What is the proper way to calculate accumulative user for everyday?

user_5258884 0
Viewed 127

I have a MySQL table, named transaction, which has 5 columns, id(int), from(int), to(int), value(float), time(datetime).

And I need to calculate the accumulative user (the number of unique "from") for some specific receiver ("to") everyday.

For example:

+-----+------+-----+-------+----------------------------+
| id  | from | to  | value | time                       |
+-----+------+-----+-------+----------------------------+
| 1   |  1   | 223 |     1 | 2019-01-01 01:11:30.000000 |
| 2   |  1   | 224 |     2 | 2019-01-01 21:37:30.000000 |
| 3   |  2   |  25 |   0.1 | 2019-01-02 03:05:30.000000 |
| 4   |  2   | 223 |   0.2 | 2019-01-02 13:26:30.000000 |
| 5   |  3   |  26 |     3 | 2019-01-02 19:29:30.000000 |
| 6   |  3   | 227 |     4 | 2019-01-03 21:37:30.000000 |
| 7   |  1   | 224 |     5 | 2019-01-05 22:03:30.000000 |
| 8   |  4   | 224 |     1 | 2019-01-05 23:48:30.000000 |
| 9   |  5   | 223 |     2 | 2019-01-06 05:41:30.000000 |
| 10  |  6   |  28 |     2 | 2019-01-06 20:19:30.000000 |
+-----+------+-----+-------+----------------------------+

And the specific to is [223, 224, 227]

Then the expected result is:

2019-01-01: 1 # [1]
2019-01-02: 3 # [1, 2, 3]
2019-01-03: 3 # [1, 2, 3]
2019-01-04: 3 # [1, 2, 3]
2019-01-05: 4 # [1, 2, 3, 4]
2019-01-05: 5 # [1, 2, 3, 4, 5]

The direct way is using SQL

SELECT COUNT(DISTINCT(`From`))
FROM `transaction`
FORCE INDEX (to_time_from)
WHERE `time` < '2019-01-0X'
AND `to` IN (223, 224, 227)

But the problem is, transaction table is big (1 million per day, about 2 years), and to list is about 1000. The above SQL is very slow, even though I have created an index on [to, time, from] and force use it.

Besides, although daily transactions amount reaches about 1 million, the daily active user is only about 10,000. So I'm considering to store DAU list in No-SQL, like 

2019-01-01: [1]
2019-01-02: [2, 3]
2019-01-03: [3]
2019-01-04: []
2019-01-05: [1, 4]
2019-01-05: [5]

And when given a date d, I just retrieve all the DAU list no later than d and make a union to get the accumulative user. Something like: len(set([dau_list1]+[dau_list2]+[dau_list3]...))

But I have no idea which No-SQL to use.

  1. Redis will load everything into memory, but I only need these data when I query.
  2. MongoDB
    1. it seems I need to create a collection for every date because I need to create a unique index on from. Am I right?
    2. I know I can use an array field and $addToSet operation. But it is O(n), very slow.

So, what is the proper way to make it?

nosql mysql database
1 Answers

In MySQL, use something like (no redis, no MongoDB):

SELECT  DATE(`time`),
        COUNT(*),
        GROUP_CONCAT(`from`)
    FROM  tbl
    WHERE  `to` IN (...)
    GROUP BY  1;    -- shorthand for "DATE(time)"

INDEX(`to`, `from`, `time`)  -- if applying to entire table
INDEX(`to`, `time`, `from`)  -- if you have `AND time ...`

Plus some formatting. (Such could be done with a messy CONCAT, or left for the application code.)

Since this seems to be a 'scaling' question, too, perhaps you need a "Summary Table" that is updated daily with the previous day's entries, thereby making the queries much faster.

CREATE TABLE Daily (
    `day` DATE NOT NULL,
    `from` ... NOT NULL,
    `to` ... NOT NULL,
    `ct` SMALLINT UNSIGNED NOT NULL,
    PRIMARY KEY(`to`, `day`, `from`)
) ENGINE=InnoDB;

and the query becomes

SELECT  `day`,
        SUM(ct),
        GROUP_CONCAT(DISTINCT `from`)
    FROM Daily
    WHERE  `to` IN (...)`
    GROUP BY `day`;

(It could help you provided CREATE TABLE and INSERTs to build a test case from.)

user_17668310
Related