I have data date and a in this table my table
How I can get b where the value is a subtract by a from previous date?
You may use the LAG analytic function here:
SELECT date, a, a - LAG(a) OVER (ORDER BY date) AS b
FROM yourTable
ORDER BY date;
Related
I have trouble integrating a simple window function into my query. I work with this avocado dataset from Kaggle. I started off with a simple query:
SELECT
date,
SUM(Total_Bags) as weekly_bags,
FROM
`course.avocado`
WHERE
EXTRACT(year FROM date) = 2015
GROUP BY
date
ORDER BY
date
And it works just fine. Next, I want to add the rolling sum to the query to display along the weekly sum. I tried the following:
SELECT
date,
SUM(Total_Bags) as weekly_bags,
SUM(Total_Bags) OVER(
PARTITION BY date
ORDER BY date
ROWS BETWEEN 4 PRECEDING AND CURRENT ROW
)
FROM
`course.avocado`
WHERE
EXTRACT(year FROM date) = 2015
GROUP BY
date
ORDER BY
date
but im getting the common error:
SELECT list expression references column Total_Bags which is neither grouped nor aggregated at [4:7]
and im confused. Total_Bags in the first query was aggregated yet when it's introduced again in the second query, it's not aggregated anymore. How do I fix this query? Thanks.
In your query, which returns 2 columns: date and aggregate SUM(Total_Bags), the window function SUM() is evaluated after the aggregation when there is no column Total_Bags and this is why you can't use it inside the window function.
However, you can do want you want, without group by, by using only window functions and DISTINCT:
SELECT DISTINCT date,
SUM(Total_Bags) OVER(PARTITION BY date) AS weekly_bags,
SUM(Total_Bags) OVER(
PARTITION BY date
ORDER BY date
ROWS BETWEEN 4 PRECEDING AND CURRENT ROW
)
FROM course.avocado
WHERE EXTRACT(year FROM date) = 2015
ORDER BY date;
or, use window function on the the aggregated result:
SELECT date,
SUM(Total_Bags) AS weekly_bags,
SUM(SUM(Total_Bags)) OVER(
ORDER BY date
ROWS BETWEEN 4 PRECEDING AND CURRENT ROW
)
FROM course.avocado
WHERE EXTRACT(year FROM date) = 2015
GROUP BY date
ORDER BY date;
I tried to approach it from a different angle and seems I have figured it out, the results seem just right. Here's the code:
WITH daily_bags AS
(SELECT
Date,
CAST(SUM(Total_Bags) as int64) as all_bags
FROM
`course.avocado`
WHERE
EXTRACT(year from Date) = 2015
GROUP BY
Date
ORDER BY
Date)
SELECT
Date,
all_bags,
SUM(all_bags) OVER(
ROWS BETWEEN 4 PRECEDING AND CURRENT ROW
) as rolling_sum
FROM
daily_bags
Thanks everyone for your help.
I have a table of accident date I want to calculate the maximum between the difference of date i and date i + 1 which are in the same column. when we declare an accident date, I want to find the record of days without accidents.
You can use lag(). Assuming a table structure like mytable(dt), where dt is of a date-like datatype, you would do:
select max(diff)
from (select dt - lag(dt) over(order by dt) diff from mytable) t
I have a sample data, i want to get the Difference in month over month data 'Lag' column for only row B
If there always is just one row per month and id, then just use lag(). You can wrap this in a case expression so it only applies to id 'B'.
select
id,
date,
data,
case when id = 'B'
then data - lag(data) over(partition by id order by date)
end lag_diff
from mytable
I wan this type of calculated date value in SQL.
Is any way to get this type of calculated data ?
I think that you want:
select
t.*,
dateadd(
day,
sum(t.duration) over(order by autoid),
first_value(t.date) over(order by autoid)
) date
from mytable t
Starting from the first value in the date column (which, as I understand, is the only non-null value in that column), this incrementally adds the number of days in the duration column.
You seem to want the cumulative duration. I would do this just by subtracting the earliest:
select t.*,
datediff(day, min(date) over (), date) as total_duration
from t;
I have a date column I need to do date difference with previous row date using hive query?
Below would be the query -
select date, datediff(date, prev_date) from (select date, lag(date) over (partition by <colName> order by <Name>) as prev_date from tableName) t1;
If you can provide the sample data with tableName Can help you with the exact query