Pandas: Calculate Discrete Differences with diff()

Compute discrete differences between current and prior elements across Series and DataFrames using the Pandas diff() method. Measure period-over-period changes, calculate velocity or acceleration in time series data, and configure custom lag periods.

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Introduction to diff() 🔝

The diff() method calculates the absolute difference between an element and a prior element (specified by the periods parameter). By default, it subtracts the immediately preceding row (`periods=1`). The first row naturally evaluates to `NaN`.

Create numerical sample series 🔝

We start with a numerical Series representing sequential measurements or stock prices.

import pandas as pd

s = pd.Series([10, 15, 12, 18], name="VAL")
print(s)
assert s.shape == (4,)

Expected output

0    10
1    15
2    12
3    18
Name: VAL, dtype: int64

Basic discrete difference calculation 🔝

Call diff() on the Series to compute row-over-row differences.

res = s.diff()
print(res)
assert res.iloc[1] == 5.0 and res.iloc[2] == -3.0

Expected output

0    NaN
1    5.0
2   -3.0
3    6.0
Name: VAL, dtype: float64

Configuring lag periods 🔝

Use the periods parameter to calculate differences across multiple rows (e.g., comparing against two periods prior).

res_lag = s.diff(periods=2)
print(res_lag)
assert res_lag.iloc[2] == 2.0

Expected output

0    NaN
1    NaN
2    2.0
3    3.0
Name: VAL, dtype: float64

Differencing DataFrames across rows or columns 🔝

When applied to a DataFrame, diff() computes row-wise differences (axis=0) by default. Set axis=1 to calculate column-wise differences.

df = pd.DataFrame({
    "A": [10, 20, 30],
    "B": [15, 25, 40]
})
df_diff = df.diff(axis=1)
print(df_diff)
assert df_diff.loc[0, "B"] == 5.0

Expected output

     A    B
0  NaN  5.0
1  NaN  5.0
2  NaN  10.0

Exercise: compute daily temperature changes 🔝

Given a Series temps = pd.Series([30, 32, 31, 35], name="TEMP"), compute the discrete difference using diff() and verify that the difference at index 3 equals 4.0.

Exercise solution 🔝

Solution creating the Series and applying diff().

temps = pd.Series([30, 32, 31, 35], name="TEMP")
answer = temps.diff()
print(answer)
assert answer.iloc[3] == 4.0

Expected output

0    NaN
1    2.0
2   -1.0
3    4.0
Name: TEMP, dtype: float64

Practice in Google Colab 🔝

Interactive Colab Notebook

Practice diff() examples interactively directly in your browser:

View on GitHub

Reference: Pandas diff() documentation.

Subhendu Mohapatra

Author & Instructor at plus2net

I write and maintain practical tutorials on Python, PHP, SQL, JavaScript, HTML, jQuery, and web development at plus2net. The tutorials focus on clear explanations, working examples, and code that readers can test and adapt while learning.

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