December 18, 2017 at 6:42 pm
Hello all,
Hoping someone out there can help with this because I'm just not seeing the answer for some reason, and it seems like it should be a fairly easy one. If I can just get some ideas on the approach, I can work through it....
I have one table with trip information, i.e. tripID, pickup address, dropoff address for each trip for a day. For a typical rider, they will have 2 trips, one outbound and one returning, and for the return trip generally the dropoff address and pickup address will just be swapped from the outbound trip. Most, but not all. I need to find the distinct trips between pickup address and dropoff address --even if they're reversed as dropoff to pickup. Trying to identify just the unique trips between two addresses (regardless of whether they are pickup or dropoff address).
In other words, John Doe shows up in this table/dataset with two records, one trip with pickup address of 1234 5th ave s and dropoff address of 789 10th ave se, and one record with pickup address of 789 10th ave se and dropoff address of 1234 5th ave s. I am only looking for the distinct combination of these two addresses once.
Tried concatenating the pickup and dropoff address into another field in one temp table, swapping the dropoff and pickup to concatenate again for another temp table, and then get distinct but that didn't quite work as expected.
I'm not sure why I'm not seeing the answer here.
Anyone have any ideas?
December 19, 2017 at 5:29 am
Tracey,
I think you will need to post DDL and data for us to help you.Otherwise the short answer is "Because they don't match so are not distinct"
Could you build a CTE that includes a RANK() OVER(Partition by Rider, Date ORDER BY TripID )
You can then join the CTE to itself
SELECT
CTE1.*
,CTE2.*
FROM
CTE CTE1
LEFT JOIN
CTE CTE2 ON CTE2.Rider = CTE1.Rider
AND CTE2.Date = CTE1.Date
AND CTE2.Rank > CTE1.Rank
AND CTE2.Pickup = CTE1.Dropoff
AND CTE2.Dropoff = CTE1.Pickup
ORDER BY
CTE1.Rider
,CTE1.Date
,CTE1.Rank
This will give you all of the trips and the return trip if it exists
You will get multiple joins if the rider can do the same round trip more than once in a day but this should at least indicate where the round trips are and why they are not displaying correctly in your distinct
December 19, 2017 at 5:50 am
How do you identify a return trip of the second address is different? You didn't mention it, but do you also have a RiderID or something along those lines? Otherwise, let's say two riders come along. RiderA goes from LocationA to LocationB. RiderB goes from LocationC to LocationB. Now, RiderA goes home, LocationB to LocationA. RiderB follows RiderA home, LocationB to LocationA. How would we differentiate those two? It would look like a single outbound with two inbound trips and the other outbound would have no inbound.
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December 19, 2017 at 9:16 am
Isn't it as simple as a CASE expression?
SELECT CASE WHEN pickup_address < dropoff_address THEN pickup_address ELSE dropoff_address END AS address1,
CASE WHEN pickup_address < dropoff_address THEN dropoff_address ELSE pickup_address END AS address2
FROM YourTable
Drew
J. Drew Allen
Business Intelligence Analyst
Philadelphia, PA
December 19, 2017 at 9:22 am
Thanks for the info. Maybe this will clarify:
My ultimate goal is to only find the distinct combination of two addresses in this table. (Regardless of the passenger or TripID or whether the addresses show up as outbound trip or return trip.)
I tried concatenating ((PAddress + ' ' + DAddress) [Trip]) to make a new column ("Trip") which should have distinct values but that doesn't help me to narrow down things as it will of course still show two records for each "trip" between two addresses. I only want to see one.
Here's some sample data: (sorry for the weird format spacing, not sure what's going on here)
CREATE TABLE #TRIPS
(TRIPID INT
,PICKUP_ADDRESS VARCHAR(50)
,DROPOFF_ADDRESS VARCHAR(50)
)
INSERT INTO #TRIPS
VALUES
(23084416
,'100 MELROSE AVE E'
,'915 NW 45TH ST')
INSERT INTO #TRIPS
VALUES
(23079056
,'1001 5TH AVE'
,'18100 NE 95TH ST')
INSERT INTO #TRIPS
VALUES
(23084870
,'3252 64TH AVE SW'
,'1524 S 328TH ST')
INSERT INTO #TRIPS
VALUES
(23059443
,'915 NW 45TH ST'
,'100 MELROSE AVE E')
INSERT INTO #TRIPS
VALUES
(23081731
,'7100 CALIFORNIA AVE SW'
,'1001 5TH AVE')
INSERT INTO #TRIPS
VALUES
(23086885
,'18100 NE 95TH ST'
,'1001 5TH AVE')
INSERT INTO #TRIPS
VALUES
(23064853
,'1524 S 328TH ST'
,'3252 64TH AVE SW')
INSERT INTO #TRIPS
VALUES
(23084523
,'1001 5TH AVE'
,'7100 CALIFORNIA AVE SW')
SELECT *
FROM #TRIPS
DROP TABLE #TRIPS
What I'm hoping to get as the resultset is something like this:
CREATE TABLE #EXPECTEDRESULTS
(PICKUP_ADDRESS VARCHAR(50)
,DROPOFF_ADDRESS VARCHAR(50)
)
INSERT INTO #EXPECTEDRESULTS
VALUES
('1001 5TH AVE'
,'7100 CALIFORNIA AVE SW')
INSERT INTO #EXPECTEDRESULTS
VALUES
('100 MELROSE AVE E'
,'915 NW 45TH ST')
INSERT INTO #EXPECTEDRESULTS
VALUES
('1001 5TH AVE'
,'18100 NE 95TH ST')
INSERT INTO #EXPECTEDRESULTS
VALUES
('3252 64TH AVE SW'
,'1524 S 328TH ST')
Notice that doesn't return every combination of the addresses, since most of these have two (outbound and return).
