Interview Questions

CHAPTER 6: INTERVIEW QUESTIONS WITH SOLUTIONS

🧠 Yeh questions 90% interviews mein poochhe jaate hain. Har ek ka code + theory DONO yaad rakho. "Approach batao" bole toh theory batao, "Code likho" bole toh code likho.

Q1. How do you detect and handle outliers?

🧠 Outlier kya hai? Wo data point jo baaki sab se bahut alag ho. Jaise class mein sabke marks 50-80 hain, ek bande ke 2 hain — woh outlier hai. IQR method se pehchaante hain.

def detect_outliers_iqr(df, column):
    """Detect outliers using the IQR (Interquartile Range) method."""
    Q1 = df[column].quantile(0.25)  # 25th percentile
    Q3 = df[column].quantile(0.75)  # 75th percentile
    IQR = Q3 - Q1                   # Interquartile Range
    
    lower_bound = Q1 - 1.5 * IQR
    upper_bound = Q3 + 1.5 * IQR
    
    # Identify outlier rows
    outliers = df[(df[column] < lower_bound) | (df[column] > upper_bound)]
    
    print(f"Column: {column}")
    print(f"  Q1={Q1}, Q3={Q3}, IQR={IQR}")
    print(f"  Valid range: [{lower_bound}, {upper_bound}]")
    print(f"  Outliers found: {len(outliers)}")
    
    return outliers

# Detect
outliers = detect_outliers_iqr(df, 'revenue')

# Option 1: Remove outliers
df_clean = df[~df.index.isin(outliers.index)]

# Option 2: Cap outliers (winsorization)
Q1 = df['revenue'].quantile(0.25)
Q3 = df['revenue'].quantile(0.75)
IQR = Q3 - Q1
df['revenue_capped'] = df['revenue'].clip(lower=Q1-1.5*IQR, upper=Q3+1.5*IQR)

Theory: An outlier is a data point that is significantly different from other observations. The IQR method defines outliers as values below Q1 - 1.5×IQR or above Q3 + 1.5×IQR. This is the same logic behind box plot whiskers.

Q2. What is the difference between apply() and map()?

Method	Works On	Use Case
map()	Series only	Simple element-wise transformation
apply()	Series or DataFrame	Complex functions, can access entire row
applymap()	DataFrame only	Element-wise on entire DataFrame

# map — simple substitution
df['segment_code'] = df['segment'].map({
    'Corporate': 'C', 
    'Consumer': 'N', 
    'Home Office': 'H'
})

# apply on Series — custom function
df['revenue_log'] = df['revenue'].apply(lambda x: np.log(x) if x > 0 else 0)

# apply on DataFrame — access entire row
df['summary'] = df.apply(
    lambda row: f"{row['name']} from {row['city']} spent ₹{row['revenue']}", 
    axis=1  # axis=1 means apply to each ROW
)

Q3. How do you read a large CSV file that doesn't fit in memory?

# Method 1: Read in chunks
chunk_size = 10000
results = []
for chunk in pd.read_csv('huge_file.csv', chunksize=chunk_size):
    # Process each chunk
    filtered = chunk[chunk['amount'] > 1000]
    results.append(filtered)
df = pd.concat(results)

# Method 2: Read only needed columns
df = pd.read_csv('huge_file.csv', usecols=['customer_id', 'amount', 'date'])

# Method 3: Specify data types to reduce memory
df = pd.read_csv('huge_file.csv', dtype={
    'customer_id': 'int32',     # Instead of int64
    'amount': 'float32',        # Instead of float64
    'category': 'category'      # For low-cardinality strings
})

Q4. Find the top 3 customers by revenue in each city.

# Method: rank within groups
df['rank'] = df.groupby('city')['revenue'].rank(method='dense', ascending=False)
top3 = df[df['rank'] <= 3].sort_values(['city', 'rank'])

# OR using nlargest within groups
top3 = df.groupby('city').apply(
    lambda x: x.nlargest(3, 'revenue')
).reset_index(drop=True)

Q5. Calculate month-over-month revenue growth.

monthly = df.groupby(df['order_date'].dt.to_period('M'))['revenue'].sum()
monthly_growth = monthly.pct_change() * 100  # percentage change
print(monthly_growth.round(2))

Q6. Write a function to generate a summary report for any DataFrame.

def data_summary(df):
    """Generate a comprehensive summary of any DataFrame."""
    print("=" * 60)
    print("DATASET SUMMARY REPORT")
    print("=" * 60)
    
    print(f"\n📊 Shape: {df.shape[0]} rows × {df.shape[1]} columns")
    
    print(f"\n📋 Column Types:")
    for dtype in df.dtypes.unique():
        cols = df.select_dtypes(include=[dtype]).columns.tolist()
        print(f"  {dtype}: {len(cols)} columns → {cols}")
    
    print(f"\n❌ Missing Values:")
    missing = df.isnull().sum()
    missing_pct = (missing / len(df) * 100).round(2)
    for col in df.columns:
        if missing[col] > 0:
            print(f"  {col}: {missing[col]} ({missing_pct[col]}%)")
    if missing.sum() == 0:
        print("  None! ✅")
    
    print(f"\n🔢 Numerical Summary:")
    print(df.describe().round(2))
    
    print(f"\n📝 Categorical Summary:")
    for col in df.select_dtypes(include='object').columns:
        print(f"  {col}: {df[col].nunique()} unique values, "
              f"top = '{df[col].mode()[0]}'")
    
    print(f"\n🔁 Duplicates: {df.duplicated().sum()}")

# Usage
data_summary(df)

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CHAPTER 7: COMMON PYTHON INTERVIEW CODE SNIPPETS

Reverse a string

s = "DecisionTree"
s[::-1]  # "eerTnoisiceD"

Check if a string is a palindrome

def is_palindrome(s):
    s = s.lower().replace(" ", "")
    return s == s[::-1]

is_palindrome("madam")  # True

Find the most frequent element in a list

from collections import Counter
nums = [1, 3, 2, 1, 4, 1, 3, 3, 3]
Counter(nums).most_common(1)  # [(3, 4)] → 3 appears 4 times

Remove duplicates from a list while preserving order

def remove_dupes(lst):
    seen = set()
    result = []
    for item in lst:
        if item not in seen:
            seen.add(item)
            result.append(item)
    return result

remove_dupes([1, 3, 2, 1, 4, 3])  # [1, 3, 2, 4]

Flatten a nested list

nested = [[1, 2], [3, 4], [5, 6]]
flat = [item for sublist in nested for item in sublist]
# [1, 2, 3, 4, 5, 6]

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Practice: Work through Kaggle's Pandas exercises and try the Titanic dataset for a real EDA practice run. Aim to complete 3–5 EDA projects before the interview.