pandas拆分point列触发Columns must be same length as key错误如何解决

报错原因 #

该报错的核心诱因是point列存在None值：你设置的逻辑是地址解析失败时，point直接赋值为None。当你把point列转为列表构造DataFrame时，None无法被拆分为长度为3的lat/lon/altitude序列，导致生成的DataFrame列数和你要赋值的3个列长度不匹配，最终触发报错。

修复方案 #

调整point列的生成逻辑，确保每一行的point值都是长度为3的元组，解析失败的位置用None填充三个维度即可，修改后的完整代码如下：

import tabula
import pandas as pd
import re
import numpy as np
# 补充缺失的geopy初始化代码
from geopy.geocoders import Nominatim
geolocator = Nominatim(user_agent="condo_geo_parse") # user_agent可自定义

### 数据爬取与清洗
dsf = tabula.read_pdf('/content/drive/MyDrive/Topcondoimage 11-22-2021.pdf', pages='all',lattice=True)
df = dsf[0]
df.columns = df.iloc[0]
df = df.drop(df.index[0])
df = df.iloc[: , 1:]
df = df.replace(np.nan, 'Not Available', regex=True)
df['geo_Address'] = df['Building / Address / City']
df['geo_Address'] = df['geo_Address'].map(lambda x: re.sub(r'\r', ' ', x))
df['loc'] = df['geo_Address'].apply(geolocator.geocode, timeout=10)
# 调整point生成逻辑，确保每个值都是长度为3的元组
df['point'] = df['loc'].apply(lambda loc: tuple(loc.point) if loc else (None, None, None))

df = df.rename(columns={'Building / Address / City': 'building_address_city','Days on\rMarket':'days_on_market','Price /\rSq. Ft.':'price_per_sqft'})
df.reset_index(drop=True, inplace=True)

# 此时生成的DataFrame列数固定为3，不会再报长度不匹配错误
df[['lat','lon','altitude']] = pd.DataFrame(df['point'].to_list(),index=df.index)

补充说明 #

后续如果不需要地址解析失败的行，可以直接用df = df.dropna(subset=['lat'])过滤掉无效数据即可。

内容的提问来源于stack exchange，提问作者Adam