Data Science Python 面试完全指南 2026：Pandas/NumPy/Sklearn + 真题解析

1import pandas as pd
2
3# 真题：计算每个用户的月均消费金额，并找出消费最高的前 10% 用户
4df = pd.read_csv('transactions.csv')  # columns: user_id, date, amount
5
6# 解法
7df['month'] = pd.to_datetime(df['date']).dt.to_period('M')
8monthly_avg = (
9    df.groupby(['user_id', 'month'])['amount']
10    .sum()
11    .reset_index()
12    .groupby('user_id')['amount']
13    .mean()
14    .reset_index()
15    .rename(columns={'amount': 'monthly_avg_spend'})
16)
17
18# 找出前 10% 用户
19threshold = monthly_avg['monthly_avg_spend'].quantile(0.9)
20top_users = monthly_avg[monthly_avg['monthly_avg_spend'] >= threshold]

1# 真题：计算用户 7 日滚动留存率
2df['date'] = pd.to_datetime(df['date'])
3df = df.sort_values(['user_id', 'date'])
4
5# 计算每个用户的首次登录日期
6df['first_login'] = df.groupby('user_id')['date'].transform('min')
7df['days_since_first'] = (df['date'] - df['first_login']).dt.days
8
9# 7 日留存：首次登录后第 7 天是否有活跃记录
10retention = df[df['days_since_first'] == 7]['user_id'].nunique() / df['user_id'].nunique()

1# 真题：处理一张包含 30% 缺失值的用户特征表
2# 面试官考察的不是代码，而是你的决策逻辑
3
4# 1. 先分析缺失值的分布
5print(df.isnull().sum() / len(df))  # 缺失率
6print(df.isnull().mean())  # 同上，更简洁
7
8# 2. 根据业务逻辑选择填充策略
9# 数值型：用中位数（对异常值鲁棒）
10df['age'].fillna(df['age'].median(), inplace=True)
11
12# 类别型：用众数或单独的 'Unknown' 类别
13df['city'].fillna('Unknown', inplace=True)
14
15# 时间序列：用前向填充（forward fill）
16df['price'].fillna(method='ffill', inplace=True)
17
18# 3. 对缺失率 > 50% 的列，考虑删除
19cols_to_drop = df.columns[df.isnull().mean() > 0.5]
20df.drop(columns=cols_to_drop, inplace=True)

1from sklearn.pipeline import Pipeline
2from sklearn.compose import ColumnTransformer
3from sklearn.preprocessing import StandardScaler, OneHotEncoder
4from sklearn.impute import SimpleImputer
5from sklearn.ensemble import RandomForestClassifier
6from sklearn.model_selection import cross_val_score
7
8# 定义特征类型
9numeric_features = ['age', 'income', 'tenure']
10categorical_features = ['city', 'product_type', 'channel']
11
12# 数值特征处理：填充缺失值 + 标准化
13numeric_transformer = Pipeline(steps=[
14    ('imputer', SimpleImputer(strategy='median')),
15    ('scaler', StandardScaler())
16])
17
18# 类别特征处理：填充缺失值 + One-Hot 编码
19categorical_transformer = Pipeline(steps=[
20    ('imputer', SimpleImputer(strategy='constant', fill_value='missing')),
21    ('onehot', OneHotEncoder(handle_unknown='ignore', sparse_output=False))
22])
23
24# 合并特征处理器
25preprocessor = ColumnTransformer(
26    transformers=[
27        ('num', numeric_transformer, numeric_features),
28        ('cat', categorical_transformer, categorical_features)
29    ]
30)
31
32# 完整 Pipeline
33pipeline = Pipeline(steps=[
34    ('preprocessor', preprocessor),
35    ('classifier', RandomForestClassifier(n_estimators=100, random_state=42))
36])
37
38# 交叉验证
39scores = cross_val_score(pipeline, X_train, y_train, cv=5, scoring='roc_auc')
40print(f'CV AUC: {scores.mean():.4f} (+/- {scores.std()*2:.4f})')

1import shap
2
3# 用 SHAP 解释 RandomForest 的预测
4explainer = shap.TreeExplainer(pipeline.named_steps['classifier'])
5X_processed = pipeline.named_steps['preprocessor'].transform(X_test)
6shap_values = explainer.shap_values(X_processed)
7
8# 可视化特征重要性
9shap.summary_plot(shap_values[1], X_processed, feature_names=feature_names)

1from scipy import stats
2import numpy as np
3
4# 真题：给定两组用户的购买金额数据，判断两组是否存在显著差异
5group_a = np.array([45, 52, 48, 61, 39, 55, 47, 58, 43, 51])
6group_b = np.array([52, 61, 58, 67, 49, 63, 55, 70, 48, 59])
7
8# 1. 先检验正态性（Shapiro-Wilk 检验）
9_, p_normal_a = stats.shapiro(group_a)
10_, p_normal_b = stats.shapiro(group_b)
11
12if p_normal_a > 0.05 and p_normal_b > 0.05:
13    # 数据近似正态分布，使用 t 检验
14    t_stat, p_value = stats.ttest_ind(group_a, group_b)
15    print(f't-statistic: {t_stat:.4f}, p-value: {p_value:.4f}')
16else:
17    # 数据不符合正态分布，使用非参数检验（Mann-Whitney U）
18    u_stat, p_value = stats.mannwhitneyu(group_a, group_b, alternative='two-sided')
19    print(f'U-statistic: {u_stat:.4f}, p-value: {p_value:.4f}')
20
21# 2. 计算效应量（Effect Size）
22cohens_d = (group_b.mean() - group_a.mean()) / np.sqrt(
23    ((len(group_a)-1)*group_a.std()**2 + (len(group_b)-1)*group_b.std()**2) /
24    (len(group_a) + len(group_b) - 2)
25)
26print(f"Cohen's d: {cohens_d:.4f}")