Five Plus One Data Moves

Meet the data

Here are two small tables. You will act on them one move at a time. Every move is shown three ways: an instruction in plain English, the line of Python an AI tool would write, and the table before and after.

`schools`
school	district	students	absent
Adams	North	400	60
Baker	North	250	50
Cole	North	150	45
Diaz	South	600	60
Evans	South	100	30
Ford	East	300	75
Grant	East	200	NaN
Hale	NaN	350	70

NaN is how Python shows a missing value ("not a number"). Here, Grant's absences and Hale's district are missing.

In schools, what does one row represent?

`funding`
district	year	per_pupil
North	2025	12,000
South	2024	9,500
South	2025	9,800
East	2025	10,200

In funding, what does one row represent?

Part 1 of 2 · Five moves on one tableMost of what an analysis does to a table is one of these five moves.

1 · Keep rows One table

“Keep only schools with at least 250 students.”

schools[schools["students"] >= 250]

How to read this line

`schools[ ... ]`	Square brackets with a condition inside: keep the rows where it is true.
`schools["students"]`	The `students` column. Quoted name in brackets = a column.
`>= 250`	The condition: at least 250.

Before

school	district	students	absent
Adams	North	400	60
Baker	North	250	50
Cole	North	150	45
Diaz	South	600	60
Evans	South	100	30
Ford	East	300	75
Grant	East	200	NaN
Hale	NaN	350	70

Predict: how many rows will come out?

After 8 → 5 rows

school	district	students	absent
Adams	North	400	60
Baker	North	250	50
Diaz	South	600	60
Ford	East	300	75
Hale	NaN	350	70

What can go wrong: the question quietly narrows. Any average you compute next describes schools with at least 250 students, not schools. Nothing in the output table says so.

2 · Pick columns One table

“Show just each school's name and number of students.”

schools[["school", "students"]]

How to read this line

[["school", "students"]] A list of column names inside the brackets (note the double brackets): keep those columns.

Before

school	district	students	absent
Adams	North	400	60
Baker	North	250	50
Cole	North	150	45
Diaz	South	600	60
Evans	South	100	30
Ford	East	300	75
Grant	East	200	NaN
Hale	NaN	350	70

Predict: how many rows will come out?

After 8 → 8 rows

school	students
Adams	400
Baker	250
Cole	150
Diaz	600
Evans	100
Ford	300
Grant	200
Hale	350

What changed: only what you see. The data behind the table is untouched.

3 · Sort One table

“List the schools from largest to smallest.”

schools.sort_values("students", ascending=False)

How to read this line

`schools.sort_values(...)`	A dot means do this to the thing on the left: sort `schools`.
`"students"`	Sort by this column.
`ascending=False`	A named option: biggest first. Leave it out and the default (smallest first) applies.

Before

school	district	students	absent
Adams	North	400	60
Baker	North	250	50
Cole	North	150	45
Diaz	South	600	60
Evans	South	100	30
Ford	East	300	75
Grant	East	200	NaN
Hale	NaN	350	70

Predict: which school will be in the first row?

After 8 → 8 rows

school	district	students	absent
Diaz	South	600	60
Adams	North	400	60
Hale	NaN	350	70
Ford	East	300	75
Baker	North	250	50
Grant	East	200	NaN
Cole	North	150	45
Evans	South	100	30

What changed: only the order. But the order of rows is what people read first, so it shapes the story.

4 · Add a column One table

“Calculate each school's absence rate.”

schools["absent_rate"] = schools["absent"] / schools["students"]

How to read this line

`schools["absent_rate"] = ...`	A single `=` stores a result. Here it stores a new column named `absent_rate`.
`schools["absent"] / schools["students"]`	Divide one column by another, row by row.

Before

school	district	students	absent
Adams	North	400	60
Baker	North	250	50
Cole	North	150	45
Diaz	South	600	60
Evans	South	100	30
Ford	East	300	75
Grant	East	200	NaN
Hale	NaN	350	70

Predict: what will Cole's absent_rate be?

