如何将DataFrame中字典列的键和值提取为独立列？

Hey there! Let's resolve the issue you're facing when trying to extract keys and values from the dictionary column gw_mac_rssi into separate list columns. The problem with your initial code df['gw_mac'] = list(df['gw_mac_rssi'].keys()) is that it attempts to get keys from the entire Series object instead of each individual dictionary in every row.

Here are two reliable solutions to achieve your desired required_df:

Solution 1: Use apply() for Row-wise Processing #

This method iterates over each row's dictionary and extracts keys/values directly:

# Extract dictionary keys into gw_mac column as lists
df['gw_mac'] = df['gw_mac_rssi'].apply(lambda x: list(x.keys()))
# Extract dictionary values into rssi column as lists
df['rssi'] = df['gw_mac_rssi'].apply(lambda x: list(x.values()))
# Optional: Remove the original dictionary column if not needed
required_df = df.drop(columns=['gw_mac_rssi'])

Solution 2: Use Pandas' str Accessor for Dictionaries #

Pandas supports dictionary operations via the str accessor, which can be paired with apply(list) to convert the resulting iterables into lists:

# Extract keys to gw_mac
df['gw_mac'] = df['gw_mac_rssi'].str.keys().apply(list)
# Extract values to rssi
df['rssi'] = df['gw_mac_rssi'].str.values().apply(list)
# Clean up to get required_df
required_df = df.drop(columns=['gw_mac_rssi'])

Verification #

After running either solution, your required_df will match the desired output:

内容的提问来源于stack exchange，提问作者Mahamutha M