{"id":"l-pda-4-5","courseId":"python-data-analysis","lang":"en","kind":"lesson","summary":"Rank customers by spend, measure how concentrated revenue is, and turn a continuous number into segments with cut (fixed boundaries) and qcut (equal-sized groups).","html":"

“Who are our best customers?” is a question every sales or marketing team asks, and the honest answer has two parts: a ranked list, and how much the top of that list matters. At Cartwheel the answer is striking, and it shapes how you should read every revenue number, including the Q4 jump.

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One row per customer

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Start with a customer-level table: one row per customer, with orders and spend, built with the groupby from Lesson 4.1.

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Python
import pandas as pd\n\nitems = pd.read_csv(\"order_items.csv\")\norders = pd.read_csv(\"orders.csv\", parse_dates=[\"order_date\"])\nitems[\"revenue\"] = items[\"quantity\"] * items[\"unit_price\"] * (1 - items[\"discount\"])\norders = orders.merge(items.groupby(\"order_id\", as_index=False)[\"revenue\"].sum(), on=\"order_id\", validate=\"one_to_one\")\norders = orders[orders[\"status\"] != \"cancelled\"]\n\ncust = orders.groupby(\"customer_id\").agg(orders=(\"order_id\", \"count\"), spend=(\"revenue\", \"sum\"))\nprint(len(cust), \"customers with a non-cancelled order\")\nprint(cust[\"spend\"].describe().round(0))
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503 customers. The median spent 551 dollars, the mean 888, the top one 5,370. When the mean sits far above the median, a few large values are pulling it up; Lesson 5.1 returns to what that means for reporting averages.

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Ranking

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Python
import pandas as pd\n\nitems = pd.read_csv(\"order_items.csv\")\norders = pd.read_csv(\"orders.csv\", parse_dates=[\"order_date\"])\nitems[\"revenue\"] = items[\"quantity\"] * items[\"unit_price\"] * (1 - items[\"discount\"])\norders = orders.merge(items.groupby(\"order_id\", as_index=False)[\"revenue\"].sum(), on=\"order_id\", validate=\"one_to_one\")\norders = orders[orders[\"status\"] != \"cancelled\"]\ncust = orders.groupby(\"customer_id\").agg(orders=(\"order_id\", \"count\"), spend=(\"revenue\", \"sum\"))\n\ncust[\"rank\"] = cust[\"spend\"].rank(ascending=False, method=\"min\")\ncust[\"percentile\"] = cust[\"spend\"].rank(pct=True)\nprint(cust.sort_values(\"rank\").head(5).round(2))\n\ntop20 = cust.nlargest(int(len(cust) * 0.2), \"spend\")\nprint(len(top20), \"customers =\", round(top20[\"spend\"].sum() / cust[\"spend\"].sum(), 3), \"of revenue\")
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rank gives every row its position, with 1 for the highest spend because of ascending=False. Ties are where the methods differ: "min" gives tied customers the same rank and skips the next numbers, like a sports table, "dense" never skips, "first" breaks ties by row order, and the default "average" gives them the mean rank. pct=True turns ranks into percentiles between 0 and 1, which compare across groups of different sizes.

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The last line is the concentration figure: the top 100 customers, a fifth of the base, generate 55.5% of revenue. That is less extreme than the textbook 80/20 rule, but still means losing a handful of big customers would hurt more than losing a hundred small ones.

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Ranking within groups

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Country managers want their own top customers, not a global list where the largest markets crowd everyone else out. rank works on a grouped column too, restarting at 1 inside each group:

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Python
import pandas as pd\n\nitems = pd.read_csv(\"order_items.csv\")\norders = pd.read_csv(\"orders.csv\", parse_dates=[\"order_date\"])\ncustomers = pd.read_csv(\"customers.csv\")\nitems[\"revenue\"] = items[\"quantity\"] * items[\"unit_price\"] * (1 - items[\"discount\"])\norders = orders.merge(items.groupby(\"order_id\", as_index=False)[\"revenue\"].sum(), on=\"order_id\", validate=\"one_to_one\")\norders = orders[orders[\"status\"] != \"cancelled\"]\ncust = orders.groupby(\"customer_id\", as_index=False).agg(spend=(\"revenue\", \"sum\"))\ncust = cust.merge(customers[[\"customer_id\", \"country\"]], on=\"customer_id\", validate=\"one_to_one\")\n\ncust[\"rank_in_country\"] = cust.groupby(\"country\")[\"spend\"].rank(ascending=False, method=\"first\")\nprint(cust[cust[\"rank_in_country\"] == 1].sort_values(\"spend\", ascending=False).round(0))
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Filtering on rank_in_country == 1 keeps each country’s top spender, eight rows in all; keeping ranks of 3 or less would give a top-three list per country. method="first" guarantees exactly one row per rank even when two customers tie, which is what you want for a fixed-length list. This “rank within group, then filter” pattern answers a whole family of questions: the best-selling product per category, the first order of each customer, the latest ticket per order.

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cut: bins with boundaries you choose

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Marketing wants spend bands they can name in a campaign. You choose the edges:

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Python
import numpy as np\nimport pandas as pd\n\nitems = pd.read_csv(\"order_items.csv\")\norders = pd.read_csv(\"orders.csv\", parse_dates=[\"order_date\"])\nitems[\"revenue\"] = items[\"quantity\"] * items[\"unit_price\"] * (1 - items[\"discount\"])\norders = orders.merge(items.groupby(\"order_id\", as_index=False)[\"revenue\"].sum(), on=\"order_id\", validate=\"one_to_one\")\norders = orders[orders[\"status\"] != \"cancelled\"]\ncust = orders.groupby(\"customer_id\").agg(orders=(\"order_id\", \"count\"), spend=(\"revenue\", \"sum\"))\n\ncust[\"band\"] = pd.cut(cust[\"spend\"], bins=[0, 250, 500, 1000, 2000, np.inf],\n                      labels=[\"<250\", \"250-500\", \"500-1k\", \"1k-2k\", \"2k+\"])\nprint(cust.groupby(\"band\", observed=True)[\"spend\"].agg([\"count\", \"sum\"]).round(0))
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Each edge pair defines a bin, closed on the right by default: (250, 500] includes 500 but not 250. The 57 customers in the 2k+ band, 11% of the base, brought in 173,407 dollars, more than any other band. cut returns a categorical column whose categories keep your order, so the table lists the bands from smallest to largest rather than alphabetically.

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qcut: equal-sized groups

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When you want groups of equal size rather than fixed boundaries, qcut cuts at quantiles:

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\n Top row: cut splits the spend axis at fixed values 250, 500, 1000 and 2000; the bins hold different numbers of customers. Bottom row: qcut splits at the quartiles 210, 551 and 1236 so each of the four bins holds about 126 customers.\n cut\n 143\n 92\n 118\n 93\n 57\n customers\n qcut\n 126\n 126\n 125\n 126\n 210\n 551\n 1236\n quartile edges\n
cut uses fixed boundaries; qcut uses equal counts
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Python
import pandas as pd\n\nitems = pd.read_csv(\"order_items.csv\")\norders = pd.read_csv(\"orders.csv\", parse_dates=[\"order_date\"])\nitems[\"revenue\"] = items[\"quantity\"] * items[\"unit_price\"] * (1 - items[\"discount\"])\norders = orders.merge(items.groupby(\"order_id\", as_index=False)[\"revenue\"].sum(), on=\"order_id\", validate=\"one_to_one\")\norders = orders[orders[\"status\"] != \"cancelled\"]\ncust = orders.groupby(\"customer_id\").agg(orders=(\"order_id\", \"count\"), spend=(\"revenue\", \"sum\"))\n\ncust[\"quartile\"] = pd.qcut(cust[\"spend\"], q=4, labels=[\"Q1 low\", \"Q2\", \"Q3\", \"Q4 high\"])\nshare = cust.groupby(\"quartile\", observed=True)[\"spend\"].sum() / cust[\"spend\"].sum()\nprint(share.round(3))\nprint(pd.qcut(cust[\"orders\"], q=4, duplicates=\"drop\").value_counts().sort_index())
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The top quartile of customers brings 63.2% of revenue, the bottom quartile 2.8%. The last line shows qcut’s limit: a third of customers placed exactly one order, so the two lowest quartile edges are both 1, and plain qcut raises ValueError: Bin edges must be unique. duplicates="drop" merges the repeated edges, leaving fewer, unequal groups. For counts with many ties, cut with hand-picked edges is usually clearer.

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In the exercise you segment customers yourself. That completes the analysis toolkit; Section 5 turns it on real questions, starting with what an average can and cannot tell you.

\n","toc":[{"id":"sec-1","text":"One row per customer","level":2},{"id":"sec-2","text":"Ranking","level":2},{"id":"sec-3","text":"Ranking within groups","level":3},{"id":"sec-4","text":"cut: bins with boundaries you choose","level":2},{"id":"sec-5","text":"qcut: equal-sized groups","level":2}],"takeaways":["rank(ascending=False, method="min") numbers rows by value; the method decides how ties are numbered, and pct=True gives percentile ranks.","pd.cut assigns values to bins with boundaries you choose; use np.inf as the last edge so nothing falls outside.","pd.qcut makes bins with (nearly) equal numbers of rows, such as quartiles; when many values tie, pass duplicates="drop".","Both return a categorical column, so group by it with observed=True and sort by the category order, not alphabetically.","A concentration figure such as “the top 20% of customers bring 55% of revenue” is often the most useful single number you can give a business."],"further":[{"title":"pandas.cut","url":"https://pandas.pydata.org/docs/reference/api/pandas.cut.html"},{"title":"pandas.qcut","url":"https://pandas.pydata.org/docs/reference/api/pandas.qcut.html"},{"title":"pandas.Series.rank","url":"https://pandas.pydata.org/docs/reference/api/pandas.Series.rank.html"}],"quiz":[{"q":"

pd.cut(spend, bins=[0, 250, 500, 1000]) leaves some customers as NaN. Which ones?

","options":[{"text":"Customers who spent exactly 250","why":"With the default right=True, 250 falls in the (0, 250] bin, so it is labelled."},{"text":"Customers who spent more than 1,000","why":"Correct. Values above the last edge belong to no bin. End the edges with np.inf to catch everything."},{"text":"Customers with very small spend, such as 18 dollars","why":"18 is inside (0, 250]. Only a spend of exactly 0 would fall outside, because the left edge is excluded."},{"text":"Customers who placed only one order","why":"cut looks only at the values you pass, here spend; the number of orders plays no part."}],"answer":1},{"q":"

You want four customer groups with the same number of customers each. Which tool fits?

","options":[{"text":"pd.cut(spend, 4)","why":"An integer bins with cut makes four ranges of equal width, which for skewed spend puts most customers in the first bin."},{"text":"spend.rank(pct=True)","why":"Percentile ranks are a step in the right direction, but they are numbers, not four labelled groups."},{"text":"pd.qcut(spend, 4)","why":"Correct. qcut cuts at the quartiles, so each group holds about a quarter of the customers."}],"answer":2},{"q":"

Two customers tie for second place by spend. With rank(ascending=False, method="min"), what ranks do they and the next customer get?

