I have a table (table1) with 3 column (ID,V,R), I want to divide it to some table based on ID. for example suppose it is a table1:
ID V R
1 1 T
1 2 F
2 1 T
2 3 T
3 4 F
then because I have 3 type of ID (which it could be more or less) I want 3 different table like this:
table1_1:
ID V R
1 1 T
1 2 F
table 1_2:
ID V R
2 1 T
2 3 T
table 1_3:
ID V R
3 4 F
is there any solution to do it in Laravel?
I apologize in advance if my question is so simple.
Related
Suppose I have a table as follows:
id=`A`B`A`B`B`B`A
item= 10 1 1 3 5 10 6
t=table(id,item)
id item
-- ----
A 10
B 1
A 1
B 3
B 5
B 10
A 6
For example, I want to sort the table with two conditions: first, by the most commonly occurring item in column item, then by the highest number in column item.
How can I sort like this:
id item
--- ----
A 10
B 10
A 1
B 1
A 6
B 5
B 3
Is there any way to go about this? Thanks!
Try this:
t1=table(id,item);
update t1 set count=count(item) context by item;
select * from t1 order by count desc, item desc;
I want to filter a pivot table in the following set up:
My Table:
Key Value1 Value2
1 23 a
2 33 b
3 1 c
4 5 d
My pivot table (simplified):
Key SUM of Value1 COUNTA of Value2
1 23 1
2 33 1
3 1 1
4 5 1
Grand Total 62 4
I now want to filter the pivot table by the values in this list:
Keys
1
2
4
So the resulting pivot table should look like this:
Key SUM of Value1 COUNTA of Value2
1 23 1
2 33 1
4 5 1
Grand Total 61 3
I thought this should be possible by using a custom formula in the pivot filter but it seems I have no way of using the current cell in the pivot e.g. to make a lookup.
I created a simple example of this setup here: https://docs.google.com/spreadsheets/d/1GlQDYtW8v8ri5L68RhryTZxwTikV_NXZQlccSI6_7pU/edit?usp=sharing
paste this formula in Filters!B1:
=ARRAYFORMULA(IFERROR(VLOOKUP(A1:A, Table!A1:C, {2,3}, 0), ))
and create a resulting pivot table from there:
demo spreadsheet
I have the following mock up table
#n a b group
1 1 1 1
2 1 2 1
3 2 2 1
4 2 3 1
5 3 4 2
6 3 5 2
7 4 5 2
I am using SAS for this problem. In column group, the rows that are interconnected through a and b are grouped. I will try to explain why these rows are in the same group
row 1 to 2 are in group 2 since they both have a = 1
row 3 is in group 2 since b = 2 in row 2 and 3 and row 2 is in group 1
row 3 and 4 are in group 1 since a = 2 in both rows and row 3 is in group 1
The overall logic is that if a row x contains the same value of a or b as row y, row x also belongs to the same group as y is a part of.
Following the same logic, row 5,6 and 7 are in group 2.
Is there any way to make an algorithm to find these groups?
Case I:
Grouping defined as to be item linkage within contiguous rows.
Use the LAG function to examine both variables prior values. Increase the group value if both have changed. For example
group + ( a ne lag(a) and b ne lag(b) );
Case II:
Grouping determined from pair item slot value linkages over all data.
From grouping pairs by either key
General statement of problem:
-----------------------------
Given: P = p{i} = (p{i,1},p{i,2}), a set of pairs (key1, key2).
Find: The distinct groups, G = g{x}, of P,
such that each pair p in a group g has this property:
key1 matches key1 of any other pair in g.
-or-
key2 matches key2 of any other pair in g.
Demonstrates
… an iterative way using hashes.
Two hashes maintain the groupId assigned to each key value.
Two additional hashes are used to maintain group mapping paths.
When the data can be passed without causing a mapping, then the groups have
been fully determined.
A final pass is done, at which point the groupIds are assigned to each
pair and the data is output to a table.
Could you please explain to me how to do a NATURAL JOIN on these two relations (one having 5 and the other one 3 rows?
1st relation
A C
3 3
6 4
2 3
3 5
7 1
2nd relation
B C D
5 1 6
1 5 8
4 3 9
In your question you have two separate relations, which have one attribute (i.e. column) in common: C.
A natural join will combine all tuples in both relations with that attribute in common. You will end up with the results:
A B C D
7 5 1 6
3 4 3 9
2 4 3 9
3 1 5 8
This can be performed in SQL by using the code #Matthew posted.
Something like:
SELECT * FROM 1stRelation NATURAL JOIN 2ndReleation
It will do the same thing and an inner join using the explicit column names. I.e.:
SELECT * from 1stRelation as x INNER JOIN 2ndRelation as z ON x.C=z.C
Personally - I prefer not to use them except in the possible case where I am not aware of the table structure in advance but know they should be able to be joined.
Basicly you do a CROSS JOIN, i. e. you combine every row from the 1st relation with every row of the 2nd relation. Then you have two C columns. Now you eliminate every row where the two C are not equal and merge them as only one column C.
I have a table with these columns
create table demo (
ID integer not null,
INDEX integer not null,
DATA varchar2(10)
)
Example data:
ID INDEX DATA
1 1 A
1 3 B
2 1 C
2 1 D
2 5 E
I want:
ID INDEX DATA
1 1 A
1 2 B -- 3 -> 2
2 1 C or D -- 1 -> 1 or 2
2 2 D or C -- 1 -> 2 or 1
2 3 E -- 5 -> 3 (closing the gap)
Is there a way to renumber the INDEX per ID using a single SQL update? Note that the order of items should be preserved (i.e. the item "E" should still be after the items "C" and "D" but the order of "C" and "D" doesn't really matter.
The goal is to be able to create a primary key over ID and INDEX.
If that matters, I'm on Oracle 10g.
MERGE
INTO demo d
USING (
SELECT rowid AS rid, ROW_NUMBER() OVER (PARTITION BY id ORDER BY index) AS rn
FROM demo
) d2
ON (d.rowid = d2.rid)
WHEN MATCHED THEN
UPDATE
SET index = rn