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How to extract data?

2 messages · james.holtman@convergys.com, Uwe Ligges

#
By 'ignore', can we delete those from the list of data?  I would then
assume that if you have a sequence of +0+0+ that you would want the last
"+" for the increase of three.

If that is the case, then do a 'diff' and delete the entries that are 0.
Then create a new 'diff' and then use 'rle' to see what the length of the
sequences are:
[1] 1 2 2 3 3 4 3 3 2 2 2 1
[1]  1  0  1  0  1 -1  0 -1  0  0 -1
[1] 1 2 3 4 3 2 1
[1]  1  1  1 -1 -1 -1
Run Length Encoding
  lengths: int [1:2] 3 3
  values : num [1:2] 1 -1
you can check the results of 'rle' to determine where the changes are.
__________________________________________________________
James Holtman        "What is the problem you are trying to solve?"
Executive Technical Consultant  --  Office of Technology, Convergys
james.holtman at convergys.com
+1 (513) 723-2929


                                                                                                                                           
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I appreciate if anyone can help me,
I have a table as follow,
DATE VALUE
1   1997-01-10  5.30
2   1997-01-17  5.30
3   1997-01-24  5.28
4   1997-01-31  5.30
5   1997-02-07  5.29
6   1997-02-14  5.26
7   1997-02-21  5.24
8   1997-02-28  5.26
9   1997-03-07  5.30
10  1997-03-14  5.30
.    ......     ...
.    ......     ...
.    ......     ...
I want to extract the DATE(s) on which the VALUE has
already dropped twice and the DATE(s) that VALUE has
already increased for three times,( ignore where
VALUE(i+1)-VALUE(i)=0),I try to use diff() function,
however that works only for one increase or decrease.

Sincerely,

Sean

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#
james.holtman at convergys.com wrote:

            
Additionally, if there are differences are > 1, you might want apply 
sign() before rle().

Uwe Ligges