
149:
Read 148 items
> rle(dur)
Run Length Encoding
lengths: int [1:51] 3 2 1 1 3 1 2 3 3 2 ...
values : num [1:51] 7 9 5 7 5 9 5 6 1 2 ...
> mean(rle(dur)$lengths)
[1] 2.901961
--
Just a guess, mind you. Your problem description seemed vague in both
the input ahd the output sides.
David Winsemius, MD
West Hartford, CT
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