如何在Python循环中向DataFrame追加函数返回的变量？

问题 #

目标生成如下格式的输出表格：

campaign | mean | left | right 
A        | 5%   | 0%   | 10% 
B        | 8%   | 2%   | 6% etc.

使用Python循环调用bootstrap_ci函数获取campaign、mean、left、right四个变量，但追加新DataFrame时失败，最终返回空DataFrame。相关代码如下：

bootstrap_ci函数代码 #

import numpy as np
def bootstrap_ci(df, variable, classes, repetitions = 1000, alpha = 0.05, random_state=None):
    df = df[[variable, classes]]
    bootstrap_sample_size = len(df)

    mean_diffs = []

    for i in range(repetitions):
        bootstrap_sample = df.sample(n = bootstrap_sample_size, replace = True, random_state = random_state)
        mean_diff = bootstrap_sample.groupby(classes).mean().iloc[1,0] - bootstrap_sample.groupby(classes).mean().iloc[0,0]
        mean_diffs.append(mean_diff)

        left = np.percentile(mean_diffs, alpha/2*100)*(-1)
        right = np.percentile(mean_diffs, 100-alpha/2*100)*(-1)
        mean = -(df.groupby(classes).mean().iloc[1,0] - df.groupby(classes).mean().iloc[0,0])

循环调用及追加代码 #

df = pd.DataFrame([]) # 初始化新df

for i in campaigns:     
    print(f'Lift for: {i}')     
    bootstrap_ci(df[df['campaign_name']==i],'conversion','group')
    df = df.append(i, mean, left, right) # 将相关变量追加到df

解决建议 #

1. 修复bootstrap_ci函数的返回值与计算逻辑 #

• 原函数无返回值，调用后无法获取mean、left、right变量
• 原函数中left、right的计算缩进错误，应放在循环结束后（否则每次循环都会重复计算百分位数，效率低且结果不准确）

修改后的函数：

import numpy as np
import pandas as pd

def bootstrap_ci(df, variable, classes, repetitions=1000, alpha=0.05, random_state=None):
    df = df[[variable, classes]]
    bootstrap_sample_size = len(df)
    mean_diffs = []

    for _ in range(repetitions):
        bootstrap_sample = df.sample(n=bootstrap_sample_size, replace=True, random_state=random_state)
        group_means = bootstrap_sample.groupby(classes).mean().iloc[:, 0]
        mean_diff = group_means.iloc[1] - group_means.iloc[0]
        mean_diffs.append(mean_diff)
    
    # 循环结束后统一计算置信区间与均值
    left = np.percentile(mean_diffs, alpha/2 * 100) * (-1)
    right = np.percentile(mean_diffs, 100 - alpha/2 * 100) * (-1)
    original_group_means = df.groupby(classes).mean().iloc[:, 0]
    mean = -(original_group_means.iloc[1] - original_group_means.iloc[0])
    
    # 返回计算结果
    return mean, left, right

2. 修复DataFrame的追加逻辑 #

• df.append()已被Pandas弃用，推荐先收集所有结果为字典列表，最后一次性转为DataFrame，效率更高且不易出错
• 原代码未接收bootstrap_ci的返回值，导致mean、left、right未定义

修改后的循环代码：

# 初始化列表存储所有campaign的结果
results = []

for campaign in campaigns:     
    print(f'Lift for: {campaign}')     
    # 接收函数返回的三个计算值
    mean, left, right = bootstrap_ci(df[df['campaign_name'] == campaign], 'conversion', 'group')
    # 将当前campaign的结果存入字典，追加到列表
    results.append({
        'campaign': campaign,
        'mean': f'{mean*100:.1f}%', # 转为百分比格式
        'left': f'{left*100:.1f}%',
        'right': f'{right*100:.1f}%'
    })

# 将列表转为目标格式的DataFrame
result_df = pd.DataFrame(results)
print(result_df)

3. 额外优化点 #

• 给random_state传入固定值（如42），保证结果可复现
• 如果group列的分组顺序不确定，建议用分组名索引替代iloc，避免索引错误，例如假设分组为control和treatment：

group_means = bootstrap_sample.groupby(classes).mean().loc[:, variable]
mean_diff = group_means['treatment'] - group_means['control']

内容的提问来源于stack exchange，提问作者Lakem