I don't see any options for doing anything other than an exact match...
That wouldn't be possible with a merge join. The Merge Join algorithm relies on having two sorted input sets and proceeds as follows.
get first row R1 from input 1
get first row R2 from input 2
while not at the end of either input
begin
if R1 joins with R2
begin
return (R1, R2)
get next row R2 from input 2
end
else if R1 < R2
get next row R1 from input 1
else
get next row R2 from input 2
end
If the input sets were
input 1 input 2
------ ------
1 7
2 8
3 9
Then the join input2.value > input1.value would return 9 rows (every permutation). However that wouldn't be possible with the algorithm above which does a single pass through each set.
Related
As we all known, the window function "last_value" returns the last value within an ordered group of values.
In the following example, group by field "A" and sort by field "B" in positive order.
In the group of "A = 1", the last value is returned, which is, the C value 4 when B = 2.
However, in the group of "A = 2", the values of field "B" are the same.
At this time, instead of the last value, which is, the C value 4 in line 6, the first C value 1 in B = 2 is returned.
This puzzles me why the last value within an ordered group of values is not returned when I encounter the value I want to use for sorting.
Example
row_number
A
B
C
LAST_VALUE(C) IGNORE NULLS OVER (PARTITION BY A ORDER BY B ASC)
1
1
1
2
4
2
1
1
1
4
3
1
1
3
4
4
1
2
4
4
5
2
2
1
1
6
2
2
4
1
This puzzles me why the last value within an ordered group of values is not returned when I encounter the value I want to use for sorting.
For partition A equals 2 and column B, there is a tie:
The sort is NOT stable. To achieve stable sort a column or a combination of columns in ORDER BY clause must be unique.
To ilustrate it:
SELECT C
FROM tab
WHERE A = 2
ORDER BY B
LIMIT 1;
It could return either 1 or 4.
If you sort by B within A then any duplicate rows (same A and B values) could appear in any order and therefore last_value could give any of the possible available values.
If you want a specific row, based on some logic, then you would need to sort by all columns within the group to reflect that logic. So in your case you would need to sort by B and C
Good day Bill!
Right, the sorting is not stable and it will return different output each time.
To get stable results, we can run something like below
select
column1,
column2,
column3,
last_value(column3) over (partition by column1 order by
column2,column3) as column2_last
from values
(1,1,2), (1,1,1), (1,1,3),
(1,2,4), (2,2,1), (2,2,4)
order by column1;
I have a list of players and a list of their scores in a game. I'd like to compare the two columns, and return a value of 1 or 0. 1 = Highest number, 0 = Lowest number. If the two values are equal, I'd like to return two 1's.
How can I do this using Excel or SQL Server 2012?
If I understood you correctly, you want the player with the highest score in a team to be marked with 1 and the lowest with 0. In that case:
SELECT t.team_id,t.player_id,t.score,
case when t.score = s.max_score then 1
when t.score = s.min_score then 0
end as pro_ind
FROM YourTable t
INNER JOIN(select team_id,max(score) as max_score,min(score) as min_score
FROM YourTable
GROUP BY team_id) s
ON (t.team_id = s.team_id)
Of course I guessed the names of the columns, so you will have to adjust it.
My question is that I want to find the "Balie" with the least number of "Maatschappijen" booked on it. So far I got this query wich displays all "Balies" and all the "Maatschappijen" with them. The wanted result is one "balienummer" record with the least number of "maatschappijen" booked on it.
Query
SELECT [Balie].[balienummer], [IncheckenBijMaatschappij].[balienummer], [IncheckenBijMaatschappij].[maatschappijcode]
FROM [Balie]
JOIN [IncheckenBijMaatschappij]
ON [Balie].[balienummer] = [IncheckenBijMaatschappij].[balienummer]
Query result
balienummer balienummer maatschappijcode
1 1 BA
1 1 TR
2 2 AF
2 2 NZ
3 3 KL
4 4 KL
LRS: https://www.dropbox.com/s/f2l9a874d5witpt/LRS_CasusGelreAirport.pdf
SELECT [Balie].[balienummer], count([IncheckenBijMaatschappij].[maatschappijcode])
FROM [Balie]
JOIN [IncheckenBijMaatschappij]
ON [Balie].[balienummer] = [IncheckenBijMaatschappij].[balienummer]
GROUP BY [Balie].[balienummer]
ORDER BY count([IncheckenBijMaatschappij].[maatschappijcode])
First record should be your answer.
I have a query analogous to:
update x
set x.y = (
select sum(x2.y)
from mytable x2
where x2.y < x.y
)
from mytable x
the point being, I'm iterating over rows and updating a field based on a subquery over those fields which are changing.
What I'm seeing is the subquery is being executed for each row before any updates occur, so the changed values for each row are not being picked up.
How can I force the subquery to be re-evaluated for each row of the update?
Is there a suitable table hint or something?
As an aside, I was doing the below and it did work, however since modifying my query somewhat (for logic purposes, not to try and solve this issue) this trick no longer works :(
declare #temp int
update x
set #temp = (
select sum(x2.y)
from mytable x2
where x2.y < x.y
),
x.y = #temp
from mytable x
I'm not particularly concerned about performance, this is a background task run over a few rows
It looks like task is incorrect or other rules should apply.
