Question

I need to show distinct users per week. I have a date-visit column, and a user id, it is a big table with 1 billion rows.
I can change the date column from the CSVs to year,month, day columns. but how do I deduce the week from that in the query.
I can calculate the week from the CSV, but this is a big process step.

I also need to show how many distinct users visit day after day, looking for workaround as there is no date type.

any ideas?

Answer 1

If you have your date as a timestamp (i.e microseconds since the epoch) you can use the UTC_USEC_TO_DAY/UTC_USEC_TO_WEEK functions. Alternately, if you have an iso-formatted date string (e.g. “2012/03/13 19:00:06 -0700”) you can call PARSE_UTC_USEC to turn the string into a timestamp and then use that to get the week or day.

To see an example, try:

SELECT LEFT((format_utc_usec(day)),10) as day, cnt 
FROM (
    SELECT day, count(*) as cnt 
    FROM (
        SELECT UTC_USEC_TO_DAY(PARSE_UTC_USEC(created_at)) as day 
        FROM [publicdata:samples.github_timeline])
    GROUP BY day 
    ORDER BY cnt DESC)

To show week, just change UTC_USEC_TO_DAY(...) to UTC_USEC_TO_WEEK(..., 0) (the 0 at the end is to indicate the week starts on Sunday). See the documentation for the above functions at https://developers.google.com/bigquery/docs/query-reference for more information.