Does that make more sense?
December 19, 2017 at 9:43 am
Drew.allen - I think that might be it!
I dumped the CASE data into a temp table and added a SELECT DISTINCT ADDRESS1, ADDRESS2 to get the distinct list and I think this works. I used this query against the sample data I posted and it returns the same address sets as my ExpectedResults table.
I never would have thought to use < to compare.
Here's another challenge on this, should you all choose to accept it.....
What if there were additional columns in the table, such as City, Lat, and Lon for each record? So expand the original table from TripID, Pickup_Address, Dropoff_Address to TripID, Pickup_Address, PickupCity, Pickup_State, Pickup_Lat, Pickup_Lon, Dropoff_Address, Dropoff_City, Dropoff_State, Dropoff_Lat, Dropoff_Lon.
I can't imaging that the < compare would still work here, would it?
December 19, 2017 at 10:25 am
tacy.highland - Tuesday, December 19, 2017 9:43 AMDrew.allen - I think that might be it!I dumped the CASE data into a temp table and added a SELECT DISTINCT ADDRESS1, ADDRESS2 to get the distinct list and I think this works. I used this query against the sample data I posted and it returns the same address sets as my ExpectedResults table.
I never would have thought to use < to compare.
Here's another challenge on this, should you all choose to accept it.....
What if there were additional columns in the table, such as City, Lat, and Lon for each record? So expand the original table from TripID, Pickup_Address, Dropoff_Address to TripID, Pickup_Address, PickupCity, Pickup_State, Pickup_Lat, Pickup_Lon, Dropoff_Address, Dropoff_City, Dropoff_State, Dropoff_Lat, Dropoff_Lon.
I can't imaging that the < compare would still work here, would it?
Yes, it does. You just have to make sure that you are using the same field(s) for all of the comparisons. Since you have latitude and longitude, I would be inclined to use those, since they would be less subject to variability. That is, the same address can be given several different ways, and it might be hard to determine that they are actually the same, but a latitude and longitude will always be the same within a certain degree of accuracy.
Drew
J. Drew Allen
Business Intelligence Analyst
Philadelphia, PA
December 19, 2017 at 10:46 am
How would the CASE statement look? Would it be a case for each column, something like this?:
SELECT CASE WHEN pickup_address < dropoff_address THEN pickup_address ELSE dropoff_address END AS address1,
CASE WHEN pickup_address < dropoff_address THEN dropoff_address ELSE pickup_address END AS address2,
CASE WHEN pickup_city < dropoff_city THEN pickup_city ELSE dropoff_city END as city1,
CASE WHEN pickup_city < dropoff_city THEN dropoff_city ELSE pickup_city END as city2,
CASE WHEN pickup_Lat < dropoff_Lat THEN pickup_Lat ELSE dropoff_Lat END as Lat1,
........
FROM YourTable
December 19, 2017 at 11:47 am
tacy.highland - Tuesday, December 19, 2017 9:22 AMThanks for the info. Maybe this will clarify:
My ultimate goal is to only find the distinct combination of two addresses in this table. (Regardless of the passenger or TripID or whether the addresses show up as outbound trip or return trip.)
I tried concatenating ((PAddress + ' ' + DAddress) [Trip]) to make a new column ("Trip") which should have distinct values but that doesn't help me to narrow down things as it will of course still show two records for each "trip" between two addresses. I only want to see one.
