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How to get repeatable partition as new one after another partition?
How to get a list of user accounts using the command line in MySQL?How to get the sizes of the tables of a MySQL database?Group OHLC-Stockmarket Data into multiple timeframes - MysqlReason for Column is invalid in the select list because it is not contained in either an aggregate function or the GROUP BY clauseIdentifying repeated fields in SQL queryOracle Partition By to get data from multiple groupsHow do I select each data set from a Row_Number Over Partition by table based on the Row_Number Over Partition by column?MySQL how to select first row each group with countOracle - Get min & max date from group of rowsWindow functions: PARTITION BY one column after ORDER BY another
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"Partition by" in OVER clause groups all of the values as unique, just like "Distinct" or "Group by" do.
This is how it works in my query with row_number():
id st t row_number
-------------------
1 1 1 1
1 1 2 2
1 1 3 3
2 1 3 1
1 2 4 1
1 1 10 4
This is what I want:
id st t uniq_row_number
------------------
1 1 1 1
1 1 2 2
1 1 3 3
2 1 3 1
1 2 4 1
1 1 10 1
No matter if new string already was before, it's read as new partition after every change of partition.
If partition repeats, so uniq_row_number gets +1. If new partition comes with new string: boom, it gets uniq_row_number 1.
My SQL query:
SELECT id, st, t,
row_number() OVER (PARTITION BY id, st ORDER BY id, st) cat_num,
min(t) over (PARTITION BY id, st) min_t,
max(t) over (PARTITION BY id, st) max_t
FROM tabl ORDER BY t;
SQL code is here: http://sqlfiddle.com/#!18/d4290/2
mysql sql window-functions
asked Mar 24 at 16:27
Andrey YashinAndrey Yashin
32
add a comment |
"Partition by" in OVER clause groups all of the values as unique, just like "Distinct" or "Group by" do.
This is how it works in my query with row_number():
id st t row_number
-------------------
1 1 1 1
1 1 2 2
1 1 3 3
2 1 3 1
1 2 4 1
1 1 10 4
This is what I want:
id st t uniq_row_number
------------------
1 1 1 1
1 1 2 2
1 1 3 3
2 1 3 1
1 2 4 1
1 1 10 1
No matter if new string already was before, it's read as new partition after every change of partition.
If partition repeats, so uniq_row_number gets +1. If new partition comes with new string: boom, it gets uniq_row_number 1.
My SQL query:
SELECT id, st, t,
row_number() OVER (PARTITION BY id, st ORDER BY id, st) cat_num,
min(t) over (PARTITION BY id, st) min_t,
max(t) over (PARTITION BY id, st) max_t
FROM tabl ORDER BY t;
SQL code is here: http://sqlfiddle.com/#!18/d4290/2
mysql sql window-functions
asked Mar 24 at 16:27
Andrey YashinAndrey Yashin
32
"if new string already was before" - To say that one row was before the other, you need a column (or a set of columns) that determines an order. So what are these columns? Why is (1, 2, 4) before (1, 1, 10)?
– Paul Spiegel
Mar 24 at 17:05
@paul-spiegel, in this example I codedrow_number() OVER (PARTITION BY id, st) row_number, two columns. So in last row I got '4' near '1 1 10', because '1 1' was already in first, second and third rows. I want not '4' but '1' near ' 1 1 10', because between third and sixth lines there is another partition in.
– Andrey Yashin
Mar 24 at 18:49
add a comment |
"Partition by" in OVER clause groups all of the values as unique, just like "Distinct" or "Group by" do.
This is how it works in my query with row_number():
id st t row_number
-------------------
1 1 1 1
1 1 2 2
1 1 3 3
2 1 3 1
1 2 4 1
1 1 10 4
This is what I want:
id st t uniq_row_number
------------------
1 1 1 1
1 1 2 2
1 1 3 3
2 1 3 1
1 2 4 1
1 1 10 1
No matter if new string already was before, it's read as new partition after every change of partition.
If partition repeats, so uniq_row_number gets +1. If new partition comes with new string: boom, it gets uniq_row_number 1.
