我认为您列出的输入与您期望的输出并不匹配。如果您希望获取单个用户的数据，下面的查询可能是一个不错的起点。它使用了公用表表达式（CTE）来使查询更易读和编写。您可以遵循相同的模式，向结果中添加更多数据。
有一点我不太确定：您是需要某一天的数据总和，还是每天发生的行为次数？例如，您是想了解在 2019-10-31 这一天，该用户在其所有在站期间总共发布了 20 篇帖子，还是想了解在 2019-10-31 这一天该用户恰好发布了 3 篇帖子？当前查询的结构返回的是后一种结果。
--[params]
-- string :username
-- date :start_date
WITH target_user AS (SELECT id FROM users WHERE username = :username),
days AS (
SELECT day::date
FROM GENERATE_SERIES(:start_date, NOW()::date, INTERVAL '1 day') AS day
),
topics_created AS(
SELECT
tu.id,
day,
COUNT(tu.id) AS topics
FROM topics t
JOIN target_user tu
ON tu.id = t.user_id
JOIN days
ON t.created_at::date = day
WHERE t.archetype = 'regular'
AND t.deleted_at IS NULL
GROUP BY day, tu.id
),
posts_created AS(
SELECT
tu.id,
day,
COUNT(tu.id) AS posts,
SUM(array_length(regexp_split_to_array(raw, '\s'), 1)) AS word_count
FROM posts p
JOIN target_user tu
ON tu.id = p.user_id
JOIN days
ON p.created_at::date = day
WHERE p.post_type = 1
AND p.deleted_at IS NULL
GROUP BY day, tu.id
),
likes_received AS (
SELECT tu.id,
day,
COUNT(tu.id) AS likes_received_count
FROM user_actions ua
JOIN target_user tu
ON tu.id = ua.user_id
JOIN days
ON ua.created_at::date = day
WHERE ua.action_type = 2
GROUP BY day, tu.id
)
SELECT
d.day,
COALESCE(topics, 0) AS topics,
COALESCE(posts, 0) AS posts,
COALESCE(word_count, 0) AS word_count,
COALESCE(likes_received_count, 0) AS likes_received
FROM days d
LEFT JOIN topics_created tc
ON tc.day = d.day
LEFT JOIN posts_created pc
ON pc.day = d.day
LEFT JOIN likes_received lc
ON lc.day = d.day
ORDER BY d.day