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Unit 4 • Lesson 3

Selecting Columns and Rows

Overview

loc goes by label, iloc goes by position, and the difference matters.

Intermediate 20–25 min
← Previous Next: Adding and Dropping Columns

What You Will Learn in This Lesson

By the end of this lesson, you will be able to:

  • Take one column as a Series and several as a smaller table.
  • Say what loc and iloc each read.
  • Explain why they stop agreeing after a filter or a sort.
  • Find a row by what it contains rather than by where it sits.

Why This Matters

Almost every "why is this the wrong row" question comes from reaching for a position when you meant a label, or the reverse. The two look interchangeable on a freshly loaded table and stop being interchangeable the moment you sort or filter it — which is to say, immediately.

Step 1: Columns by name

One name gives a Series; a list of names gives a smaller DataFrame. The double brackets are not a typo — the inner pair is the list.

1

One column, or several

Single brackets with one name give the column itself. Wrap the name in a list and you get a table back, with one column in it.

import pandas as pd

df = pd.DataFrame({"name": ["Ada", "Grace"], "score": [92, 88]})
print(type(df["score"]).__name__)
print(type(df[["score"]]).__name__)
print(df[["name", "score"]])

Mini Practice #1: Series or table

Try It Yourself

Print the type of each of these three. Two of them are the same and one is not.

Press Run to see output

Step 2: Rows by label or by position

loc reads the index label; iloc reads the position. They agree on a default index and disagree the moment it changes.

1

The two accessors

Both take a row and optionally a column. loc names them; iloc counts them.

import pandas as pd

df = pd.DataFrame({"score": [92, 88, 79]}, index=["a", "b", "c"])
print(df.loc["b", "score"])    # by label
print(df.iloc[1]["score"])     # by position
2

Where they part company

Sort the table and the labels travel with their rows while the positions do not. After this sort, position 0 and label 0 are different rows.

import pandas as pd

df = pd.DataFrame({"score": [70, 99, 80]})
best = df.sort_values("score", ascending=False)
print(best)
print("iloc[0]:", best.iloc[0]["score"])   # the top score
print("loc[0]:", best.loc[0]["score"])     # still the original row 0

Neither is wrong. They are answering different questions, and a filtered or sorted table is where the difference becomes visible.

3

One more asymmetry

A loc slice includes its end label; an iloc slice excludes its end position, like every other slice in Python. This is worth knowing once rather than debugging twice.

import pandas as pd

df = pd.DataFrame({"score": [10, 20, 30, 40]})
print(len(df.loc[0:2]))    # 3 -- inclusive
print(len(df.iloc[0:2]))   # 2 -- exclusive

Mini Practice #2: After a sort they disagree

Try It Yourself

Sort this table by score descending, then print iloc[0] and loc[0]. Say which one is "the best score" before you run it.

Press Run to see output

Step 3: Finding a row by what is in it

Usually you do not know either the label or the position — you know a value. Mask first, then read off what survives.

1

Mask, then take

The mask keeps the rows you meant. .iloc[0] on the survivors takes the first of them, which is safe because you are indexing into the filtered result, not the original table.

import pandas as pd

df = pd.DataFrame({"name": ["Zoe", "Ada"], "score": [10, 92]})
match = df[df["name"] == "Ada"]
print(match)
print(int(match["score"].iloc[0]))

Step 4: When to Use It

loc for labels

Rows by index label and columns by name, with slices that include the end.

df.loc[df["score"] >= 60, ["name", "score"]]

iloc for positions

The first n rows, the last column — anything defined by position rather than name.

When Not to Use It

Do not chain selections and then assign, as in df[mask]["score"] = 0. It may change a temporary copy and leave df alone; do it in one .loc call.

End-of-Lesson Exercises

Exercise 1: The score belonging to a name

Write score_for(df, name) that returns the score belonging to name, where the table has a name column and a score column.

Mask on the name column, then read the score off what survives. Assuming a position is the bug the second hidden case is looking for.

Press Run to try it, then Check my work to mark it.

Exercise 2: The top row, by position

Write best_name(df) that sorts the table by score descending and returns the name on the top row. Take the row by position on the sorted table, not by label.

sort_values("score", ascending=False) then .iloc[0]["name"]. Using .loc[0] would give you the original row 0, which the first hidden case is checking.

Press Run to try it, then Check my work to mark it.
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