Peter Szabo a écrit :
I see now. anyway, it is really easy after you follow me. Leave @loc function (the last line i wrote in my previous letter), then you only have to use some if-then cycles to get how many occurence are.let s2=if conseq eq 2 then 1; let howmany2=s2[i=@sum] !and so on... so howmany2 will give you the number of events of 2 consecutive values.or use the @evnt.
Sorry again Peter but this not the right answer.
Using your code, you get 7 events of 2 consecutives values of 1.
You should find 4.
yes? use v1.nc
yes? let conseq=v1[i=@cda:1]
yes? let s2=if conseq eq 2 then 1
yes? list v1,conseq,s2
yes? let howmany2=s2[i=@sum]
yes? list howmany2
VARIABLE : S2[I=@SUM]
FILENAME : v1.nc
X : 0.5 to 52.5
7.000
Publish on mailling list if you can find the exact result which are for
the file attached v1.nc
5 events of 1 consecutive values at 1 4 events of 2 consecutive values at 1 1 event of 3 consecutive values at 1 3 events of 4 consecutive values at 1 Patrick
On Fri, Jun 20, 2008 at 3:21 PM, Brockmann Patrick <Patrick.Brockmann@xxxxxx <mailto:Patrick.Brockmann@xxxxxx>> wrote:Peter Szabo a écrit : Hi Patrick, try something like this in the beginning of your script (the leftover is not necessary): let b = If a le 0.45 then 1 !i changed it intentionally. you will have missing values instead of 1 at the right points let conseq=b[i=@cda:1] !with @cda you will have the data in a format like ,,5,4,3,2,1,,,, instead of ,,1,1,1,1,1,,,, let whereis5=conseq[i=@loc:5] !the @loc finds the start of the consecutive events. in this case 5. Is this what you wanted? Not at all. I would like the result as follows 5 events of 1 consecutive values at 1 4 events of 2 consecutive values at 1 1 event of 3 consecutive values at 1 3 events of 4 consecutive values at 1 Ok thank you anyway. Good try. Patrick-- LSCE/IPSL, Laboratoire CEA-CNRS-UVSQData Analysis and Visualization Engineer IPSL Global Climate Modelling Group --
Attachment:
v1.nc
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