如何将Pandas DataFrame中A-Z类别的字符列转换为整数？

Here are a few straightforward methods to turn your uppercase letter-based 'class' column into integers, depending on whether you want alphabetical ordering or just unique integer identifiers:

Method 1: Map to Alphabetical Position (0 or 1-indexed) #

This method directly converts each letter to its position in the alphabet (e.g., 'A' → 0, 'B' →1, ..., 'Z'→25, or start at 1 if you prefer).

import pandas as pd

# Sample DataFrame based on your example
data = {
    'x1': [180.9863129, 52.20132828, -17.17127419, 37.28710938, -132.2395782, -12.52811623],
    'x2': [-0.266379416, 28.93587875, 29.97013283, -69.96691132, 27.02541733, -87.90951538],
    'km': [24,16,17,3,15,22],
    'gmm': [19,14,16,6,18,5],
    'class': ['T','I','D','N','G','S']
}
df = pd.DataFrame(data)

# Convert to 0-indexed alphabetical integers
df['class'] = df['class'].apply(lambda x: ord(x) - ord('A'))

# Or convert to 1-indexed (A=1, B=2, ... Z=26)
# df['class'] = df['class'].apply(lambda x: ord(x) - ord('A') + 1)

After running this, your 'class' column will have values like: 'T' →19, 'I'→8, 'D'→3, etc. (for 0-indexed).

Method 2: Use pd.factorize() for Unique Identifiers (Order of Appearance) #

If alphabetical order doesn't matter and you just want a unique integer for each distinct letter (based on their first occurrence in the DataFrame), use factorize():

df['class'] = pd.factorize(df['class'])[0]

In your example, this would assign:

• 'T' →0, 'I'→1, 'D'→2, 'N'→3, 'G'→4, 'S'→5

Method 3: Custom Mapping Dictionary #

You can also create an explicit dictionary to map each letter to an integer, which gives you full control over the mapping:

# Create a dictionary for A-Z to 0-25
letter_map = {chr(ord('A') + idx): idx for idx in range(26)}

# Apply the mapping
df['class'] = df['class'].map(letter_map)

This works the same as Method 1 but uses a pre-defined map instead of a lambda function. It's handy if you need to adjust the mapping later (e.g., assign specific integers to certain letters).

Quick Notes: #

• If your 'class' column has lowercase letters, add .upper() to the lambda or mapping to ensure consistency: lambda x: ord(x.upper()) - ord('A')
• All methods assume your 'class' values are only A-Z; if there are other characters, you might want to add error handling (like try-except blocks) to avoid issues.

内容的提问来源于stack exchange，提问作者Jadu Sen