如何将嵌套列表字典转换为指定结构的pandas DataFrame？

Why Your Current Code Isn't Working #

When you use pd.DataFrame.from_dict(dict1, orient='index'), pandas treats the dictionary key (808) as the index and the nested list as separate columns. This results in a single row where each column holds an inner list—definitely not the flattened structure you're aiming for. Let's fix that with two straightforward approaches.

Method 1: Manual Row Construction (Intuitive for Beginners) #

We'll loop through the dictionary to build a list of complete rows, where each row starts with the memberid followed by the elements from each sublist. Then we convert this list to a DataFrame with your desired column names.

import pandas as pd

dict1 = {808: [['a', 5.4, 'b'], ['c', 4.1, 'b'], ['d', 3.7, 'f']]}

# Build a list of full rows
rows = []
for memberid, entries in dict1.items():
    for entry in entries:
        rows.append([memberid] + entry)

# Create the DataFrame with specified columns
df = pd.DataFrame(rows, columns=['memberid', 'userid', 'score', 'related'])
print(df)

Output: #

memberid userid  score related
0       808      a    5.4       b
1       808      c    4.1       b
2       808      d    3.7       f

Method 2: Pandas-Native Approach (Using Explode) #

If you prefer a more pandas-focused workflow, you can use explode() to expand the nested list into separate rows, then split the exploded column into individual columns.

import pandas as pd

dict1 = {808: [['a', 5.4, 'b'], ['c', 4.1, 'b'], ['d', 3.7, 'f']]}

# Convert dict to initial DataFrame and rename columns
df = pd.DataFrame.from_dict(dict1, orient='index').reset_index()
df.columns = ['memberid', 'data']

# Explode the list into individual rows
df = df.explode('data')

# Split the 'data' column into separate columns
df[['userid', 'score', 'related']] = pd.DataFrame(df['data'].tolist(), index=df.index)

# Drop the temporary 'data' column
df = df.drop('data', axis=1)
print(df)

This will produce the exact same output as Method 1.

内容的提问来源于stack exchange，提问作者Wendy D.