Here's some sample data: (sorry for the weird format spacing, not sure what's going on here)
CREATE TABLE #TRIPS
(TRIPID INT
,PICKUP_ADDRESS VARCHAR(50)
,DROPOFF_ADDRESS VARCHAR(50)
)
INSERT INTO #TRIPS
VALUES
(23084416
,'100 MELROSE AVE E'
,'915 NW 45TH ST')
INSERT INTO #TRIPS
VALUES
(23079056
,'1001 5TH AVE'
,'18100 NE 95TH ST')
INSERT INTO #TRIPS
VALUES
(23084870
,'3252 64TH AVE SW'
,'1524 S 328TH ST')
INSERT INTO #TRIPS
VALUES
(23059443
,'915 NW 45TH ST'
,'100 MELROSE AVE E')
INSERT INTO #TRIPS
VALUES
(23081731
,'7100 CALIFORNIA AVE SW'
,'1001 5TH AVE')
INSERT INTO #TRIPS
VALUES
(23086885
,'18100 NE 95TH ST'
,'1001 5TH AVE')
INSERT INTO #TRIPS
VALUES
(23064853
,'1524 S 328TH ST'
,'3252 64TH AVE SW')
INSERT INTO #TRIPS
VALUES
(23084523
,'1001 5TH AVE'
,'7100 CALIFORNIA AVE SW')
SELECT *
FROM #TRIPS
DROP TABLE #TRIPS
What I'm hoping to get as the resultset is something like this:
CREATE TABLE #EXPECTEDRESULTS
(PICKUP_ADDRESS VARCHAR(50)
,DROPOFF_ADDRESS VARCHAR(50)
)
INSERT INTO #EXPECTEDRESULTS
VALUES
('1001 5TH AVE'
,'7100 CALIFORNIA AVE SW')
INSERT INTO #EXPECTEDRESULTS
VALUES
('100 MELROSE AVE E'
,'915 NW 45TH ST')
INSERT INTO #EXPECTEDRESULTS
VALUES
('1001 5TH AVE'
,'18100 NE 95TH ST')
INSERT INTO #EXPECTEDRESULTS
VALUES
('3252 64TH AVE SW'
,'1524 S 328TH ST')
Notice that doesn't return every combination of the addresses, since most of these have two (outbound and return).
Does that make more sense?
So, something like this:
CREATE TABLE #TRIPS
(TRIPID INT
,PICKUP_ADDRESS VARCHAR(50)
,DROPOFF_ADDRESS VARCHAR(50)
);
INSERT INTO #TRIPS ([TRIPID],[PICKUP_ADDRESS],[DROPOFF_ADDRESS])
VALUES
(23084416,'100 MELROSE AVE E','915 NW 45TH ST')
, (23079056,'1001 5TH AVE','18100 NE 95TH ST')
, (23084870,'3252 64TH AVE SW','1524 S 328TH ST')
, (23059443,'915 NW 45TH ST','100 MELROSE AVE E')
, (23081731,'7100 CALIFORNIA AVE SW','1001 5TH AVE')
, (23086885,'18100 NE 95TH ST','1001 5TH AVE')
, (23064853,'1524 S 328TH ST','3252 64TH AVE SW')
, (23084523,'1001 5TH AVE','7100 CALIFORNIA AVE SW');
SELECT DISTINCT
[ca1].[Address1]
,[ca1].[Address2]
FROM
#TRIPS t
CROSS APPLY (SELECT MIN(dt.[Address]), MAX(dt.[Address]) FROM (VALUES (t.[PICKUP_ADDRESS]),(t.[DROPOFF_ADDRESS]))dt([Address]))ca1([Address1],[Address2]);
DROP TABLE #TRIPS;
December 20, 2017 at 9:17 am
Hello? Any feedback?
December 20, 2017 at 9:30 am
My apologies, Lynn. Thank you for your input here.
The problem got a little more complex with additional columns added for each address. So now, instead of just getting the unique combinations of address1 and address2, I need those and also address1_city, address1_state, address1_lat, address1_lon, and same for address2. I've not used CROSS APPLY before so not sure how to adapt that to your query.
I had added a modification (question) to the other query that drew provided to add those columns but never heard anything back on whether that was quite correct.
December 20, 2017 at 9:44 am
Could you post an expand example to work with?
December 20, 2017 at 9:48 am
tacy.highland - Wednesday, December 20, 2017 9:30 AMMy apologies, Lynn. Thank you for your input here.The problem got a little more complex with additional columns added for each address. So now, instead of just getting the unique combinations of address1 and address2, I need those and also address1_city, address1_state, address1_lat, address1_lon, and same for address2. I've not used CROSS APPLY before so not sure how to adapt that to your query.
I had added a modification (question) to the other query that drew provided to add those columns but never heard anything back on whether that was quite correct.
I didn't respond to the modification, because you should have been able to figure out for yourself whether it would work. You learn a lot more by trying things for yourself than having someone tell you what to do.
Drew
J. Drew Allen
Business Intelligence Analyst
Philadelphia, PA
December 21, 2017 at 5:15 am
How about concatenating the address pieces, and then comparing the two resulting strings?
Remember to handle any pieces of the address that are null. And you might consider replacing all spaces, so you don't get mismatches simply because one string has a double space and the other has a single space.
As with your single address compare before, your results will only be as good as the data input. Is there something that standardizes the addresses? For example, "Street" <> "St" <> "St." - if someone enters 123 Main St and someone else enters 123 Main Street, they won't match.
December 21, 2017 at 10:15 am
Thanks everyone for your help with this!
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