After 8 → 8 rows, one new column

school	district	students	absent	absent_rate
Adams	North	400	60	0.15
Baker	North	250	50	0.20
Cole	North	150	45	0.30
Diaz	South	600	60	0.10
Evans	South	100	30	0.30
Ford	East	300	75	0.25
Grant	East	200	NaN	NaN
Hale	NaN	350	70	0.20

What can go wrong: this is where definitions live. Divide by something else (the district's students, say, instead of the school's) and every rate changes, with nothing in the output to tell you which definition you got.

5 · Collapse One table

“Total students in each district.”

schools.groupby("district")["students"].sum()

How to read this line

`.groupby("district")`	Split the rows into one pile per district. On its own this produces nothing you can see.
`["students"].sum()`	Apply a sum to the students column in each pile, then combine the answers into a new table.

Before (coloured by district)

school	district	students	absent
Adams	North	400	60
Baker	North	250	50
Cole	North	150	45
Diaz	South	600	60
Evans	South	100	30
Ford	East	300	75
Grant	East	200	NaN
Hale	NaN	350	70

Predict: how many rows will come out?

Split → apply → combine

East

Ford · 300

Grant · 200

sum = 500

North

Adams · 400

Baker · 250

Cole · 150

sum = 800

South

Diaz · 600

Evans · 100

sum = 700

NaN

Hale · 350

sum = dropped

After 8 → 3 rows

district	students
East	500
North	800
South	700

Compare: the same sum without the grouping

schools["students"].sum()

2,350

What changed: one row now represents a district, not a school. The three districts add up to 2,000 students, but the whole table adds up to 2,350. The missing 350 is Hale, and nothing in the output says so.

Part 2 of 2 · Combining two tablesSo far every move used one table. Often what you need is spread across two.

6 · Match Two tables

“Add each district's funding per pupil to the schools table.”

schools.merge(funding, on="district", how="left")

How to read this line

`schools.merge(funding, ...)`	Match rows of `schools` to rows of `funding`.
`on="district"`	Rows match when their district is the same.
`how="left"`	Keep every school, even one with no match in `funding`.

schools

school	district	students
Adams	North	400
Baker	North	250
Cole	North	150
Diaz	South	600
Evans	South	100
Ford	East	300
Grant	East	200
Hale	NaN	350

funding

district	year	per_pupil
North	2025	12,000
South	2024	9,500
South	2025	9,800
East	2025	10,200

Predict: how many rows will come out?

After 8 → 10 rows

school	district	students	absent	year	per_pupil
Adams	North	400	60	2025	12,000
Baker	North	250	50	2025	12,000
Cole	North	150	45	2025	12,000
Diaz	South	600	60	2024	9,500
Diaz	South	600	60	2025	9,800
Evans	South	100	30	2024	9,500
Evans	South	100	30	2025	9,800
Ford	East	300	75	2025	10,200
Grant	East	200	NaN	2025	10,200
Hale	NaN	350	70	NaN	NaN

What can go wrong: a match can multiply rows. Total students in this table: 3,050, up from 2,350. Any sum or average from here counts Diaz and Evans twice. The question to ask of any match: did the row count change?

Recap: five plus one data moves

	Move	Python you saw
Five moves on one table	Keep rows	`schools[schools["students"] >= 250]`
	Pick columns	`schools[["school", "students"]]`
	Sort	`schools.sort_values("students")`
	Add a column	`schools["absent_rate"] = ...`
	Collapse	`schools.groupby("district")["students"].sum()`
Combining two tables	Match	`schools.merge(funding, on="district")`

In code an AI tool writes, the table usually has a generic name like df (short for “data frame”, Python's word for a table) or data. It's just a name: df[df["students"] >= 250] is the same Keep-rows move on a table called df.

Along the way, the table changed without any warning: Hale vanished in the Collapse, and Diaz and Evans doubled in the Match. Neither showed up as an error.

For any analysis you did not run, build two habits. After each step, ask what does one row represent now? and how many rows went in, and how many came out?