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Finding the intersection of two date lists


I've got two lists that look like this


list1={{"1/01/2010 6:15", 0.0565625}, {"11/06/2010 0:15",0}, {"11/06/2010 0:30", 0},{"11/06/2010 0:45",0}, {"11/06/2010 1:00", 0}}


list2={{"01/01/2010 06:15", 0.04375}, {"01/01/2010 06:30",0.04375}, {"01/01/2010 06:45", 0.04375}, {"01/01/2010 07:00",0.04375}, {"01/01/2010 07:15", 0.04375}}

What I want to do is create a list that looks like this;


list3={{"1/01/2010 6:15", 0.0565625, 0.04375},{{"01/01/2010 06:30",,0.04375},......}

Either list may have gaps in it. It's not important that all the gaps be filled (if the date is missing from both list1&2 then it can be missing from list3). What is important is that it's fast, the data set is about 100,000 records.



Answer



I confess to being allergic to database operations that require data to be in a particular sort order: it's too easy for huge mistakes to creep in. What is needed here is to turn the source list (say, list1) into a lookup table so it reliably returns its value (second element in the list) when given its key (first element in the list).


To assure reliability of key matches, let's first convert strings into bona fide dates:


l1 = {DateList[First[#]], Last[#]} & /@  list1;

l2 = {DateList[First[#]], Last[#]} & /@  list2;

Now convert the source list into a rule dispatch table:


rules = Dispatch[Flatten[({First[#] -> Sequence @@ #} & /@  l1), 1]]

(This can take some time with 100,000 entries: my system requires almost ten seconds. I consider the wait to be worthwhile. But be careful if the source list is much longer than this!)


These rules can repeatedly be applied to any number of target lists:


l2 /. rules // MatrixForm



$\left( \begin{array}{c} \{\{2010,1,1,6,15,0.\},0.0565625,0.04375\} \\ \{\{2010,1,1,6,30,0.\},0.04375\} \\ \{\{2010,1,1,6,45,0.\},0.04375\} \\ \{\{2010,1,1,7,0,0.\},0.04375\} \\ \{\{2010,1,1,7,15,0.\},0.04375\} \end{array} \right)$



When the target has $100000$ elements and there are $100000$ source elements, this last step still takes only $0.25$ seconds.


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