BigQuery: A Clever Solution for a Difficult Query

Introduction

BigQuery is a powerful data analytics service offered by Google Cloud Platform. It allows users to process and analyze large datasets using SQL-like queries. However, sometimes, queries can be challenging due to the complexity of the data or the requirements of the analysis. In this article, we’ll explore a difficult query related to live chat services, where conversations consist of multiple messages with timestamps, and channels determine the difficulty of the inquiry.

Background

A conversation in our context consists of multiple messages, each with a timestamp registered. Conversations are held in one or more channels, depending on the difficulty of the inquiry. The initial difficulty is determined at row 5 based on the content of the messages. Rows 2 till 4 should have the same difficulty as these messages form a group. After that, it gets easier, as all messages between two difficulties belong to the one earlier in time.

The Problem

The task is to write a query that can assign the correct difficulty to rows 5 till 29 using the same logic and a subquery. Additionally, we need to find a way to assign the correct difficulty to messages before the first channel ID. This seems like a daunting task at first glance.

Initial Solution

Initially, I attempted to solve this problem using the following query:

WITH channelMessage AS (
  SELECT conversation, channel, timestamp
  FROM MessagesTable
  WHERE channel IS NOT NULL
  AND channel != ''
),
messages AS (
  SELECT a.message, b.channel, a.conversation, a.timestamp,
         CASE WHEN b.timestamp IS NULL THEN 0 ELSE TIMESTAMP_DIFF(b.timestamp, a.timestamp, MILLISECOND) END AS diff,
         CASE WHEN b.timestamp IS NULL THEN 0 ELSE MAX(TIMESTAMP_DIFF(b.timestamp, a.timestamp, MILLISECOND)) OVER (PARTITION BY a.message ORDER BY a.timestamp) END AS diff2
  FROM MessagesTable AS a
  LEFT JOIN (SELECT * FROM channelMessage) AS b
  ON a.conversation = b.conversation AND a.timestamp >= b.timestamp
)
SELECT timestamp, conversation, message, channel, diff, diff2,
FROM messages 
WHERE diff = diff2
ORDER BY timestamp;

This query first identifies the channels for each message that doesn’t already have one. Then, it calculates the difference in timestamps between consecutive messages and determines which group they belong to.

Alternative Solution

However, upon further analysis, I realized that this initial solution might not be optimal. The task is to assign the correct difficulty to rows 5 till 29 using the same logic and a subquery. Additionally, we need to find a way to assign the correct difficulty to messages before the first channel ID.

A more straightforward approach is to use the following query:

SELECT a.timestamp, a.conversation,
  COALESCE(
    -- Message already has a channel
    a.channel,
    -- Channel from most recent earlier message
    (SELECT MAX(c.channel) FROM Messages c
     WHERE c.conversation = a.conversation
     AND c.timestamp =
       (SELECT MAX(c2.timestamp) FROM Messages c2
        WHERE c2.conversation = a.conversation
        AND c2.channel IS NOT NULL
        AND c2.timestamp < a.timestamp)),
    -- Channel of earliest message
    (SELECT MAX(c.channel) FROM Messages c
     WHERE c.conversation = a.conversation
     AND c.timestamp =
       (SELECT MIN(c2.timestamp) FROM Messages c2
        WHERE c2.conversation = a.conversation
        AND c2.channel IS NOT NULL))) AS channel
FROM Messages a;

This query identifies the channels for each message that doesn’t already have one, similar to the initial solution. However, it uses a more efficient approach to determine which group a message belongs to.

Recursive CTE Solution

Another possible solution involves assigning a number to each row of the original table and using a recursive CTE (Common Table Expression) to find the previous non-NULL value. This approach can be used to assign the correct difficulty to rows before the first channel ID. However, this solution is more complex and requires careful consideration of how to handle rows 2-4.

Conclusion

In conclusion, the task of assigning the correct difficulty to messages in a live chat service is challenging but solvable. The initial query uses a subquery to identify channels for each message that doesn’t already have one. However, an alternative solution provides a more straightforward approach using COALESCE to determine which group a message belongs to. Additionally, a recursive CTE solution can be used to assign the correct difficulty to rows before the first channel ID. By understanding the logic behind these solutions and how they apply to BigQuery queries, developers can effectively tackle similar challenges in their own projects.

Recommendations

Use COALESCE to determine which group a message belongs to.
Assign a number to each row of the original table and use a recursive CTE to find the previous non-NULL value.
Consider using a subquery or JOINs to identify channels for each message that doesn’t already have one.

Resources

Note: The above code snippets are written in SQL and can be used as a starting point for your own queries.

Last modified on 2023-05-17