My SQL query:
SELECT id, st, t,
row_number() OVER (PARTITION BY id, st ORDER BY id, st) cat_num,
min(t) over (PARTITION BY id, st) min_t,
max(t) over (PARTITION BY id, st) max_t
FROM tabl ORDER BY t;
SQL code is here: http://sqlfiddle.com/#!18/d4290/2
mysql sql window-functions
asked Mar 24 at 16:27
Andrey YashinAndrey Yashin
32
"Partition by" in OVER clause groups all of the values as unique, just like "Distinct" or "Group by" do.
This is how it works in my query with row_number():
id st t row_number
-------------------
1 1 1 1
1 1 2 2
1 1 3 3
2 1 3 1
1 2 4 1
1 1 10 4
This is what I want:
id st t uniq_row_number
------------------
1 1 1 1
1 1 2 2
1 1 3 3
2 1 3 1
1 2 4 1
1 1 10 1
No matter if new string already was before, it's read as new partition after every change of partition.
If partition repeats, so uniq_row_number gets +1. If new partition comes with new string: boom, it gets uniq_row_number 1.
My SQL query:
SELECT id, st, t,
row_number() OVER (PARTITION BY id, st ORDER BY id, st) cat_num,
min(t) over (PARTITION BY id, st) min_t,
max(t) over (PARTITION BY id, st) max_t
FROM tabl ORDER BY t;
SQL code is here: http://sqlfiddle.com/#!18/d4290/2
mysql sql window-functions
mysql sql window-functions
asked Mar 24 at 16:27
Andrey YashinAndrey Yashin
32
asked Mar 24 at 16:27
Andrey YashinAndrey Yashin
32
asked Mar 24 at 16:27
Andrey YashinAndrey Yashin
32
asked Mar 24 at 16:27
Andrey YashinAndrey Yashin
32
asked Mar 24 at 16:27
Andrey YashinAndrey Yashin
32
32
"if new string already was before" - To say that one row was before the other, you need a column (or a set of columns) that determines an order. So what are these columns? Why is (1, 2, 4) before (1, 1, 10)?
– Paul Spiegel
Mar 24 at 17:05
@paul-spiegel, in this example I codedrow_number() OVER (PARTITION BY id, st) row_number, two columns. So in last row I got '4' near '1 1 10', because '1 1' was already in first, second and third rows. I want not '4' but '1' near ' 1 1 10', because between third and sixth lines there is another partition in.
– Andrey Yashin
Mar 24 at 18:49
add a comment |
"if new string already was before" - To say that one row was before the other, you need a column (or a set of columns) that determines an order. So what are these columns? Why is (1, 2, 4) before (1, 1, 10)?
– Paul Spiegel
Mar 24 at 17:05
@paul-spiegel, in this example I codedrow_number() OVER (PARTITION BY id, st) row_number, two columns. So in last row I got '4' near '1 1 10', because '1 1' was already in first, second and third rows. I want not '4' but '1' near ' 1 1 10', because between third and sixth lines there is another partition in.
– Andrey Yashin
Mar 24 at 18:49
"if new string already was before" - To say that one row was before the other, you need a column (or a set of columns) that determines an order. So what are these columns? Why is (1, 2, 4) before (1, 1, 10)?
– Paul Spiegel
Mar 24 at 17:05
"if new string already was before" - To say that one row was before the other, you need a column (or a set of columns) that determines an order. So what are these columns? Why is (1, 2, 4) before (1, 1, 10)?
– Paul Spiegel
Mar 24 at 17:05
@paul-spiegel, in this example I coded
row_number() OVER (PARTITION BY id, st) row_number, two columns. So in last row I got '4' near '1 1 10', because '1 1' was already in first, second and third rows. I want not '4' but '1' near ' 1 1 10', because between third and sixth lines there is another partition in.– Andrey Yashin
Mar 24 at 18:49
@paul-spiegel, in this example I coded
row_number() OVER (PARTITION BY id, st) row_number, two columns. So in last row I got '4' near '1 1 10', because '1 1' was already in first, second and third rows. I want not '4' but '1' near ' 1 1 10', because between third and sixth lines there is another partition in.– Andrey Yashin
Mar 24 at 18:49
add a comment |
1 Answer
1
active
oldest
votes
This is called a "gaps-and-islands" problem. You need to define a group for each "island" of similar values. Then you can use row_number().
