1. Lighthouse

Level 5 · Find duplicate customer records

The request

From Tyler Morgan, Junior Data Engineer

“I think some customers signed up twice with the same email, just typed differently. Can you find the duplicates? For each email, I need how many records there are and which customer_id was created first, so we can merge into that one.”

Your goal

From customers, treat emails as equal after trimming spaces and lowercasing. Return one row per email that appears on more than one customer record, with: email (trimmed and lowercased), records (number of customer rows), first_customer_id (lowest customer_id) and last_customer_id (highest customer_id).

  • Read from the customers tablenot done yet
  • Show exactly email records first_customer_id last_customer_idnot done yet
  • 18 rowsnot done yet

Your query

SQL

Loading the editor…

    ResultPreview of customers · press Run to see yours
    Preview of customers
    customer_idfirst_namelast_nameemailcity
    1OwenHudsonowen.hudson@icloud.comSan Francisco
    2DouglasMendozadmendoza69@gmail.comAtlanta
    3HenryBrookshbrooks@lakesidecafe.comLos Angeles
    4BrianGrahambgraham@bluecedarhomes.comSeattle

    Mote checks your result

    Press Run. Each row that matches the goal becomes a firefly in your lantern.

    Tables

    Harbor & Pine store data, ready to query.

    customers2,018 rows · 9 columns
    • customer_idinteger
    • first_namevarchar
    • last_namevarchar
    • emailvarchar
    • cityvarchar
    • statevarchar
    • signup_datedate
    • segmentvarchar
    • referral_sourcevarchar
    First 5 rows of customers
    customer_idfirst_namelast_nameemailcitystatesignup_datesegmentreferral_source
    1OwenHudsonowen.hudson@icloud.comSan FranciscoCA2022-06-02consumersearch
    2DouglasMendozadmendoza69@gmail.comAtlantaGA2022-06-02consumerNULL
    3HenryBrookshbrooks@lakesidecafe.comLos AngelesCA2022-06-04businessNULL
    4BrianGrahambgraham@bluecedarhomes.comSeattleWA2022-06-06businesspinterest
    5KyleJimenezkjimenez27@yahoo.comIndianapolisIN2022-06-07consumerNULL
    products120 rows · 8 columns
    • product_idinteger
    • namevarchar
    • categoryvarchar
    • subcategoryvarchar
    • unit_pricedecimal
    • unit_costdecimal
    • launched_ondate
    • is_activeboolean
    First 5 rows of products
    product_idnamecategorysubcategoryunit_priceunit_costlaunched_onis_active
    101Vale Velvet LoveseatFurnitureSofas1659.00NULL2023-09-11true
    102Alder Oak Coffee TableFurnitureTables919.00503.502024-05-16true
    103Orchard Rattan Dining ChairFurnitureChairs789.00429.872023-05-27true
    104Upland Oak BookcaseFurnitureStorage429.00214.702022-12-20true
    105Avery Leather Bed FrameFurnitureBeds449.00232.702021-08-31true
    orders12,000 rows · 7 columns
    • order_idinteger
    • customer_idinteger
    • order_datetimestamp
    • statusvarchar
    • channelvarchar
    • shipping_feedecimal
    • coupon_codevarchar
    First 5 rows of orders
    order_idcustomer_idorder_datestatuschannelshipping_feecoupon_code
    10011252023-01-01 20:51:19completedweb0.00NULL
    10021562023-01-02 07:58:46completedweb0.00NULL
    1003222023-01-02 10:25:27completedmarketplace4.99WELCOME10
    1004482023-01-02 11:21:51completedmarketplace4.99WELCOME10
    1005822023-01-02 11:43:36completedweb0.00WELCOME10
    order_items30,338 rows · 6 columns
    • order_item_idinteger
    • order_idinteger
    • product_idinteger
    • quantityinteger
    • unit_pricedecimal
    • discount_pctdecimal
    First 5 rows of order_items
    order_item_idorder_idproduct_idquantityunit_pricediscount_pct
    11001121931.990.00
    21001154311.990.00
    31002154411.990.00
    410021941149.000.00
    51002193219.990.00
    returns1,673 rows · 5 columns
    • return_idinteger
    • order_item_idinteger
    • returned_ondate
    • reasonvarchar
    • refund_amountdecimal
    First 5 rows of returns
    return_idorder_item_idreturned_onreasonrefund_amount
    1142023-01-10damaged83.69
    2422023-01-10wrong_item19.99
    31052023-01-14wrong_item10.99
    4242023-01-16changed_mind26.99
    51262023-01-16late20.99
    employees60 rows · 8 columns
    • employee_idinteger
    • first_namevarchar
    • last_namevarchar
    • departmentvarchar
    • titlevarchar
    • manager_idinteger
    • hire_datedate
    • salaryinteger
    First 5 rows of employees
    employee_idfirst_namelast_namedepartmenttitlemanager_idhire_datesalary
    1CatherineHaleExecutiveChief Executive OfficerNULL2016-03-01285000
    2RobertKimExecutiveChief Operating Officer12017-01-09235000
    3MeganCollinsSalesHead of Sales12018-04-16172000
    4ChloeGutierrezSalesAccount Executive32020-01-2172000
    5FrankBellSalesAccount Executive32023-05-2173000
    web_sessions40,000 rows · 7 columns
    • session_idinteger
    • customer_idinteger
    • started_attimestamp
    • devicevarchar
    • sourcevarchar
    • pages_viewedinteger
    • convertedboolean
    First 5 rows of web_sessions
    session_idcustomer_idstarted_atdevicesourcepages_viewedconverted
    1NULL2023-01-01 23:02:24mobilepaid1false
    2182023-01-01 15:07:49desktoppaid1false
    3NULL2023-01-01 16:55:46desktopdirect7false
    4NULL2023-01-01 12:43:28mobileorganic3false
    5NULL2023-01-01 13:06:21mobilesocial2false
    marketing_spend3,276 rows · 3 columns
    • spend_datedate
    • channelvarchar
    • amountdecimal
    First 5 rows of marketing_spend
    spend_datechannelamount
    2023-01-01paid492.99
    2023-01-01email37.38
    2023-01-01social200.53
    2023-01-02paid709.13
    2023-01-02email64.13

