map() for element-wise transformation in Pandas - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of map() with dictionary mapping

What is the output of the following code snippet?

Pandas

import pandas as pd

s = pd.Series(['cat', 'dog', 'bird', 'cat'])
mapping = {'cat': 'feline', 'dog': 'canine'}
result = s.map(mapping)
print(result)

A

0    feline
1    canine
2    None
3    feline
dtype: object

B

0    cat
1    dog
2    bird
3    cat
dtype: object

C

0    feline
1    canine
2    bird
3    feline
dtype: object

D

0    feline
1    canine
2      NaN
3    feline
dtype: object

💡 Hint

Think about what happens when a value is not found in the mapping dictionary.

The map() function replaces values found in the dictionary. Values not found become NaN (missing). Here, 'bird' is not in the dictionary, so it becomes NaN.

❓ data_output

intermediate

Result of map() with a function

Given the code below, what is the resulting Series after applying map()?

Pandas

import pandas as pd

s = pd.Series([1, 2, 3, 4])
result = s.map(lambda x: x**2)
print(result)

A

0     1
1     4
2     9
3    16
dtype: int64

B

0     2
1     3
2     4
3     5
dtype: int64

C

0     1
1     8
2    27
3    64
dtype: int64

D

0     1
1     2
2     3
3     4
dtype: int64

💡 Hint

The lambda function squares each element.

map() applies the lambda function to each element, squaring the numbers: 1²=1, 2²=4, 3²=9, 4²=16.

🔧 Debug

advanced

Identify the error in map() usage

What error does the following code raise?

Pandas

import pandas as pd

s = pd.Series(['apple', 'banana', 'cherry'])
result = s.map(['fruit', 'yellow', 'red'])
print(result)

A TypeError: mapping must be a callable, dict, or Series

B TypeError: unhashable type: 'list'

C ValueError: Mapping correspondence is not valid

D No error, outputs a Series with mapped values

💡 Hint

Check what types map() accepts as its argument.

map() expects a function, dictionary, or Series. Passing a list causes a TypeError.

🚀 Application

advanced

Using map() to replace values with a default

You want to replace values in a Series using a dictionary but keep original values if not found in the dictionary. Which code achieves this?

Pandas

import pandas as pd

s = pd.Series(['red', 'blue', 'green', 'yellow'])
mapping = {'red': '#FF0000', 'blue': '#0000FF'}

A s.map(mapping).fillna('unknown')

B s.map(lambda x: mapping.get(x))

C s.map(mapping).fillna(s)

D s.map(mapping).replace({None: s})

💡 Hint

Think about how to keep original values when mapping returns NaN.

map() replaces known values and returns NaN for unknown. fillna(s) replaces NaN with original values.

🧠 Conceptual

expert

Understanding map() behavior with mixed types

Consider a Series with mixed types. What will be the output of map() when mapping integers to strings but the Series contains strings and integers?

Pandas

import pandas as pd

s = pd.Series([1, '2', 3, '4'])
mapping = {1: 'one', 2: 'two', 3: 'three', 4: 'four'}
result = s.map(mapping)
print(result)

A

0      one
1      two
2    three
3      four
dtype: object

B

0      one
1      NaN
2    three
3      NaN
dtype: object

C

0      NaN
1      NaN
2      NaN
3      NaN
dtype: object

D

0      one
1      2
2    three
3      4
dtype: object

💡 Hint

map() matches keys exactly, including type.

The keys in mapping are integers. The Series has strings '2' and '4' which do not match integer keys, so those map to NaN.

Practice

1. What does the map() function do when used on a pandas Series?

easy

A. It combines two Series into one.

B. It applies a transformation to each element individually.

C. It filters out elements based on a condition.

D. It sorts the Series in ascending order.

Solution

1. Step 1: Understand the purpose of map()

   map() applies a function or dictionary mapping to each element in a pandas Series one by one.

2. Step 2: Compare with other operations

   Combining, filtering, or sorting are different operations; map() specifically transforms elements individually.

3. Final Answer:

   It applies a transformation to each element individually. -> Option B

4. Quick Check:

   map() = element-wise transform [OK]

Hint: Remember: map changes each item one by one [OK]

Common Mistakes:

• Confusing map() with merge or join
• Thinking map() filters data
• Assuming map() sorts data

2. Which of the following is the correct syntax to replace values in a pandas Series s using a dictionary d with map()?

easy

A. map(s, d)

B. s.map([d])

C. s.map(d)

D. map(d, s)

Solution

1. Step 1: Recall the method call format

   In pandas, map() is a Series method called as s.map(), where s is the Series.

