pct_change() for percentage change in Pandas - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to calculate the percentage change of the 'sales' column in the DataFrame.

Pandas

df['pct_change'] = df['sales'].[1]()

Drag options to blanks, or click blank then click option'

Adiff

Bmean

Ccumsum

Dpct_change

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'diff()' which calculates difference, not percentage change.

Using 'cumsum()' which calculates cumulative sum.

Using 'mean()' which calculates average.

The pct_change() function calculates the percentage change between the current and prior element in the column.

2fill in blank

medium

Complete the code to calculate the percentage change with a period of 2 rows.

Pandas

df['pct_change_2'] = df['revenue'].[1](periods=2)

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Apct_change

Bdiff

Cshift

Drolling

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'diff()' which does not accept 'periods' for percentage change.

Using 'shift()' which shifts data but does not calculate percentage change.

Using 'rolling()' which is for moving window calculations.

Setting periods=2 in pct_change() calculates the percentage change compared to 2 rows before.

3fill in blank

hard

Fix the error in the code to correctly calculate percentage change of the 'price' column.

Pandas

df['price_change'] = df['price'].[1](periods=1)

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Apctchange

Bpct_change

Cpct_changes

Dpercent_change

Attempts:

3 left

💡 Hint

Common Mistakes

Misspelling the function name.

Using 'period' instead of 'periods' parameter.

The correct function name is pct_change() with an underscore and plural 'periods' parameter.

4fill in blank

hard

Fill both blanks to create a dictionary comprehension that maps each word to its length only if the length is greater than 3.

Pandas

{word: [1] for word in words if [2]

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Alen(word)

Bword

Clen(word) > 3

Dword > 3

Attempts:

3 left

💡 Hint

Common Mistakes

Using the word itself as the value instead of its length.

Checking if the word is greater than 3 instead of its length.

The dictionary comprehension maps each word to its length using len(word). The condition filters words with length greater than 3 using len(word) > 3.

5fill in blank

hard

Fill all three blanks to create a dictionary comprehension that maps each uppercase word to its value only if the value is greater than 0.

Pandas

{ [1]: [2] for k, v in data.items() if v [3] 0 }

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Ak.upper()

Bv

C>

Dk

Attempts:

3 left

💡 Hint

Common Mistakes

Using the original key without uppercase.

Using incorrect comparison operators.

Swapping keys and values.

The key is the uppercase version of k using k.upper(). The value is v. The condition filters values greater than 0 using >.