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从列表元素的每一行创建DataFrame:假设列表内DataFrame结构一致

Combine DataFrames from a List of Identical-Structure DataFrames

Got it, since you have a list of DataFrames with exactly the same structure (like your final01_1, final02_1, final03_1 examples), here are the most practical ways to create a single consolidated DataFrame from all their rows:

This is the standard, efficient way to stack all rows from your DataFrame list into one.

import pandas as pd

# Assume your list of DataFrames is defined like this
df_list = [final01_1, final02_1, final03_1]

# Combine all DataFrames into one
combined_df = pd.concat(df_list, ignore_index=True)

# Preview the result
print(combined_df.head())
  • Key Note: The ignore_index=True parameter resets the index of the final DataFrame, avoiding duplicate index values from the original individual DataFrames.
  • Optional: If you want to track which original DataFrame each row came from, add a source identifier column first:
    # Add a 'source' column to each DataFrame in the list
    for index, df in enumerate(df_list, start=1):
        df['source'] = f'final{index:02d}_1'
    
    # Now combine them
    combined_df = pd.concat(df_list, ignore_index=True)
    

Method 2: Build from Individual Rows

If you specifically need to extract each row first before creating the final DataFrame (e.g., for custom processing per row), you can do this:

import pandas as pd

df_list = [final01_1, final02_1, final03_1]
all_rows = []

# Extract every row from each DataFrame as a dictionary
for df in df_list:
    all_rows.extend(df.to_dict('records'))

# Create the final DataFrame from the collected rows
combined_df = pd.DataFrame(all_rows)

This works because df.to_dict('records') converts each row of the DataFrame into a dictionary (with column names as keys), and we collect all these dictionaries before building the new DataFrame.

Verification

After running either method, your combined_df will have the same column structure as your original DataFrames, and will contain all rows from every DataFrame in your list.

内容的提问来源于stack exchange,提问作者rook1996

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最近更新时间:2026.05.25 07:23:47