2. Step 2: Check argument type

   The argument to map() can be a dictionary for replacement, so s.map(d) is correct syntax.

3. Final Answer:

   s.map(d) -> Option C

4. Quick Check:

   Series.map(dict) = s.map(d) [OK]

Hint: Use Series.map(dict) syntax, not standalone map() [OK]

Common Mistakes:

• Using map() as a standalone function
• Passing dictionary inside a list
• Swapping argument order

3. Given the code:

import pandas as pd
s = pd.Series(['cat', 'dog', 'bird'])
mapping = {'cat': 'feline', 'dog': 'canine'}
result = s.map(mapping)

What is the value of result?

medium

A. 0 feline
1 canine
2 NaN
dtype: object

B. 0 feline
1 canine
2 canine
dtype: object

C. 0 cat
1 dog
2 bird
dtype: object

D. 0 feline
1 canine
2 bird
dtype: object

Solution

1. Step 1: Apply dictionary mapping with map()

   Values 'cat' and 'dog' are replaced by 'feline' and 'canine' respectively. 'bird' is not in the dictionary, so it becomes NaN.

2. Step 2: Check the output Series

   The resulting Series has 'feline', 'canine', and NaN for the third element.

3. Final Answer:

   0 feline 1 canine 2 NaN dtype: object -> Option A

4. Quick Check:

   Missing keys map to NaN [OK]

Hint: Unmapped values become NaN by default [OK]

Common Mistakes:

• Expecting unmapped values to stay unchanged
• Confusing map() with replace() behavior
• Assuming map fills missing keys automatically

4. What is wrong with this code snippet?

import pandas as pd
s = pd.Series([1, 2, 3])
result = s.map({1: 'one', 2: 'two', 3: 'three'})
print(result)

It raises a KeyError. Why?

medium

A. The dictionary keys match Series values, so no error should occur.

B. The dictionary keys do not match Series values type.

C. The dictionary keys are integers but Series has strings.

D. map() cannot accept dictionaries as arguments.

Solution

1. Step 1: Check Series and dictionary key types

   Series contains integers 1, 2, 3 and dictionary keys are also integers 1, 2, 3.

2. Step 2: Understand map() behavior with matching keys

   Since keys match values, map() replaces all values correctly without error.

3. Final Answer:

   The dictionary keys match Series values, so no error should occur. -> Option A

4. Quick Check:

   Matching keys cause no error [OK]

Hint: Matching keys and values avoid KeyError in map() [OK]

Common Mistakes:

• Assuming map() raises error on missing keys
• Confusing map() with indexing behavior
• Thinking map() cannot use dictionaries

5. You have a pandas Series s = pd.Series(['apple', 'banana', 'cherry', 'date']). You want to replace 'banana' and 'date' with their uppercase versions using map(), but keep other fruits unchanged. Which code achieves this?

hard

A. s.map({'banana': 'BANANA', 'date': 'DATE'})

B. s.map({'banana': 'BANANA', 'date': 'DATE'}, na_action='ignore')

C. s.map(lambda x: x if x in ['banana', 'date'] else x.upper())

D. s.map(lambda x: x.upper() if x in ['banana', 'date'] else x)

Solution

1. Step 1: Understand dictionary map behavior

   Using a dictionary with map replaces only keys found; others become NaN, so s.map({'banana': 'BANANA', 'date': 'DATE'}) loses 'apple' and 'cherry'.

2. Step 2: Check lambda function usage

   s.map(lambda x: x.upper() if x in ['banana', 'date'] else x) uses a lambda to uppercase only 'banana' and 'date', else keeps original, preserving all values correctly.

3. Step 3: Analyze other options

   Options A and B produce NaN for 'apple' and 'cherry' since dictionary map() sets NaN for missing keys even with na_action='ignore'; s.map(lambda x: x if x in ['banana', 'date'] else x.upper()) reverses the condition and uppercases wrong fruits.

4. Final Answer:

   s.map(lambda x: x.upper() if x in ['banana', 'date'] else x) -> Option D

5. Quick Check:

   Use lambda for selective transformation [OK]

Hint: Use lambda in map() for conditional changes [OK]

Common Mistakes:

• Using dictionary map without fillna loses data
• Confusing condition in lambda
• Assuming map() changes all values automatically