Question

我将绘制存储在MySQL数据库中的netflow数据，我需要一种有效的方法来获取相关的数据点。自纪元以来，它们的记录以日期作为int存储了几秒钟。我想能够像：

Select SUM(bytes) from table where stime > x and stime < Y  
group by (10 second intervals)

有没有这样做？或者，在python本地处理它会更快吗？即使是500K行表？

修改我的错误，时间存储为无符号双精度而不是INT。我目前正在使用GROUP BY (FLOOR(stime / I))，其中我是所需的间隔。

Answer 1

您可以使用整数除法来执行此操作。不确定性能。

让我成为你想要的间隔秒。

SELECT SUM(bytes), ((stime - X) DIV I) as interval
FROM table
WHERE (stime > X) and (stime < Y)
GROUP BY interval

Example, let X = 1500 and I = 10
stime = 1503 -> (1503 - 1500) DIV 10 = 0 
stime = 1507 -> (1507 - 1500) DIV 10 = 0
stime = 1514 -> (1514 - 1500) DIV 10 = 1
stime = 1523 -> (1523 - 1500) DIV 10 = 2

Answer 2

您是否尝试过以下操作？只需将tyiem列除以10并将结果向下舍入。

SELECT    SUM(bytes) 
FROM      table 
WHERE     stime > x 
AND       stime < Y
GROUP BY  ROUND(stime/10, -1)

我不知道ROUND（）函数和函数调用分组在MySQL中工作，但上面是T-SQL。

Answer 3

小组中的

FLOOR有时会失败。它有时将不同的时间分组为一个值，例如当您将值除以3但是当您除以4时它不会相同，尽管这两个值之间的差异远大于3或4它应该分组为两个不同的群体。最好将它投射到地板上的无符号，其工作方式如下：

CAST(FLOOR(UNIX_TIMESTAMP(time_field)/I) AS UNSIGNED INT)

问题：

与GROUP BY FLOOR(UNIX_TIMESTAMP(time_field)/3)相比，有时候GROUP BY FLOOR(UNIX_TIMESTAMP(time_field)/4)提供的群组较少，这在数学上是不可能的。

Answer 4

SELECT sec_to_time(time_to_sec(datefield)- time_to_sec(datefield)%(10)) as intervals,SUM(bytes) 
FROM table
WHERE where stime > x and stime < Y
group by intervals

Answer 5

我使用了答案和同事的建议。最终结果如下：

Select FROM_UNIXTIME(stime), bytes 
from argusTable_2009_10_22 
where stime > (UNIX_TIMESTAMP()-600)
group by floor(stime /10)

我也尝试了舍入解决方案，但结果不一致。

可能有

Answer 6

我之前做过这个，所以我创建了一些函数（使用sql server，但我认为它几乎相同）：

首先我创建了一个标量函数，它根据间隔和日期部分（分钟，小时，日，蛾，年）返回日期的ID：

CREATE FUNCTION [dbo].[GetIDDate]
(
    @date datetime,
    @part nvarchar(10),
    @intervalle int
)
RETURNS int
AS
BEGIN
    -- Declare the return variable here
    DECLARE @res int
    DECLARE @date_base datetime
    SET @date_base = convert(datetime,'01/01/1970',103)

    set @res = case @part 
                WHEN 'minute' THEN datediff(minute,@date_base,@date)/@intervalle
                WHEN 'hour' THEN datediff(hour,@date_base,@date)/@intervalle
                WHEN 'day' THEN datediff(day,@date_base,@date)/@intervalle
                WHEN 'month' THEN datediff(month,@date_base,@date)/@intervalle
                WHEN 'year' THEN datediff(year,@date_base,@date)/@intervalle
                ELSE datediff(minute,@date_base,@date)/@intervalle END



