如何避免字典为空或无匹配键时Pandas脚本报错？

Hey there, let's sort out this error you're hitting. The issue almost certainly comes from trying to access a dictionary key that doesn't exist (or working with an empty dictionary) using direct bracket notation like my_dict[row['Key']]—that'll throw a KeyError every time. Let's walk through safe, efficient solutions tailored to your DataFrame.

First, let's start with a corrected version of your DataFrame (I fixed the admit column to match the length of other columns, which would cause an error otherwise):

import pandas as pd

df = pd.DataFrame({
    'Key': ['10003', '10003', '10003', '10003', '10003','10003','10034'], 
    'Num1': [12,13,30,12,13,13,16], 
    'Num2': [121,122,122,124,125,126,127], 
    'admit': [2015019, 20150124, 20150206,20150211,20150215,20150220,20150225]
})

Solution 1: Safe apply() Handling #

If you want to stick with apply(), the key is to check if the key exists in the dictionary first (this automatically handles empty dictionaries too, since key in empty_dict returns False).

Define a helper function that safely checks for the key and returns your marker:

# Example target dictionary (use {} to test empty case)
target_dict = {'10003': 'mark_me'}

def mark_row(row):
    # Return '1' if Key is in the dictionary, else '0'
    return '1' if row['Key'] in target_dict else '0'

# Add the new marked column to your DataFrame
df['marked_column'] = df.apply(mark_row, axis=1)

This will never throw an error: if the dictionary is empty, all rows get '0'; if a row's Key isn't in the dictionary, it also gets '0'.

Solution 2: Faster Vectorized Operation (Recommended) #

apply() is slow for large DataFrames because it processes rows one by one. A better approach is to use pandas' built-in vectorized methods, which are optimized for speed:

# Use isin() to check if Key is in the dictionary's keys, then map to '1'/'0'
df['marked_column'] = df['Key'].isin(target_dict.keys()).map({True: '1', False: '0'})

# Alternative: Convert boolean to int then string
df['marked_column'] = df['Key'].isin(target_dict.keys()).astype(int).astype(str)

This does the same job but runs orders of magnitude faster on big datasets.

Why Your Original Code Threw Errors #

If you were using something like return target_dict[row['Key']] directly, Python will raise a KeyError as soon as it hits a Key that's not in the dictionary (or if the dictionary is empty). By checking row['Key'] in target_dict first, we avoid that entirely.

内容的提问来源于stack exchange，提问作者CandleWax