The difference of row numbers is a convenient way to define the islands:
select t.*,
row_number() over (partition by id, seqnum_t - seqnum_it
order by t
) as uniq_row_number
from (select t.*,
row_number() over (order by t) as seqnum_t,
row_number() over (partition by id order by t) as seqnum_it,
from t
) t;
The best way to understand how this works is to look at the results of the subquery. You should be able to see how the difference of row numbers defines the groups that you care about.
answered Mar 24 at 16:39
Gordon LinoffGordon Linoff
817k37330437
thank you! This worked foe me!
– Andrey Yashin
Mar 24 at 18:42
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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active
oldest
votes
active
oldest
votes
This is called a "gaps-and-islands" problem. You need to define a group for each "island" of similar values. Then you can use row_number().
The difference of row numbers is a convenient way to define the islands:
select t.*,
row_number() over (partition by id, seqnum_t - seqnum_it
order by t
) as uniq_row_number
from (select t.*,
row_number() over (order by t) as seqnum_t,
row_number() over (partition by id order by t) as seqnum_it,
from t
) t;
The best way to understand how this works is to look at the results of the subquery. You should be able to see how the difference of row numbers defines the groups that you care about.
answered Mar 24 at 16:39
Gordon LinoffGordon Linoff
817k37330437
thank you! This worked foe me!
– Andrey Yashin
Mar 24 at 18:42
add a comment |
This is called a "gaps-and-islands" problem. You need to define a group for each "island" of similar values. Then you can use row_number().
The difference of row numbers is a convenient way to define the islands:
select t.*,
row_number() over (partition by id, seqnum_t - seqnum_it
order by t
) as uniq_row_number
from (select t.*,
row_number() over (order by t) as seqnum_t,
row_number() over (partition by id order by t) as seqnum_it,
from t
) t;
The best way to understand how this works is to look at the results of the subquery. You should be able to see how the difference of row numbers defines the groups that you care about.
answered Mar 24 at 16:39
Gordon LinoffGordon Linoff
817k37330437
thank you! This worked foe me!
– Andrey Yashin
Mar 24 at 18:42
add a comment |
This is called a "gaps-and-islands" problem. You need to define a group for each "island" of similar values. Then you can use row_number().
The difference of row numbers is a convenient way to define the islands:
select t.*,
row_number() over (partition by id, seqnum_t - seqnum_it
order by t
) as uniq_row_number
from (select t.*,
row_number() over (order by t) as seqnum_t,
row_number() over (partition by id order by t) as seqnum_it,
from t
) t;
The best way to understand how this works is to look at the results of the subquery. You should be able to see how the difference of row numbers defines the groups that you care about.
answered Mar 24 at 16:39
Gordon LinoffGordon Linoff
817k37330437
This is called a "gaps-and-islands" problem. You need to define a group for each "island" of similar values. Then you can use row_number().
The difference of row numbers is a convenient way to define the islands:
select t.*,
row_number() over (partition by id, seqnum_t - seqnum_it
order by t
) as uniq_row_number
from (select t.*,
row_number() over (order by t) as seqnum_t,
row_number() over (partition by id order by t) as seqnum_it,
from t
) t;
The best way to understand how this works is to look at the results of the subquery. You should be able to see how the difference of row numbers defines the groups that you care about.
answered Mar 24 at 16:39
Gordon LinoffGordon Linoff
817k37330437
answered Mar 24 at 16:39
Gordon LinoffGordon Linoff
817k37330437
answered Mar 24 at 16:39
Gordon LinoffGordon Linoff
817k37330437
answered Mar 24 at 16:39
Gordon LinoffGordon Linoff
817k37330437
817k37330437
thank you! This worked foe me!
– Andrey Yashin
Mar 24 at 18:42
add a comment |
thank you! This worked foe me!
– Andrey Yashin
Mar 24 at 18:42
thank you! This worked foe me!
– Andrey Yashin
Mar 24 at 18:42
thank you! This worked foe me!
– Andrey Yashin
Mar 24 at 18:42
add a comment |
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"if new string already was before" - To say that one row was before the other, you need a column (or a set of columns) that determines an order. So what are these columns? Why is (1, 2, 4) before (1, 1, 10)?
– Paul Spiegel
Mar 24 at 17:05
@paul-spiegel, in this example I coded
row_number() OVER (PARTITION BY id, st) row_number, two columns. So in last row I got '4' near '1 1 10', because '1 1' was already in first, second and third rows. I want not '4' but '1' near ' 1 1 10', because between third and sixth lines there is another partition in.– Andrey Yashin
Mar 24 at 18:49