如何用Pandas中列值作为嵌套字典的键新增列？

Let’s start with sample data matching your structure to make this concrete:

import pandas as pd

# Your nested dictionary
dict1 = {
    'key1': {'key2': 'value12', 'key3': 'value13'},
    'key4': {'key5': 'value45', 'key6': 'value46'}
}

# Your DataFrame with A and B columns
df = pd.DataFrame({
    'A': ['key1', 'key4', 'key1'],
    'B': ['key2', 'key5', 'key3']
})

Why Your Previous Attempts Failed #

• dict1[df['A']][df['C']]: Dictionaries can’t be indexed with a pandas Series (which is what df['A'] is)—they require single scalar keys, so this throws an error.
• The apply call with lambda x,y: When using apply on a DataFrame, each row is passed as a single argument to the lambda, not two separate values. Your lambda expects two inputs but only gets one, leading to a mismatch.

Efficient, Loop-Free Solutions #

Method 1: Use apply with axis=1 (Simple & Readable)

This approach avoids explicit loops like itertuples() or iterrows() and is easy to understand. We pass each row to a lambda that performs the nested lookup:

df['C'] = df.apply(lambda row: dict1[row['A']][row['B']], axis=1)

Method 2: Flatten the Dict for Vectorized Lookup (Faster for Large Data)

For big DataFrames, a fully vectorized approach will perform better. First flatten your nested dict into a multi-index Series, then map the (A,B) tuples from your DataFrame to it:

# Flatten the nested dict into a Series with (outer_key, inner_key) as index
flat_lookup = pd.Series({(outer, inner): val 
                         for outer, inner_dict in dict1.items() 
                         for inner, val in inner_dict.items()})

# Map the (A,B) pairs from the DataFrame to the flattened Series
df['C'] = df.set_index(['A', 'B']).index.map(flat_lookup.get)
df = df.reset_index(drop=True)  # Restore original index if needed

Method 3: Two-Step map (Clearer Breakdown)

You can split the lookup into two steps for better readability: first map column A to get the inner dictionaries, then extract values using column B:

# Step 1: Retrieve the inner dict for each row using column A
df['inner_dict'] = df['A'].map(dict1)

# Step 2: Look up the value from the inner dict using column B
df['C'] = df.apply(lambda row: row['inner_dict'][row['B']], axis=1)

# Clean up the temporary column
df.drop('inner_dict', axis=1, inplace=True)

Final Result #

After applying any of these methods, your DataFrame will have the desired C column:

内容的提问来源于stack exchange，提问作者formicaman