    -- Return the result of the function
    RETURN @res

END

然后我创建了一个表函数，它返回了日期范围内的所有id：

CREATE FUNCTION [dbo].[GetTableDate] 
(   
    -- Add the parameters for the function here
    @start_date datetime, 
    @end_date datetime,
    @interval int,
    @unite varchar(10)
)
RETURNS @res TABLE (StartDate datetime,TxtStartDate nvarchar(50),EndDate datetime,TxtEndDate nvarchar(50),IdDate int)
AS
begin
    declare @current_date datetime 
    declare @end_date_courante datetime
    declare @txt_start_date nvarchar(50)
    declare @txt_end_date nvarchar(50)
    set @current_date = case @unite 
                WHEN 'minute' THEN dateadd(minute, datediff(minute,0,@start_date),0)
                WHEN 'hour' THEN dateadd(hour, datediff(hour,0,@start_date),0)
                WHEN 'day' THEN dateadd(day, datediff(day,0,@start_date),0)
                WHEN 'month' THEN dateadd(month, datediff(month,0,@start_date),0)
                WHEN 'year' THEN dateadd(year, datediff(year,0,dateadd(year,@interval,@start_date)),0)
                ELSE dateadd(minute, datediff(minute,0,@start_date),0) END

    while @current_date < @end_date
    begin
        set @end_date_courante = 
            case @unite 
                WHEN 'minute' THEN dateadd(minute, datediff(minute,0,dateadd(minute,@interval,@current_date)),0)
                WHEN 'hour' THEN dateadd(hour, datediff(hour,0,dateadd(hour,@interval,@current_date)),0)
                WHEN 'day' THEN dateadd(day, datediff(day,0,dateadd(day,@interval,@current_date)),0)
                WHEN 'month' THEN dateadd(month, datediff(month,0,dateadd(month,@interval,@current_date)),0)
                WHEN 'year' THEN dateadd(year, datediff(year,0,dateadd(year,@interval,@current_date)),0)
                ELSE dateadd(minute, datediff(minute,0,dateadd(minute,@interval,@current_date)),0) END
        SET @txt_start_date = case @unite 
                WHEN 'minute' THEN CONVERT(VARCHAR(20), @current_date, 100)
                WHEN 'hour' THEN CONVERT(VARCHAR(20), @current_date, 100)
                WHEN 'day' THEN REPLACE(CONVERT(VARCHAR(11), @current_date, 106), ' ', '-')
                WHEN 'month' THEN REPLACE(RIGHT(CONVERT(VARCHAR(11), @current_date, 106), 8), ' ', '-')
                WHEN 'year' THEN CONVERT(VARCHAR(20), datepart(year,@current_date))
                ELSE CONVERT(VARCHAR(20), @current_date, 100) END
        SET @txt_end_date = case @unite 
                WHEN 'minute' THEN CONVERT(VARCHAR(20), @end_date_courante, 100)
                WHEN 'hour' THEN CONVERT(VARCHAR(20), @end_date_courante, 100)
                WHEN 'day' THEN REPLACE(CONVERT(VARCHAR(11), @end_date_courante, 106), ' ', '-')
                WHEN 'month' THEN REPLACE(RIGHT(CONVERT(VARCHAR(11), @end_date_courante, 106), 8), ' ', '-')
                WHEN 'year' THEN CONVERT(VARCHAR(20), datepart(year,@end_date_courante))
                ELSE CONVERT(VARCHAR(20), @end_date_courante, 100) END
        INSERT INTO @res (
StartDate,
EndDate,
TxtStartDate,
TxtEndDate,
IdDate) values(
@current_date,
@end_date_courante,
@txt_start_date,
@txt_end_date,
dbo.GetIDDate(@current_date,@unite,@interval)
)
        set @current_date = @end_date_courante

    end
    return
end

因此，如果我想计算每个33分钟间隔内添加的所有用户：

SELECT count(id_user) , timeTable.StartDate
FROM user
INNER JOIn dbo.[GetTableDate]('1970-01-01',datedate(),33,'minute') as timeTable
ON dbo.getIDDate(user.creation_date,'minute',33) = timeTable.IDDate

GROUP BY dbo.getIDDate（user.creation_date，'minute'，33） ORDER BY timeTable.StartDate

：）

MySQL组按日期范围内的间隔

6 个答案: