The cut decides the segment

There are two families of knife and pandas has a function for each. pd.cut takes edges you choose: the mean, the median, a round R$ 300, a 180-day line. The shares that land in each bin are whatever the data makes them, and on a skewed column they are lopsided. pd.qcut takes counts you choose: thirds, fifths, tenths. It then puts the edges wherever it has to so the bins come out equal, which on a skewed column means the edges crowd into the dense part and one bin at the top spans everything from "a bit above typical" to the largest order in the file.

Two mechanics to hold onto. Both functions build left-open, right-closed intervals by default, so a customer sitting exactly on an edge falls into the lower bin; if your threshold is "R$ 300 or more counts as big", pass right=False or shift the edge. And recency is scored in reverse: fewer days is better, so the labels for R run labels=[5, 4, 3, 2, 1] while F and M run [1, 2, 3, 4, 5]. Forgetting the reversal is the most common RFM bug in the wild, and it produces a Champions segment made of people who have not bought in a year.

The five placements tonight, on the spend column: the mean (R$ 165.20, 29.3 percent above it), the median (R$ 107.78, 50 percent above it by construction), rank thirds (the top 33 percent by construction), rank fifths (the top 20 percent by construction), and a merchant threshold, a round number the owner would actually write on a whiteboard. On Olist a R$ 300 line takes roughly the top tenth of customers. Five knives, five shares, one column.

If spend were symmetric the mean and the median would sit on the same number and "above the mean" would be half the customers. On Olist the mean is R$ 165 and the median R$ 108. The gap is the tail: a minority of big baskets pulls the arithmetic average up while leaving the middle customer where she was. The top 10 percent of customers by spend hold 38.3 percent of revenue, and every one of those large numbers pushes the mean right.

The consequence for the knife: "above the mean" on a skewed column is not "the top half", it is a minority, 29.3 percent here, and the more skewed the column the smaller that minority gets. The median is immune to the tail by construction. Neither is wrong. The mean answers "who spends more than the average customer's share of the money"; the median answers "who is in the upper half". Write down which question you are asking before choosing the knife, and expect the two to disagree more on spend than on recency, because money has a longer tail than time does.

Run pd.qcut(rfm["F"], 5) on Olist and pandas raises ValueError: Bin edges must be unique: [1, 1, 1, 1, 1, 15]. Read the edges: the 20th, 40th, 60th and 80th percentiles of frequency are all 1, because 97 percent of the column is 1. There is no honest place to put four knives. The error is pandas telling you the truth about your data.

The fix every tutorial reaches for is to rank first: pd.qcut(rfm["F"].rank(method="first"), 5). method="first" breaks ties by order of appearance, so the 90,557 identical one-time buyers get 90,557 different ranks, the edges become unique, and the error disappears. What has actually happened is that a column with almost no information has been given five equal bins by row order. On Olist that scores 18,672 identical one-time buyers a 1 and 15,871 identical one-time buyers a 5. The 2,801 repeat buyers all land in the top fifth too, which is correct, but they share it with fifteen thousand people who are no different from the ones in the bottom fifth.

Every downstream Champion built on that score is decided by that row order. It is the whole reason the vendor maps in Part 3 crown one-time buyers.

The rank trick is not wrong in itself. It exists for spiky columns with real variety: a frequency column where most values are distinct but a few hundred customers share a value that happens to sit on a quintile edge. There, ranking breaks a tie among a few people and moves an edge by a few rows, which is a fair price for equal bins. The trick becomes arbitrary only when the tie is most of the column, as it is on any one-time-buyer base, which is most retail data.

The test before applying it: rfm["F"].value_counts(normalize=True).head(). If one value holds more than a fifth of the rows, fifths on that column are not fifths, whatever pandas returns, and the honest move is a cut with chosen edges (once, twice, three or more) or a different dimension altogether. The full quintile build, the personas it feeds, and the persona-migration crosstab between two snapshots live in the retention course, Retention a4; the mechanics are taught there and not repeated here. Tonight is about what the mechanics do to a base like this one.

Most RFM tutorials, including the one many analysts learn from, do not put the knife at the mean. They rank each column into fifths with pd.qcut, then collapse: a score of 4 or 5 becomes a 1, a score of 1 to 3 becomes a 0, and the three bits are read as the eight groups of session 1. It looks like the same taxonomy with a tidier scoring step in front of it.

It is a different knife. Collapsing at 3 is a rank split at the top two fifths of each axis, by construction, so 40 percent of customers are "high" on every column regardless of the column's shape. On frequency that is impossible to do honestly: 97 percent of the column is the value 1, so the fifths come from row order (Part 2), and the collapse then promotes about two fifths of identical one-time buyers to "frequent". On Olist the route calls 6.92 percent of customers High value, and 84.8 percent of them bought exactly once; its Key win-back group is 9.8 percent of customers and 86.3 percent of those bought once. The mean split on the same people: 1.28 percent High value with nobody who bought once, 0.79 percent Key win-back and every one a repeat buyer.

Neither is the wrong taxonomy. The eight names are identical. What differs is where the knife sits and whether the column can bear it, and that is a decision to write down, not inherit. Session 3 runs both on the same 4,000 people side by side.

The mean split's High value corner is small because "frequent" means above the mean of 1.033 orders, which is two or more, and only 3.0 percent of Olist customers ever got there. Zero percent of High value bought once, by construction. It is a tiny group, and it is a true one.

Klaviyo's rule scores recency by calendar (3 if the last order was within 180 days, 2 within 365, else 1) and frequency and spend by rank thirds. Rank thirds on a 97-percent-ones column is the Part 2 row-order problem with three bins instead of five: two thirds of the one-time buyers are handed an F of 2 or 3 by row order. Champions are the codes 333, 332 and 323, which lands on 13.9 percent of customers, and 91.0 percent of them bought once. Putler's table takes fifths on all three axes, averages F and M, and calls R 4-5 with F+M 4-5 Champions: 16.1 percent, 92.1 percent of them one-time buyers. On this base the two vendor maps agree with each other and disagree with the data.

Replication note. Neither Klaviyo nor Putler documents how its live product breaks ties on a column that is 97 percent the same value. The figures above use the pandas rank(method="first") convention, which is what every tutorial that "fixes" the duplicate-edge error does. They are a property of that convention on this base, not a claim about either product's output on your account. Ask your vendor the tie question directly; the honest answer is worth more than the segment tile.

Klaviyo (help centre article 17797937793179). Recency: 3 if the customer bought within the last 180 days, 2 if within 365 days, 1 otherwise. Frequency and monetary: each ranked into thirds, 3 for the top third. The 27 three-digit codes collapse into six groups:

On Olist that produces Recent 21.6 percent, Loyal 21.5 percent, At risk 17.3 percent, Needs attention 15.5 percent, Champions 13.9 percent. Needs attention, the natural win-back audience, is 96.7 percent one-time buyers.

Putler (putler.com/rfm-analysis). Each of R, F and M ranked into fifths, 5 best; F and M averaged into one score; eleven segments by ranges of R and F+M: Champions R 4-5 and F+M 4-5; Loyal R 2-5 and F+M 3-5; Potential loyalist R 3-5 and F+M 1-3; New R 4-5 and F+M 0-1; Promising R 3-4 and F+M 0-1; Needs attention R 2-3 and F+M 2-3; About to sleep R 2-3 and F+M 0-2; At risk R 0-2 and F+M 2-5; Can't lose R 0-1 and F+M 4-5; Hibernating R 1-2 and F+M 1-2; Lost R 0-2 and F+M 0-2. The published ranges overlap, so a customer can match more than one; the replication takes the first match reading top to bottom, and the build-along says so in the code.

On Olist that produces Loyal 31.6 percent, Potential loyalist 24.3 percent, At risk 19.2 percent, Champions 16.1 percent, and four of the eleven segments empty: New, Promising, Can't lose and Lost never fill because averaging F and M fifths never reaches the corner values when 97 percent of F is tied. At risk plus Can't lose, the win-back audience, is 97.6 percent one-time buyers.

Say the shop has 300 customers in its community group. Fifths put 60 in each bin, thirds put 100. A "Champions" segment of 5 percent is 15 people. Percentages below that are not segments; they are a short list of names the owner could recite. Any rank-based cut on 300 rows manufactures a precision the pile does not have, and the tie problem from Part 2 is worse, not better, because a community shop's frequency column is often a handful of values shared by dozens.

Then the household. One member buys for an entire compound, so her spend is many times anyone else's. The mean of the spend column is dragged toward her, and "above the mean" is suddenly a group of a few dozen people who happen to be on the near side of one buyer's total. The median does not move; a threshold the owner would write down ("regulars are anyone over 500 yuan a quarter") does not move either. The honest cut at this size is a median or a merchant threshold, plus owner knowledge: the owner already knows which twenty customers matter, and the model's job is to confirm and organise that, not to overrule it with fifths.

Your numbers will differ, and no output is printed for this shop anywhere on the page, because it is invented. What transfers is the check: divide your N by the number of bins before you choose the knife, and if the answer is a room of people rather than a population, cut with edges you can defend out loud.