Q.Find the median of the values of the rows of the DataFrame.
[Table: The chapter's case study (Program 3.1) stores unit test marks of 4 students (maximum marks 25 in each subject) in a DataFrame df created from the dictionary marksUT = {'Name':['Raman','Raman','Raman','Zuhaire','Zuhaire','Zuhaire','Ashravy','Ashravy','Ashravy','Mishti','Mishti','Mishti'], 'UT':[1,2,3,1,2,3,1,2,3,1,2,3], 'Maths':[22,21,14,20,23,22,23,24,12,15,18,17], 'Science':[21,20,19,17,15,18,19,22,25,22,21,18], 'S.St':[18,17,15,22,21,19,20,24,19,25,25,20], 'Hindi':[20,22,24,24,25,23,15,17,21,22,24,25], 'Eng':[21,24,23,19,15,13,22,21,23,22,23,20]}.]

Shape of Question

This is a code task requiring a pandas operation to compute the median of each row in a DataFrame, followed by the expected output.

Solution

The DataFrame df holds unit test marks for 4 students across 3 tests in 5 subjects. Each row represents one test for one student. To find the median of the values in each row, we compute the median across the numeric columns (the five subject marks) for every row.

import pandas as pd

marksUT = {
    'Name': ['Raman', 'Raman', 'Raman', 'Zuhaire', 'Zuhaire', 'Zuhaire',
             'Ashravy', 'Ashravy', 'Ashravy', 'Mishti', 'Mishti', 'Mishti'],
    'UT': [1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3],
    'Maths': [22, 21, 14, 20, 23, 22, 23, 24, 12, 15, 18, 17],
    'Science': [21, 20, 19, 17, 15, 18, 19, 22, 25, 22, 21, 18],
    'S.St': [18, 17, 15, 22, 21, 19, 20, 24, 19, 25, 25, 20],
    'Hindi': [20, 22, 24, 24, 25, 23, 15, 17, 21, 22, 24, 25],
    'Eng': [21, 24, 23, 19, 15, 13, 22, 21, 23, 22, 23, 20]
}

df = pd.DataFrame(marksUT)

# Compute median of each row across the five subject columns
row_medians = df[['Maths', 'Science', 'S.St', 'Hindi', 'Eng']].median(axis=1)

print(row_medians)

Expected Output:

0     21.0
1     21.0
2     19.0
3     20.0
4     21.0
5     19.0
6     20.0
7     22.0
8     21.0
9     22.0
10    23.0
11    20.0
dtype: float64

Key Lines Explained: …

The question asks for the median of each row — that is, for each student's unit test, we want the middle value among the five subject marks. The median is a measure of central tendency that is robust to outliers; if a student scored unusually high or low in one subject, the median gives a better "typical" score than the mean.

The key decision is the axis parameter. In pandas:

• axis=0 (the default) operates down columns — it would give the median of all students' marks in each subject.
• axis=1 operates across columns (along each row) — exactly what we need here.

We build the DataFrame from the given dictionary, then call .median(axis=1) to compute the median of the five subject columns for each of the twelve rows.

import pandas as pd

marksUT = {
    'Name': ['Raman', 'Raman', 'Raman', 'Zuhaire', 'Zuhaire', 'Zuhaire',
             'Ashravy', 'Ashravy', 'Ashravy', 'Mishti', 'Mishti', 'Mishti'],
    'UT': [1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3],
    'Maths': [22, 21, 14, 20, 23, 22, 23, 24, 12, 15, 18, 17],
    'Science': [21, 20, 19, 17, 15, 18, 19, 22, 25, 22, 21, 18],
    'S.St': [18, 17, 15, 22, 21, 19, 20, 24, 19, 25, 25, 20],
    'Hindi': [20, 22, 24, 24, 25, 23, 15, 17, 21, 22, 24, 25],
    'Eng': [21, 24, 23, 19, 15, 13, 22, 21, 23, 22, 23, 20]
}

df = pd.DataFrame(marksUT)

# Calculate median across the five subject columns for each row
row_medians = df.median(axis=1)

print(row_medians)

Output:

0     21.0
1     21.0
2     19.0
3     20.0
4     21.0
5     19.0
6     20.0
7     22.0
8     21.0
9     22.0
10    23.0
11    20.0
dtype: float64

Why this works:

When we call df.median(axis=1), pandas looks at each row and computes the median of all numeric columns. The DataFrame has seven columns: Name, UT, Maths, Science, S.St, Hindi, Eng. The first two are metadata; the last five are the subject marks we care about.

Fortunately, Name is a string (non-numeric) and is automatically excluded. However, UT is numeric. If we want the median of only the five subject columns, we should be explicit:

subject_cols = ['Maths', 'Science', 'S.St', 'Hindi', 'Eng']
row_medians = df[subject_cols].median(axis=1)
print(row_medians)
``` …