","options":[{"text":"2, 2 and 4","why":"Correct. min gives tied rows the lowest rank of the group, and the next rank skips to account for the tie, like sports rankings."},{"text":"2, 3 and 4","why":"That is method="first", which breaks ties by the order rows appear in."},{"text":"2, 2 and 3","why":"That is method="dense", which never leaves gaps after ties."},{"text":"2.5, 2.5 and 4","why":"That is the default method="average", which gives tied rows the mean of the ranks they span."}],"answer":0}],"exercise":{"kind":"python","files":["order_items.csv","orders.csv"],"starter":"import numpy as np\nimport pandas as pd\n\nitems = pd.read_csv(\"order_items.csv\")\norders = pd.read_csv(\"orders.csv\", parse_dates=[\"order_date\"])\nitems[\"revenue\"] = items[\"quantity\"] * items[\"unit_price\"] * (1 - items[\"discount\"])\norders = orders.merge(items.groupby(\"order_id\", as_index=False)[\"revenue\"].sum(), on=\"order_id\", validate=\"one_to_one\")\norders = orders[orders[\"status\"] != \"cancelled\"]\n\n# 1. cust: one row per customer_id with columns orders (count) and spend (sum of revenue)\ncust = None\n\n# 2. Column \"rank\": 1 for the highest spender, ties share the lowest rank (\"min\")\n\n# 3. Column \"band\": spend bands with edges 0, 250, 500, 1000, 2000, infinity\n# and labels \"<250\", \"250-500\", \"500-1k\", \"1k-2k\", \"2k+\"\n\n# 4. Share of all revenue from the top 10% of customers by spend (top 50), rounded to 3 decimals\ntop10_share = None\n","solution":"import numpy as np\nimport pandas as pd\n\nitems = pd.read_csv(\"order_items.csv\")\norders = pd.read_csv(\"orders.csv\", parse_dates=[\"order_date\"])\nitems[\"revenue\"] = items[\"quantity\"] * items[\"unit_price\"] * (1 - items[\"discount\"])\norders = orders.merge(items.groupby(\"order_id\", as_index=False)[\"revenue\"].sum(), on=\"order_id\", validate=\"one_to_one\")\norders = orders[orders[\"status\"] != \"cancelled\"]\n\ncust = orders.groupby(\"customer_id\").agg(orders=(\"order_id\", \"count\"), spend=(\"revenue\", \"sum\"))\n\ncust[\"rank\"] = cust[\"spend\"].rank(ascending=False, method=\"min\")\n\ncust[\"band\"] = pd.cut(cust[\"spend\"], bins=[0, 250, 500, 1000, 2000, np.inf],\n labels=[\"<250\", \"250-500\", \"500-1k\", \"1k-2k\", \"2k+\"])\n\ntop = cust.nlargest(50, \"spend\")\ntop10_share = round(top[\"spend\"].sum() / cust[\"spend\"].sum(), 3)\nprint(cust[\"band\"].value_counts().sort_index(), top10_share)\n","checks":[{"id":"cust","test":"isinstance(cust, pd.DataFrame) and len(cust) == 503 and {'orders', 'spend'} <= set(cust.columns)"},{"id":"rank","test":"'rank' in cust.columns and int(cust['rank'].min()) == 1 and int((cust.set_index('customer_id') if 'customer_id' in cust.columns else cust).loc[331, 'rank']) == 2"},{"id":"band","test":"'band' in cust.columns and int(cust['band'].isna().sum()) == 0 and cust['band'].astype(str).value_counts().to_dict() == {'<250': 143, '500-1k': 118, '1k-2k': 93, '250-500': 92, '2k+': 57}"},{"id":"top10","test":"round(float(top10_share), 3) == np.float64(0.355)"}],"title":"Segment Cartwheel's customers","prompt":"

Build cust, one row per customer_id, with orders (number of orders) and spend (total revenue). Add rank (1 for the highest spender, ties sharing the lowest rank), and band, made with pd.cut using the edges 0, 250, 500, 1000, 2000 and infinity and the labels "<250", "250-500", "500-1k", "1k-2k", "2k+". Finally set top10_share to the share of all revenue that comes from the 50 highest-spending customers (about the top 10%), rounded to 3 decimals.

\n","hints":["orders.groupby("customer_id").agg(orders=("order_id", "count"), spend=("revenue", "sum")).","cust["spend"].rank(ascending=False, method="min"); for the bands, pd.cut(cust["spend"], bins=[0, 250, 500, 1000, 2000, np.inf], labels=[...]).","cust.nlargest(50, "spend")["spend"].sum() divided by the total spend, then round(..., 3)."],"explanation":"

The customer table is a groupby over the order-level table, so cancelled orders are already excluded. method="min" matters only for ties, and ascending=False makes the biggest spender rank 1 (customer 36, with 5,370 dollars). Ending the bins at np.inf leaves no customer unlabelled, which the check confirms. The top 50 customers, a tenth of the base, account for 35.5% of revenue. Concentration figures like this belong in any report about customers: they tell you whether to worry about the many or the few.

\n","checkLabels":{"cust":"cust has 503 customers with orders and spend","rank":"rank starts at 1, and customer 331 ranks 2nd","band":"band has no gaps and the expected counts per band","top10":"top10_share is the top 50 customers’ share of revenue"}},"words":680,"readMinutes":3}