The request

From Priya Shah, Head of Data
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
customersDataFramenot done yet - Store a number in
resultnot done yet
Your code
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Reference solution
| customer_id | first_name | last_name | city | |
|---|---|---|---|---|
| 1 | Owen | Hudson | owen.hudson@icloud.com | San Francisco |
| 2 | Douglas | Mendoza | dmendoza69@gmail.com | Atlanta |
| 3 | Henry | Brooks | hbrooks@lakesidecafe.com | Los Angeles |
| 4 | Brian | Graham | bgraham@bluecedarhomes.com | Seattle |
Mote checks your result
Press Run. Each row that matches the goal becomes a firefly in your lantern.
What Priya gets, and why
DataFrames
Loaded for you as pandas DataFrames, plus pd and np.
customers2,018 rows · 9 columns
customer_idintegerfirst_namevarcharlast_namevarcharemailvarcharcityvarcharstatevarcharsignup_datedatesegmentvarcharreferral_sourcevarchar
| customer_id | first_name | last_name | city | state | signup_date | segment | referral_source | |
|---|---|---|---|---|---|---|---|---|
| 1 | Owen | Hudson | owen.hudson@icloud.com | San Francisco | CA | 2022-06-02 | consumer | search |
| 2 | Douglas | Mendoza | dmendoza69@gmail.com | Atlanta | GA | 2022-06-02 | consumer | NaN |
| 3 | Henry | Brooks | hbrooks@lakesidecafe.com | Los Angeles | CA | 2022-06-04 | business | NaN |
| 4 | Brian | Graham | bgraham@bluecedarhomes.com | Seattle | WA | 2022-06-06 | business | |
| 5 | Kyle | Jimenez | kjimenez27@yahoo.com | Indianapolis | IN | 2022-06-07 | consumer | NaN |
products120 rows · 8 columns
product_idintegernamevarcharcategoryvarcharsubcategoryvarcharunit_pricedecimalunit_costdecimallaunched_ondateis_activeboolean
| product_id | name | category | subcategory | unit_price | unit_cost | launched_on | is_active |
|---|---|---|---|---|---|---|---|
| 101 | Vale Velvet Loveseat | Furniture | Sofas | 1659.00 | NaN | 2023-09-11 | true |
| 102 | Alder Oak Coffee Table | Furniture | Tables | 919.00 | 503.50 | 2024-05-16 | true |
| 103 | Orchard Rattan Dining Chair | Furniture | Chairs | 789.00 | 429.87 | 2023-05-27 | true |
| 104 | Upland Oak Bookcase | Furniture | Storage | 429.00 | 214.70 | 2022-12-20 | true |
| 105 | Avery Leather Bed Frame | Furniture | Beds | 449.00 | 232.70 | 2021-08-31 | true |
orders12,000 rows · 7 columns
order_idintegercustomer_idintegerorder_datetimestampstatusvarcharchannelvarcharshipping_feedecimalcoupon_codevarchar
| order_id | customer_id | order_date | status | channel | shipping_fee | coupon_code |
|---|---|---|---|---|---|---|
| 1001 | 125 | 2023-01-01 20:51:19 | completed | web | 0.00 | NaN |
| 1002 | 156 | 2023-01-02 07:58:46 | completed | web | 0.00 | NaN |
| 1003 | 22 | 2023-01-02 10:25:27 | completed | marketplace | 4.99 | WELCOME10 |
| 1004 | 48 | 2023-01-02 11:21:51 | completed | marketplace | 4.99 | WELCOME10 |
| 1005 | 82 | 2023-01-02 11:43:36 | completed | web | 0.00 | WELCOME10 |
order_items30,338 rows · 6 columns
order_item_idintegerorder_idintegerproduct_idintegerquantityintegerunit_pricedecimaldiscount_pctdecimal
| order_item_id | order_id | product_id | quantity | unit_price | discount_pct |
|---|---|---|---|---|---|
| 1 | 1001 | 121 | 9 | 31.99 | 0.00 |
| 2 | 1001 | 154 | 3 | 11.99 | 0.00 |
| 3 | 1002 | 154 | 4 | 11.99 | 0.00 |
| 4 | 1002 | 194 | 1 | 149.00 | 0.00 |
| 5 | 1002 | 193 | 2 | 19.99 | 0.00 |
returns1,673 rows · 5 columns
return_idintegerorder_item_idintegerreturned_ondatereasonvarcharrefund_amountdecimal
| return_id | order_item_id | returned_on | reason | refund_amount |
|---|---|---|---|---|
| 1 | 14 | 2023-01-10 | damaged | 83.69 |
| 2 | 42 | 2023-01-10 | wrong_item | 19.99 |
| 3 | 105 | 2023-01-14 | wrong_item | 10.99 |
| 4 | 24 | 2023-01-16 | changed_mind | 26.99 |
| 5 | 126 | 2023-01-16 | late | 20.99 |
employees60 rows · 8 columns
employee_idintegerfirst_namevarcharlast_namevarchardepartmentvarchartitlevarcharmanager_idintegerhire_datedatesalaryinteger
| employee_id | first_name | last_name | department | title | manager_id | hire_date | salary |
|---|---|---|---|---|---|---|---|
| 1 | Catherine | Hale | Executive | Chief Executive Officer | NaN | 2016-03-01 | 285000 |
| 2 | Robert | Kim | Executive | Chief Operating Officer | 1 | 2017-01-09 | 235000 |
| 3 | Megan | Collins | Sales | Head of Sales | 1 | 2018-04-16 | 172000 |
| 4 | Chloe | Gutierrez | Sales | Account Executive | 3 | 2020-01-21 | 72000 |
| 5 | Frank | Bell | Sales | Account Executive | 3 | 2023-05-21 | 73000 |
web_sessions40,000 rows · 7 columns
session_idintegercustomer_idintegerstarted_attimestampdevicevarcharsourcevarcharpages_viewedintegerconvertedboolean
| session_id | customer_id | started_at | device | source | pages_viewed | converted |
|---|---|---|---|---|---|---|
| 1 | NaN | 2023-01-01 23:02:24 | mobile | paid | 1 | false |
| 2 | 18 | 2023-01-01 15:07:49 | desktop | paid | 1 | false |
| 3 | NaN | 2023-01-01 16:55:46 | desktop | direct | 7 | false |
| 4 | NaN | 2023-01-01 12:43:28 | mobile | organic | 3 | false |
| 5 | NaN | 2023-01-01 13:06:21 | mobile | social | 2 | false |
marketing_spend3,276 rows · 3 columns
spend_datedatechannelvarcharamountdecimal
| spend_date | channel | amount |
|---|---|---|
| 2023-01-01 | paid | 492.99 |
| 2023-01-01 | 37.38 | |
| 2023-01-01 | social | 200.53 |
| 2023-01-02 | paid | 709.13 |
| 2023-01-02 | 64.13 |
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=Falseand 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.
Lantern lit!
Practised: duplicateddrop duplicatessubsetkeepnormalisation
- Solved
- No hints
- First run
