Python：使用字典、函数和lambda为DataFrame新增条件列

Hey there! Let's get your DataFrame sorted out with that new column you need. Your initial approach with a dictionary is a good start, but we can refine it to work directly with pandas without manual looping, which is way more efficient.

First, let's fix the core issue with your current code #

Your loop is iterating over the dictionary keys instead of the rows in your DataFrame, and it's just printing results instead of assigning them to the new column. Also, there's a typo (df['state'] should be df['id']) and the condition logic isn't targeting each row's value.

Here are 3 clean ways to achieve your desired result: #

1. Dictionary Mapping + map() + fillna() (Most Efficient)

This is the pandas-idiomatic way—use a pre-defined mapping for the known IDs, then fill in the rest with your default string:

import pandas as pd

df = pd.DataFrame({
    'id': ['id1', 'id2', 'id3', 'id8', 'id9']
})

# Define the exact strings we want for matching IDs
id_to_new = {
    'id1': 'id1 is cat 1',
    'id2': 'id2 is cat 2',
    'id3': 'id3 is cat 3'
}

# Map the IDs, then fill missing values with the default pattern
df['new'] = df['id'].map(id_to_new).fillna(df['id'].apply(lambda x: f"{x} is cat 0"))

2. apply() with Nested Ternary Logic (Straightforward)

If you prefer explicit condition checks for each case, use apply() with a lambda function:

df['new'] = df['id'].apply(
    lambda x: f"{x} is cat 1" if x == 'id1'
    else f"{x} is cat 2" if x == 'id2'
    else f"{x} is cat 3" if x == 'id3'
    else f"{x} is cat 0"
)

3. numpy.select() (Great for Multiple Conditions)

For more readability with multiple conditions, use numpy.select():

import pandas as pd
import numpy as np

df = pd.DataFrame({
    'id': ['id1', 'id2', 'id3', 'id8', 'id9']
})

# List of conditions to check
conditions = [
    df['id'] == 'id1',
    df['id'] == 'id2',
    df['id'] == 'id3'
]

# Corresponding values for each condition
choices = [
    'id1 is cat 1',
    'id2 is cat 2',
    'id3 is cat 3'
]

# Apply conditions, use default for everything else
df['new'] = np.select(conditions, choices, default=df['id'].apply(lambda x: f"{x} is cat 0"))

Result #

All three methods will give you exactly the output you want:

id              new
0  id1   id1 is cat 1
1  id2   id2 is cat 2
2  id3   id3 is cat 3
3  id8   id8 is cat 0
4  id9   id9 is cat 0

内容的提问来源于stack exchange，提问作者shsh