Understand the idea
replace and map both use a lookup dictionary, but handle unlisted values differently.
A small example
For ["A", "B"] and {"A": 1}: replace gives [1, "B"]; map gives [1, missing].
Follow the code
Apply the idea to the supplied table. Read from top to bottom; the final line displays the result.
df["flavour"] = df["flavour"].replace({"fruity": "fruit"})
dfWhat each part does
replace({"old": "new"})- preserve other values
Your inputs
The editable setup on the right creates df. Run executes the setup and your work from top to bottom.
| candy | flavour | price | rating | shelf |
|---|---|---|---|---|
| Gummy Bear | fruity | 1.2 | 4.1 | A |
| Choco Pop | chocolate | 2.1 | 4.6 | B |
| Mint Bite | mint | 1.5 | 3.8 | A |
| Berry Loop | fruity | 2.8 | 4.4 | B |
| Cocoa Cube | chocolate | 3.4 | 4.9 | A |
| Lemon Drop | fruity | 1.8 | 4 | B |
Your task · Follow
- The supplied catalogue df is adopting the label fruit in place of fruity.
- Update flavour while retaining all other labels and columns; display df.
- Use: replace().
Hint
replace updates the named category without erasing the others.
Reveal solution
One way to do it. Keep any supplied setup in the editor and use this in the Your work section.
df["flavour"] = df["flavour"].replace({"fruity": "fruit"})
dfOptional stretch
Try map with the same dictionary. Use isna().sum() to see the difference.