shift() for lagging data in Pandas - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of shift() with default parameters

What is the output of the following code snippet using pandas shift() with default lag?

Pandas

import pandas as pd

df = pd.DataFrame({'value': [10, 20, 30, 40]})
df['lagged'] = df['value'].shift()
print(df)

A

   value  lagged
0     10     NaN
1     20    10.0
2     30    20.0
3     40    30.0

B

   value  lagged
0     10    10.0
1     20    20.0
2     30    30.0
3     40    40.0

C

   value  lagged
0     10    20.0
1     20    30.0
2     30    40.0
3     40     NaN

D

   value  lagged
0     10     NaN
1     20     NaN
2     30     NaN
3     40     NaN

💡 Hint

Remember that shift() moves data down by 1 row by default, introducing NaN at the top.

The shift() function moves the data down by one row, so the first row in the new column is NaN, and each subsequent row contains the previous row's value.

❓ data_output

intermediate

Result of shift() with negative periods

What is the resulting DataFrame after applying shift() with periods=-1 on the 'value' column?

Pandas

import pandas as pd

df = pd.DataFrame({'value': [5, 15, 25, 35]})
df['lead'] = df['value'].shift(periods=-1)
print(df)

A

   value  lead
0      5  35.0
1     15  25.0
2     25  15.0
3     35   NaN

B

   value  lead
0      5   NaN
1     15   5.0
2     25  15.0
3     35  25.0

C

   value  lead
0      5  15.0
1     15  25.0
2     25  35.0
3     35   NaN

D

   value  lead
0      5   NaN
1     15   NaN
2     25   NaN
3     35   NaN

💡 Hint

Negative periods shift data up, so the last row becomes NaN.

Using periods=-1 shifts the data up by one row, so the last row in the new column is NaN, and each row contains the next row's value.

🔧 Debug

advanced

Identify the error in shift() usage

What error will this code raise and why? import pandas as pd df = pd.DataFrame({'score': [100, 200, 300]}) df['shifted'] = df['score'].shift(periods='2')

Pandas

import pandas as pd

df = pd.DataFrame({'score': [100, 200, 300]})
df['shifted'] = df['score'].shift(periods='2')

A TypeError: unsupported operand type(s) for +: 'int' and 'str'

B SyntaxError: invalid syntax

C No error, code runs successfully

D ValueError: periods must be an integer

💡 Hint

Check the type of the 'periods' argument in shift().

The periods argument must be an integer. Passing a string like '2' causes a ValueError.

🚀 Application

advanced

Using shift() to calculate daily returns

Given a DataFrame with daily stock prices, which code correctly calculates the daily return as percentage change using shift()?

Pandas

import pandas as pd

df = pd.DataFrame({'price': [100, 105, 110, 120]})

A df['return'] = df['price'] - df['price'].shift(1) * 100

B df['return'] = (df['price'] - df['price'].shift(1)) / df['price'].shift(1) * 100

C df['return'] = (df['price'] / df['price'].shift(1)) * 100

D df['return'] = (df['price'].shift(1) - df['price']) / df['price'] * 100

💡 Hint

Daily return is (today's price - yesterday's price) divided by yesterday's price.

Option B correctly computes the percentage change from the previous day using shift(1).

🧠 Conceptual

expert

Effect of shift() on time series with missing dates

If a time series DataFrame has missing dates and you apply shift() on the value column, what is the effect on the lagged data?

A The lagged data aligns with the previous row regardless of missing dates, so gaps remain unfilled.

B shift() automatically fills missing dates with NaN before shifting, so lagged data aligns with actual previous dates.

C shift() fills missing dates with the last available value before shifting, so lagged data has no gaps.

D shift() removes rows with missing dates before shifting, so lagged data only contains continuous dates.

💡 Hint

shift() works row-wise and does not modify the index or fill missing dates.

shift() moves data down by rows without considering the index values, so missing dates cause gaps in lagged data.