如何用Python正则findall结合月份列表批量匹配文本日期

Got it, let's sort this out. The issue with your initial approach is that you're trying to loop through each month and call findall separately, which won't work as a single line (and would be inefficient anyway). Instead, we can convert your months list into a regex alternation group (using | to separate options) so we can match all month variants in one go.

Step-by-Step Explanation #

1. Convert the months list to a regex pattern: We'll join all month names/abbreviations with |, escaping each one to avoid any unexpected regex behavior (though your months don't have special characters, this is a safe practice).
2. Build the full date regex: Attach the pattern for day and year to the month group, making sure to capture each part (month, day, year) as separate groups.
3. Run findall once: Apply the combined regex to your DataFrame column to extract all matching dates in one line.

Working Code Example #

import pandas as pd
import re

# Your month list
months = ['January', 'February', 'March', 'April', 'May', 'June', 'July', 'August', 'September', 'October','November', 'December', 'Jan', 'Feb', 'Mar', 'Apr', 'Jun', 'Jul', 'Aug', 'Sep', 'Oct', 'Nov', 'Dec']

# Sample DataFrame
df = pd.DataFrame({
    'text': ['last day of the championship July 28, 1983 Mar 11, 1990 record of the first division April 27, 1982 record of played matches']
})

# Build the regex pattern
month_pattern = '|'.join(re.escape(month) for month in months)
date_regex = rf'({month_pattern})\s(\d?\d),\s(\d{{4}})'

# Extract all matches
df['extracted_dates'] = df['text'].str.findall(date_regex)

# View the result
print(df['extracted_dates'].iloc[0])

Expected Output #

[('July', '28', '1983'), ('Mar', '11', '1990'), ('April', '27', '1982')]

Why This Works #

• The month_pattern becomes a string like January|February|...|Dec, so regex will match any of these month variants.
• The full date_regex captures three groups: the month (full or abbreviated), the day (1 or 2 digits), and the 4-digit year.
• str.findall() returns all non-overlapping matches as tuples, exactly the format you're looking for.

If you want to condense this into a single line (as you asked), you can combine the pattern building and findall call:

df['extracted_dates'] = df['text'].str.findall(r'(' + '|'.join(re.escape(m) for m in months) + r')\s(\d?\d),\s(\d{4})')

内容的提问来源于stack exchange，提问作者Joe