    The lesson

    Duplicate records are one of the most common data quality issues an analyst meets. Customers sign up twice, a sync job runs twice, a CRM import repeats a batch. Each duplicate inflates “customer count” and splits one person’s history across two IDs, which deflates lifetime value and repeat rates.

    Step 1: choose the key

    What makes two rows “the same customer”? Here it is the email, but only after normalizing it. 'Owen.Hudson@iCloud.com ' and 'owen.hudson@icloud.com' are different strings and the same person. So the key is LOWER(TRIM(email)). Picking the key is a business decision; write down what you chose.

    Step 2: group and keep groups larger than one

    SELECT
      LOWER(TRIM(email)) AS email,
      COUNT(*) AS records,
      MIN(customer_id) AS first_customer_id,
      MAX(customer_id) AS last_customer_id
    FROM customers
    GROUP BY LOWER(TRIM(email))
    HAVING COUNT(*) > 1;

    HAVING COUNT(*) > 1 is the classic duplicate finder. MIN(customer_id) gives the surviving record for a merge, assuming IDs increase over time, which Tyler can confirm with signup_date.

    Step 3: see the full rows

    To inspect every duplicate row, not just one line per email, join back to the table:

    WITH dupes AS (
      SELECT LOWER(TRIM(email)) AS email_key
      FROM customers
      GROUP BY 1
      HAVING COUNT(*) > 1
    )
    SELECT c.*
    FROM customers AS c
    JOIN dupes AS d ON d.email_key = LOWER(TRIM(c.email))
    ORDER BY d.email_key, c.customer_id;

    Counting unique people

    Until the records are merged, report customers as COUNT(DISTINCT LOWER(TRIM(email))) rather than COUNT(*), and say why. Harbor & Pine has 2,018 customer rows but 2,000 distinct people. A small gap here, but in a real CRM duplicates can reach 5 to 10 percent of rows.

    Keeping one row per group

    Finding duplicates is half the job. To pick one row per key with all its columns, the standard tool is ROW_NUMBER(), covered in the next chapter:

    QUALIFY ROW_NUMBER() OVER (PARTITION BY LOWER(TRIM(email)) ORDER BY customer_id) = 1

    QUALIFY works in DuckDB, BigQuery and Snowflake; in Postgres and MySQL you wrap the window function in a CTE and filter in the outer query.

    Common mistakes

    • Grouping by the raw email and concluding “no duplicates”.
    • Using SELECT DISTINCT *, which only removes rows that are identical in every column. Duplicate customers usually differ in ID and signup date, so DISTINCT removes nothing.
    • Deleting duplicates without moving their orders to the surviving ID. That is an engineering decision; an analyst’s job is to find and size the problem.