c# - Reactive Extension : 3 Consecutive occurance of a Record -


i have hot observable -

class tablerecord

{     string propertya;     datetime dt;     dictionary<string,int> checkthreshholdvalues; }

i need check each value present in dictionary, if 3 consecutive values exceeding threshold or not. if yes, need generate new event property. (here each reacord come @ different of 15 minutes ie 4 records in 1 hour.

class generatedrecord {   string propertya;   datetime lasttimeobserved;   string dictionarypropertyname;   int lastvalue }

i know have use groupbyuntil buffer, somehow unable think , how split dictionary individual parts still keeping condition appreciated.

example : lets have propertya value in {"towna","townb","townc"} lets properties in dictionary {"pollution","temperature","pressure"}

lets say, recv values per 15 minutes, data -

{"towna" ,"6/14/2015 10:00",{ {"pollution",61},{"temperature",48},{"pressure",40}}}  {"towna" ,"6/14/2015 10:15",{ {"pollution",63},{"temperature",63},{"pressure",40}}}  {"towna" ,"6/14/2015 10:30",{ {"pollution",49},{"temperature",64},{"pressure",40}}} {"towna" ,"6/14/2015 10:45",{ {"pollution",70},{"temperature",65},{"pressure",40}}}

say threshhold value 60. above example, temperature satisfies condition has value of {63,64,65 } . pollution 2 consecutive values above 60

regards

you should able overlapping windows couldn't work out syntax start new window every time element received , close window when contains 3 items. here conceptual solution based on observable.create , maintaining queue. in case use interval , simple threshold (>3):

observable.create<string>(observer => {     var queue = new queue<long>(3);        return observable.interval(timespan.fromseconds(1)).subscribe(value =>     {         if (queue.count == 3)         {             queue.dequeue();         }          queue.enqueue(value);          if (queue.count == 3)         {             var values = queue.toarray();             if (values.all(v => v > 3))             {                 observer.onnext("all above 3. " + values[0] + " " + values[1] + " " + values[2]);             }         }     }); })

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