1. Lighthouse

Level 3 · Find duplicate customers

The request

From Priya Shah, Head of Data

“We think some people signed up twice. Sometimes it's the same email, sometimes the same email typed with capitals or a stray space. How many extra customer records are there?”

Your goal

Normalise email by stripping spaces and lowercasing it. Set result to the number of customer rows (an int) whose normalised email already appeared on an earlier row, that is, the rows a deduplication would remove.

  • Use the customers DataFramenot done yet
  • Store a number in resultnot done yet

Your code

Python

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.

    DataFrames

    Loaded for you as pandas DataFrames, plus pd and np.

    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.comAtlantaGA2022-06-02consumerNaN
    3HenryBrookshbrooks@lakesidecafe.comLos AngelesCA2022-06-04businessNaN
    4BrianGrahambgraham@bluecedarhomes.comSeattleWA2022-06-06businesspinterest
    5KyleJimenezkjimenez27@yahoo.comIndianapolisIN2022-06-07consumerNaN
    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 LoveseatFurnitureSofas1659.00NaN2023-09-11true
    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 20:51:19completedweb0.00NaN
    10021562023-01-02 07:58:46completedweb0.00NaN
    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 OfficerNaN2016-03-01285000
    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
    1NaN2023-01-01 23:02:24mobilepaid1false
    2182023-01-01 15:07:49desktoppaid1false
    3NaN2023-01-01 16:55:46desktopdirect7false
    4NaN2023-01-01 12:43:28mobileorganic3false
    5NaN2023-01-01 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

    Exact duplicates are the easy part

    duplicated() marks rows that repeat an earlier row:

    customers.duplicated().sum()                    # whole-row duplicates
    customers.duplicated(subset=["email"]).sum()    # same email, any other columns

    Whole-row duplicates are rare in real data, because a second signup gets a new ID and a new date. What you need is a business key: the column (or columns) that should identify a person uniquely. For customers, that’s the email.

    Normalise the key first

    Six of Harbor & Pine’s duplicates share the email exactly. Twelve more differ only by case or a leading space, ' owen.hudson@icloud.com' versus 'owen.hudson@icloud.com'. A raw comparison misses them, so build a normalised key:

    email_key = customers["email"].str.strip().str.lower()
    email_key.duplicated().sum()

    Compare that with customers["email"].duplicated().sum() to see how many duplicates the normalisation uncovers. That gap is the point of this level.

    keep: which copy counts as the original?

    duplicated takes keep=:

    • keep="first" (default): the first occurrence is the original; later ones are True.
    • keep="last": the last occurrence is the original.
    • keep=False: every row in a duplicate group is True. Use it to look at all copies side by side.
    dupes = customers[email_key.duplicated(keep=False)]
    dupes.assign(email_key=email_key).sort_values(["email_key", "customer_id"])

    Eyeballing the groups before removing anything is good practice. Here you’d see the same name and city, a later signup date and a higher ID on the second record.

    drop_duplicates

    To remove the repeats, keep one row per key. Sort first so the row you want to keep comes first:

    deduped = (
        customers.assign(email_key=email_key)
        .sort_values("customer_id")
        .drop_duplicates(subset="email_key", keep="first")
    )

    Keeping the lowest ID keeps the original account, which is the one with the order history. Always decide which copy survives on purpose. “Whichever came first in the file” is not a rule.

    Duplicates and joins

    Duplicate keys cause the fan-out you met in the merge chapter: an order matched to two copies of its customer is counted twice. If you merge orders with customers on customer_id, these duplicates are harmless, because each has its own ID. If you match on email (say, to join a newsletter list), they double count. Deduplicate before joining on a business key.

    Don’t delete, flag

    In production, it’s often better to add a flag (is_duplicate) or a mapping from duplicate ID to original ID than to drop rows. Other tables may reference the duplicate IDs, and finance may need to know they existed.

    Common mistakes

    • Checking duplicates on the raw email and concluding there are only six.
    • Using keep=False and then dropping, which removes the originals too.
    • Deduplicating on all columns, which finds nothing because IDs differ.

    Interview note

    “How would you find duplicate customers?” is a classic. A strong answer names the business key, normalises it, inspects the groups with keep=False and explains which record survives.