Let's see on example. Let's say you have values 4, 1, 2, 3, 1, 2
Sql will update rows based on original values. I.e. during single update statement newly calculated values is NOT mixing with original values:
-- only original values used
4 -> 9 (1+2+3+1+2)
1 -> null
2 -> 2 (1+1)
3 -> 6 (1+2+1+2)
1 -> null
2 -> 2 (1+1)
Based on your request you wants that update of each rows will count newly calculated values. (Note, that SQL does not guarantees the sequence in which rows will be processed.)
Let's do this calculation by processing rows from top to bottom:
-- from top
4 -> 9 (1+2+3+1+2)
1 -> null
2 -> 1 (1)
3 -> 4 (1+1+2)
1 -> null
2 -> 1 (1)
Do the same in other sequence - from bottom to top:
-- from bottom
4 -> 3 (2+1)
1 -> null
2 -> 1 (1)
3 -> 5 (2+2+1)
1 -> null
2 -> 2 (1+1)
How you can see your expected result is inconsistent. To make it right you need to correct the calculation rule - for instance define strong sequence of the rows to process (date, id, ...)
Also, if you want to do some recursive processing look at the common_table_expression:
http://technet.microsoft.com/en-us/library/ms186243(v=sql.105).aspx
I have a struct mpc with the following structure:
num type col3 col4 ...
mpc.bus = 1 2 ... ...
2 2 ... ...
3 1 ... ...
4 3 ... ...
5 1 ... ...
10 2 ... ...
99 1 ... ...
to from col3 col4 ...
mpc.branch = 1 2 ... ...
1 3 ... ...
2 4 ... ...
10 5 ... ...
10 99 ... ...
What I need to do is:
1: Re-order the rows of mpc.bus, such that all rows of type 1 are first, followed by 2 and at last, 3. There is only one element of type 3, and no other types (4 / 5 etc.).
2: Make the numbering (column 1 of mpc.bus, consecutive, starting at 1.
3: Change the numbers in the to-from columns of mpc.branch, to correspond to the new numbering in mpc.bus.
4: After running simulations, reverse the steps above to turn up with the same order and numbering as above.
It is easy to update mpc.bus using find.
type_1 = find(mpc.bus(:,2) == 1);
type_2 = find(mpc.bus(:,2) == 2);
type_3 = find(mpc.bus(:,2) == 3);
mpc.bus(:,:) = mpc.bus([type1; type2; type3],:);
mpc.bus(:,1) = 1:nb % Where nb is the number of rows of mpc.bus
The numbers in the to/from columns in mpc.branch corresponds to the numbers in column 1 in mpc.bus.
It's OK to update the numbers on the to, from columns of mpc.branch as well.
However, I'm not able to find a non-messy way of retracing my steps. Can I update the numbering using some simple commands?
For the record: I have deliberately not included my code for re-numbering mpc.branch, since I'm sure someone has a smarter, simpler solution (that will make it easier to redo when the simulations are finished).
Edit: It might be easier to create normal arrays (to avoid woriking with structs):
bus = mpc.bus;
branch = mpc.branch;
Edit #2: The order of things:
Re-order and re-number.
Columns (3:end) of bus and branch are changed. (Not part of this question)
Restore original order and indices.
Thanks!
I'm proposing this solution. It generates a n x 2 matrix, where n corresponds to the number of rows in mpc.bus and a temporary copy of mpc.branch:
function [mpc_1, mpc_2, mpc_3] = minimal_example
mpc.bus = [ 1 2;...
2 2;...
3 1;...
4 3;...
5 1;...
10 2;...
99 1];
mpc.branch = [ 1 2;...
1 3;...
2 4;...
10 5;...
10 99];
mpc.bus = sortrows(mpc.bus,2);
mpc_1 = mpc;
mpc_tmp = mpc.branch;
for I=1:size(mpc.bus,1)
PAIRS(I,1) = I;
PAIRS(I,2) = mpc.bus(I,1);
mpc.branch(mpc_tmp(:,1:2)==mpc.bus(I,1)) = I;
mpc.bus(I,1) = I;
end
mpc_2 = mpc;
% (a) the following mpc_tmp is only needed if you want to truly reverse the operation
mpc_tmp = mpc.branch;
%
% do some stuff
%
for I=1:size(mpc.bus,1)
% (b) you can decide not to use the following line, then comment the line below (a)
mpc.branch(mpc_tmp(:,1:2)==mpc.bus(I,1)) = PAIRS(I,2);
mpc.bus(I,1) = PAIRS(I,2);
end
% uncomment the following line, if you commented (a) and (b) above:
% mpc.branch = mpc_tmp;
mpc.bus = sortrows(mpc.bus,1);
mpc_3 = mpc;
The minimal example above can be executed as is. The three outputs (mpc_1, mpc_2 & mpc_3) are just in place to demonstrate the workings of the code but are otherwise not necessary.
1.) mpc.bus is ordered using sortrows, simplifying the approach and not using find three times. It targets the second column of mpc.bus and sorts the remaining matrix accordingly.
2.) The original contents of mpc.branch are stored.
3.) A loop is used to replace the entries in the first column of mpc.bus with ascending numbers while at the same time replacing them correspondingly in mpc.branch. Here, the reference to mpc_tmp is necessary so ensure a correct replacement of the elements.
4.) Afterwards, mpc.branch can be reverted analogously to (3.) - here, one might argue, that if the original mpc.branch was stored earlier on, one could just copy the matrix. Also, the original values of mpc.bus are re-assigned.
5.) Now, sortrows is applied to mpc.bus again, this time with the first column as reference to restore the original format.