Python TypeError:to_append应该是一个序列或序列的列表/元组，got DataFrame

代码运行得更早，然后刚刚停止

我不明白为什么，如果有人能帮我解释一下这个问题，那就太棒了

import urllib.request
from bs4 import BeautifulSoup as bs
import re
import pandas as pd
import requests


locations = ['las-vegas-nv','denver-co']
for location in locations:
    headers = {
        'User-Agent' : 'Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/90.0.4430.93 Safari/537.36'
    }



    url = f'https://www.apartments.com/{location}/1-bedrooms/'
    page = requests.get(url, headers=headers)
    soup = bs(page.text, 'lxml')
    table = soup.find("table", class_="rentTrendGrid")
    rows = []

    for tr in table.find_all('tr'):
        rows.append([td.text for td in tr.find_all(['th', 'td'])])

#header_row = rows[0]
    rows = list(zip(*rows[0:])) # tranpose the table
    df = pd.DataFrame(rows[1:], columns=rows[0])
    df['City'] = location
    dfs.append(df)
    df = pd.concat(dfs).reset_index(drop = True)
print(df)

---

结果我只需要从for循环中取出最后一部分

import urllib.request
from bs4 import BeautifulSoup as bs
import re
import pandas as pd
import requests


locations = ['las-vegas-nv','denver-co', 'lehi-ut']

dfs = []
for location in locations:
    headers = {
        'User-Agent' : 'Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/90.0.4430.93 Safari/537.36'
    }



    url = f'https://www.apartments.com/{location}/1-bedrooms/'
    page = requests.get(url, headers=headers)
    soup = bs(page.text, 'lxml')
    table = soup.find("table", class_="rentTrendGrid")
    rows = []

    for tr in table.find_all('tr'):
        rows.append([td.text for td in tr.find_all(['th', 'td'])])

#header_row = rows[0]
    rows = list(zip(*rows[0:])) # tranpose the table
    df = pd.DataFrame(rows[1:], columns=rows[0])
    df['City'] = location
    dfs.append(df)
df = pd.concat(dfs).reset_index(drop = True)
print(df)

---

什么是dfs？我看不到它在任何地方被定义。它只是让我在代码中乱搞…我知道它没有被定义，但是当dfs等于df时，它仍然抛出相同的错误