如何将getEmails函数的多次输出合并为统一列表并存入DataFrame

当前代码的核心问题：每次匹配到符合条件的发件人时，都会在逻辑块内新建临时列表results，仅存储当前匹配结果后打印，临时列表会随循环迭代被销毁，且函数无返回值，外部无法拿到所有聚合结果。

推荐实现方案（函数返回法，无全局变量污染） #

import os
import pickle
import base64
import re
from googleapiclient.discovery import build
from google_auth_oauthlib.flow import InstalledAppFlow
from google.auth.transport.requests import Request
from bs4 import BeautifulSoup
import pandas as pd

# 请自行替换为你的SCOPES配置
SCOPES = ['https://www.googleapis.com/auth/gmail.readonly']

def getEmails():
    # 函数内部初始化总列表，收集所有匹配结果
    all_sites = []
    creds = None

    if os.path.exists('token.pickle'):
        with open('token.pickle', 'rb') as token:
            creds = pickle.load(token)
    if not creds or not creds.valid:
        if creds and creds.expired and creds.refresh_token:
            creds.refresh(Request())
        else:
            flow = InstalledAppFlow.from_client_secrets_file('credentials.json', SCOPES)
            creds = flow.run_local_server(port=0)
        with open('token.pickle', 'wb') as token:
            pickle.dump(creds, token)

    service = build('gmail', 'v1', credentials=creds)
    result = service.users().messages().list(userId='me').execute()
    messages = result.get('messages')

    for msg in messages:
        txt = service.users().messages().get(userId='me', id=msg['id']).execute()
        try:
            payload = txt['payload']
            headers = payload['headers']
            sender = None
            for d in headers:
                if d['name'] == 'From':
                    sender = d['value']
            
            parts = payload.get('parts')[0]
            data = parts['body']['data'].replace("-","+").replace("_","/")
            decoded_data = base64.b64decode(data)
            soup = BeautifulSoup(decoded_data , "lxml")
            page = soup.find('p', text=re.compile('has been listed on ')).getText()
            
            site = None
            if page:
                match = re.search(r'has been listed on (\b\w+\b) ', str(page))
                if match:
                    site = match.group(1)
            
            if sender == 'Cryptocurrency Alerting <alerts@mail.cryptocurrencyalerting.com>' and site:
                # 匹配到结果直接追加到总列表
                all_sites.append([site])
                print([site])
        except:
            pass
    # 函数执行完成后返回全量聚合列表
    return all_sites

# 调用函数接收聚合后的合法列表
results_list = getEmails()
# 可直接生成DataFrame
df = pd.DataFrame(results_list, columns=['exchange'])

补充说明 #

• 原冗余的Convert函数已移除，直接通过正则分组提取交易所名称即可，无需额外拆分字符串
• 此前尝试的k = print(...)写法无效，因为print仅负责将内容输出到控制台，本身无返回值，变量k会一直为None，无法通过该方式收集结果
• 如果坚持要用外部变量收集，可在函数外定义总列表后，在函数内声明global 总列表名后追加结果，该方式存在全局变量污染风险，不推荐使用

内容的提问来源于stack exchange，提问作者crispYpickle